1.1.
Solve:
1.2.
Solve:
1.3.
Solve:
1.4. Solve: (a) The basic idea of the particle model is that we will treat...
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1.1.
Solve:
1.2.
Solve:
1.3.
Solve:
1.4. Solve: (a) The basic idea of the particle model is that we will treat an object as if all its mass is concentrated into a single point. The size and shape of the object will not be considered. This is a reasonable approximation of reality if (i) the distance traveled by the object is large in comparison to the size of the object, and (ii) rotations and internal motions are not significant features of the object’s motion. The particle model is important in that it allows us to simplify a problem. Complete reality—which would have to include the motion of every single atom in the object—is too complicated to analyze. By treating an object as a particle, we can focus on the most important aspects of its motion while neglecting minor and unobservable details. (b) The particle model is valid for understanding the motion of a satellite or a car traveling a large distance. (c) The particle model is not valid for understanding how a car engine operates, how a person walks, how a bird flies, or how water flows through a pipe.
1.5. Solve: (a) An operational definition defines a concept or an idea in terms of a procedure, or a set of operations, that is used to identify or measure the concept. G (b) The displacement Δr of an object is a vector found by drawing an arrow from the object’s initial location to G G G G its final location. Mathematically, Δr = rf − ri . The average velocity v of an object is a vector that points in the G G same direction as the displacement Δr and has length, or magnitude, Δr / Δt , where Δt = tf − ti is the time interval during which the object moves from its initial location to its final location.
1.6.
Solve: The player starts from rest and moves faster and faster (accelerates).
1.7. Solve: The particle starts with an initial velocity but as it slides it moves slower and slower till coming to rest. This is a case of negative acceleration because it is an acceleration opposite to the positive direction of motion.
G Solve: The acceleration of an object is a vector formed by finding the ratio of Δv , the change in the G object’s velocity, to Δt , the time in which the change occurs. The acceleration vector a points in the direction G of Δv , which is found by vector subtraction.
1.8.
G (a) Acceleration is found by the method of Tactics Box 1.3. Let v0 be the velocity vector G between points 0 and 1 and v1 be the velocity vector between points 1 and 2.
1.9.
Solve:
(b) Speed v1 is greater than speed v0 because more distance is covered in the same interval of time.
G Solve: (a) Acceleration is found by the method of Tactics Box 1.3. Let v0 be the velocity vector G between points 0 and 1 and v1 be the velocity vector between points 1 and 2.
1.10.
(b) Speed v1 is greater than speed v0 because more distance is covered in the same interval of time.
1.11. (a)
Solve: (b)
1.12. (a)
Solve: (b)
1.13.
Model:
Represent the car as a particle.
Visualize: The dots are equally spaced until brakes are applied to the car. Equidistant dots indicate constant average speed. On braking, the dots get closer as the average speed decreases.
1.14.
Model:
Represent the (child + sled) system as a particle.
Visualize: The dots in the figure are equally spaced until the sled encounters a rocky patch. Equidistant dots indicate constant average speed. On encountering a rocky patch, the average speed decreases and the sled comes to a stop. This part of the motion is indicated by a separation between the dots that becomes smaller and smaller.
1.15.
Model: Represent the tile as a particle. Visualize: The tile falls from the roof with an acceleration equal to a = g = 9.8 m/s2. Starting from rest, its velocity increases until the tile hits the water surface. This part of the motion is represented by dots with increasing separation, indicating increasing average velocity. After the tile enters the water, it settles to the bottom at roughly constant speed.
1.16.
Model: Represent the tennis ball as a particle. Visualize: The particle falls freely for the three stories under the acceleration of gravity. It strikes the ground and very quickly decelerates to zero (while decompresses) and finally travels upward with negative acceleration under gravity to zero velocity at a height of two stories. The downward and upward motions of the ball are shown in the figure. The increasing length between the dots during downward motion indicates increasing average velocity or downward acceleration. On the other hand, the decreasing length between the dots during upward motion indicates acceleration in a direction opposite to its motion; that is, in the downward direction. Assess: For a free-fall motion, acceleration due to gravity is always vertically downward.
1.17.
Model: Represent the toy car as a particle. Visualize: As the toy car rolls down the ramp, its average speed increases. This is indicated by the increasing G length of the velocity arrows. That is, motion down the ramp is under an acceleration a. At the bottom of the ramp, the toy car continues with the speed obtained with no change in velocity.
1.18. Solve: (a)
Dot 1 2 3 4 5 6 7 8 9
Time (s) 0 2 4 6 8 10 12 14 16
x (m) 0 30 95 215 400 510 600 670 720
(b)
1.19. Solve: A forgetful physics professor goes for a walk on a straight country road. Walking at a constant speed, he covers a distance of 300 m in 300 s. He then stops and watches the sunset for 100 s. Finding that it was getting dark, he walks faster back to his house covering the same distance in 200 s.
1.20. Solve: Forty miles into a car trip north from his home in El Dorado, an absent-minded English professor stopped at a rest area one Saturday. After staying there for one hour, he headed back home thinking that he was supposed to go on this trip on Sunday. Absent-mindedly he missed his exit and stopped after one hour of driving at another rest area 20 miles south of El Dorado. After waiting there for one hour, he drove back very slowly, confused and tired as he was, and reached El Dorado in two hours.
Visualize: The bicycle is moving with an acceleration of 1.5 m/s2. Thus, the velocity will increase by 1.5 m/s each second of motion.
1.21.
G Visualize: The particle moves upward with a constant acceleration a . The final velocity is 200 m/s and is reached at a height of 1000 m.
1.22.
⎛ 10−6 s ⎞ −6 Solve: (a) 9.12 μ s = (9.12 μs) ⎜ ⎟ = 9.12 × 10 s ⎝ 1 μs ⎠ ⎛ 103 m ⎞ 3 (b) 3.42 km = (3.42 km) ⎜ ⎟ = 3.42 × 10 m ⎝ 1 km ⎠
1.23.
−2 ⎛ cm ⎞ ⎛ 10 m ⎞ ⎛ 1 ms ⎞ 2 (c) 44 cm/ms = 44 ⎜ ⎟ ⎜ −3 ⎟ = 4.4 × 10 m/s ⎟⎜ ⎝ ms ⎠ ⎝ 1 cm ⎠ ⎝ 10 s ⎠ 3 ⎛ km ⎞ ⎛ 10 m ⎞ ⎛ 1 hour ⎞ (d) 80 km/hour = 80 ⎜ ⎟⎜ ⎟⎜ ⎟ = 22 m/s ⎝ hour ⎠ ⎝ 1 km ⎠ ⎝ 3600 s ⎠
1.24.
−2 ⎛ 2.54 cm ⎞ ⎛ 10 m ⎞ Solve: (a) 8.0 inches = 8.0 (inch) ⎜ ⎟ = 0.20 m ⎟⎜ ⎝ 1 inch ⎠ ⎝ 1 cm ⎠
1m ⎛ feet ⎞⎛ 12 inch ⎞⎛ ⎞ (b) 66 feet /s = 66 ⎜ ⎟⎜ ⎟⎜ ⎟ = 20 m/s ⎝ s ⎠⎝ 1 foot ⎠⎝ 39.37 inch ⎠ 3 ⎛ miles ⎞⎛ 1.609 km ⎞ ⎛ 10 m ⎞ ⎛ 1 hour ⎞ (c) 60 mph = 60 ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ = 27 m/s ⎝ hour ⎠⎝ 1 mile ⎠ ⎝ 1 km ⎠ ⎝ 3600 s ⎠
2
1m ⎛ ⎞ −3 (d) 14 square inches = 14 (inches) 2 ⎜ ⎟ = 9.0 × 10 square meter ⎝ 39.37 inches ⎠
1.25.
Solve: (a)
⎛ 3600 s ⎞ 3 1 hour = 1(hour) ⎜ ⎟ = 3600 s = 3.60 × 10 s 1 hour ⎝ ⎠
⎛ 24 hours ⎞ ⎛ 3600 s ⎞ 4 (b) 1 day = 1 (day) ⎜ ⎟⎜ ⎟ = 8.64 × 10 s ⎝ 1 day ⎠ ⎝ 1 hour ⎠ ⎛ 365.25 days ⎞ ⎛ 8.64 × 104 s ⎞ 7 (c) 1 year = 1 (year) ⎜ ⎟ = 3.16 × 10 s ⎟⎜ ⎝ 1 year ⎠⎝ 1 day ⎠ 1m ⎛ ft ⎞⎛ 12 inch ⎞⎛ ⎞ 2 (d) 32 ft /s 2 = 32 ⎜ 2 ⎟⎜ ⎟⎜ ⎟ = 9.75 m/s ⎝ s ⎠⎝ 1 ft ⎠⎝ 39.37 inch ⎠
1.26.
Solve: (a)
⎛1 m ⎞ 20 ft = 20 ( ft ) ⎜ ⎟ = 7.0 m ⎝ 3 ft ⎠
⎛ 1 km ⎞⎛ 1000 m ⎞ 5 (b) 60 miles = 60(miles) ⎜ ⎟⎜ ⎟ = 1.0 × 10 m ⎝ 0.6 miles ⎠⎝ 1 km ⎠
⎛ 1 m/s ⎞ (c) 60 mph = 60(mph) ⎜ ⎟ = 30 m/s ⎝ 2 mph ⎠
⎛ 1 cm ⎞ ⎛ 10−2 m ⎞ (d) 8 in = 8(in) ⎜ ⎟⎜ ⎟ = 0.16 m ⎝ 1/2 in ⎠⎝ 1 cm ⎠
1.27.
Solve: ⎛ 4 in ⎞ (a) (30 cm ) ⎜ ⎟ = 12 in ⎝ 10 cm ⎠
⎛ 2 mph ⎞ (b) ( 25 m/s ) ⎜ ⎟ = 50 mph ⎝ 1 m/s ⎠ ⎛ 0.6 mi ⎞ (c) (5 km ) ⎜ ⎟ = 3 mi ⎝ 1 km ⎠ ⎛1 ⎞ in ⎛1 ⎞⎜ 2 ⎟ 1 (d) ⎜ cm ⎟ ⎜ ⎟ = in ⎝2 ⎠ ⎝ 1 cm ⎠ 4
Solve: (a) 33.3 × 25.4 = 846 (b) 33.3 − 25.4 = 7.9
1.28.
(c) 33.3 = 5.77 (d) 333.3 ÷ 25.4 = 13.1
1.29.
Solve: (a) (33.3) 2 = 1.109 × 103. For numbers starting with “1” an extra digit is kept.
(b) 33.3 × 45.1 = 1.50 × 103 Scientific notation is an easy way to establish significance. (c) 22.2 − 1.2 = 3.5 (d) 1/ 44.4 = 0.0225
1.30.
Solve: The length of a typical car is 15 ft. Or
1m ⎛ 12 inch ⎞⎛ ⎞ 15(ft) ⎜ ⎟⎜ ⎟ = 4.6 m 1 ft 39.37 inch ⎝ ⎠⎝ ⎠ This length of 15 ft is approximately two-and-a-half times my height.
1.31.
Solve: The height of a telephone pole is estimated to be around 50 ft or 15 m. This height is approximately 8 times my height.
1.32.
Solve: I typically take 15 minutes in my car to cover a distance of approximately 6 miles from home to campus. My average speed is ⎛ 0.447 m/s ⎞ 6 miles ⎛ 60 min ⎞ ⎟ = 11 m/s ⎜ ⎟ = 24 mph = 24(mph) ⎜ 15 min ⎝ 1 hour ⎠ ⎝ 1 mph ⎠
1.33.
Solve: My barber trims about an inch of hair when I visit him every month for a haircut. The rate of hair growth is 1(inch) ⎛ 2.54 cm ⎞ ⎛ 10−2 m ⎞⎛ 1 month ⎞ ⎛ 1 day ⎞⎛ 1 h ⎞ −9 ⎟⎜ ⎟⎜ ⎜ ⎟⎜ ⎟⎜ ⎟ = 9.8 × 10 m/s (month) ⎝ 1 inch ⎠ ⎝ 1 cm ⎠ ⎝ 30 days ⎠ ⎝ 24 h ⎠⎝ 3600 s ⎠ 6 ⎛ m ⎞ ⎛ 10 μ m ⎞ ⎛ 3600 s ⎞ = 9.8 × 10−9 ⎜ ⎟ ⎜ ⎟⎜ ⎟ = 35 μ m/h ⎝ s ⎠⎝ 1 m ⎠⎝ 1 h ⎠
1.34.
Model: Visualize:
Represent the Porsche as a particle for the motion diagram.
1.35.
Model: Visualize:
Represent the watermelon as a particle for the motion diagram.
1.36. Model: Visualize:
Represent (Sam + car) as a particle for the motion diagram.
1.37.
Model: Visualize:
Represent the speed skater as a particle for the motion diagram.
1.38.
Model: Visualize:
Represent the wad as a particle for the motion diagram.
1.39.
Model: Visualize:
Represent the ball as a particle for the motion diagram.
1.40.
Model: Visualize:
Represent the ball as a particle for the motion diagram.
1.41.
Model: Visualize:
Represent the motorist as a particle for the motion diagram.
1.42.
Model: Visualize:
Represent Bruce and the puck as particles for the motion diagram.
1.43.
Model: Visualize:
Represent Fred and yourself as particles for the motion diagram.
1.44.
Solve: Rahul was coasting on interstate highway I-35 from Wichita to Kansas City at 65 mph. Seeing an accident at a distance of 200 feet in front of him, he braked his car to a stop with steady deceleration.
1.45. Solve: A car starts coasting at an initial speed of 30.0 m/s up a 10° incline. 230 m up the incline the road levels out to a flat road and the car continues coasting at a reduced speed along the road.
1.46.
Solve: A skier starts from rest down a 25° slope with very little friction. At the bottom of the 100 m slope the skier moves to a flat area and continues at constant velocity.
1.47.
Solve: A ball is dropped from a height to check its rebound properties. It rebounds to 80% of its original height.
1.48.
Solve: Two boards lean against each other at equal angles to the vertical direction. A ball rolls up the incline, over the peak, and down the other side.
1.49.
Solve:
(a)
(b) Sue passes 3rd Street doing 40 mph, slows steadily to the stop sign at 4th Street, stops for 1 s, then speeds up and reaches her original speed as she passes 5th Street. If the blocks are 50 m long, how long does it take Sue to drive from 3rd Street to 5th Street? (c)
1.50.
Solve:
(a)
(b) A train moving at 100 km/hour slows down in 10 s to a speed of 60 km/hour as it enters a tunnel. The driver maintains this constant speed for the entire length of the tunnel that takes the train a time of 20 s to traverse. Find the length of the tunnel. (c)
1.51. Solve: (a)
(b) Jeremy has perfected the art of steady acceleration and deceleration. From a speed of 60 mph he brakes his car to rest in 10 seconds with a constant deceleration. Then he turns into an adjoining street. Starting from rest, Jeremy accelerates with exactly the same magnitude as his earlier deceleration and reaches the same speed of 60 mph over the same distance in exactly the same time. Find the car’s acceleration or deceleration. (c)
1.52. Solve: (a)
(b) A coyote (A) sees a rabbit and begins to run toward it with an acceleration of 3.0 m/s2. At the same instant, the rabbit (B) begins to run away from the coyote with an acceleration of 2.0 m/s2. The coyote catches the rabbit after running 40 m. How far away was the rabbit when the coyote first saw it? (c)
1.53. Solve: Since area equals length × width, the smallest area will correspond to the smaller length and the smaller width. Similarly, the largest area will correspond to the larger length and the larger width. Therefore, the smallest area is (64 m)(100 m) = 6.4 × 103 m2 and the largest area is (75 m)(110 m) = 8.3 × 103 m2.
1.54.
Solve: (a) We need kg/m3. There are 100 cm in 1 m. If we multiply by 3
⎛ 100 cm ⎞ 3 ⎜ ⎟ = (1) ⎝ 1m ⎠ we do not change the size of the quantity, but only the number in terms of the new unit. Thus, the mass density of aluminum is 3
⎛ kg ⎞⎛ 100 cm ⎞ 3 kg 2.7 × 10−3 ⎜ 3 ⎟⎜ ⎟ = 2.7 × 10 3 m ⎝ cm ⎠⎝ 1 m ⎠ (b) Likewise, the mass density of alcohol is
⎛ g ⎞⎛ 100 cm ⎞ 0.81⎜ 3 ⎟⎜ ⎟ ⎝ cm ⎠⎝ 1 m ⎠
3
⎛ 1 kg ⎞ kg ⎜ ⎟ = 810 3 1000 g m ⎝ ⎠
1.55.
Model:
The car is represented by the particle model as a dot.
Solve: (a) Time t (s) 0 10 20 30 40 50 60 70 80 90
Position x (m) 1200 975 825 750 700 650 600 500 300 0
(b)
1.56. Solve: Susan enters a classroom, sees a seat 40 m directly ahead, and begins walking toward it at a constant leisurely pace, covering the first 10 m in 10 seconds. But then Susan notices that Ella is heading toward the same seat, so Susan walks more quickly to cover the remaining 30 m in another 20 seconds, beating Ella to the seat. Susan stands next to the seat for 10 seconds to remove her backpack.
1.57. Solve: A crane operator holds a ton of bricks 30 m above the ground. Four seconds after he is told to lower the bricks, he takes four seconds to lower them 15 m at a constant rate before stopping the bricks to make an eight-second safety check. He then continues lowering the bricks the remaining 15 m, taking four more seconds.
1-1
2.1. Model: We will consider the car to be a particle that occupies a single point in space. Visualize:
Solve:
Since the velocity is constant, we have xf = xi + vx Δt. Using the above values, we get
x1 = 0 m + (10 m/s)(45 s) = 450 m Assess: 10 m/s ≈ 22 mph and implies a speed of 0.4 miles per minute. A displacement of 450 m in 45 s is reasonable and expected.
2.2.
Model: Visualize:
Solve:
We will consider Larry to be a particle.
Since Larry’s speed is constant, we can use the following equation to calculate the velocities:
vs =
sf − si tf − ti
(a) For the interval from the house to the lamppost:
v1 =
200 yd − 600 yd = −200 yd/min 9:07 − 9:05
For the interval from the lamppost to the tree: 1200 yd − 200 yd v2 = = +333 yd/min 9:10 − 9:07 (b) For the average velocity for the entire run: 1200 yd − 600 yd vavg = = +120 yd/min 9:10 − 9:05
2.3.
Model: Visualize:
Solve:
Cars will be treated by the particle model.
Beth and Alan are moving at a constant speed, so we can calculate the time of arrival as follows: Δx x1 − x0 x −x v= = ⇒ t1 = t0 + 1 0 Δt t1 − t0 v
Using the known values identified in the pictorial representation, we find: x −x 400 mile tAlan 1 = tAlan 0 + Alan 1 Alan 0 = 8:00 AM + = 8:00 AM + 8 hr = 4:00 PM v 50 miles/hour x −x 400 mile tBeth 1 = tBeth 0 + Beth 1 Beth 0 = 9:00 AM + = 9:00 AM + 6.67 hr = 3:40 PM v 60 miles/hour (a) Beth arrives first. (b) Beth has to wait tAlan 1 − tBeth 1 = 20 minutes for Alan. Assess: Times of the order of 7 or 8 hours are reasonable in the present problem.
Solve: (a) The time for each segment is Δt1 = 50 mi/40 mph = 5/4 hr and Δt2 = 50 mi/60 mph = 5/6hr. The average speed to the house is 100 mi = 48 mph 5/6 h + 5/4 h
2.4.
(b) Julie drives the distance Δx1 in time Δt1 at 40 mph. She then drives the distance Δx2 in time Δt2 at 60 mph. She spends the same amount of time at each speed, thus Δt1 = Δt2 ⇒ Δx1 /40 mph = Δx2 /60 mph ⇒ Δx1 = (2/3)Δx2
But Δx1 + Δx2 = 100 miles, so (2 / 3)Δx2 + Δx2 = 100 miles. This means Δx2 = 60 miles and Δx1 = 40 miles. Thus, the times spent at each speed are Δt1 = 40 mi/40 mph = 1.00 h and Δt2 = 60 mi/60 mph = 1.00 h. The total time for her return trip is Δt1 + Δt2 = 2.00 h. So, her average speed is 100 mi/2 h = 50 mph.
2.5.
Model: The bicyclist is a particle. Visualize: Please refer to Figure EX2.5. Solve: The slope of the position-versus-time graph at every point gives the velocity at that point. The slope at t = 10 s is Δs 100 m − 50 m v= = = 2.5 m/s 20 s Δt The slope at t = 25 s is v=
100 m − 100 m = 0 m/s 10 s
v=
0 m − 100 m = −10 m/s 10 s
The slope at t = 35 s is
2.6. Solve:
Visualize: Please refer to Figure EX2.6. (a) We can obtain the values for the velocity-versus-time graph from the equation v = Δs/Δt.
(b) There is only one turning point. At t = 3 s the velocity changes from +5 m/s to −20 m/s, thus reversing the direction of motion. At t = 1 s, there is an abrupt change in motion from rest to +5 m/s, but there is no reversal in motion.
Visualize: Please refer to Figure EX2.7. The particle starts at x0 = 10 m at t0 = 0. Its velocity is initially in the –x direction. The speed decreases as time increases during the first second, is zero at t = 1 s, and then increases after the particle reverses direction. Solve: (a) The particle reverses direction at t = 1 s, when vx changes sign. (b) Using the equation xf = x0 + area of the velocity graph between t1 and tf ,
2.7.
x2 s = 10 m − (area of triangle between 0 s and 1 s) + (area of triangle between 1 s and 2 s) 1 1 (4 m/s)(1 s) + (4 m/s)(1 s) = 10 m 2 2 x3 s = 10 m + area of trapazoid between 2 s and 3 s = 10 m −
1 = 10 m + (4 m/s + 8 m/s)(3 s − 2 s) = 16 m 2 x4 s = x3 s + area between 3 s and 4 s 1 = 16 m + (8 m/s + 12 m/s)(1 s) = 26 m 2
2.8. Visualize: Please refer to Figure EX2.8. Solve: A constant velocity from t = 0 s to t = 2 s means zero acceleration. On the other hand, a linear increase in velocity between t = 2 s and t = 4 s implies a constant positive acceleration.
2.9. Visualize: Please refer to Figure EX2.9. Solve: (a) The acceleration of the train at t = 3.0 s is the slope of the v vs t graph at t = 3 s. Thus a = ( 2 m/s − ( − 2 m/s) ) (8 s ) = 0.5 m/s 2 . (b)
2.10. Solve:
Visualize: Please refer to Figure EX2.10. (a) At t = 2.0 s, the position of the particle is
x2 s = 2.0 m + area under velocity graph from t = 0 s to t = 2.0 s 1 = 2.0 m + (4 m/s)(2.0 s) = 6 m 2 (b) From the graph itself at t = 2.0 s, v = 4 m/s. (c) The acceleration is ax =
Δvx vfx − vix 6 m/s − 0 m/s = = = 2 m/s 2 3s Δt Δt
2.11. Solve:
Visualize: Please refer to Figure EX2.11. (a) Using the equation
xf = xi + area under the velocity-versus-time graph between ti and tf we have
x (at t = 1 s) = x (at t = 0 s) + area between t = 0 s and t = 1 s
= 2.0 m + (4 m/s)(1 s) = 6 m
Reading from the velocity-versus-time graph, vx (at t = 1 s) = 4 m/s. Also, ax = slope = Δv/Δt = 0 m/s 2 . (b) x (at t = 3.0 s) = x(at t = 0 s) + area between t = 0 s and t = 3 s = 2.0 m + 4 m/s × 2 s + 2 m/s × 1 s + (1/2) × 2 m/s × 1 s = 13.0 m
Reading from the graph, vx (t = 3 s) = 2 m/s. The acceleration is ax (t = 3 s) = slope =
vx (at t = 4 s) − vx (at t = 2 s) = −2 m/s 2 2s
2.12.
Model: Visualize:
Solve:
Represent the jet plane as a particle.
(a) Since we don’t know the time of acceleration, we will use
v12 = v02 + 2a ( x1 − x0 ) ⇒a=
v12 − v02 (400 m/s) 2 − (300 m/s) 2 = = 8.75 m/s 2 2 x1 2(4000 m)
(b) The acceleration of the jet is approximately equal to g, the acceleration due to gravity.
2.13.
Model: We are using the particle model for the skater and the kinematics model of motion under constant acceleration. Solve: Since we don’t know the time of acceleration we will use
vf2 = vi2 + 2a ( xf − xi ) ⇒a= Assess:
v −v (6.0 m/s) 2 − (8.0 m/s) 2 = = −2.8 m/s 2 2( xf − xi ) 2(5.0 m) 2 f
2 i
A deceleration of 2.8 m/s2 is reasonable.
2.14.
Model: Visualize:
Solve:
We are assuming both cars are particles.
The Porsche’s time to finish the race is determined from the position equation
1 xP1 = xP0 + vP0 (tP1 − tP0 ) + aP (tP1 − tP0 ) 2 2 1 ⇒ 400 m = 0 m + 0 m + (3.5 m/s 2 )(tP1 − 0 s)2 ⇒ tP1 = 15.1 s 2 The Honda’s time to finish the race is obtained from Honda’s position equation as
1 xH1 = xH0 + vH0 (tH1 − tH0 ) + aH0 (tH1 − tH0 ) 2 2 1 400 m = 50 m + 0 m + (3.0 m/s 2 )(tH1 − 0 s)2 ⇒ tH1 = 15.3 s 2 So, the Porsche wins.
2.15.
Model: Visualize:
Solve:
Represent the spherical drop of molten metal as a particle.
(a) The shot is in free fall, so we can use free fall kinematics with a = − g . The height must be such that
the shot takes 4 s to fall, so we choose t1 = 4 s. Then,
1 1 1 y1 = y0 + v0 (t1 − t0 ) − g (t1 − t0 ) 2 ⇒ y0 = gt12 = (9.8 m/s 2 )(4 s) 2 = 78.4 m 2 2 2 (b) The impact velocity is v1 = v0 − g (t1 − t0 ) = − gt1 = −39.2 m/s. G Assess: Note the minus sign. The question asked for velocity, not speed, and the y-component of v is negative because the vector points downward.
2.16.
Model: Visualize:
Solve:
Assume the ball undergoes free-fall acceleration and that the ball is a particle.
(a) We will use the kinematic equations v = v0 + a (t − t0 ) and
1 y = y0 + v0 (t − t0 ) + a (t − t0 ) 2 2
as follows:
v(at t = 1.0 s) = 19.6 m/s + (−9.8 m/s 2 )(1.0 s − 0 s) = 9.8 m/s y (at t = 1.0 s) = 0 m + (19.6 m/s)(1.0 s − 0 s) + 1/2(−9.8 m/s 2 )(1.0 s − 0 s) 2 = 14.7 m
v(at t = 2.0 s) = 19.6 m/s + (−9.8 m/s 2 )(2.0 s − 0 s) = 0 m/s y (at t = 2.0 s) = 0 m + (19.6 m/s)(2.0 s − 0 s) + 1/2(−9.8 m/s 2 )(2.0 s − 0 s) 2 = 19.6 m v(at t = 3.0 s) = 19.6 m/s + (−9.8 m/s 2 )(3 s − 0 s) = 9.8 m/s
y (at t = 3.0 s) = 0 m + (19.6 m/s)(3.0 s − 0 s) + 1/2(−9.8 m/s 2 )(3.0 s − 0 s)2 = 14.7 m v(at t = 4.0 s) = 19.6 m/s + (−9.8 m/s 2 )(4.0 s − 0 s) = −19.6 m/s y (at t = 4.0 s) = 0 m + (19.6 m/s)(4.0 s − 0 s) + 1/2(−9.8 m/s 2 )(4.0 s − 0 s) 2 = 0 m (b)
Assess: (a) A downward acceleration of 9.8 m/s 2 on a particle that has been given an initial upward velocity of +19.6 m/s will reduce its speed to 9.8 m/s after 1 s and then to zero after 2 s. The answers obtained in this solution are consistent with the above logic.
(b) Velocity changes linearly with a negative uniform acceleration of 9.8 m/s 2 . The position is symmetrical in time around the highest point which occurs at t = 2 s.
2.17.
Model: Visualize:
We represent the ball as a particle.
Solve: Once the ball leaves the student’s hand, the ball undergoes free fall and its acceleration is equal to the acceleration due to gravity that always acts vertically downward toward the center of the earth. According to the constant-acceleration kinematic equations of motion 1 y1 = y0 + v0 Δt + a Δt 2 2
Substituting the known values
−2 m = 0 m + (15 m/s)t1 + (1/2)(−9.8 m/s 2 )t12 The solution of this quadratic equation gives t1 = 3.2 s. The other root of this equation yields a negative value for t1 , which is not valid for this problem. Assess: A time of 3.2 s is reasonable.
2.18.
Model: Visualize:
Solve:
We will use the particle model and the constant-acceleration kinematic equations.
(a) Substituting the known values into y1 = y0 + v0 Δt + 12 a Δt 2 , we get
1 −10 m = 0 m + 20 (m/s)t1 + (−9.8 m/s 2 )t12 2 One of the roots of this equation is negative and is not relevant physically. The other root is t1 = 4.53 s, which is
the answer to part (b). Using v1 = v0 + aΔt , we obtain
v1 = 20(m/s) + (−9.8 m/s 2 )(4.53 s) = −24 m/s (b) The time is 4.5 s. Assess: A time of 4.5 s is a reasonable value. The rock’s velocity as it hits the bottom of the hole has a negative sign because of its downward direction. The magnitude of 24 m/s compared to 20 m/s, when the rock was tossed up, is consistent with the fact that the rock travels an additional distance of 10 m into the hole.
2.19.
Model: Visualize:
We will represent the skier as a particle.
Note that the skier’s motion on the horizontal, frictionless snow is not of any interest to us. Also note that the acceleration parallel to the incline is equal to g sin 10°. Solve: Using the following constant-acceleration kinematic equations,
vf2x = vi2x + 2ax ( xf − xi ) ⇒ (15 m / s) 2 = (3.0 m / s) 2 + 2(9.8 m / s 2 )sin10°( x1 − 0 m) ⇒ x1 = 64 m vfx = vix + ax (tf − ti ) ⇒ (15 m / s) = (3.0 m / s) + (9.8 m / s 2 )(sin10°)t ⇒ t = 7.1 s Assess:
A time of 7.1 s to cover 64 m is a reasonable value.
2.20.
Model: Visualize:
Represent the car as a particle.
Solve: Note that the problem “ends” at a turning point, where the car has an instantaneous speed of 0 m/s before rolling back down. The rolling back motion is not part of this problem. If we assume the car rolls without friction, then we have motion on a frictionless inclined plane with an accleration a = − g sin θ = − g sin 20° = −3.35 m/s 2 . Constant acceleration kinematics gives v12 = v02 + 2a ( x1 − x0 ) ⇒ 0 m 2 /s 2 = v02 + 2ax1 ⇒ x1 = −
v02 (30 m/s) 2 =− = 134 m 2a 2( −3.35 m/s 2 )
Notice how the two negatives canceled to give a positive value for x1. G Assess: We must include the minus sign because the a vector points down the slope, which is in the negative x-direction.
Solve: (a) The position t = 2 s is x2 s = [2(2) 2 − 2 + 1] m = 7 m. (b) The velocity is the derivative v = dx / dt and the velocity at t = 2 s is calculated as follows:
2.21.
v = (4t 2 − 1) m/s ⇒ v2 s = [4(2) − 1] m/s = 7 m/s (c) The acceleration is the derivative a = dv / dt and the acceleration at t = 2 s is calculated as follows:
a = (4) m/s 2 ⇒ a2 s = 4 m/s 2
2.22.
Solve: The formula for the particle’s position along the x-axis is given by tf
xf = xi + ∫ vx dt ti
Using the expression for vx we get 2 xf = xi + ⎡⎣tf3 − ti3 ⎤⎦ 3
ax =
dvx d = (2t 2 m/s) = 4t m/s 2 dt dt
(a) The particle’s position at t = 1 s is x1 s = 1 m + 23 m = 53 m. (b) The particle’s speed at t = 1 s is v1 s = 2 m/s. (c) The particle’s acceleration at t = 1 s is a1 s = 4 m/s 2 .
2.23.
Solve: The formula for the particle’s velocity is given by
vf = vi + area under the acceleration curve between ti and tf For t = 4 s, we get 1 v4 s = 8 m/s + (4 m/s 2 )4 s = 16 m/s 2 Assess: The acceleration is positive but decreases as a function of time. The initial velocity of 8.0 m/s will therefore increase. A value of 16 m/s is reasonable.
2.24.
Solve: (a) Time (s) 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
Position (m) −4 −2 0 1.75 3 4 5 6 7
(b)
(c) Δs = s (at t = 1 s) − s (at t = 0 s) = 0 m − (−4 m) = 4 m. (d) Δs = s (at t = 4 s) − s (at t = 2 s) = 7 m − 3 m = 4 m. (e) From t = 0 s to t = 1 s, vs = Δs / Δt = 4 m/s. (f) From t = 2 s to t = 4 s, vs = Δs / Δt = 2 m/s. (g) The average acceleration is a=
Δv 2 m/s − 4 m/s = = −2 m/s 2 Δt 2 s −1 s
2.25.
Solve: (a)
(b) To be completed by student. dx = vx = 2t − 4 ⇒ vx (at t = 1 s) = [2 m/s 2 (1 s) − 4 m/s] = −2 m/s (c) dt (d) There is a turning point at t = 2 s. (e) Using the equation in part (c), vx = 4 m/s = (2t − 4) m/s ⇒ t = 4
Since x = (t 2 − 4t + 2) m, x = 2 m.
(f)
2.26.
Visualize: Please refer to Figure P2.26. Solve: The graph for particle A is a straight line from t = 2 s to t = 8 s. The slope of this line is −10 m/s, which is the velocity at t = 7.0 s. The negative sign indicates motion toward lower values on the x-axis. The velocity of particle B at t = 7.0 s can be read directly from its graph. It is −20 m/s, The velocity of particle C can be obtained from the equation vf = vi + area under the acceleration curve between ti and tf
This area can be calculated by adding up three sections. The area between t = 0 s and t = 2 s is 40 m/s, the area between t = 2 s and t = 5 s is 45 m/s, and the area between t = 5 s and t = 7 s is −20 m/s. We get (10 m/s) + (40 m/s) + (45 m/s) − (20 m/s) = 75 m/s.
2.27. Solve:
Visualize: Please refer to Figure P2.27. (a) We can calculate the position of the particle at every instant with the equation
xf = xi + area under the velocity-versus-time graph between ti and tf The particle starts from the origin at t = 0 s, so xi = 0 m. Notice that the each square of the grid in Figure P2.27 has “area” (5 m/s) × (2 s) = 10 m. We can find the area under the curve, and thus determine x, by counting squares. You can see that x = 35 m at t = 4 s because there are 3.5 squares under the curve. In addition, x = 35 m at t = 8 s because the 5 m represented by the half square between 4 and 6 s is cancelled by the –5 m represented by the half square between 6 and 8 s. Areas beneath the axis are negative areas. The particle passes through x = 35 m at t = 4 s and again at t = 8 s. (b) The particle moves to the right for 0 s ≥ t ≥ 6 s, where the velocity is positive. It reaches a turning point at x = 40 m at t = 6 s. The motion is to the left for t > 6 s. This is shown in the motion diagram below.
2.28. Visualize:
Solve: We will determine the object’s velocity using graphical methods first and then using calculus. Graphically, v(t ) = v0 + area under the acceleration curve from 0 to t. In this case, v0 = 0 m/s. The area at each time t requested is a triangle. t =0 s
v(t = 0 s) = v0 = 0 m/s
t=2s
1 v(t = 2 s) = (2 s)(5 m/s) = 5 m/s 2
t=4s
1 v(t = 4 s) = (4 s)(10 m/s) = 20 m/s 2
t =6 s
1 v(t = 6 s) = (6 s)(10 m/s) = 30 m/s 2
t =8 s
v(t = 8 s) = v (t = 6 s) = 30 m/s
The last result arises because there is no additional area after t = 6 s. Let us now use calculus. The acceleration function a(t) consists of three pieces and can be written: 0 s≤t ≤4 s ⎧ 2.5t ⎪ a (t ) = ⎨−5t + 30 4 s ≤ t ≤ 6 s ⎪ 0 6 s≤t ≤8 s ⎩ These were determined by the slope and the y-intercept of each of the segments of the graph. The velocity function is found by integration as follows: For 0 ≤ t ≤ 4 s, t
t
v(t ) = v(t = 0 s) + ∫ a (t )dt = 0 + 2.5 0
This gives
t2 = 1.25t 2 20
t =0 s
v(t = 0 s) = 0 m / s
t=2s
v(t = 2 s) = 5 m / s
t=4s
v(t = 4 s) = 20 m / s
For 4 s ≤ t ≤ 6 s, t
t ⎡ −5t 2 ⎤ v(t ) = v(t = 4 s) + ∫ a(t )dt = 20 m/s + ⎢ + 30t ⎥ = −2.5t 2 + 30t − 60 4 ⎣ 2 ⎦4
This gives:
t=6s
v(t = 6 s) = 30 m/s
For 6 s ≤ t ≤ 8 s, t
v(t ) = v(t = 6 s) + ∫ a (t )dt = 30 m/s + 0 m/s = 30 m/s 6
This gives: t=8s
v(t = 8 s) = 30 m/s
Assess: graphs.
The same velocities are found using calculus and graphs, but the graphical method is easier for simple
2.29.
Visualize: Please refer to Figure P2.29. Solve: (a) The velocity-versus-time graph is the derivative with respect to time of the distance-versus-time graph. The velocity is zero when the slope of the position-versus-time graph is zero, the velocity is most positive when the slope is most positive, and the velocity is most negative when the slope is most negative. The slope is zero at t = 0, 1 s, 2 s, 3 s, . . . ; the slope is most positive at t = 0.5 s, 2.5 s, . . ; and the slope is most negative at t = 1.5 s, 3.5 s, . . . (b)
2.30. Solve:
Visualize: Please refer to Figure P2.30. (a) We can determine the velocity as follows: vx = vx (at t = ti ) + area under the acceleration-versus-time graph from t = ti to tf
vx (at t = 4 s) = 0 m/s + (−1 m/s 2 )(4 s) = −4 m/s vx (at t = 8 s) = −4 m/s + (3 m/s 2 )(4 s) m/s = 8 m/s vx (at t = 10 s) = 8 m/s + (−2 m/s 2 )(4 s) = 4 m/s
(b) If vx (at t = 0 s) = 2.0 m/s, the entire velocity-versus-time graph will be displaced upward by 2.0 m/s.
2.31.
Solve: (a) The velocity-versus-time graph is given by the derivative with respect to time of the position function: vx =
dx = (6t 2 − 18t )m/s dt
For vx = 0 m/s, there are two solutions to the quadratic equation: t = 0 s and t = 3 s. (b) At the first of these solutions,
x(at t = 0 s) = 2(0 s)3 − 9(0 s) 2 + 12 = 12 m The acceleration is the derivative of the velocity function: ax =
dvx = (12t − 18) m/s 2 ⇒ a (at t = 0 s) = −18 m/s 2 dt
At the second solution,
x(at t = 3 s) = 2(3 s)3 − 9(3 s) 2 + 12 = −15 m
ax (at t = 3 s) = 12(3 s) − 18 = +18 m/s 2
2.32.
Model: Represent the object as a particle. Solve: (a) Known information: x0 = 0 m, v0 = 0 m/s, x1 = 40 m, v1 = 11 m/s, t1 = 5 s. If the acceleration is uniform (constant a), then the motion must satisfy the three equations 1 x1 = at12 ⇒ a = 3.20 m/s 2 2
v1 = at1 ⇒ a = 2.20 m/s 2
v12 = 2ax1 ⇒ a = 1.51 m/s 2
But each equation gives a different value of a. Thus the motion is not uniform acceleration. (b) We know two points on the velocity-versus-time graph, namely at t0 = 0 and t1 = 5 s. What shape does the function have between these two points? If the acceleration was uniform, which it’s not, then the graph would be a straight line. The area under the graph is the displacement Δx. From the figure you can see that Δx = 27.5 m for a straight-line graph. But we know that, in reality, Δx = 40 m. To get a larger Δx, the graph must bulge upward above the straight line. Thus the graph is curved, and it is concave downward.
2.33.
Solve: The position is the integral of the velocity. t1
t1
t1
t0
0
0
x1 = x0 + ∫ vx dt = x0 + ∫ kt 2 dt = x0 + 13 kt 3 = x0 + 13 kt13 We’re given that x0 = −9.0 m and that the particle is at x1 = 9.0 m at t1 = 3.0 s. Thus
9.0 m = ( −9.0 m ) + 13 k (3.0 s)3 = ( −9.0 m ) + k ( 9.0 s3 )
Solving for k gives
k = 2.0 m/s3 .
2.34.
Solve: (a) The velocity is the integral of the acceleration.
v1x = v0 x + ∫ ax dt = 0 m/s + ∫ (10 − t ) dt = (10t − 12 t 2 ) = 10t1 − 12 t12 t1
t1
t1
t0
0
0
The velocity is zero when
v1x = 0 m/s = (10t1 − 12 t12 ) = (10 − 12 t1 ) × t1 ⇒ t1 = 0 s or t1 = 20 s The first solution is the initial condition. Thus the particle’s velocity is again 0 m/s at t1 = 20 s. (b) Position is the integral of the velocity. At t1 = 20 s, and using x0 = 0 m at t0 = 0 s, the position is t1
20
t0
0
x1 = x0 + ∫ vx dt = 0 m + ∫
(10t − t ) dt = 5t 1 2 2
2 20 0
− 16 t 3
20 0
= 667 m
2.35.
Model: Represent the ball as a particle. Visualize: Please refer to Figure P2.35. Solve: In the first and third segments the acceleration as is zero. In the second segment the acceleration is negative and constant. This means the velocity vs will be constant in the first two segments and will decrease linearly in the third segment. Because the velocity is constant in the first and third segments, the position s will increase linearly. In the second segment, the position will increase parabolically rather than linearly because the velocity decreases linearly with time.
2.36.
Model: Represent the ball as a particle. Visualize: Please refer to Figure P2.36. The ball rolls down the first short track, then up the second short track, and then down the long track. s is the distance along the track measured from the left end (where s = 0). Label t = 0 at the beginning, that is, when the ball starts to roll down the first short track. Solve: Because the incline angle is the same, the magnitude of the acceleration is the same on all of the tracks.
Assess: Note that the derivative of the s versus t graph yields the vs versus t graph. And the derivative of the vs versus t graph gives rise to the as versus t graph.
2.37.
Model: Represent the ball as a particle. Visualize: Please refer to Figure P2.37. The ball moves to the right along the first track until it strikes the wall, which causes it to move to the left on a second track. The ball then descends on a third track until it reaches the fourth track, which is horizontal. Solve:
Assess: Note that the time derivative of the position graph yields the velocity graph, and the derivative of the velocity graph gives the acceleration graph.
2.38. Solve:
Visualize:
Please refer to Figure P2.38.
2.39. Solve:
Visualize:
Please refer to Figure P2.39.
2.40.
Visualize: Please refer to Figure P2.40. There are four frictionless tracks. Solve: For the first track, as is negative and constant and vs is decreasing linearly. This is consistent with a ball rolling up a straight track but not so far that vs goes to zero. For the second track, as is zero and vs is constant but greater than zero. This is consistent with a ball rolling on a horizontal track. For the third track, as is positive and constant and vs is increasing linearly. This is consistent with a ball rolling down a straight track. For the fourth track, as is zero and vs is constant. This is again consistent with a ball rolling on a horizontal track. The as on the first track has the same absolute value as the as on the third track. This means the slope of the first track up is the same as the slope of the third track down.
2.41.
Model: The plane is a particle and the constant-acceleration kinematic equations hold. Solve: (a) To convert 80 m/s to mph, we calculate 80 m/s × 1 mi/1609 m × 3600 s/h = 179 mph. (b) Using as = Δv / Δt , we have, as (t = 0 to t = 10 s) =
23 m/s − 0 m/s = 2.3 m/s 2 10 s − 0 s
as (t = 20 s to t = 30 s) =
69 m/s − 46 m/s = 2.3 m/s 2 30 s − 20 s
For all time intervals a is 2.3 m/s2. (c) Using kinematics as follows:
vfs = vis + a (tf − ti ) ⇒ 80 m/s = 0 m/s + (2.3 m/s 2 )(tf − 0 s) ⇒ tf = 35 s (d) Using the above values, we calculate the takeoff distance as follows:
1 1 sf = si + vis (tf − ti ) + as (tf − ti ) 2 = 0 m + (0 m/s)(35 s) + (2.3 m/s 2 )(35 s) 2 = 1410 m 2 2 For safety, the runway should be 3 ×1410 m = 4230 m or 2.6 mi. This is longer than the 2.4 mi long runway, so the takeoff is not safe.
2.42. Solve:
Model: (a)
The automobile is a particle.
The acceleration is not constant because the velocity-versus-time graph is not a straight line. (b) Acceleration is the slope of the velocity graph. You can use a straightedge to estimate the slope of the graph at t = 2 s and at t = 8 s. Alternatively, we can estimate the slope using the two data points on either side of 2 s and 8 s. Either way, you need the conversion factor 1 mph = 0.447 m/s from Table 1.4.
ax ( at 2 s ) ≈
vx ( at 4 s ) − vx ( at 0 s ) mph 0.447 m/s = 11.5 × = 5.1 m/s 2 4 s−0 s s 1 mph
ax ( at 8 s ) ≈
vx ( at 10 s ) − vx ( at 6 s ) mph 0.447 m/s = 4.5 × = 2.0 m/s 2 10 s − 6 s s 1 mph
(c) The displacement, or distance traveled, is
Δx = x ( at 10 s ) − x ( at 0 s ) = ∫
10 s 0s
vx dx = area under the velocity curve from 0 s to 10 s
We can approximate the area under the curve as the area of five rectangular steps, each with width Δt = 2 s and height equal to the average of the velocities at the beginning and end of each step. 0 s to 2 s
vavg = 14 mph = 6.26 m/s
area = 12.5 m
2 s to 4 s
vavg = 37 mph = 16.5 m/s
area = 33.0 m
4 s to 6 s 6 s to 8 s
vavg = 53 mph = 23.7 m/s
area = 47.4 m
vavg = 65 mph = 29.1 m/s
area = 58.2 m
8 s to 10 s
vavg = 74 mph = 33.1 m/s
area = 66.2 m
The total area under the curve is ≈ 217 m , so the distance traveled in 10 s is ≈ 217 m .
2.43. Solve:
Model: Represent the car as a particle. (a) First, we will convert units:
60
miles 1 hour 1610 m × × = 27 m/s hour 3600 s 1 mile
The motion is constant acceleration, so v1 = v0 + aΔt ⇒ a =
v1 − v0 (27 m/s − 0 m/s) = = 2.7 m/s 2 10 s Δt
(b) The fraction is a/g = 2.7 / 9.8 = 0.28. So a is 28% of g. (c) The distance is calculated as follows:
1 1 x1 = x0 + v0 Δt + a (Δt ) 2 = a (Δt ) 2 = 1.3 × 102 m = 4.3 × 102 feet 2 2
2.44.
Model: Represent the spaceship as a particle. Solve: (a) The known information is: x0 = 0 m, v0 = 0 m/s, t0 = 0 s, a = g = 9.8 m/s 2 , and v1 = 3.0 × 108 m/s. Constant acceleration kinematics gives v1 = v0 + aΔt ⇒ Δt = t1 =
v1 − v0 = 3.1 × 107 s a
The problem asks for the answer in days, so we need a conversion: t1 = (3.1× 107 s) ×
1 hour 1 day × = 3.6 × 102 days 3600 s 24 hour
(b) The distance traveled is 1 1 x1 − x0 = v0 Δt + a (Δt ) 2 = at12 = 4.6 × 1015 m 2 2
(c) The number of seconds in a year is
1 year = 365 days ×
24 hours 3600 s × = 3.15 × 107 s 1 day 1 hour
In one year light travels a distance
1 light year = (3.0 × 108m/s)(3.15 × 107 s) = 9.46 × 1015 m The distance traveled by the spaceship is 4.6 × 1015m/9.46 × 1015m = 0.49 of a light year. Assess:
Note that x1 gives “Where is it?” rather than “How far has it traveled?” “How far” is represented by
x1 − x0 . They happen to be the same number in this problem, but that isn’t always the case.
2.45.
Model: Visualize:
Solve:
The car is a particle, and constant-acceleration kinematic equations hold.
This is a two-part problem. During the reaction time, we can use
1 x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 2 = 0 m + (20 m/s)(0.50 s − 0 s) + 0 m = 10 m During deceleration,
v22 = v12 + 2a1 ( x2 − x1 )
0 = (20 m/s) 2 + 2(−6.0 m/s 2 )( x2 − 10 m) ⇒ x2 = 43 m
She has 50 m to stop, so she can stop in time.
2.46.
Model: Visualize:
Solve:
The car is a particle and constant-acceleration kinematic equations hold.
(a) This is a two-part problem. During the reaction time,
x1 = x0 + v0 (t1 − t0 ) + 1/2a0 (t1 − t0 ) 2 = 0 m + (20 m/s)(0.50 s − 0 s) + 0 m = 10 m After reacting, x2 − x1 = 110 m − 10 m = 100 m, that is, you are 100 m away from the intersection. (b) To stop successfully,
v22 = v12 + 2a1 ( x2 − x1 ) ⇒ (0 m/s) 2 = (20 m/s) 2 + 2a1 (100 m) ⇒ a1 = −2 m/s 2 (c) The time it takes to stop once the brakes are applied can be obtained as follows:
v2 = v1 + a1 (t2 − t1 ) ⇒ 0 m/s = 20 m/s + (−2 m/s 2 )(t2 − 0.50 s) ⇒ t2 = 11 s The total time to stop since the light turned red is 11.5 s.
2.47.
Model: Visualize:
Solve:
(a)
To
We will use the particle model and the constant-acceleration kinematic equations.
find
x2 ,
we
first
need
to
determine
x1.
Using
x1 = x0 + v0 (t1 − t0 ),
we
get
x1 = 0 m + (20 m/s)(0.50 s − 0 s) = 10 m. Now,
v22 = v12 + 2a1 ( x2 − x1 ) ⇒ 0 m 2 /s 2 = (20 m/s) 2 + 2(−10 m/s 2 )( x2 − 10 m) ⇒ x2 = 30 m The distance between you and the deer is ( x3 − x2 ) or (35 m − 30 m) = 5 m. (b) Let us find v0 max such that v2 = 0 m/s at x2 = x3 = 35 m. Using the following equation,
v22 − v02 max = 2a1 ( x2 − x1 ) ⇒ 0 m 2 /s 2 − v02 max = 2(−10 m/s 2 )(35 m − x1 ) Also, x1 = x0 + v0 max (t1 − t0 ) = v0 max (0.50 s − 0 s) = (0.50 s)v0 max . Substituting this expression for x1 in the above equation yields
−v02 max = (−20 m/s 2 )[35 m − (0.50 s) v0 max ] ⇒ v02 max + (10 m/s)v0 max − 700 m 2 /s 2 = 0 The solution of this quadratic equation yields v0 max = 22 m/s. (The other root is negative and unphysical for the present situation.) Assess: An increase of speed from 20 m/s to 22 m/s is very reasonable for the car to cover an additional distance of 5 m with a reaction time of 0.50 s and a deceleration of 10 m/s2.
2.48.
Model: Visualize:
The car is represented as a particle.
Solve: (a) This is a two-part problem. First, we need to use the information given to determine the acceleration during braking. Second, we need to use that acceleration to find the stopping distance for a different initial velocity. First, the car coasts at constant speed before braking:
x1 = x0 + v0 (t1 − t0 ) = v0t1 = (30 m/s)(0.5 s) = 15 m Then, the car brakes to a halt. Because we don’t know the time interval, use
v22 = 0 = v12 + 2a1 ( x2 − x1 ) v12 (30 m/s) 2 =− = −10 m/s 2 2( x2 − x1 ) 2(60 m − 15 m) G We used v1 = v0 = 30 m/s. Note the minus sign, because a1 points to the left. ⇒ a1 = −
We can repeat these steps now with v0 = 40 m/s. The coasting distance before braking is x1 = v0t1 = (40 m/s)(0.5 s) = 20 m
The position x2 after braking is found using
v22 = 0 = v12 + 2a1 ( x2 − x1 ) ⇒ x2 = x1 −
v12 (40 m/s) 2 = 20 m − = 100 m 2a1 2(−10 m/s 2 )
(b) The car coasts at a constant speed for 0.5 s, traveling 20 m. The graph will be a straight line with a slope of 40 m/s. For t ≥ 0.5 the graph will be a parabola until the car stops at t2.We can find t2 from
v2 = 0 = v1 + a1 (t2 − t1 ) ⇒ t2 = t1 −
v1 = 4.5 s a1
The parabola will reach zero slope (v = 0 m/s) at t = 4.5 s. This is enough information to draw the graph shown in the figure.
2.49.
Model: Visualize:
The rocket is represented as a particle.
Solve: (a) There are three parts to the motion. Both the second and third parts of the motion are free fall, with a = − g . The maximum altitude is y2.. In the acceleration phase:
1 1 1 y1 = y0 + v0 (t1 − t0 ) + a (t1 − t0 ) 2 = at12 = (30 m/s 2 )(30 s) 2 = 13,500 m 2 2 2 v1 = v0 + a (t1 − t0 ) = at1 = (30 m/s 2 )(30 s) = 900 m/s In the coasting phase,
v22 = 0 = v12 − 2 g ( y2 − y1 ) ⇒ y2 = y1 +
v12 (900 m/s) 2 = 13,500 m + = 54,800 m = 54.8 km 2g 2(9.8 m/s 2 )
The maximum altitude is 54.8 km (≈ 33 miles). (b) The rocket is in the air until time t3 . We already know t1 = 30 s. We can find t2 as follows:
v2 = 0 m/s = v1 − g (t2 − t1 ) ⇒ t2 = t1 +
v1 = 122 s g
Then t3 is found by considering the time needed to fall 54,800 m:
1 1 2 y2 y3 = 0 m = y2 + v2 (t3 − t2 ) − g (t3 − t2 ) 2 = y2 − g (t3 − t2 ) 2 ⇒ t3 = t2 + = 228 s 2 2 g (c) The velocity increases linearly, with a slope of 30 (m/s)/s, for 30 s to a maximum speed of 900 m/s. It then begins to decrease linearly with a slope of −9.8(m/s)/s. The velocity passes through zero (the turning point at
y2 ) at t2 = 122 s. The impact velocity at t3 = 228 s is calculated to be v3 = v2 − g (t3 − t2 ) = −1040 m/s.
Assess: In reality, friction due to air resistance would prevent the rocket from reaching such high speeds as it falls, and the acceleration upward would not be constant because the mass changes as the fuel is burned, but that is a more complicated problem.
2.50.
Model: Visualize:
Solve:
We will model the rocket as a particle. Air resistance will be neglected.
(a) Using the constant-acceleration kinematic equations,
v1 = v0 + a0 (t1 − t0 ) = 0 m/s + a0 (16 s − 0 s) = a0 (16 s) 1 1 y1 = y0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 = a0 (16 s − 0 s) 2 = a0 (128 s 2 ) 2 2 1 y2 = y1 + v1 (t2 − t1 ) + a1 (t2 − t1 ) 2 2 1 2 ⇒ 5100 m = 128 s a0 + 16 s a0 (20 s − 16 s) + ( − 9.8 m/s 2 )(20 s − 16 s) 2 ⇒ a0 = 27 m/s 2 2 (b) The rocket’s speed as it passes through a cloud 5100 m above the ground can be determined using the kinematic equation:
v2 = v1 + a1 (t2 − t1 ) = (16 s) a0 + (−9.8 m/s 2 )(4 s) = 4.0 × 102 m/s Assess: 400 m/s ≈ 900 mph, which would be the final speed of a rocket that has been accelerating for 20 s at a rate of approximately 20 m/s2 or 66 ft/s2.
2.51.
Model: equations. Visualize:
We will model the lead ball as a particle and use the constant-acceleration kinematic
Note that the particle undergoes free fall until it hits the water surface. Solve: The kinematics equation y1 = y0 + v0 (t1 − t0 ) + 12 a0 (t1 − t0 ) 2 becomes
1 −5.0 m = 0 m + 0 m + (−9.8 m/s 2 )(t1 − 0) 2 ⇒ t1 = 1.01 s 2 Now, once again,
1 y2 = y1 + v1 (t2 − t1 ) + a1 (t2 − t1 ) 2 2 ⇒ y2 − y1 = v1 (3.0 s − 1.01 s) + 0 m/s = 1.99 v1 v1 is easy to determine since the time t1 has been found. Using v1 = v0 + a0 (t1 − t0 ) , we get
v1 = 0 m/s − (9.8 m/s 2 )(1.01 s − 0 s) = − 9.898 m/s With this value for v1 , we go back to: y2 − y1 = 1.99v1 = (1.99)(−9.898 m/s) = −19.7 m y2 − y1 is the displacement of the lead ball in the lake and thus corresponds to the depth of the lake. The negative sign shows the direction of the displacement vector. Assess: A depth of about 60 ft for a lake is not unusual.
2.52.
Model: Visualize:
Solve:
The elevator is a particle moving under constant-acceleration kinematic equations.
(a) To calculate the distance to accelerate up:
(v1 ) 2 = v02 + 2a0 ( y0 − y0 ) ⇒ (5 m/s) 2 = (0 m/s) 2 + 2(1 m/s 2 )( y1 − 0 m) ⇒ y1 = 12.5 m (b) To calculate the time to accelerate up:
v1 = v0 + a0 (t1 − t0 ) ⇒ 5 m/s = 0 m/s + (1 m/s 2 )(t1 − 0 s) ⇒ t1 = 5 s To calculate the distance to decelerate at the top:
v32 = v22 + 2a2 ( y3 − y2 ) ⇒ (0 m/s) 2 = (5 m/s) 2 + 2(−1 m/s 2 )( y3 − y2 ) ⇒ y3 − y2 = 12.5 m To calculate the time to decelerate at the top: v3 = v2 + a2 (t3 − t2 ) ⇒ 0 m/s = 5 m/s + (−1 m/s 2 )(t3 − t2 ) ⇒ t3 − t2 = 5 s The distance moved up at 5 m/s is y2 − y1 = ( y3 − y0 ) − ( y3 − y2 ) − ( y1 − y0 ) = 200 m − 12.5 m − 12.5 m = 175 m The time to move up 175 m is given by 1 y2 − y1 = v1 (t2 − t1 ) + a1 (t2 − t1 ) 2 ⇒ 175 m = (5 m/s)(t2 − t1 ) ⇒ (t2 − t1 ) = 35 s 2 To total time to move to the top is (t1 − t0 ) + (t2 − t1 ) + (t3 − t2 ) = 5 s + 35 s + 5 s = 45 s Assess: To cover a distance of 200 m at 5 m/s (ignoring acceleration and deceleration times) will require a time of 40 s. This is comparable to the time of 45 s for the entire trip as obtained above.
2.53.
Model: Visualize:
The car is a particle moving under constant-acceleration kinematic equations.
Solve: This is a three-part problem. First the car accelerates, then it moves with a constant speed, and then it decelerates. First, the car accelerates:
v1 = v0 + a0 (t1 − t0 ) = 0 m/s + (4.0 m/s 2 )(6 s − 0 s) = 24 m/s 1 1 x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 = 0 m + (4.0 m/s 2 )(6 s − 0 s) 2 = 72 m 2 2 Second, the car moves at v1: 1 x2 − x1 = v1 (t2 − t1 ) + a1 (t2 − t1 ) 2 = (24 m/s)(8 s − 6 s) + 0 m = 48 m 2 Third, the car decelerates: v3 = v2 + a2 (t3 − t2 ) ⇒ 0 m/s = 24 m/s + (−3.0 m/s 2 )(t3 − t2 ) ⇒ (t3 − t2 ) = 8 s 1 1 x3 = x2 + v2 (t3 − t2 ) + a2 (t3 − t2 ) 2 ⇒ x3 − x2 = (24 m/s)(8 s) + (−3.0 m/s 2 )(8 s) 2 = 96 m 2 2 Thus, the total distance between stop signs is:
x3 − x0 = ( x3 − x2 ) + ( x2 − x1 ) + ( x1 − x0 ) = 96 m + 48m + 72 m = 216 m Assess: A distance of approximately 600 ft in a time of around 10 s with an acceleration/deceleration of the order of 7 mph/s is reasonable.
2.54.
Model: Visualize:
Solve:
The car is a particle moving under constant linear acceleration.
Using the kinematic equation for position:
1 1 x2 = x1 + v1 (t2 − t1 ) + a1 (t2 − t1 ) 2 ⇒ x1 + 30 m = x1 + v1 (5.0 s − 4.0 s) + (2 m/s 2 )(5.0 s − 4.0 s) 2 2 2 1 ⇒ 30 m = v1 (1.0 s) + (2 m/s 2 )(1.0 s) 2 ⇒ v1 = 29 m/s 2 And 4.0 seconds before:
v1 = v0 + a0 (t1 − t0 ) ⇒ 29 m/s = v0 + (2 m/s 2 )(4.0 s − 0 s) ⇒ v0 = 21 m/s Assess:
21 m/s ≈ 47 mph and is a reasonable value.
2.55.
Model: Santa is a particle moving under constant-acceleration kinematic equations. Visualize: Note that our x-axis is positioned along the incline.
Solve:
Using the following kinematic equation,
v12 = v02 + 2a|| ( x1 − x2 ) = (0 m/s) 2 + 2(4.9 m/s 2 )(10 m − 0 m) ⇒ v1 = 9.9 m/s Assess:
Santa’s speed of 20 mph as he reaches the edge is reasonable.
2.56.
Model: Visualize:
The cars are represented as particles.
Solve: (a) Ann and Carol start from different locations at different times and drive at different speeds. But at time t1 they have the same position. It is important in a problem such as this to express information in terms of positions (that is, coordinates) rather than distances. Each drives at a constant velocity, so using constant velocity kinematics gives xA1 = xA0 + vA (t1 − tA0 ) = vA (t1 − tA0 )
xC1 = xC0 + vC (t1 − tC0 ) = xC0 + vCt1
The critical piece of information is that Ann and Carol have the same position at t1 , so xA1 = xC1. Equating these two expressions, we can solve for the time t1 when Ann passes Carol: vA (t1 − tA0 ) = xC0 + vCt1
⇒ (vA − vC )t1 = xC0 + vAtA0
⇒ t1 =
xC0 + vAtA0 2.4 mi + (50 mph)(0.5 h) = = 2.0 h vA − vC 50 mph − 36 mph
(b) Their position is x1 = xA1 = xC1 = xC0 + vCt1 = 74 miles. (c) Note that Ann’s graph doesn’t start until t = 0.5 hours, but her graph has a steeper slope so it intersects Carol’s graph at t ≈ 2.0 hours.
2.57.
Model: Visualize:
We will use the particle model for the puck.
We can view this problem as two one-dimensional motion problems. The horizontal segments do not affect the motion because the speed does not change. So, the problem “starts” at the bottom of the uphill ramp and “ends” at the bottom of the downhill ramp. At the top of the ramp the speed does not change along the horizontal section. The final speed from the uphill roll (first problem) becomes the initial speed of the downhill roll (second problem). Because the axes point in different directions, we can avoid possible confusion by calling the downhill axis the z-axis and the downhill velocities u. The uphill axis as usual will be denoted by x and the uphill velocities as v. Note that the height information, h = 1 m, has to be transformed into information about positions along the two axes. Solve: (a) The uphill roll has a0 = − g sin 30D = −4.90 m/s 2 . The speed at the top is found from
v12 = v02 + 2a0 ( x1 − x0 )
⇒ v1 = v02 + 2a0 x1 = (5 m/s) 2 + 2(−4.90 m/s 2 )(2.00 m) = 2.32 m/s (b) The downward roll starts with velocity u1 = v1 = 2.32 m/s and a1 = + g sin 20D = 3.35 m/s 2 . Then,
u22 = u12 + 2a1 ( z2 − z1 ) = (2.32 m/s) 2 + 2(3.35 m/s 2 )(2.92 m − 0 m) ⇒ u2 = 5.00 m/s (c) The final speed is equal to the initial speed, so the percentage change is zero! Assess: This result may seem surprising, but can be more easily understood after we introduce the concept of energy. For now, imagine this is a one dimensional vertical problem. The total vertical change in height of the puck is zero. We have already seen how an object with an initial velocity upward has the same velocity in the opposite direction as it passes through that height going down.
2.58.
Model: Visualize:
Solve:
We will model the toy train as a particle.
Using kinematics, 1 x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 = 2 m + (2.0 m/s)(2.0 s − 0 s) + 0 m = 6.0 m 2
The acceleration can now be obtained as follows:
v22 = v12 + 2a1 ( x2 − x1 ) ⇒ 0 m 2 /s 2 = (2.0 m/s)2 + 2a1 (8.0 m − 6.0 m) ⇒ a1 = −1.0 m/s 2 Assess:
A deceleration of 1 m/s 2 in bringing the toy train to a halt over a distance of 2.0 m is reasonable.
2.59.
Model: Visualize:
Solve:
We will use the particle model and the kinematic equations at constant-acceleration.
To find x2 , let us use the kinematic equation
v22 = v12 + 2a1 ( x2 − x1 ) = (0 m/s) 2 = (50 m/s) 2 + 2(−10 m/s 2 )( x2 − x1 ) ⇒ x2 = x1 + 125 m Since the nail strip is at a distance of 150 m from the origin, we need to determine x1 : x1 = x0 + v0 (t1 − t0 ) = 0 m + (50 m/s)(0.60 s − 0.0 s) = 30 m Therefore, we can see that x2 = (30 + 125) m = 155 m. That is, he can’t stop within a distance of 150 m. He is in jail. Assess: Bob is driving at approximately 100 mph and the stopping distance is of the correct order of magnitude.
2.60.
Model: Visualize:
Solve:
We will use the particle model with constant-acceleration kinematic equations.
The acceleration, being the same along the incline, can be found as
v12 = v02 + 2a ( x1 − x0 ) ⇒ (4.0 m/s)2 = (5.0 m/s)2 + 2a (3.0 m − 0 m) ⇒ a = −1.5 m/s 2 We can also find the total time the puck takes to come to a halt as v2 = v0 + a(t2 − t0 ) ⇒ 0 m/s = (5.0 m/s) + (−1.5 m/s 2 )t2 ⇒ t2 = 3.3 s Using the above obtained values of a and t2 , we can find x2 as follows: 1 1 x2 = x0 + v0 (t2 − t0 ) + a (t2 − t0 ) 2 = 0 m + (5.0 m/s)(3.3 s) + (−1.5 m/s 2 )(3.3 s) 2 = 8.3 m 2 2 That is, the puck goes through a displacement of 8.3 m. Since the end of the ramp is 8.5 m from the starting position x0 and the puck stops 0.2 m or 20 cm before the ramp ends, you are not a winner.
2.61.
Model: Visualize:
We will use the particle model for the skier’s motion ignoring air resistance.
Solve: (a) As discussed in the text, acceleration along a frictionless incline is a = g sin 25°, where g is the acceleration due to gravity. The acceleration of the skier on snow therefore is a = (0.90) g sin 25°. Also since
h/x1 = sin 25°, x = 200 m/ sin 25°. The final velocity can now be determined using kinematics ⎛ 200 m ⎞ v12 = v02 + 2a ( x1 − x0 ) = (0 m/s) 2 + 2(0.90)(9.8 m/s 2 )sin 25° ⎜ − 0 m⎟ sin 25 ° ⎝ ⎠ ⎛ 1 km ⎞⎛ 3600 s ⎞ ⇒ v1 = 59.397 m/s = (59.397 m/s) ⎜ ⎟⎜ ⎟ = 214 km/h ⎝ 1000 m ⎠⎝ 1 h ⎠ (b) The speed lost to air resistance is (214 − 180)/214 × 100% = 16%. Assess: A record of 180 km/h on such a slope in the presence of air resistance makes the obtained speed of 214 km/h (without air resistance) physical and reasonable.
2.62.
Model: We will use the particle model for the motion of the rocks, which move according to constantacceleration kinematic equations. Visualize:
Solve:
(a) For Heather,
1 yH1 = yH0 + vH0 (tH1 − tH0 ) + a0 (tH1 − tH0 ) 2 2 1 ⇒ 0 m = (50 m) + (−20 m/s)(tH1 − 0 s) + (−9.8 m/s 2 )(tH1 − 0 s) 2 2 2 2 ⇒ 4.9 m/s tH1 + 20 m/s tH1 − 50 m = 0 The two mathematical solutions of this equation are −5.83 s and +1.75 s. The first value is not physically acceptable since it represents a rock hitting the water before it was thrown, therefore, tH1 = 1.75 s. For Jerry, 1 yJ1 = yJ0 + vJ0 (tJ1 − tJ0 ) + a0 (tJ1 − tJ0 ) 2 2 1 ⇒ 0 m = (50 m) + (+20 m/s)(tJ1 − 0 s) + (−9.8 m/s 2 )(tJ1 − 0 s) 2 2 Solving this quadratic equation will yield tJ1 = −1.75 s and +5.83 s. Again only the positive root is physically meaningful. The elapsed time between the two splashes is tJ1 − tH1 = 5.83 s − 1.75 s = 4.08 s ≈ 4.1 s. (b) Knowing the times, it is easy to find the impact velocities:
vH1 = vH0 + a0 (tH1 − tH0 ) = (−20 m/s) + (−9.8 m/s)(1.75 s − 0 s) = −37.1 m/s vJ1 = vJ0 + a0 (tJ1 − tJ0 ) = (+20 m/s) + (−9.8 m/s 2 )(5.83 s − 0 s) = −37.1 m/s Assess: The two rocks hit water with equal speeds. This is because Jerry’s rock has the same downward speed as Heather’s rock when it reaches Heather’s starting position during its downward motion.
2.63.
Model: The ball is a particle that exhibits freely falling motion according to the constant-acceleration kinematic equations. Visualize:
Solve: Using the known values, we have v12 = v02 + 2a0 ( y1 − y0 ) ⇒ (−10 m/s) 2 = v02 + 2(−9.8 m/s 2 )(5.0 m − 0 m) ⇒ v0 = 14 m/s
2.64.
Model: Visualize:
Solve:
The car is a particle that moves with constant linear acceleration.
The reaction time is 1.0 s, and the motion during this time is
x1 = x0 + v0 (t1 − t0 ) = 0 m + (20 m/s)(1.0 s) = 20 m During slowing down, 1 x2 = x1 + v1 (t2 − t1 ) + a1 (t2 − t1 ) 2 = 200 m 2 1 = 20 m + (20 m/s)(15 s − 1.0 s) + a1 (15 s − 1.0 s)2 ⇒ a1 = −1.02 m/s 2 2 The final speed v2 can now be obtained as
v2 = v1 + a1 (t2 − t1 ) = (20 m/s) + (−1.02 m/s 2 )(15 s − 1 s) = 5.7 m/s
2.65. Solve: (a) The quantity
2 P 2(3.6 × 104 W) = = 60 m 2 /s3 . Thus m 1200 kg
vx =
At t = 10 s, vx = (b) With vx =
(60 m s )t 2
3
(60 m s ) (10 s) = 24 m/s ( ≈ 50 mph), and at t = 20 s, v = 35 m/s (≈ 75 mph). 2
3
x
2 P 12 t , we have m
ax = (c) At t = 1 s, ax =
dvx 2 P 1 −12 P = × t = dt m 2 2mt
P (3.6 × 104 W) = = 3.9 m/s 2 . Similarly, at t = 10 s, ax = 1.2 m/s 2 . 2mt 2(1200 kg)(1 s)
(d) Consider the limiting case of very short times. Note that ax → ∞ as t → 0. This is physically impossible for the Alfa Romeo. dx 2 P 12 t and the initial (e) We can use the relationship that vx = and integrate to find x(t ). We have vx = dt m condition xi = 0 at ti = 0. Thus
∫
x
0
and x =
dx =
2 P t 1/ 2 t dt m ∫0
2P t 3 / 2 2 2P 3 / 2 = t m 3/ 2 3 m
(f) Time to travel a distance x is found by solving the above equation for t.
⎡3 m ⎤ t=⎢ x⎥ ⎢⎣ 2 2 P ⎥⎦ For x = 402 m, t = 18.2 s.
2/3
2.66.
Model: Visualize:
Solve:
Both cars are particles that move according to the constant-acceleration kinematic equations.
(a) David’s and Tina’s motions are given by the following equations:
1 xD1 = xD0 + vD0 (tD1 − tD0 ) + aD (tD1 − tD0 ) 2 = vD0tD1 2 1 1 2 xT1 = xT0 + vT0 (tT1 − tT0 ) + aT (tT1 − tT0 ) 2 = 0 m + 0 m + aTtT1 2 2 When Tina passes David the distances are equal and tD1 = tT1 , so we get
1 2 1 2v 2(30 m/s) xD1 = xT1 ⇒ vD0tD1 = aTtT1 ⇒ vD0 = aTtT1 ⇒ tT1 = D0 = = 30 s aT 2 2 2.0 m/s 2 Using Tina’s position equation, 1 2 1 xT1 = aTtT1 = (2.0 m/s 2 )(30 s) 2 = 900 m 2 2
(b) Tina’s speed vT1 can be obtained from
vT1 = vT0 + aT (tT1 − tT0 ) = (0 m/s) + (2.0 m/s 2 )(30 s − 0 s) = 60 m/s Assess: This is a high speed for Tina (~134 mph) and so is David’s velocity (~67 mph). Thus the large distance for Tina to catch up with David (~0.6 miles) is reasonable.
2.67.
Model: Visualize:
We will represent the dog and the cat in the particle model.
Solve: We will first calculate the time tC1 the cat takes to reach the window. The dog has exactly the same time to reach the cat (or the window). Let us therefore first calculate tC1 as follows:
1 xC1 = xC0 + vC0 (tC1 − tC0 ) + aC (tC1 − tC0 ) 2 2 1 2 ⇒ 3.0 m = 1.5 m + 0 m + (0.85 m/s 2 )tC1 ⇒ tC1 = 1.879 s 2 In the time tD1 = 1.879 s, the dog’s position can be found as follows: 1 xD1 = xD0 + vD0 (tD1 − tD0 ) + aD (tD1 − tD0 ) 2 2 1 = 0 m + (1.50 m/s)(1.879 s) + (−0.10 m/s 2 )(1.879 s) 2 = 2.6 m 2 That is, the dog is shy of reaching the cat by 0.4 m. The cat is safe.
2.68.
Model: Visualize:
Solve:
We use the particle model for the large train and the constant-acceleration equations of motion.
Your position after time tY1 is
1 xY1 = xY0 + vY0 (tY1 − tY0 ) + aY (tY1 − tY0 ) 2 2 = 0 m + (8.0 m/s)(tY1 − 0 s) + 0 m = 8.0 tY1 The position of the train, on the other hand, after time tT1 is 1 xT1 = xT0 + vT0 (tT1 − tT0 ) 2 + aT (tT1 − tT0 ) 2 2 1 2 = 30 m + 0 m + (1.0 m/s 2 )(tT1 ) 2 = 30 + 0.5tT1 2 The two positions xY1 and xT1 will be equal at time tY1 ( = tT1 ) if you are able to jump on the back step of the train. That is, 2 2 30 + 0.5tY1 = 8.0tY1 ⇒ tY1 − (16 s)tY1 + 60 s 2 = 0 ⇒ tY1 = 6 s and 10 s
The first time, 6 s, is when you will overtake the train. If you continue to run alongside, the accelerating train will then pass you at 10 s. Let us now see if the first time tY1 = 6.0 s corresponds to a distance before the barrier. From the position equation for you, xY1 = (8.0 m/s)(6.0 s) = 48.0 m. The position equation for the train will yield the same number. Since the barrier is at a distance of 50 m from your initial position, you can just catch the train before crashing into the barrier.
2.69.
Model: Jill and the grocery cart will be treated as particles that move according to the constantacceleration kinematic equations. Visualize:
Solve:
The final position of Jill when the cart is caught is given by 1 1 1 xJ1 = xJ0 + vJ0 (tJ1 − tJ0 ) + aJ0 (tJ1 − tJ0 ) 2 = 0 m + 0 m + aJ0 (tJ1 − 0 s) 2 = (2.0 m/s 2 )tJ12 2 2 2
The cart’s position when it is caught is
1 1 xC1 = xC0 + vC0 (tC1 − tC0 ) + aC0 (tC1 − tC0 ) 2 = 50 m + 0 m + (0.5 m/s 2 )(tC1 − 0 s) 2 2 2 2 = 50 m + (0.25 m/s 2 )tC1 Since xJ1 = xC1 and tJ1 = tC1 , we get
1 2 2 (2.0)tJ12 = 50 s 2 + 0.25tC1 ⇒ 0.75tC1 = 50 s 2 ⇒ tC1 = 8.20 s 2 2 ⇒ xC1 = 50 m + (0.25 m/s 2 )tC1 = 50 m + (0.25 m/s 2 )(8.2 s) 2 = 67.2 m So, the cart has moved 17.2 m.
2.70.
Model: The watermelon and Superman will be treated as particles that move according to constantacceleration kinematic equations. Visualize:
Solve:
The watermelon’s and Superman’s position as they meet each other are
1 yW1 = yW0 + vW0 (tW1 − tW0 ) + aW0 (tW1 − tW0 ) 2 2 1 yS1 = yS0 + vS0 (tS1 − tS0 ) + aS0 (tS1 − tS0 ) 2 2 1 ⇒ yW1 = 320 m + 0 m + (−9.8 m/s 2 )(tW1 − 0 s) 2 2 ⇒ yS1 = 320 m + (−35 m/s)(tS1 − 0 s) + 0 m Because tS1 = t W1 , 2 yW1 = 320 m − (4.9 m/s 2 ) t W1
yS1 = 320 m − (35 m/s) t W1
Since yW1 = yS1 , 2 320 m − (4.9 m/s 2 )tW1 = 320 m − (35 m/s)tW1 ⇒ tW1 = 0 s and 7.1 s
Indeed, t W1 = 0 s corresponds to the situation when Superman arrives just as the watermelon is dropped off the Empire State Building. The other value, tW1 = 7.1 s, is the time when the watermelon will catch up with Superman. The speed of the watermelon as it passes Superman is vW1 = vW0 + aW0 (tW1 − t W0 ) = 0 m/s + (−9.8 m/s 2 )(7.1 s − 0 s) = −70 m/s Note that the negative sign implies a downward velocity. Assess: A speed of 140 mph for the watermelon is understandable in view of the significant distance (250 m) involved in the free fall.
2.71. Model: equations. Visualize:
Treat the car and train in the particle model and use the constant acceleration kinematics
Solve: In the particle model the car and train have no physical size, so the car has to reach the crossing at an infinitesimally sooner time than the train. Crossing at the same time corresponds to the minimum a1 necessary to avoid a collision. So the problem is to find a1 such that x2 = 45 m when y2 = 60 m. The time it takes the train to reach the intersection can be found by considering its known constant velocity.
v0 y = v2 y = 30 m/s =
y2 − y0 60 m = ⇒ t2 = 2.0 s t 2 − t0 t2
Now find the distance traveled by the car during the reaction time of the driver. x1 = x0 + v0 x ( t1 − t0 ) = 0 + ( 20 m/s )( 0.50 s ) = 10 m The kinematic equation for the final position at the intersection can be solved for the minimum acceleration a1. 1 2 x2 = 45 m = x1 + v1x ( t2 − t1 ) + a1 ( t2 − t1 ) 2 1 2 = 10 m + ( 20 m/s )(1.5 s ) + a1 (1.5 s ) 2 ⇒ a1 = 4.4 m/s 2 Assess: The acceleration of 4.4 m/s2 = 2.0 miles/h/s is reasonable for an automobile to achieve. However, you should not try this yourself! Always pay attention when you drive! Train crossings are dangerous locations, and many people lose their lives at one each year.
2.72.
Solve: A comparison of the given equation with the constant-acceleration kinematics equation
1 x1 = x0 + v0 (t1 − t0 ) + ax (t1 − t0 ) 2 2 yields the following information: x0 = 0 m, x1 = 64 m, t0 = 0, t1 = 4 s, and v0 = 32 m/s. (a) After landing on the deck of a ship at sea with a velocity of 32 m/s, a fighter plane is observed to come to a complete stop in 4.0 seconds over a distance of 64 m. Find the plane’s deceleration. (b)
1 (c) 64 m = 0 m + (32 m/s)(4 s − 0 s) + ax (4 s − 0 s) 2 2
64 m = 128 m + (8 s 2 ) ax ⇒ ax = −8 m/s 2
2.73.
Solve: (a) A comparison of this equation with the constant-acceleration kinematic equation 2 (v1y ) 2 = v0y + 2(ay )( y1 − y0 )
yields the following information: y0 = 0 m, y1 = 10 m, a y = −9.8 m/s 2 , and v1 y = 10 m/s. It is clearly a problem of
free fall. On a romantic Valentine’s Day, John decided to surprise his girlfriend, Judy, in a special way. As he reached her apartment building, he found her sitting in the balcony of her second floor apartment 10 m above the first floor. John quietly armed his spring-loaded gun with a rose, and launched it straight up to catch her attention. Judy noticed that the flower flew past her at a speed of 10 m/s. Judy is refusing to kiss John until he tells her the initial speed of the rose as it was released by the spring-loaded gun. Can you help John on this Valentine’s Day? (b)
2 (c) (10 m/s) 2 = v0y − 2(9.8 m/s 2 )(10 m − 0 m) ⇒ v0y = 17.2 m/s
Assess: The initial velocity of 17.2 m/s, compared to a velocity of 10 m/s at a height of 10 m, is very reasonable.
2.74.
Solve: A comparison with the constant-acceleration kinematics equation
(v1x ) 2 = (v0x ) 2 + 2ax ( x1 − x0 ) yields the following quantities: x0 = 0 m, v0 x = 5 m/s, v1x = 0 m/s, and ax = −(9.8 m/s 2 )sin10D. (a) A wagon at the bottom of a frictionless 10° incline is moving up at 5 m/s. How far up the incline does it move before reversing direction and then rolling back down? (b)
(c) (0 m/s) 2 = (5 m/s) 2 − 2(9.8 m/s 2 )sin10°( x1 − 0 m) ⇒ 25(m/s) 2 = 2(9.8 m/s 2 )(0.174) x1 ⇒ x1 = 7.3 m
2.75.
Solve: (a) From the first equation, the particle starts from rest and accelerates for 5 s. The second equation gives a position consistent with the first equation. The third equation gives a subsequent position following the second equation with zero acceleration. A rocket sled accelerates from rest at 20 m/s2 for 5 sec and then coasts at constant speed for an additional 5 sec. Draw a graph showing the velocity of the sled as a function of time up to t = 10 s. Also, how far does the sled move in 10 s? (b)
1 (c) x1 = (20 m/s 2 )(5 s)2 = 250 m 2
v1 = 20 m/s 2 (5 s) = 100 m/s
x2 = 250 m + (100 m/s)(5 s) = 750 m
2.76. Model: Visualize:
Solve:
The masses are particles.
The rigid rod forms the hypotenuse of a right triangle, which defines a relationship between x2 and y1 :
x22 + y12 = L2 . Taking the time derivative of both sides yields 2 x2
dx2 dx + 2 y1 1 = 0 dt dt
dx2 dy and v1y = 1 to write x2v2 x + y1v1 y = 0 . dt dt ⎛ y1 ⎞ y Thus v2 x = − ⎜ ⎟ v1 y . But from the figure, 1 = tan θ ⇒ v2 x = −v1 y tan θ . x 2 ⎝ x2 ⎠
We can now use v2 x =
Assess:
As x2 decreases (v2 x < 0), y1 increases (v1 y > 0), and vice versa.
2.77.
Model: Visualize:
The rocket and the bolt will be represented as particles to investigate their motion.
The initial velocity of the bolt as it falls off the side of the rocket is the same as that of the rocket, that is, vB0 = vR1 and it is positive since the rocket is moving upward. The bolt continues to move upward with a deceleration equal to g = 9.8 m/s 2 before it comes to rest and begins its downward journey. Solve:
To find aR we look first at the motion of the rocket: 1 yR1 = yR0 + vR0 (tR1 − tR0 ) + aR (tR1 − tR0 ) 2 2 1 = 0 m + 0 m/s + aR (4.0 s − 0 s) 2 = 8aR 2
To find aR we must determine the magnitude of yR1 or yB0 . Let us now look at the bolt’s motion: 1 yB1 = yB0 + vB0 (tB1 − tB0 ) + aB (tB1 − tB0 ) 2 2 1 0 = yR1 + vR1 (6.0 s − 0 s) + (−9.8 m/s 2 )(6.0 s − 0 s) 2 2
⇒ yR1 = 176.4 m − (6.0 s) vR1 Since vR1 = vR0 + aR (tR1 − tR0 ) = 0 m/s + 4 aR = 4 aR the above equation for yR1 yields yR1 = 176.4 − 6.0(4aR ). We know from the first part of the solution that yR1 = 8aR . Therefore, 8aR = 176.4 − 24.0aR and hence aR = 5.5 m/s 2 .
2.78.
Model: motion. Visualize:
Solve:
The rocket car is a particle that moves according to the constant-acceleration equations of
This is a two-part problem. For the first part,
1 1 1 x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 = 0 m + 0 m + a0 (9.0 s − 0 s) 2 = (81 s 2 ) a0 2 2 2 v1 = v0 + a0 (t1 − t0 ) = 0 m/s + a0 (9.0 s − 0 s) = (9.0 s) a0 During the second part of the problem,
1 x2 = x1 + v1 (t2 − t1 ) + a1 (t2 − t1 ) 2 2 1 1 ⇒ 990 m = (81 s 2 )a0 + (9.0 s)a0 (12 s − 9.0 s) + (−5.0 m/s 2 )(12 s − 9.0 s) 2 2 2 ⇒ a0 = 15 m/s 2 This leads to:
v1 = (9.0 s)a0 = (9.0 s)(15 m/s 2 ) = 135 m/s Using this value of v1 , we can now calculate v2 as follows:
v2 = v1 + a1 (t2 − t1 ) = (135 m/s) + (−5.0 m/s 2 )(12 s − 9.0 s) = 120 m/s That is, the car’s speed as it passes the judges is 120 m/s. Assess: This is a very fast motion (~250 mph), but the acceleration is large and the long burn time of 9 s yields a high velocity.
2.79.
Model: Visualize:
Solve:
Use the particle model.
(a) Substituting into the constant-acceleration kinematic equation
1 10 ⎞ ⎛ x2 = x1 + v1 (t2 − t1 ) + a1 (t2 − t1 ) 2 ⇒ 100 m = x1 + v1 ⎜ t2 − ⎟ + 0 m 2 3⎠ ⎝ 100 − x1 10 t2 = + v1 3 Let us now find v1 and x1 as follows: ⎛ 10 ⎞ s − 0 s ⎟ = 12 m/s v1 = v0 + a0 (t1 − t0 ) = 0 m/s + (3.6 m/s 2 ) ⎜ 3 ⎝ ⎠ 2
1 1 ⎛ 10 ⎞ x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 = 0 m + 0 m + (3.6 m/s 2 ) ⎜ s − 0 s ⎟ = 20 m 2 2 ⎝ 3 ⎠ The expression for t2 can now be solved as
t2 =
100 m − 20 m 10 s + = 10 s 12 m/s 3
(b) The top speed = 12 m/s which means v1 = 12 m/s. To find the acceleration so that the sprinter can run the 100-meter dash in 9.9 s, we use
v1 = v0 + a0 (t1 − t0 ) ⇒ 12 m/s = 0 m/s + a0t1 ⇒ t1 =
12 m/s a0
1 1 1 x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 = 0 m + 0 m + a0t12 = a0t12 2 2 2 Since x2 = x1 + v1 (t2 − t1 ) + 12 a1 (t2 − t1 ) 2 , we get 1 100 m = a0t12 + (12 m/s) (9.9 s − t1 ) + 0 m 2
Substituting the above equation for t1 in this equation, 2
⎛ 12 m/s ⎞ ⎛ 1 ⎞ ⎛ 12 m/s ⎞ 2 100 m = ⎜ ⎟ a0 ⎜ ⎟ + (12 m/s ) ⎜ 9.9 s − ⎟ ⇒ a0 = 3.8 m/s a ⎝ 2 ⎠ ⎝ a0 ⎠ 0 ⎝ ⎠ (c) We see from parts (a) and (b) that the acceleration has to be increased from 3.6 m/s2 to 3.8 m/s2 for the sprint time to be reduced from 10 s to 9.9 s, that is, by 1%. This decrease of time by 1% corresponds to an increase of acceleration by
3.8 − 3.6 × 100% = 5.6% 3.6
2.80.
Solve:
(a) The acceleration is the time derivative of the velocity. dv d ax = x = ⎡⎣ a (1 − e − bt ) ⎤⎦ = abe − bt dt dt With a = 11.81 m/s and b = 0.6887 s −1 , ax = 8.134e −0.6887 t m/s 2 . At the times t = 0 s, 2 s, and 4 s, this has the values 8.134 m/s2, 2.052 m/s2, and 0.5175 m/s2. dx dx (b) Since vx = = a − ae −bt and the initial condition , the position x is the integral of the velocity. With vx = dt dt that xi = 0 m at ti = 0 s ,
∫
x o
t
t
o
o
dx = ∫ a dt − ∫ ae − bt dt
Thus t
a a a t x = at o + e − bt = at + e − bt − b b b o This can be written a little more neatly as a x = ( bt + e− bt − 1) b = 17.15 ( 0.6887t + e−0.6887 t − 1) m (c) By trial and error, t = 9.92 s yields x = 100.0 m. Assess: Lewis’s actual time was 9.93 s.
2.81.
Model: Visualize:
Solve:
We will use the particle-model to represent the sprinter and the equations of kinematics.
Substituting into the constant-acceleration kinematic equations,
1 1 1 1 x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 = 0 m + 0 m + a0 (4 s − 0 s) 2 = a0t12 = a0 (4.0 s) 2 2 2 2 2 ⇒ x1 = (8 s 2 ) a0 v1 = v0 + a0 (t1 − t0 ) = 0 m/s + a0 (4.0 s − 0 s) ⇒ v1 = (4.0 s) a0 From these two results, we find that x1 = (2 s)v1. Now,
1 x2 = x1 + v1 (t2 − t1 ) + a1 (t2 − t1 ) 2 2 ⇒ 100 m = (2 s)v1 + v1 (10 s − 4 s) + 0 m ⇒ v1 = 12.5 m/s Assess: Using the conversion 2.24 mph = 1 m/s, v1 = 12.5 m/s = 28 mph. This speed as the sprinter reaches the finish line is physically reasonable.
2.82. Model: Visualize:
Solve:
The balls are particles undergoing constant acceleration.
(a) The positions of each of the balls at t1 is found from kinematics.
1 2 1 gt1 = v0t1 − gt12 2 2 1 1 ( y1 ) B = ( y0 )B + ( v0 y )B t1 − gt12 = h − gt12 2 2 In the particle model the balls have no physical extent, so they meet when ( y1 ) A = ( y1 ) B . This means
( y1 ) A = ( y0 ) A + ( v0 y ) A t1 −
Thus the collision height is ycoll
1 1 h v0t1 − gt12 = h − gt12 ⇒ t1 = 2 2 v0 1 gh 2 . = h − gt12 = h − 2 2v0 2
(b) We need the collision to occur while ycoll ≥ 0 . Thus h−
So hmax =
gh gh 2 2v 2 ⇒ h≤ 0 ≥ 0 ⇒ 1≥ 2 2 2v0 g 2v0
2v0 2 . g
(c) Ball A is at its highest point when its velocity ( v1 y ) = 0 . A
(v ) = (v ) 1y A
0y A
v0 g
h h v0 v2 . Equating these, = ⇒h= 0 . v0 v0 g g Interestingly, the height at which a collision occurs while Ball A is at its highest point is exactly half of
In (a) we found that the collision occurs at t1 = Assess: hmax .
− gt1 ⇒ 0 = v0 − gt1 ⇒ t1 =
2.83.
Model: Visualize:
The space ships are represented as particles.
Solve: The difficulty with this problem is how to describe “barely avoid.” The Klingon ship is moving with constant speed, so its position-versus-time graph is a straight line from xK0 = 100 km. The Enterprise will be decelerating, so its graph is a parabola with decreasing slope. The Enterprise doesn’t have to stop; it merely has to slow quickly enough to match the Klingon ship speed at the point where it has caught up with the Klingon ship. (You do the same thing in your car when you are coming up on a slower car; you decelerate to match its speed just as you come up on its rear bumper.) Thus the parabola of the Enterprise will be tangent to the straight line of the Klingon ship, showing that the two ships have the same speed (same slopes) when they are at the same position. Mathematically, we can say that at time t1 the two ships will have the same position ( xE1 = xK1 ) and the
same velocity (vE1 = vK1 ). Note that we are using the particle model, so the ships have zero length. At time t1, xK1 = xK0 + vK0t1 1 xE1 = vE0t1 + at12 2
vK1 = vK0 vE1 = vE0 + at1
Equating positions and velocities at t1: 1 xK0 + vK0t1 = vE0t1 + at12 2
vK0 = vE0 + at1
We have two simultaneous equations in the two unknowns a and t1. From the velocity equation, t1 = (vK0 − vE0 )/ a Substituting into the position equation gives 2
xK0 = −(vK0 − vE0 ) ⋅
(vK0 − vE0 ) 1 ⎛ (vK0 − vE0 ) ⎞ (vK0 − vE0 ) 2 + a⋅⎜ ⎟ =− a 2 ⎝ a 2a ⎠
⇒a=−
(vK0 − vE0 ) 2 (20,000 m/s − 50,000 m/s) 2 =− = − 4500 m/s 2 2 xK0 2(100,000 m)
Assess: The magnitude of the deceleration is 4500 m/s2, which is a rather extreme ≈460g. Fortunately, the Enterprise has other methods to keep the crew from being killed.
3.1.
Visualize:
G G G G G Solve: (a) To find A + B , we place the tail of vector B on the tip of vector A and connect the tail of vector A G with the tip of vector B. G G G G G G (b) Since A − B = A + (− B) , we place the tail of the vector (− B ) on the tip of vector A and then connect the tail G G of vector A with the tip of vector (− B ) .
3.2.
Visualize:
G G G G G Solve: (a) To find A + B , we place the tail of vector B on the tip of vector A and then connect vector A’s G tail with vector B’s tip. G G G G G G G G (b) To find A − B , we note that A − B = A + (− B) . We place the tail of vector − B on the tip of vector A and G G then connect vector A’s tail with the tip of vector − B.
3.3.
Solve:
Visualize:
G Vector E points to the left and up, so the components Ex and E y are negative and positive,
respectively, according to the Tactics Box 3.1. (a) Ex = − E cosθ
and
E y = E sin θ .
(b) Ex = − E sin φ
and
E y = E cos φ .
Assess: Note that the role of sine and cosine are reversed because we are using a different angle. θ and φ are complementary angles.
G Visualize: The position vector r whose magnitude r is 10 m has an x-component of 6 m. It makes an angle θ with the + x -axis in the first quadrant.
3.4.
Using trigonometry, rx = r cosθ , or 6 m = (10 m)cosθ . This gives θ = 53.1°. Thus the y-component of G the position vector r is ry = r sin θ = (10 m)sin 53.1° = 8 m.
Solve:
Assess: The y-component is positive since the position vector is in the first quadrant.
3.5.
Solve:
Visualize:
The figure shows the components vx and vy, and the angle θ.
We have, v y = −v sin 40°, or −10 m/s = −v sin 40°, or v = 15.56 m/s.
Thus the x-component is vx = v cos 40° = (15.56 m/s ) cos 40° = 12 m/s. Assess: The x-component is positive since the position vector is in the fourth quadrant.
3.6.
Visualize:
We will follow rules in Tactics Box 3.1. G Solve: (a) Vector r points to the right and down, so the components rx and ry are positive and negative, respectively:
rx = r cosθ = (100 m)cos 45° = 70.7 m
ry = − r sin θ = −(100 m)sin 45° = −70.7 m
G (b) Vector v points to the right and up, so the components vx and v y are both positive:
vx = v cosθ = (300 m /s) cos 20° = 282 m/s G (c) Vector a has the following components:
ax = −a cosθ = −(5.0 m/s 2 )cos90° = 0 m/s 2
v y = v sin θ = (300 m/s)sin 20° = 103 m/s
a y = − a sin θ = −(5.0 m/s 2 )sin 90° = −5.0 m/s 2
Assess: The components have same units as the vectors. Note the minus signs we have manually inserted according to Tactics Box 3.1.
3.7.
Visualize:
We will follow the rules given in Tactics Box 3.1. Solve: v y = (5 cm/s)cos90° = 0 cm/s (a) vx = −(5 cm/s)sin 90° = −5 cm/s (b) ax = −(10 m/s 2 )sin 40° = −6.4 m/s 2 (c) Fx = (50 N)sin 36.9° = 30 N
a y = −(10 m/s 2 )cos 40° = −7.7 m/s 2
Fy = (50 N)cos36.9° = 40 N
Assess: The components have the same units as the vectors. Note the minus signs we have manually inserted according to Tactics Box 3.1.
3.8.
Visualize:
G G The components of the vector C and D, and the angles θ are shown. K K Solve: For C we have C x = −(3.15 m)cos15° = −3.04 m and C y = (3.15 m)sin15° = 0.815 m. For D we
have Dx = 25.6sin 30° = 12.8 and Dy = −25.67cos30° = −22.2. G G Assess: The components of the vector C have the same units as C itself. Dx and Dy G are unitless because D is without units. Note the minus signs we have manually inserted
following rules of Tactics Box 3.1.
3.9.
Visualize:
Solve: The magnitude of the vector is E = ( Ex ) 2 + ( E y ) 2 = (125 V/m) 2 + ( −250 V/m) 2 = 280 V/m. In the G K expression for E , the − ˆj and +iˆ means that E is in quadrant IV. The angle θ is below the positive x-axis. We have:
θ = tan −1 Assess:
| Ey | Ex
⎛ 250 V/m ⎞ −1 = tan −1 ⎜ ⎟ = tan 2 = 63.4° 125 V/m ⎝ ⎠
Since | E y | > | Ex | , the angle θ made with the +x-axis is larger than 45°. θ = 45° for | E y | = | Ex | .
3.10.
Visualize:
Solve:
(a) Using the formulas for the magnitude and direction of a vector, we have:
4 4
θ = tan −1 = tan −1 1 = 45°
B = ( − 4) 2 + (4) 2 = 5.7 (b) r = (−2 cm) 2 + (−1 cm) 2 = 2.2 cm
1 2
θ = tan −1 = tan −1 0.5 = 26.6°
100 = tan −1 10 = 84.3° 10 10 (d) a = (10 m/s 2 ) 2 + (20 m/s 2 ) 2 = 22.4 m/s 2 θ = tan −1 = tan −1 0.5 = 26.6° 20 Assess: Note that θ ≤ 45° when | E y | ≤ | Ex |, where θ is the angle made with the x-axis. On the other hand, (c) v = (−10 m/s) 2 + (−100 m/s) 2 = 100.5 m/s
θ > 45° when | E y | > | Ex | .
θ = tan −1
3.11.
Visualize:
Solve:
(a) Using the formulas for the magnitude and direction of a vector, we have: 6 4
θ = tan −1 = tan −1 1.5 = 56.3°
A = (4) 2 + ( −6) 2 = 7.21 (b) r = (50 m) 2 + (80 m) 2 = 94.3 m
⎛ ry ⎞ −1 ⎛ 80 m ⎞ ⎟ = tan ⎜ ⎟ = 58.0° r ⎝ 50 m ⎠ ⎝ x⎠
θ = tan −1 ⎜
40 = tan −1 2 = 63.4° 20 2 (d) a = (2 m/s 2 ) 2 + (−6 m/s 2 ) 2 = 6.3 m/s 2 θ = tan −1 = tan −1 0.33 = 18.4° 6 Assess: Note that the angle θ made with the x-axis is smaller than 45° whenever | E y | < | Ex | , θ = 45° for
(c) v = (−20 m/s) 2 + (40 m/s) 2 = 44.7 m/s
θ = tan −1
| E y | = | Ex |, and θ > 45° for | E y | > | Ex | . In part (d), θ is with the y-axis, where the opposite of this rule applies.
3.12.
Visualize:
G G G G G G G K K We have C = A − B or C = A + (− B), where − B = ( B, direction opposite B ). Look back at Tactics Box 1.2, which shows how to perform vector subtraction graphically. G G G G G Solve: To obtain vector C from A and B, we place the tail of − B on the tip of A, and then use the tip-totail rule of graphical addition.
3.13.
Visualize:
G G G G G The vectors A, B, and C = A + B are shown.
G G G G G (a) We have A = 5iˆ + 2 ˆj and B = −3iˆ − 5 ˆj. Thus, C = A + B = (5iˆ + 2 ˆj ) + (−3iˆ − 5 ˆj ) = 2iˆ − 3 ˆj. G G G (b) Vectors A, B, and C are shown with their tails together. G K (c) Since C = 2iˆ − 3 ˆj = C xiˆ + C y ˆj , Cx = 2, and C y = −3. Therefore, the magnitude and direction of C are Solve:
C = (2) 2 + (−3) 2 = 3.6
θ = tan −1
| Cy |
⎛3⎞ = tan −1 ⎜ ⎟ = 56° below the + x-axis Cx ⎝2⎠
G Assess: The vector C is to the right and down, thus implying a negative y-component and positive xcomponent, as obtained above. Also θ > 45° since | C y | > | Cx | .
3.14.
Visualize:
G G G G G K (a) We have A = 5iˆ + 2 ˆj, B = −3iˆ − 5 ˆj, and − B = +3iˆ + 5 ˆj. Thus, D = A + ( − B) = 8iˆ + 7 ˆj. G G G (b) Vectors A, B and D are shown in the above figure. G G (c) Since D = 8iˆ + 7 ˆj = Dxiˆ + Dy ˆj , Dx = 8 and Dy = 7. Therefore, the magnitude and direction of D are Solve:
D = (8) 2 + (7) 2 = 10.6 Assess:
⎛ Dy ⎞ −1 ⎛ 7 ⎞ ⎟ = tan ⎜ ⎟ = 41° ⎝8⎠ ⎝ Dx ⎠
θ = tan −1 ⎜
Since | Dy | < | Dx | , the angle θ is less than 45°, as it should be.
3.15.
Visualize:
G G G G Solve: (a) We have A = 5iˆ + 2 ˆj and B = −3iˆ − 5 ˆj. This means 2 A = 10iˆ + 4 ˆj and 3B = −9iˆ − 15 ˆj. Hence, G G G E = 2 A + 3B = 1iˆ − 11 ˆj. G G G (b) Vectors A , B, and E are shown in the above figure. G G (c) From the E vector, Ex = 1 and E y = −11 . Therefore, the magnitude and direction of E are
E = (1) 2 + (−11) 2 = 11.05
⎛ Ex ⎞ ⎛1⎞ = tan −1 ⎜ ⎟ = 5.19° right of the − y -axis ⎜ | E | ⎟⎟ ⎝ 11 ⎠ ⎝ y ⎠
φ = tan −1 ⎜
Assess: Note that φ is the angle made with the y-axis, and that is why φ = tan −1 ( Ex /| E y |) rather than
tan −1 (| E y |/ Ex ), which would be the case if φ were the angle made with the x-axis.
3.16.
Visualize:
G G Solve: (a) We have A = 5iˆ + 2 ˆj and B = −3iˆ − 5 ˆj. G G G F = A − 4 B = 17iˆ + 22 ˆj = Fxiˆ + Fy ˆj with Fx = 17 and Fy = 22. G G G (b) The vectors A, B, and F are shown in the above figure. G (c) The magnitude and direction of F are
This
means
F = Fx2 + Fy2 = (17) 2 + (22) 2 = 27.8 ⎛ Fy ⎞ −1 ⎛ 22 ⎞ ⎟ = tan ⎜ ⎟ = 52.3° ⎝ 17 ⎠ ⎝ Fx ⎠
θ = tan −1 ⎜ Assess:
Fy > Fx implies θ > 45°, as is observed.
G − 4 B = +12iˆ + 20 ˆj.
Hence,
3.17.
Solve: A different coordinate system can only mean a different orientation of the grid and a different origin of the grid. (a) False, because the size of a vector is fixed. (b) False, because the direction of a vector in space is independent of any coordinate system. (c) True, because the orientation of the vector relative to the axes can be different.
3.18.
Visualize:
G G In coordinate system I, A = −(4 m) ˆj , so Ax = 0 m and Ay = −4 m. The vector B makes an angle of G 60° counterclockwise from vertical, which makes it have an angle of θ = 30° with the –x-axis. Since B points to the left and up, it has a negative x-component and a positive y-component. That is, G Bx = −(5.0 m)cos30° = −4.3 m and By = + (5.0 m)sin 30° = 2.5 m. Thus, B = −(4.3 m)iˆ + (2.5 m) ˆj. G In coordinate system II, A points to the left and down, and makes an angle of 30° with the –y-axis. Therefore, Ax = −(4.0 m)sin 30° = −2.0 m and Ay = −(4.0)cos30° = −3.5 m. This implies G ˆ ˆ A = −(2.0 m)i − (3.5 m) j. The vector G B makes an angle of 30° with the +y-axis and is to the left and up. This means we have to manually insert a x-component. Bx = − B sin 30° = −(5.0 m)sin 30° = −2.5 m, and minus sign with the G ˆ ˆ By = + B cos30° = (5.0 m)cos30° = 4.3 m. Thus B = −(2.5 m)i + (4.3 m) j. Solve:
3.19.
Visualize:
G Refer to Figure EX3.19. The velocity vector v points west and makes an angle of 30° with
the G –x-axis. v points to the left and up, implying that vx is negative and v y is positive. Solve:
We
have
vx = −v cos30°
v y = + v sin 30° = (100 m/s)sin 30° = 50.0 m/s. G Assess: vx and v y have the same units as v .
= −(100 m/s)cos30°
= −86.6 m/s
and
3.20.
Visualize:
(a)
G G G Solve: (b) The components of the vectors A, B, and C are Ax = ( 3.0 m ) cos 20° = 2.8 m and Ay = − ( 3.0 m ) sin 20°= −1.0 m ; Bx = 0 m and By = 2 m;
Cx = − ( 5.0 m ) cos70° = −1.71 m and C y = − ( 5.0 m ) sin 70°= −4.7 m. This means the vectors can be written, G G G A = (2.8 m)iˆ − (1.0 m) ˆj B = (2.0 m) ˆj C = (−1.71 m)iˆ − (4.7 m) ˆj G G G G (c) We have D = A + B + C = (1.09 m)iˆ − (3.7 m) ˆj. This means D = (1.09 m) 2 + (3.7 m) 2 = 3.9 m
θ = tan −1
G The direction of D is south of east, 7 4 ° b elow the positive x-axis.
3.9 = tan −1 3.58 = 74° 1.09
3.21.
Visualize:
G G G Solve: Using the method of tail-to-tip graphical addition, the diagram shows the resultant for D + E + F in (a), G G G G G the resultant for D + 2 E in (b), and the resultant for D − 2 E + F in (c).
G E = Exiˆ + E y ˆj = 2iˆ + 3 ˆj ,
which means Ex = 2 and E y = 3. Also, G F = Fxiˆ + Fy ˆj = 2iˆ − 2 ˆj , which means Fx = 2 and Fy = −2. G G (a) The magnitude of E is given by E = Ex2 + E y2 = (2) 2 + (3) 2 = 3.6 and the magnitude of F is given by
3.22.
Solve: We
have
F = Fx2 + Fy2 = (2) 2 + (2) 2 = 2.8. G G G G (b) Since E + F = 4iˆ + 1 ˆj , the magnitude of E + F is (4) 2 + (1) 2 = 4.1. G G G G (c) Since − E − 2 F = −(2iˆ + 3 ˆj ) − 2(2iˆ − 2 ˆj ) = −6iˆ + 1 ˆj , the magnitude of − E − 2 F is
(−6) 2 + (1) 2 = 6.1.
G G Solve: We have r = (5iˆ + 4 ˆj )t 2 m. This means that r does not change the ratio of its components as t G G increases, that is, the direction of r is constant. The magnitude of r is given by
3.23.
r = (5t 2 ) 2 + (4t 2 ) 2 m = (6.40t 2 ) m. (a) The particle’s distance from the origin at t = 0 s, t = 2 s, and t = 5 s is 0 m, 25.6 m, and 160 m. G dt 2 G dr (b) The particle’s velocity is v = = (5iˆ + 4 ˆj ) m/s = (5iˆ + 4 ˆj )2t m/s = (10iˆ + 8 ˆj )t m/s dt dt (c) The magnitude of the particle’s velocity is given by v = (10 t ) 2 + (8t ) 2 = 12.8t m/s. The particle’s speed at t = 0 s, t = 2 s, and t = 5 s is 0 m/s, 25.6 m/s , and 64.0 m/s .
3.24.
Visualize:
G G G Solve: (a) Vector C is the sum of vectors A and B, which is obtained using the tip-to-tip rule of graphical addition. Its magnitude is measured to be 4.7 and its angle made with the +x-axis is measured to be 33°. (b) Using the law of cosines, C 2 = A2 + B 2 − 2 AB cos φ , and the geometry of parallelograms, which shows that φ = 180° − (θ B − θ A ) = 180° − (60° − 20°) = 140°, we obtain C = (3) 2 + (2) 2 − 2(3)(2)cos(140°) = 4.71
Using the law of sines: sin α sin140° ⇒ α = 15.8° = 2 4.71 Thus, θC = α + 20° = 35.8°. (c) We have:
Ax = A cos θ A = 3cos 20° = 2.82
Ay = A sin θ A = 3sin 20° = 1.03
Bx = B cos θ B = 2cos60° = 1.00
By = B sin θ B = 2sin 60° = 1.73
This means: Cx = Ax + Bx = 3.82 and C y = Ay + By = 2.76. The magnitude and direction of C are given by
C = Cx2 + C y2 = (3.82)2 + (2.76) 2 = 4.71
⎛ Cy ⎞ −1 ⎛ 2.76 ⎞ ⎟ = tan ⎜ ⎟ = 35.8° C ⎝ 3.82 ⎠ ⎝ x⎠
θC = tan −1 ⎜
Assess: Using the method of vector components and their algebraic addition to find the resultant vector yields the same results as using the graphical addition of vectors.
3.25.
Visualize: Refer to Figure P3.25 in your textbook. G G G G G G G Solve: (a) We are given that A + B + C = −2iˆ with A = 4iˆ, and C = −2 ˆj. This means A + C = 4iˆ − 2 ˆj. Thus, G G G G G G B = ( A + B + C ) − ( A + C ) = (−2iˆ) − (4iˆ − 2 ˆj ) = −6iˆ + 2 ˆj. G (b) We have B = Bxiˆ + By ˆj with Bx = −6 and By = 2. Hence, B = ( −6) 2 + (2) 2 = 6.3
θ = tan −1
By | Bx |
= tan −1
2 = 18° 6
G G Since B has a negative x-component and a positive y-component, the angle θ made by B is with the –x-axis and it is above the –x-axis. Assess: Since | By | < | Bx |, θ < 45° as is obtained above.
3.26.
Visualize:
Solve:
(a) θ E = tan −1 ⎛⎜⎝ 11 ⎞⎟⎠ = 45°
θ F = tan −1 ⎛⎜⎝ 12 ⎞⎟⎠ = 63.4° Thus φ = 180° − θ E − θ F = 71.6° G G (b) From the figure, E = 2 and F = 5. Using
G 2 = E 2 + F 2 − 2 EF cos φ = ( 2) 2 + ( 5) 2 − 2( 2)( 5)cos(180° − 71.6°) ⇒ G = 3.00. sin α sin(180° − 71.6°) = ⇒ α = 45° 2.975 5 G Since θ E = 45° , the angle made by the vector G with the +x-axis is θG = (α + θ E ) = 45° + 45° = 90°. (c) We have
Furthermore, using
Ex = +1.0,
and E y = +1.0
Fx = −1.0,
and Fy = +2.0
Gx = 0.0,
and
⇒G =
( 0.0 )
G y = 3.0 2
+ ( 3.0 ) = 3.0, 2
and θ = tan −1
|G y | |Gx |
⎛ 3.0 ⎞ = tan −1 ⎜ ⎟ = 90° ⎝ 0.0 ⎠
G That is, the vector G makes an angle of 90° with the x-axis. Assess: The graphical solution and the vector solution give the same answer within the given significance of figures.
3.27. Solve:
Visualize: Refer to Figure P3.27. From the rules of trigonometry,
we
have
Ay = 4sin 40° = 2.6. Also, G G G G Since A + B + C = 0,
Ax = 4cos 40° = 3.1 and
Bx = −2cos10° = −1.97 and By = +2sin10° = 0.35. G G G G G ˆ ˆ C = − A − B = (− A) + (− B) = (−3.1i − 2.6 j ) + (+1.97iˆ − 0.35 ˆj ) = −1.1iˆ − 3.0 ˆj.
3.28.
Visualize:
D u
G G In the tilted coordinate system, the vectors A and B are expressed as: G G A = (2sin15° m)iˆ + (2cos15° m) ˆj and B = (4cos15° m)iˆ − (4sin15° m) ˆj. G G G Therefore, D = 2 A + B = (4 m)[(sin15° + cos15°)iˆ + (cos15° − sin15°) ˆj ] = (4.9 m)iˆ + (2.9 m) ˆj. The magnitude of Solve:
this vector is D = 5.7 m, and it makes an angle of θ = tan −1 (2.9 m/4.9 m) = 31° with the +x-axis. Assess: The resultant vector can be obtained graphically by using the rule of tail-to-tip addition.
3.29.
Visualize:
G The magnitude of the unknown vector is 1 and its direction is along iˆ + ˆj . Let A = iˆ + ˆj as shown in the G G diagram. That is, A = 1iˆ + 1 ˆj and the x- and y-components of A are both unity. Since θ = tan −1 ( Ay / Ax ) = 45°, the unknown vector must make an angle of 45° with the +x-axis and have unit magnitude. G Solve: Let the unknown vector be B = Bxiˆ + By ˆj where
Bx = B cos 45° =
1 B 2
and By = B sin 45° =
G We want the magnitude of B to be 1, so we have
1 B 2
B = Bx2 + By2 = 1 2
2
⎛ 1 ⎞ ⎛ 1 ⎞ ⇒ ⎜ B⎟ + ⎜ B ⎟ = 1 ⇒ B2 = 1 ⇒ B = 1 ⎝ 2 ⎠ ⎝ 2 ⎠ Hence, Bx = By = Finally,
1 2
G 1 ˆ 1 ˆ B = Bxiˆ + By ˆj = i+ j 2 2
3.30.
Model: Carlos will be represented as a particle and the particle model will be used for motion under constant acceleration kinetic equations. Visualize:
Solve: Carlos runs at constant speed without changing direction. The total distance he travels is found from kinematics:
r1 = r0 + v0 Δt = 0 m + ( 5 m/s )( 600 s ) = 3000 m G This displacement is north of east, or θ = 25° from the +x-axis. Thus the position r1 becomes G r1 = (3000 m)(cos 25°iˆ + sin 25° ˆj ) = 2.7 km iˆ + 1.27 km ˆj
That is, Carlos ends up 1.27 km north of his starting position. Assess: The choice of our coordinate system is such that the x-component of the displacement is along the east and the y-component is along the north. The displacement of 3.0 km is reasonable for Carlos to run in 10 minutes if he is an athlete.
3.31. Visualize: The coordinate system (x,y,z) is shown here. While +x denotes east and +y denotes north, G G G G the +z-direction is vertically up. The vectors S morning (shortened as S m ), Safternoon (shortened as Sa ), and the total G G G displacement vector S total = Sa + S m are also shown.
G G Solve: S m = (2000iˆ + 3000 ˆj + 200kˆ) m, and Sa = (−1500iˆ + 2000 ˆj − 300kˆ) m. The total displacement is the sum of the individual displacements. (a) The sum of the z-components of the afternoon and morning displacements is
Sza + Szm = −300 m + 200 m = −100 m, that is, 100 m lower. G G G (b) S total = Sa + S m = (500iˆ + 5000 ˆj − 100kˆ) m, that is, (500 m east) + (5000 m north) – (100 m vertical). The magnitude of your total displacement is S total =
( 500 )
2
+ ( 5000 ) + ( −100 ) m = 5.03 km 2
2
3.32.
Visualize:
Only the minute hand is shown in the figure. G G Solve: (a) We have S8:00 = (2.0 cm) ˆj and S8:20 = (2.0 cm)cos30°iˆ − (2.0 cm)sin 30° ˆj. The displacement vector is G G G Δr = S8:20 − S8:00 = (2.0 cm)[cos30°iˆ − (sin 30° + 1) ˆj ]
= (2.0 cm)[0.87iˆ − 1.50 ˆj ] = (1.74 cm)iˆ − (3.00 cm) ˆj
G G G G G (b) We have S8:00 = (2.0 cm) ˆj and S9:00 = (2.0 cm) ˆj. The displacement vector is Δr = S9:00 − S8:00 = 0. Assess: The displacement vector in part (a) has a positive x-component and a negative y-component. The vector thus is to the right and points down, in quadrant IV. This is where the vector drawn from the tip of the 8:00 a.m. arm to the tip of the 8:20 a.m. arm will point.
3.33.
Visualize:
(a)
Note that +x is along the east and +y is along the north. G G Solve: (b) We are given A = −(200 m) ˆj , and can use trigonometry to obtain B = −(283 m)iˆ − (283 m) ˆj and G G G G G C = (100 m)iˆ + (173 m) ˆj. We want A + B + C + D = 0. This means G G G G D = −A − B − C = (200 m ˆj ) + (283 m iˆ + 283 m ˆj ) + (−100 m iˆ − 173 m ˆj ) = 183 m iˆ + 310 m ˆj K The magnitude and direction of D are
D = (183 m) 2 + (310 m) 2 = 360 m and θ = tan −1
Dy Dx
⎛ 310 m ⎞ = tan −1 ⎜ ⎟ = 59.4° ⎝ 183 m ⎠
G This means D = (360 m, 59.4° north of Geast). (c) The measuredGlength of the vector D on the graph (with a ruler) is approximately 1.75 times the measured length of vector A . Since A = 200 m, this gives D = 1.75 × 200 m = 350 m. Similarly, the angle θ measured with the protractor is close to 60°. These answers are in close agreement to part (b).
3.34.
Visualize:
(a) The figure shows Sparky’s individual displacements and his net displacement.
G G G G (b) Dnet = D1 + D2 + D3 , where individual displacements are G D1 = (50cos 45°iˆ + 50sin 45° ˆj ) m = (35.4iˆ + 35.4 ˆj ) m G D2 = −70iˆ m G D3 = −20 ˆj m G Thus Sparky’s displacement is Dnet = ( −35iˆ + 15.4 ˆj ) m. (c) As a magnitude and angle, Solve:
Dnet = ( Dnet ) 2x + ( Dnet ) 2y = 38 m,
⎛ ( Dnet ) y ⎞ ⎟ = 24° ⎝ | ( Dnet ) x | ⎠
θ net = tan −1 ⎜
Sparky’s net displacement is 38 m in a direction 24° north of west.
3.35.
Visualize:
Solve:
G G G We are given A = 5 m iˆ and C = (−1 m)kˆ. Using trigonometry, B = (3cos 45° m)iˆ − (3sin 45° m) ˆj. The G G G G G is displace-ment is r = A + B + C = (7.12 m)iˆ − (2.12 m) ˆj − (1 m)kˆ. The magnitude of r
total
r = (7.12)2 + (2.12) 2 + (1)2 m = 7.5 m. Assess: A displacement of 7.5 m is a reasonable displacement.
3.36.
Visualize:
Solve:
We
have
G v = vxiˆ + v y ˆj
= v||iˆ + v⊥ ˆj
= v cosθ iˆ + v sin θ ˆj.
Thus,
v|| = v cosθ = (100 m/s)cos30° = 86.6 m/s.
Assess:
For the small angle of 30°, the obtained value of 86.6 m/s for the horizontal component is reasonable.
3.37.
Visualize:
⎛ 2.5 m/s ⎞ (a) Since vx = v cosθ , we have 2.5 m/s = (3.0 m/s)cosθ ⇒ θ = cos −1 ⎜ ⎟ = 34°. ⎝ 3.0 m/s ⎠ (b) The vertical component is v y = v sin θ = (3.0 m/s) sin 34° = 1.7 m/s.
Solve:
3.38.
Visualize:
The coordinate system used here is tilted with x-axis along the slope. Solve: The component of the velocity parallel to the x-axis is v|| = −v cos70° = −v sin 20° = −10 m/s (0.34) = −3.4 m/s. This is the speed down the slope. The component of the velocity perpendicular v⊥ = −v sin 70° = −v cos 20° = −10 m/s (0.94) = −9.4 m/s. This is the speed toward the ground. Assess: A speed of approximately 10 m/s implies a fall time of approximately 1 second under free fall. Note that g = –9.8 m/s2. This time is reasonable for a drop of approximately 5 m, or 16 feet.
3.39.
Visualize:
Solve: (a) The river is 100 m wide. If Mary rows due north at a constant speed of 2.0 m/s, it will take her 50 s to row across. But while she’s doing so, the current sweeps her boat sideways a distance 1 m/s × 50 s = 50 m. Mary’s net displacement is the vector sum of the displacement due to her rowing plus the displacement due to the river’s current. She lands 50 m east of the point that was directly across the river from her when she started. (b) Mary’s net displacement is shown on the figure.
3.40. G
Visualize: Establish a coordinate system with origin at the tree and with the x-axis pointing east. Let G G A be a displacement vector directly from the tree to the treasure. Vector A is A = (100iˆ + 500 ˆj ) paces. This describes the displacement you would undergo by walking north 500 paces, then east 100 paces. Instead, you follow the road for 300 paces and undergo displacement G B = (300sin 60°iˆ + 300cos60° ˆj ) paces = (260iˆ + 150 ˆj ) paces
G G G G Solve: Now let C be the displacement vector from your position to the treasure. From the figure A = B + C. G G G So the displacement you need to reach the treasure is C = A − B = (−160iˆ + 350 ˆj ) paces. G If θ is the angle measured between C and the y-axis, ⎛ 160 ⎞ ⎟ = 24.6° ⎝ 350 ⎠
θ = tan −1 ⎜
You should head 24.6° west of north. You need to walk distance C = Cx2 + C y2 = 385 paces to get to the
treasure.
3.41. Visualize: A 3% grade rises 3 m for every 100 m horizontal distance. The angle of the ground is thus α = tan −1 (3/100) = tan −1 (0.03) = 1.72°. Establish a tilted coordinate system with one axis parallel to the ground and the other axis perpendicular to the ground.
G Solve: From the figure, the magnitude of the component vector of v perpendicular to the ground is v⊥ = v sin α = 15.0 m/s. But this is only the size. We also have to note that the direction of vG⊥ is down, so the
component is
v⊥ = −15.0 m/s.
3.42.
Visualize:
G G G G G The resulting velocity is given by v = vfly + vwind , where vwind = 6 m/s iˆ and vfly = −v sin θ iˆ − v cosθ ˆj. G Substituting the known values we get v = −8 m/s sin θ iˆ − 8 m/s cosθ ˆj + 6 m/s iˆ. We need to have vx = 0. This
Solve:
means 0 = −8 m/s sin θ + 6 m/s, so sin θ =
6 and θ = 48.6°. Thus the ducks should head 48.6° west of south. 8
3.43.
Model: Visualize:
Solve:
The car is treated as a particle in this problem.
(a) The tangential component is a|| = a sin 30° = (2.0 m/s 2 )(0.5) = 1.0 m/s 2 .
(b) The perpendicular component is a⊥ = a cos30° = (2.0 m/s 2 )(0.866) = 1.7 m/s 2 . Assess: Magnitudes of the tangential and perpendicular components of acceleration are reasonable.
3.44.
Model: Visualize:
We will treat the knot in the rope as a particle in static equilibrium.
G G Solve: Expressing the vectors using unit vectors, we have F1 = 3.0iˆ and F2 = −5.0sin 30°iˆ + 5.0cos30° ˆj. Since G G G G G G G G F1 + F2 + F3 = 0, we can write F3 = − F1 − F2 = −0.5iˆ − 4.33 ˆj . The magnitude of F3 is given by G F3 = (−0.5) 2 + (−4.33) 2 = 4.4 units. The angle F3 makes is θ = tan −1 (4.33/0.5) = 83° and is below the negative xaxis. Assess: The resultant vector has both components negative, and is therefore in quadrant III. Its magnitude and G direction are reasonable. Note the minus sign that we have manually inserted with the force F2 .
3.45.
Visualize:
Use a tilted coordinate system such that x-axis is down the slope. G G Solve: Expressing all three forces in terms of unit vectors, we have F1 = −(3.0 N)iˆ, F2 = + (6.0 N) ˆj, and G F3 = (5.0 N)sin θ iˆ − (5.0 N)cosθ ˆj. G (a) The component of Fnet parallel to the floor is ( Fnet ) x = −(3.0 N) + 0 N + (5.0 N)sin 30° = −0.50 N, or 0.50 N up the slope. G (b) The component of Fnet perpendicular to the floor is ( Fnet ) y = 0 N + (6.0 N) − (5.0 N)cos30° = 1.67 N. G G (c) The magnitude of Fnet is Fnet = ( Fnet ) x + ( Fnet ) y = (−0.50 N) 2 + (1.67 N) 2 = 1.74 N. The angle Fnet makes
is (F ) y ⎛ 1.67 N ⎞ = tan −1 ⎜ ⎟ = 73° |( Fnet ) x | ⎝ 0.50 N ⎠ G G Fnet is 73° above the floor on the left side of F2 . 00000
φ = tan −1 Gnet
3.46.
Visualize:
Using trigonometry to calculate θ, we get θ = tan −1 (100 cm/141 cm) = 35.3°. Expressing the three forces in unit G G G vectors, FB = −(3.0 N)iˆ, FC = −(6.0 N) ˆj , and FD = + (2.0 N)cos35.3iˆ − (2.0 N)sin 35.3 ˆj = (1.63 N)iˆ − (1.16 N) ˆj. The total G G G G G force is Fnet = FB + FC + FD = −1.37 N iˆ − 7.2 N ˆj. The magnitude of Fnet is Fnet = (1.37 N) 2 + (7.2 N) 2 = 7.3 N. Solve:
θ net = tan −1 G Fnet = (7.3 N,79° below −x in quadrant III).
|( Fnet ) y |
⎛ 7.2 N ⎞ = tan −1 ⎜ ⎟ = 79° |( Fnet ) x | ⎝ 1.37 N ⎠
3-1
4.1.
Solve:
(a)
(b) A race car slows from an initial speed of 100 mph to 50 mph in order to negotiate a tight turn. After making the 90° turn the car accelerates back up to 100 mph in the same time it took to slow down.
4.2.
Solve:
(a)
(b) A car drives up a hill, over the top, and down the other side at constant speed.
4.3.
Solve:
(a)
(b) A ball rolls along a level table at 3 m/s. It rolls over the edge and falls 1 m to the floor. How far away from the edge of the table does it land?
4.4. Solve: (a) The figure shows a motion diagram of a pendulum as it swings from one side to the other. It’s clear that the velocity at the lowest point is not zero. The velocity vector at this point is tangent to the circle. We can use the method of Tactics Box 1.3 to find the acceleration at the lowest point. The acceleration is not zero. Instead, you can see that the acceleration vector points toward the center of the circle.
(b) The end of the arc is like the highest point of a ball tossed straight up. The velocity is zero for an instant as the vector changes from pointing outward to pointing inward. However, the acceleration is not zero at this point. G The velocity is changing at the end point, and this requires an acceleration. The motion diagram shows that Δv , G and thus a , is tangent to the circle at the end of the arc.
4.5.
Model: The boat is treated as a particle whose motion is governed by constant-acceleration kinematic equations in a plane. Visualize:
Solve:
Resolving the acceleration into its x and y components, we obtain G a = ( 0.80 m/s 2 ) cos 40°iˆ + ( 0.80 m/s 2 ) sin 40° ˆj = ( 0.613 m/s 2 ) iˆ + ( 0.514 m/s 2 ) ˆj G G G From the velocity equation v1 = v0 + a ( t1 − t0 ) , G v1 = ( 5.0 m/s ) iˆ + ⎡⎣( 0.613 m/s 2 ) iˆ + ( 0.514 m/s 2 ) ˆj ⎤⎦ ( 6 s − 0 s ) = ( 8.68 m/s ) iˆ + ( 3.09 m/s ) ˆj G The magnitude and direction of v are
( 8.68 m/s )
v=
2
+ ( 3.09 m/s ) = 9.21 m/s 2
⎛ v1 y ⎞ −1 ⎛ 3.09 m/s ⎞ ⎟ = tan ⎜ ⎟ = 20° north of east v ⎝ 8.68 m/s ⎠ ⎝ 1x ⎠
θ = tan −1 ⎜ Assess:
An increase of speed from 5.0 m/s to 9.21 m/s is reasonable.
G Solve: (a) At t = 0 s, x = 0 m and y = 0 m, or r = (0iˆ + 0ˆj ) m. At t = 4 s, x = 0 m and y = 0 m, or G r = (0iˆ + 0ˆj ) m. In other words, the particle is at the origin at both t = 0 s and at t = 4 s. From the expressions for x and y,
4.6.
⎤ dy ˆ ⎡⎛ 3 2 G dx ⎞ v = iˆ + j = ⎢⎜ t − 4t ⎟ iˆ + ( t − 2 ) ˆj ⎥ m/s dt dt ⎠ ⎣⎝ 2 ⎦ G G At t = 0 s, v = −2ˆj m/s, v = 2 m/s. At t = 4 s, v = 8iˆ + 2ˆj m/s, v = 8.3 m/s. G (b) At t = 0 s, v is along − ˆj , or 90° south of + x. At t = 4 s,
(
)
⎛ 2 m/s ⎞ ⎟ = 14° north of +x ⎝ 8 m/s ⎠
θ = tan −1 ⎜
4.7. Solve:
Visualize: Refer to Figure EX4.7. From the figure, identify the following:
x1 = 0 m
y1 = 0 m
x2 = 2000 m
y2 = 1000 m
v1x = 0 m/s
v1 y = 200 m/s
v2 x = 200 m/s
v2 y = −100 m/s
The components of the acceleration can be found by applying v22 = v12 + 2aΔs for the x and y directions. Thus v22x − v12x ( 200 m/s ) − ( 0 m/s ) = = 10.00 m/s 2 2Δx 2 ( 2000 m − 0 m ) 2
ax =
( −100 m/s ) − ( 200 m/s ) 2 (1000 m − 0 m ) 2
ay =
2
2
= −15.00 m/s 2
G So a = (10.00iˆ − 15.00 ˆj ) m/s 2 . Assess: A time of 20 s is needed to change v1x = 0 m/s to v2 x = 200 m/s at ax = 10 m/s 2 . This is the same time
needed to change v1 y to v2 y at a y = −15 m/s 2 .
4.8.
Model: The puck is a particle and follows the constant-acceleration kinematic equations of motion. Visualize: Please refer to Figure EX4.8. G Solve: (a) At t = 2 s, the graphs give vx = 16 cm/s and v y = 30 cm/s. The angle made by the vector v with the
x-axis can thus be found as ⎛ vy ⎞ −1 ⎛ 30 cm/s ⎞ ⎟ = tan ⎜ ⎟ = 62° above the x-axis v ⎝ 16 cm/s ⎠ ⎝ x⎠
θ = tan −1 ⎜
(b) After t = 5 s, the puck has traveled a distance given by: 5s
x1 = x0 + ∫ vx dt = 0 m + area under vx -t curve = 12 (40 cm/s)(5 s) = 100 cm 0
5s
y1 = y0 + ∫ v y dt = 0 m + area under v y -t curve = (30 cm/s)(5 s) = 150 cm 0
⇒ r1 = x12 + y12 =
(100 cm )
2
+ (150 cm ) = 180 cm 2
4.9.
Model: Use the particle model for the puck. Visualize: Please refer to Figure EX4.9 Solve: (a) Since the vx vs t and v y vs t graphs are straight lines, the puck is undergoing constant acceleration
along the x- and y- axes. The components of the puck’s acceleration are ax =
dvx Δvx ( −10 m/s − 10 m/s) = = = −2.0 m/s 2 dt 10 s − 0 s Δt
ay =
(10 m/s − 0 m/s) = 1.0 m/s 2 (10 s − 0 s)
The magnitude of the acceleration is a = a x2 + a y2 = 2.2 m/s 2 . (b) The puck is undergoing constant acceleration in both the x and y directions. Identify from the graphs vix = 10 m/s,
viy = 0 m/s. Since the puck starts at the origin, xi = yi = 0 m, and set ti = 0 s. Using kinematics, x = 0 m + (10 m/s)t + 12 ( −2.0 m/s 2 )t 2
y = 0 m + 0 m/s + 12 (1.0 m/s 2 )t 2
The distance from the origin at time t is r = x 2 + y 2 . The table below shows the values of x, y, and r at the times t = 0, 5, 10 s. t =0 s 5s 10 s
x 0m 25 m 0m
y 0m 12.5 m 50 m
R 0m 28 m 50 m
Assess: The puck turns around at t = 5 s in the x direction, and constantly accelerates in the y direction. Traveling 50 m from the starting point in 10 s is reasonable.
4.10.
Model: Assume the particle model for the ball, and apply the constant-acceleration kinematic equations of motion in a plane. Visualize:
Solve:
G (a) We know the velocity v1 = (2.0iˆ + 2.0jˆ) m/s at t = 1 s. The ball is at its highest point at t = 2 s, so
v y = 0 m/s. The horizontal velocity is constant in projectile motion, so vx = 2.0 m/s at all times. Thus G v2 = 2.0iˆ m/s at t = 2 s. We can see that the y-component of velocity changed by Δv y = −2.0 m/s. between t = 1 s and t = 2 s. Because a y is constant, v y changes by –2.0 m/s in any 1-s interval. At t = 3 s, v y is 2.0 m/s less than its value of 0 at t = 2 s. At t = 0 s, v y must have been 2.0 m/s more than its value of 2.0 m/s at t = 1 s. Consequently, at t = 0 s,
G v0 = (2.0iˆ + 4.0 jˆ ) m/s
At t = 1 s, G v0 = (2.0iˆ + 2.0 jˆ ) m/s
At t = 2 s, G v0 = (2.0iˆ + 0.0 jˆ ) m/s
At t = 3 s, G v0 = (2.0iˆ − 2.0 jˆ ) m/s (b) Because v y is changing at the rate –2.0 m/s per s, the y-component of acceleration is a y = −2.0 m/s 2 . But
a y = − g for projectile motion, so the value of g on Exidor is g = 2.0 m/s 2 . G (c) From part (a) the components of v0 are v0 x = 2.0 m/s and v0 y = 4.0 m/s. This means
⎛ v0 y ⎞ −1 ⎛ 4.0 m/s ⎞ ⎟ = tan ⎜ ⎟ = 63.4° above +x v ⎝ 2.0 m/s ⎠ ⎝ 0x ⎠
θ = tan −1 ⎜
Assess: The y-component of the velocity vector decreases from 2.0 m/s at t = 1 s to 0 m/s at t = 2 s. This gives an acceleration of −2 m/s 2 . All the other values obtained above are also reasonable.
4.11.
Model: Visualize:
Solve:
The ball is treated as a particle and the effect of air resistance is ignored.
Using x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) , 2
50 m = 0 m + (25 m/s)(t1 − 0 s) + 0 m ⇒ t1 = 2.0 s Now, using y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) , 2
y1 = 0 m + 0 m + 12 ( −9.8 m/s 2 )(2.0 s − 0 s) 2 = −19.6 m Assess: The minus sign with y1 indicates that the ball’s displacement is in the negative y direction or downward. A magnitude of 19.6 m for the height is reasonable.
4.12.
Model: neglected. Visualize:
Solve:
The bullet is treated as a particle and the effect of air resistance on the motion of the bullet is
(a) Using y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) , we obtain 2
(−2.0 × 10−2 m) = 0 m + 0 m + 12 (−9.8 m/s 2 )(t1 − 0 s) 2 ⇒ t1 = 0.0639 s (b) Using x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) , 2
(50 m) = 0 m + v0 x (0.0639 s − 0 s) + 0 m ⇒ v0 x = 782 m/s Assess: The bullet falls 2 cm during a horizontal displacement of 50 m. This implies a large initial velocity, and a value of 782 m/s is understandable.
4.13.
Model: We will use the particle model for the food package and the constant-acceleration kinematic equations of motion. Visualize:
Solve: For the horizontal motion, x1 = x0 + v0 x (t1 − t0 ) + 12 ax (t1 − t0 ) 2 = 0 m + (150 m/s)(t1 − 0 s) + 0 m = (150 m/s)t1 We will determine t1 from the vertical y-motion as follows: y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2
⇒ 0 m = 100 m + 0 m + 12 (−9.8 m/s 2 )t12 ⇒ t1 =
200 m = 4.518 s 9.8 m/s 2
From the above x-equation, the displacement is x1 = (150 m/s)(4.518 s) = 678 m. Assess: The horizontal distance of 678 m covered by a freely falling object from a height of 100 m and with an initial horizontal velocity of 150 m/s (≈ 335 mph) is reasonable.
4.14.
Model: Visualize:
Assume the particle model for the spyglass and use the projectile motion equations.
Solve: (a) The spyglass has an initial horizontal velocity equal to that of the ship. As the spyglass falls, it and the ship move forward together at the same velocity, so the spyglass lands at the bottom of the mast directly vertically below on the ship where the sailor dropped it. (b) The time the spyglass takes to fall can be found by considering the vertical motion:
y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2 ⇒ 0 m = 15 m + 0 m + 12 (−9.8 m/s 2 )(t1 − 0 s)2 ⇒ t1 = 1.749 s Therefore, x1 = 0 m/s + (4.0 m/s)(1.749 s) = 7.0 m. The spyglass lands 7.0 m to the right of the fisherman.
4.15.
Model:
G G G G G The position vectors r and r ′ in frames S and S′ are related by the equation r = r′ + R, where
G R is the position vector of the origin of frame S′ as measured in frame S. S is Ted’s frame and S′ is Stella’s frame. Visualize:
Solve:
G G The relation between R and the velocity of Stella V is G G R = V = (100 m/s ) cos 45°iˆ − (100 m/s ) sin 45° ˆj 5s
G G G Because r = r′ + R,
G ⎡100 ˆ 100 ⇒ R = ( 5.0 m ) ⎢ i− 2 ⎣ 2
ˆj ⎤ = 353.6iˆ − 353.6 ˆj m ⎥ ⎦
(
)
G G G r ′ = r − R = ( 200 m ) iˆ − ⎡⎣( 353.6 m ) iˆ − ( 353.6 m ) ˆj ⎤⎦ = − (154 m ) iˆ + ( 354 m ) ˆj
G The vector r ′ determines the position of the exploding firecracker as seen by Stella.
4.16.
Model: Assume motion along the x-direction. Let the earth frame be S and a frame attached to the water be S′. Frame S′ moves relative to S with a velocity Vx . Visualize:
Solve: Let vx be the velocity of the boat in frame S, and v′x be the velocity of the boat in S′. Then for travel down the river, 30 km vx = v′x + Vx = = 10.0 km/h 3.0 hr For travel up the river,
⎛ 30 km ⎞ −v′x + Vx = − ⎜ ⎟ = −6.0 km/h ⎝ 5.0 hr ⎠ Adding these two equations yields Vx = 2.0 km/h. That is, the velocity of the flowing river relative to the earth is 2.0 km/h.
4.17.
Motion: Assume motion along the x-direction. Let the earth frame be S and a frame attached to the moving sidewalk be S′. Frame S′ moves relative to S with velocity Vx . Solve:
Let vx be your velocity in frame S and v′x be your velocity in S′. In the first case, when the moving
sidewalk is broken, Vx = 0 m/s and vx W =
( x1 − x0 ) 50 s
In the second case, when you stand on the moving sidewalk, v′x = 0 m/s. Therefore, using vx = v′x + Vx , we get vx S = Vx =
x1 − x0 75 s
In the third case, when you walk while riding, v′x = vx W . Using vx = v′x + Vx , we get
x1 − x0 x1 − x0 x1 − x0 = + ⇒ t = 30 s 50 s 75 s t Assess:
A time smaller than 50 s was expected.
4.18. Model: Let the earth frame be S and a frame attached to the water be S′. Frame S′ moves relative to S with velocity V. We define the x-axis along the direction of east and the y-axis along the direction of north for both frames. The frames S and S′ have their origins coincident at t = 0 s. G G G G G Solve: (a) According to the Galilean transformation of position: r = r ′ + R = r′ + tV . We need to find Mary’s G position vector r from the earth frame S. The observer in frame S′ will observe the boat move straight north G G and will find its position as r ′ = ( 2.0t ) ˆj m/s. We also know that V = ( 3.0 ) iˆ m/s. Since r ′ = 100 m = ( 2.0 m/s ) t and t = 50 s, we have G r = ( 2.0 m/s )( 50 s ) ˆj + ( 50 s )( 3.0 m/s ) t iˆ = (150 m ) iˆ + (100 m ) ˆj
Thus she lands 150 m east of the point that was straight across the river from her when she started. (b)
G G G Note that r ′ is the displacement due to rowing, R is the displacement due to the river’s motion, and r is the net displacement.
Model: Let Susan’s frame be S and Shawn’s frame be S′. S′ moves relative to S with velocity V. Both Susan and Shawn are observing the intersection point from their frames. G G G G Solve: The Galilean transformation of velocity is v = v′ + V , where v is the velocity of the intersection point G G from Susan’s reference frame, v′ is the velocity of the intersection point from Shawn’s frame S′, and V is the G velocity of S′ relative to S or Shawn’s velocity relative to Susan. Because v = − ( 60 mph ) ˆj and G G G v′ = − ( 45 mph ) iˆ, we have V = v − v′ = ( 45 mph ) iˆ − ( 60 mph ) ˆj. This means that Shawn’s speed relative to
4.19.
Susan is V=
( 45 mph )
2
+ ( −60 mph ) = 75 mph 2
Solve: (a) From t = 0 s to t = 1 s the particle does not rotate. From t = 1 s to t = 3 s, the particle rotates clockwise from the angular position 0 rad to −2π rad. Therefore, Δθ = −2π rad in two seconds, or
4.20.
ω = −π rad s. From t = 3 s to t = 4 s the particle rotates counterclockwise from the angular position −2π rad to +4π rad. Thus Δθ = 4π − ( −2π ) = 6π rad and ω = +6π rad s. (b)
4.21.
Solve: Since ω = ( dθ dt ) we have
θ f = θ i + area under the ω -versus-t graph between ti and tf From t = 0 s to t = 2 s, the area is
( 20 rad/s )( 2 s ) = 40 rad. (60 rad) × (1 rev/2π rad) = 9.55 revolutions.
1 2
( 20 rad/s )( 2 s ) = 20 rad.
From t = 2 s to t = 4 s, the area is
Thus, the area under the ω -versus-t graph during the total time interval of 4 s is 60 rad or
4.22.
Solve: Since ω = ( dθ dt ) we have
θ f = θ i + area under the ω versus t graph between ti and tf From t = 0 s to s to t = 4 s, the area is 1 2
1 2
( 20 rad/s )( 4 s ) = 40 rad.
( −10 rad/s )( 4 s ) = −20 rad. Thus, the area under the ω
rad or (20 rad) × (1 rev/2π rad) = 3.2 revolutions.
From t = 4 s to t = 8 s, the area is
versus t graph during the total time interval of 8 s is 20
4.23.
Model: Treat the record on a turntable as a particle rotating at 45 rpm. Solve: (a) The angular velocity is
ω = 45 rpm ×
1min 2π rad × = 1.5π rad/s 60 s 1 rev
(b) The period is
T=
2π rad
ω
=
2π rad = 1.33 s 1.5π rad/s
4.24.
Model: Visualize:
Solve:
The airplane is to be treated as a particle.
(a) The angle you turn through is
θ 2 − θ1 =
s 5000 miles 180° = = 1.2500 rad = 1.2500 rad × = 71.62° π rad r 4000 miles
(b) The plane’s angular velocity is
ω=
θ 2 − θ1 t2 − t1
=
1.2500 rad rad 1h = 0.13889 rad/h = 0.13889 × = 3.858 × 10−5 rad/s 9 hr h 3600 s
Assess: An angular displacement of approximately one-fifth of a complete rotation is reasonable because the separation between Kampala and Singapore is approximately one-fifth of the earth’s circumference.
Solve: Let RE be the radius of the earth at the equator. This means RE + 300 m is the radius to the top of the tower. Letting T be the period of rotation, we have
4.25.
vtop − vbottom =
2π ( RE + 300 m) 2π RE 2π (300 m) 600π m − = = = 2.18 × 10−2 m/s T T 24 h 24(3600) s
4.26.
Solve: The plane must fly as fast as the earth’s surface moves, but in the opposite direction. That is, the plane must fly from east to west. The speed is
km km 1 mile ⎛ 2π rad ⎞ 3 v = ωr = ⎜ = 1680 × = 1040 mph ⎟ (6.4 × 10 km) = 1680 h h 1.609 km ⎝ 24 h ⎠
4.27. Solve:
Model: The rider is assumed to be a particle. Since ar = v 2 / r , we have
v 2 = ar r = (98 m/s 2 )(12 m) ⇒ v = 34 m/s Assess:
34 m/s ≈ 70 mph is a large yet understandable speed.
4.28. Solve:
Model: The earth is a particle orbiting around the sun. (a) The magnitude of the earth’s velocity is displacement divided by time:
v=
2π r 2π (1.5 × 1011 m) = = 3.0 × 104 m/s 24 h 3600 s T 365 days × × 1 day 1h
(b) Since v = rω , the angular acceleration is
v r
ω= =
3.0 × 104 m/s = 2.0 × 10−7 rad/s 1.5 × 1011 m
(c) The centripetal acceleration is
ar =
v 2 (3.0 × 104 m/s) 2 = = 6.0 × 10−3 m/s 2 r 1.5 × 1011 m
Assess: A tangential velocity of 3.0 × 104 m/s or 30 km/s is large, but needed for the earth to go through a displacement of 2π (1.5 × 1011 m) ≈ 9.4 × 108 km in 1 year.
Solve: The pebble’s angular velocity ω = (3.0 rev/s)(2π rad/rev) = 18.9 rad/s. The speed of the pebble as it moves around a circle of radius r = 30 cm = 0.30 m is
4.29.
v = ω r = (18.9 rad/s)(0.30 m) = 5.7 m/s
The radial acceleration is v 2 ( 5.7 m/s ) = = 108 m/s 2 0.30 m r 2
ar =
4.30.
Model: The crankshaft is a rotating rigid body. Solve: The crankshaft at t = 0 s has an angular velocity of 250 rad/s. It gradually slows down to 50 rad/s in 2 s, maintains a constant angular velocity for 2 s until t = 4 s, and then speeds up to 200 rad/s from t = 4 s to t = 7 s. The angular acceleration (α ) graph is based on the fact that α is the slope of the ω -versus-t graph.
4.31. Model: The turntable is a rotating rigid body. The angular velocity is the area under the α -versus-t graph:
Solve:
α=
dω ⇒ ω = ∫ α ( x ) dt = ω 0 + area under the α graph dt
The values of ω at selected values of time (t) are: t (s)
0
ω (rad/s)
0.5
0 (5 + 3.75)(0.5)/2 = 2.18
1.0
(5 + 2.5)(1)/2 = 3.75
1.5
(5 + 1.25)(1.5)/2 = 4.68
2.0
(5 + 0)(2)/2 = 5.0
2.5
5.0
3.0
5.0
4.32. Model: The wheel is a rotating rigid body. Solve:
(a)
The angular acceleration (α ) is the slope of the ω -versus-t graph. (b) The car is at rest at t = 0 s. It gradually speeds up for 4 s and then slows down for 4 s. The car is at rest from t = 8 s to t = 12 s, and then speeds up again for 4 s.
4.33. Model: The angular velocity and angular acceleration graphs correspond to a rotating rigid body. Solve:
(a) The α-versus-t graph has a positive slope of 5 rad/s2 from t = 0 s to t = 2 s and a negative slope of −5 rad/s 2 from t = 2 s to t = 4 s. (b) The angular velocity is the area under the α -versus-t graph:
α=
dω ⇒ ω = ∫ α ( x ) dt = ω 0 + area under α graph. dt
4.34.
Model: Visualize:
Model the car as a particle in nonuniform circular motion.
Note that halfway around the curve, the tangent is 45° south of east. The perpendicular component of the acceleration is 45° north of east. Solve: The radial and tangential components of the acceleration are
ar = a cos 25° = ( 3.0 m s 2 ) cos 25° = 2.7 m s 2 at = a sin 25° = ( 3.0 m s 2 ) sin 25° = 1.27 m s 2
4.35.
Model: Visualize:
Model the child on the merry-go-round as a particle in nonuniform circular motion.
Solve: (a) The speed of the child is v0 = rω = (2.5 m)(1.57 rad/s) = 3.9 m/s. (b) The merry-go-round slows from 1.57 rad/s to 0 in 20 s. Thus
ω1 = 0 = ω 0 +
at rω (2.5 m)(1.57 rad/s) = –0.197 m/s 2 t1 ⇒ at = − 0 = − r t1 20 s
During these 20 s, the wheel turns through angle
θ1 = θ 0 + ω 0t1 +
at 2 0.197 m/s 2 t1 = 0 + (1.57 rad/s) (20 s) − (20 s) 2 = 15.6 rad 2r 2(2.5 m)
In terms of revolutions, θ1 = (15.6 rad)(1 rev/2π rad) = 2.49 rev.
4.36.
Model: Visualize:
Model the particle on the crankshaft as being in nonuniform circular motion.
Solve: (a) The initial angular velocity is ω 0 = 2500 rpm × (1 min/60 s) × (2π rad/rev) = 261.8 rad/s. The crankshaft slows from 261.8 rad/s to 0 in 1.5 s. Thus
ω1 = 0 = ω 0 +
at rω (0.015 m)(261.8 rad/s) = –2.618 m/s 2 = −2.6 m/s 2 t1 ⇒ at = − 0 = − r t1 1.5 s
(b) During these 1.5 s, the crankshaft turns through angle
θ1 = θ 0 + ω 0t1 +
at 2 2.618 m/s 2 t1 = 0 + (261.8 rad/s) (1.5 s) − (1.5 s) 2 = 196 rad 2r 2(0.015 m)
In terms of revolutions, θ1 = (196 rad)(1 rev/π rad) = 31.2 rev.
4.37. Model: The fan is in nonuniform circular motion. Visualize:
Solve:
⎛ min ⎞ Note 1800 rev/min ⎜ ⎟ = 30 rev/s. Thus ⎝ 60 s ⎠
ω f = ωi + αΔt ⇒ 30 rev/s = 0 rev/s + α ( 4.0 s ) ⇒ α =7.5 rev/s 2 . This can be expressed as 2π rad ⎞ 2 ⎟ = 47 rad/s . ⎝ rev ⎠ Assess: An increase in the angular velocity of a fan blade by 7.5 rev/s each second seems reasonable.
( 7.5 rev/s ) ⎛⎜
4.38. Model: The wheel is in nonuniform circular motion. Visualize:
Solve:
(a) Express ωi in rad/s:
⎛ min ⎞⎛ 2π rad ⎞ ⎟⎜ ⎟ ⇒ 5.2 rad/s ⎝ 60 s ⎠⎝ rev ⎠
ωi = ( 50 rev/min ) ⎜
After 10 s, ω f = ωi + αΔt ⇒ ω f = 5.2 rad/s + ( 0.50 rad/s 2 ) (10 s ) = 55 rad/s. Converting to rpm, 2
( 55 rad/s ) ⎛⎜
60 s ⎞⎛ rev ⎞ ⎟⎜ ⎟ = 53 rpm ⎝ min ⎠⎝ 2π rad ⎠
(b) In 10 s, the wheel has turned a number of radians
θ f = θi + ω0 Δt + 12 αΔt 2 ⇒ θ f = 0 rad/sec + (5.2 rad/s)(10 s) + 12 (0.50 rad/s 2 )(10 s) 2 = 77 radians. Converting, 77 rad = 12.3 revolutions. Assess: Making a bicycle wheel turn just over 12 revolutions in 10 s when it is initially turning almost one revolution per second to begin with seems attainable by a cyclist.
4.39.
Model: Visualize:
We will assume that constant-acceleration kinematic equations in a plane apply.
G (a) The particle’s position r0 = (9.0 ˆj ) m implies that at t0 the particle’s coordinates are x0 = 0 m and G y0 = 9.0 m. The particle’s position r1 = (20iˆ) m at time t1 implies that x1 = 20 m and y1 = 0 m. This is the
Solve:
position where the wire hoop is located. Let us find the time t1 when the particle crosses the hoop at x1 = 20 m. From the vx -versus-t curve and using the relation x1 = x0 + area of the vx -t graph, we get 20 m = 0 m + area of the vx -t graph = area of the vx -t graph From Figure P4.39 we see that the area of the vx -t graph equals 20 m when t = t1 = 3 s. (b) We can now look at the y-motion to find a y . Note that the slope of the vx -t graph (that is, a y ) is negative
and constant, and we can determine a y by substituting into y3 = y0 + v0 y (t3 − t0 ) + 12 a y (t3 − t0 ) 2 : 0 m = 9 m + 0 m + 12 a y (3 s − 0 s) 2 ⇒ a y = −2 m/s 2 Therefore, v4 y = v0 y + a y (t4 − t0 ) = 0 m/s + (−2 m/s 2 )(4 s − 0 s) = −8 m/s (c)
4.40.
Solve: From the expression for R, Rmax = v02 /g . Therefore,
R= Assess:
Rmax v02 sin 2θ 1 = ⇒ sin 2θ = ⇒ θ = 15° and 75° g 2 2
The discussion of Figure 4.22 explains why launch angles θ and (90° − θ ) give the same range.
4.41.
Model: Visualize:
Solve:
Assume particle motion in a plane and constant-acceleration kinematics for the projectile.
(a) We know that v0 y = v0 sin θ , a y = − g , and v1 y = 0 m/s. Using v1y2 = v02y + 2a y ( y1 − y0 ) ,
0 m 2 /s 2 = v02 sin 2 θ + 2 ( − g ) h ⇒ h =
v02 sin 2 θ 2g
(b) Using Equation 4.19 and the above expression for θ = 30.0° :
h=
( x2 − x0 ) =
( 33.6 m/s )
2
sin 2 30.0°
2 ( 9.8 m/s 2 )
= 14.4 m
( 33.6 m/s ) sin ( 2 × 30.0° ) = 99.8 m v02 sin 2θ = g ( 9.8 m/s2 ) 2
For θ = 45.0° : h=
( 33.6 m/s )
2
sin 2 45.0°
2 ( 9.8 m/s 2 )
( x2 − x0 ) =
( 33.6 m/s )
2
= 28.8 m
sin ( 2 × 45.0° )
( 9.8 m/s2 )
= 115.2 m
For θ = 60.0° : h=
( x2 − x0 ) = Assess:
( 33.6 m/s )
2
sin 2 60.0°
2 ( 9.8 m/s 2 )
( 33.6 m/s )
2
= 43.2 m
sin ( 2 × 60.0° )
2 ( 9.8 m/s 2 )
= 99.8 m
The projectile’s range, being proportional to sin(2θ ), is maximum at a launch angle of 45°, but the
maximum height reached is proportional to sin 2 (θ ). These dependencies are seen in this problem.
4.42.
Model: Visualize:
Solve:
Assume the particle model for the projectile and motion in a plane.
(a) Using y2 = y0 + v0 y ( t2 − t0 ) + 12 a y ( t2 − t0 ) , 2
y2 = 0 m + ( 30 m/s ) sin 60° ( 7.5 s − 0 s ) + 12 ( −9.8 m/s 2 ) ( 7.5 s − 0 s ) = −80.8 m 2
Thus the launch point is 81 m higher than where the projectile hits the ground. (b) Using v12y = v02y + 2a y ( y1 − y0 ) ,
0 m 2 /s 2 = ( 30sin 60° m/s ) + 2 ( −9.8 m/s 2 ) ( y1 − 0 m ) ⇒ y1 = 34.4 m , or y1 = 34 m 2
(c) The x-component is v2 x = v0 x = v0 cos60° = ( 30 m/s ) cos60° = 15 m/s. The y-component is
v2 y = v0 y + a y ( t2 − t0 ) = v0 sin 60° − g ( t2 − t1 ) = ( 30 m/s ) sin 60° − ( 9.8 m/s 2 ) ( 7.5 s − 0 s ) = −47.52 m/s ⇒v=
(15 m/s )
θ = tan −1
v2 y v2 x
2
+ ( −47.52 m/s ) = 50 m/s
= tan −1
2
47.52 = 73° below + x 15
Assess: An angle of 73° made with the ground, as the projectile hits the ground 81 m below its launch point, is reasonable in view of the fact that the projectile was launched at an angle of 60°.
4.43.
Model: plane. Visualize:
Solve:
Assume the particle model and motion under constant-acceleration kinematic equations in a
(a) Using y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) , 2
0 m = 1.80 m + v0 sin 40° ( t1 − 0 s ) + 12 ( −9.8 m/s 2 ) ( t1 − 0 s )
2
= 1.80 m + ( 7.713 m/s ) t1 − ( 4.9 m/s 2 ) t12 ⇒ t1 = − 0.206 s and 1.780 s The negative value of t1 is unphysical for the current situation. Using t1 = 1.780 s and x1 = x0 + v0 x ( t1 − t0 ) , we get
x1 = 0 + ( v0 cos 40° m/s )(1.780 s − 0 s ) = (12.0 m/s ) cos 40° (1.78 s ) = 16.36 m (b) We can repeat the calculation for each angle. A general result for the flight time at angle θ is
)
(
t1 = 12sin θ + 144sin 2 θ + 35.28 / 9.8 s and the distance traveled is x1 = (12.0)cosθ × t1. We can put the results in a table.
θ
t1
x1
40.0° 42.5° 45.0° 47.5°
1.780 s 1.853 s 1.923 s 1.990 s
16.36 m 16.39 m 16.31 m 16.13 m
Maximum distance is achieved at θ ≈ 42.5°. Assess: The well-known “fact” that maximum distance is achieved at 45° is true only when the projectile is launched and lands at the same height. That isn’t true here. The extra 0.03 m = 3 cm obtained by increasing the angle from 40.0° to 42.5° could easily mean the difference between first and second place in a world-class meet.
4.44.
Model: Visualize:
The golf ball is a particle following projectile motion.
(a) The distance traveled is x1 = v0 xt1 = v0 cosθ × t1. The flight time is found from the y-equation, using the fact that the ball starts and ends at y = 0 :
y1 − y0 = 0 = v0 sin θ t1 − 12 gt12 = (v0 sinθ − 12 gt1 ) t1 ⇒ t1 =
2v0 sin θ g
Thus the distance traveled is x1 = v0 cosθ ×
2v0 sin θ 2v02sin θ cosθ = g g
For θ = 30°, the distances are
( x1 )earth =
2v02sin θ cosθ 2(25 m/s) 2sin30°cos30° = = 55.2 m g earth 9.80 m/s2
( x1 ) moon =
2v02sin θ cosθ 2v02sin θ cosθ 2v02sin θ cosθ = = 6 × = 6( x1 )earth = 331.2 m 1 g moon g earth 6 g earth
The golf ball travels 331.2 m − 55.2 m = 276 m farther on the moon than on earth. (b) The flight times are (t1 )earth =
2v0 sin θ = 2.55 s g earth
(t1 ) moon =
2v0 sin θ 2v0 sinθ = 1 = 6(t1 )earth = 15.30 s g moon g 6 earth
The ball spends 15.30 s − 2.55 s = 12.75 s longer in flight on the moon.
4.45.
Model: Visualize:
Solve:
The particle model for the ball and the constant-acceleration equations of motion are assumed.
(a) Using y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) , 2
h = 0 m + ( 30 m/s ) sin 60° ( 4 s − 0 s ) + 12 ( −9.8 m/s 2 ) ( 4 s − 0 s ) = 25.5 m 2
The height of the cliff is 26 m. (b) Using ( v y2 ) = v y2 + 2a y ( ytop − y0 ) , top
0 m /s = ( v0 sin θ ) + 2 ( − g ) ( ytop ) ⇒ ytop 2
2
2
( v sinθ ) = 0 2g
2
⎡( 30 m/s ) sin 60°⎤⎦ =⎣ = 34.4 m 2 ( 9.8 m/s 2 ) 2
The maximum height of the ball is 34 m. (c) The x and y components are v1 y = v0 y + a y ( t1 − t0 ) = v0 sin θ − gt1 = ( 30 m/s ) sin 60° − ( 9.8 m/s 2 ) × ( 4.0 s ) = −13.22 m/s
v1x = v0 y = v0 cos60° = ( 30 m/s ) cos60° = 15.0 m/s ⇒ v1 = v12x + v12y = 20.0 m/s
The impact speed is 20 m/s. Assess: Compared to a maximum height of 34.4 m, a height of 25.5 for the cliff is reasonable.
4.46.
Model: assumed. Visualize:
Solve:
The particle model for the ball and the constant-acceleration equations of motion in a plane are
The initial velocity is
v0 x = v0 cos5.0° = ( 20 m/s ) cos5.0° = 19.92 m/s v0 y = v0 sin 5.0° = ( 20 m/s ) sin 5.0° = 1.743 m/s The time it takes for the ball to reach the net is
x1 = x0 + v0 x ( t1 − t0 ) ⇒ 7.0 m = 0 m + (19.92 m/s )( t1 − 0 s ) ⇒ t = 0.351 s The vertical position at t1 = 0.351 s is y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 )
2
= ( 2.0 m ) + (1.743 m/s )( 0.351 s − 0 s ) + 12 ( −9.8 m/s 2 ) ( 0.351 s − 0 s ) = 2.01 m 2
Thus the ball clears the net by 1.01 m. Assess: The vertical free fall of the ball, with zero initial velocity, in 0.351 s is 0.6 m. The ball will clear by approximately 0.4 m if the ball is thrown horizontally. The initial launch angle of 5° provides some initial vertical velocity and the ball clears by a larger distance. The above result is reasonable.
4.47. Model: The particle model for the ball and the constant-acceleration equations of motion in a plane are assumed. Visualize:
Solve:
(a) The time for the ball to fall is calculated as follows:
y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 )
2
⇒ 0 m = 4 m + 0 m + 12 ( −9.8 m/s 2 ) ( t1 − 0 s ) ⇒ t1 = 0.9035 s 2
Using this result for the horizontal velocity:
x1 = x0 + v0 x ( t1 − t0 ) ⇒ 25 m = 0 m + v0 x ( 0.9035 s − 0 s ) ⇒ v0 x = 27.7 m/s The friend’s pitching speed is 28 m/s. (b) We have v0 y = ± v0 sinθ , where we will use the plus sign for up 5° and the minus sign for down 5°. We can
write y1 = y0 ± v0 sin θ ( t1 − t0 ) −
g g 2 ( t1 − t0 ) ⇒ 0 m = 4 m ± v0 sinθ t1 − t12 2 2
Let us first find t1 from x1 = x0 + v0 x ( t1 − t0 ) : 25 m = 0 m + v0 cosθ t1 ⇒ t1 =
25 m v0 cosθ
Now substituting t1 into the y-equation above yields ⎛ 25 m ⎞ g ⎛ 25 m ⎞ 0 m = 4 m ± v0 sin θ ⎜ ⎟− ⎜ ⎟ ⎝ v0 cosθ ⎠ 2 ⎝ v0 cosθ ⎠
2
g ( 25 m ) ⎧⎪ 1 ⎪⎫ ⎨ ⎬ = 22.3 m/s and 44.2 m/s 2 2cos θ ⎩⎪ 4 m ± ( 25 m ) tan θ ⎭⎪ 2
⇒ v02 =
The range of speeds is 22 m/s to 44 m/s, which is the same as 50 mph to 92 mph. Assess: These are reasonable speeds for baseball pitchers.
4.48.
Model: Visualize:
Solve:
We will use the particle model and the constant-acceleration kinematic equations in a plane.
The x- and y-equations of the ball are
x1B = x0B + ( v0B ) x ( t1B − t0B ) + 12 ( aB ) x ( t1B − t0B ) ⇒ 65 m = 0 m + ( v0B cos30° ) t1B + 0 m 2
2 y1B = y0B + ( v0B ) y ( t1B − t0B ) + 12 ( aB ) y ( t1B − t0B ) ⇒ 0 m = 0 m + ( v0B sin 30° ) t1B + 12 ( − g ) t1B
2
From the y-equation,
v0B =
gt1B
( 2sin 30° )
Substituting this into the x-equation yields 2 g cos30° t1B 2sin 30° ⇒ t1B = 2.77 s
65 m =
For the runner:
t1R =
20 m = 2.50 s 8.0 m/s
Thus, the throw is too late by 0.27 s. Assess: The times involved in running the bases are small, and a time of 2.5 s is reasonable.
4.49.
Model: Visualize:
Use the particle model for the ball and the constant-acceleration kinematic equations.
Solve: (a) The distance from the ground to the peak of the house is 6.0 m. From the throw position this distance is 5.0 m. Using the kinematic equation v12y = v02y + 2a y ( y1 − y0 ) ,
0 m 2 /s 2 = v02y + 2 ( −9.8 m/s 2 ) ( 5.0 m − 0 m ) ⇒ v0 y = 9.899 m/s The time for up and down motion is calculated as follows: y2 = y0 + v0 y ( t2 − t0 ) + 12 a y ( t2 − t0 ) ⇒ 0 m = 0 m + ( 9.899 m/s ) t2 − 12 ( 9.8 m/s 2 ) t22 ⇒ t2 = 0 s and 2.02 s 2
The zero solution is not of interest. Having found the time t2 = 2.02 s, we can now find the horizontal velocity needed to cover a displacement of 18.0 m:
x2 = x0 + v0 x ( t2 − t0 ) ⇒ 18.0 m = 0 m + v0 x ( 2.02 s − 0 s ) ⇒ v0 x = 8.911 m/s ⇒ v0 =
(8.911 m/s )
2
+ ( 9.899 m/s ) = 13.3 m/s 2
G (b) The direction of v0 is given by
θ = tan −1
v0 y v0 x
= tan −1
9.899 = 48° 8.911
Assess: Since the maximum range corresponds to an angle of 45°, the value of 48° corresponding to a range of 18 m and at a modest speed of 13.3 m/s is reasonable.
4.50.
Model: equations. Visualize:
Solve:
We will assume a particle model for the sand, and use the constant-acceleration kinematic
Using the equation x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) , 2
x1 = 0 m + ( v0 cos15° )( t1 − 0 s ) + 0 m = ( 60 m/s ) (cos15°)t1 We can find t1 from the y-equation, but note that v0 y = −v0 sin15° because the sand is launched at an angle below
horizontal. y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) ⇒ 0 m = 3.0 m − ( v0 sin15° ) t1 − 12 gt12 2
= 3.0 m − ( 6.0 m/s ) (sin15°)t1 − 12 ( 9.8 m/s 2 ) t12
⇒ 4.9t12 + 1.55t1 − 3.0 = 0 ⇒ t1 = 0.6399 s and − 0.956 s (unphysical) Substituting this value of t1 in the x-equation gives the distance
d = x1 = ( 6.0 m/s ) cos15° ( 0.6399 s ) = 3.71 m
4.51.
Model: We will assume a particle model for the cannonball, and apply the constant-acceleration kinematic equations. Visualize:
Solve:
(a) The cannonball that was accidentally dropped can be used to find the height of the wall: 2 y1A = y0A + ( v0A ) y ( t1A − t0A ) + 12 ( aA ) y ( t1A − t0A ) ⇒ 0 m = y0A + 0 m − 12 gt1A 2
⇒ y0A =
1 2
( 9.8 m/s ) (1.5 s ) 2
2
= 11.03 m
For the cannonball that was shot: (v0S ) x = v0S cos30° = (50 m/s)cos30° = 43.30 m/s
(v0S ) y = v0s sin 30° = (50 m/s)sin 30° = 25.0 m/s We can now find the time it takes the cannonball to hit the ground: y2S = y0S + (v0S ) y (t2S − t0S ) + 12 (aS ) y (t2S − t0S ) 2
⇒ 0 m = (11.03 m) + (25.0 m/s)t2S −
(9.8 m/s 2 ) 2 t2S 2
2 ⇒ (4.9 m/s 2 )t2S − (25.0 m/s)t2S − (11.03 m) = 0 ⇒ t2S = 5.51 s
There is also an unphysical root t2S = −0.41 s. Using this time t2S , we can now find the horizontal distance from the wall as follows: x2s = x0s + (v0s ) x (t2s − t0s ) = 0 m + (43.30 m/s)(5.51 s) = 239 m The cannonball hits the ground 2.4 × 10 2 m from the castle wall. (b) At the top of the trajectory (v1S ) y = 0 m/s. Using (v1S ) 2y = (v0S ) 2y − 2 g ( y1S − y0S ),
0 m 2 /s 2 = (25.0 m/s)2 − 2(9.8 m/s 2 )( y1s − 11.03 m) ⇒ y1s = 42.9 m The maximum height above the ground is 43 m. Assess: In view of the fact that the cannonball has a speed of approximately 110 mph, a distance of 239 m for the cannonball to hit the ground is reasonable.
4.52.
Model: Visualize:
Solve:
We will use the particle model and the constant-acceleration kinematic equations for the car.
(a) The initial velocity is
v0 x = v0 cosθ = (20 m/s)cos 20° = 18.79 m/s
v0 y = v0 sin θ = (20 m/s)sin 20° = 6.840 m/s Using y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2 ,
0 m = 30 m + (6.840 m/s)(t1 − 0 s) + 12 (−9.8 m/s 2 )(t1 − 0 s)2 ⇒ 4.9t12 − 6.840t1 − 30 = 0 The positive root to this equation is t1 = 3.269 s. The negative root is physically unreasonable in the present case. Using x1 = x0 + v0 x (t1 − t0 ) + 12 ax (t1 − t0 )2 , we get
x1 = 0 m + (18.79 m/s)(3.269 s − 0 s) + 0 = 61.4 m The car lands 61 m from the base of the cliff. (b) The components of the final velocity are v1x = v0 x = 18.79 m/s and v1 y = v0 y + a y (t1 − t0 ) = 6.840 m/s − (9.8 m/s 2 )(3.269 s − 0 s) = −25.2 m/s
⇒ v = (18.79 m/s) + (−25.2 m/s) 2 = 31.4 m/s The car’s impact speed is 31 m/s. Assess: A car traveling at 45 mph and being driven off a 30-m high cliff will land at a distance of approximately 200 feet (61.4 m). This distance is reasonable.
4.53.
Model: Visualize:
Solve:
Use the particle model for the cat and apply the constant-acceleration kinematic equations.
The relative velocity of the cat from the mouse’s reference frame is
(4.0cos30° − 1.5) m/s = 1.964 m/s = v0 Thus, v0 x = v0 = 1.964 m/s and v0 y = v0 sin 30° = (4.0 m/s)sin 30° = 2.0 m/s . The time for the cat to land on the floor is found as follows: y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2 ⇒ 0 m = 0 m + (2.0 m/s)t1 + 12 (−9.8 m/s 2 )t12
⇒ t1 = 0 s (trivial solution) and 0.408 s The horizontal distance covered in time t1 is
x1 = x0 + v0 x (t1 − t0 ) + 12 ax (t1 − t0 ) 2 = 0 m + (1.964 m/s)(0.408 s) = 0.802 m That is, the cat should leap when he is 80 cm behind the mouse.
4.54.
Model: Assume motion along the x-direction. Let the earth frame be S and a frame attached to the staple gun be S′. Frame S′ moves relative to S with velocity Vx . Solve:
The velocity vx of the parts to be stapled in frame S is +3.0 m/s. Also Vx = −1.0 m/s. Using
vx = v′x + Vx , we get v′x = vx − Vx = +3.0 m/s − (−1.0 m/s) = 4.0 m/s That is, the velocity of the parts in the frame of the staple gun is 4.0 m/s. In this frame 10 staples are fired per second. That is, 4.0 m/s =
distance between two staples distance = ⇒ Distance = 0.4 m 1 time between two staples s 10
Assess: Note that the staple gun is in frame S′ and we had to find the velocity of the moving parts in this frame to solve the problem.
G Model: If a frame S′ is in motion with velocity V relative to another frame S and has a displacement G G G G G R relative to S, the positions and velocities (r and v ) in S are related to the positions and velocities (r ′ and v′) G G G G G in S ′ as r = r ′ + R and v = v′ + V . In the present case, ship A is frame S and ship B is frame S′. Both ships have a common origin at t = 0 s. The position and velocity measurements are made in S and S′ relative to their origins.
4.55.
Solve: (a) The velocity vectors of the two ships are:
G
G vA = (20 mph)[cos30°iˆ − sin 30° ˆj ] = (17.32 mph)iˆ − (10.0 mph) ˆj G vB = (25 mph)[cos 20°iˆ + sin 20° ˆj ] = (23.49 mph)iˆ + (8.55 mph) ˆj
G
Since r = v Δt ,
G
G
K
As rA = rB + R,
G G rA = vA (2 h) = (34.64 miles)iˆ − (20.0 miles) ˆj G G rB = vB (2 h) = (46.98 miles)iˆ + (17.10 miles) ˆj G G G R = rA − rB = (−12.34 miles)iˆ − (37.10 miles) ˆj ⇒ R = 39.1 miles
The distance between the ships two hours after they depart is 39 miles. G G G (b) Because vA = vB + V , G G G V = vA − vB = −(6.17 mph)iˆ − (18.55 mph) ˆj ⇒ V = 19.5 mph The speed of ship A as seen by ship B is 19.5 mph. Assess: The value of the speed is reasonable.
4.56. Model: Let the earth frame be S and a frame attached to the water be S′. Frame S′ moves relative to S with velocity V. We define the x-axis along the direction of east and the y-axis along the direction of north for both frames. Solve: (a) The kayaker’s speed of 3.0 m/s is relative to the water. Since he’s being swept toward the east, he needs to point at angle θ west of north. In frame S′, the water frame, his velocity is G v′ = (3.0 m/s, θ west of north) = (−3.0sinθ m/s)iˆ + (3.0cosθ m/s)ˆj G G G G We can find his velocity in earth frame S from the Galilean transformation v = v′ + V , with V = (2.0 m/s)iˆ. Thus G v = ((−3.0sin θ + 2.0) m/s)iˆ + (3.0cosθ m/s)ˆj In order to go straight north in the earth frame, the kayaker needs vx = 0. This will be true if sin θ =
2.0 3.0
⎛ 2.0 ⎞ ⇒ θ = sin −1 ⎜ ⎟ = 41.8° ⎝ 3.0 ⎠
Thus he must paddle in a direction 42° west of north. (b) His northward speed is v y = 3.0 cos(41.8°) m/s = 2.236 m/s. The time to cross is t=
The kayaker takes 45 s to cross.
100 m = 44.7 s 2.236 m/s
Model: Let Mike’s frame be S and Nancy’s frame be S′. Frame S′ moves relative to S with velocity V. x is the horizontal direction and y is the vertical direction for motion. The frames S and S′ coincide at t = 0 s. G G G Solve: (a) According to the Galilean transformation of velocity v = v′ + V . Mike throws the ball with velocity G G v = (22 m/s)cos63°iˆ + (22 m/s)sin63° ˆj, and V = (30 m/s)iˆ. Thus in Nancy’s frame G G G v′ = v − V = (22cos63° − 30)iˆ m/s + (22sin 63°) ˆj m/s = (−20.0iˆ + 19.6ˆj ) m/s
4.57.
θ = tan −1
v′y 19.6 m/s = tan −1 = 44.4° 20.0 m/s v′x
The direction of the angle is 44.4° above the − x′ axis (in the second quadrant). (b) In Nancy’s frame S′ the equation x′ = x0′ + v′x (t ′ − t0′ ) + 12 a′x (t ′ − t0′ ) 2 becomes x′ = −(20.0 m/s)t ′
and the equation y′ = y0′ + v′y (t ′ − t0′ ) + 12 a′y (t ′ − t0′ ) 2 becomes y′ = 0 m + (19.6 m/s)t ′ − 12 (9.8 m/s 2 )t ′2 = (19.6 m/s)t ′ − (4.9 m/s 2 )t ′2
Model: Let the earth frame be S and a frame attached to the sailboat be S′. Frame S′ moves relative to G S with velocity V = 8.0iˆ mph. G G G G Solve: (a) From Equation 6.24, v = v′ + V , where v′ = (12.0 ) cos 45°iˆ mph + (12.0 ) sin 45° ˆj mph is the velocity
4.58.
of the wind in S′, or the apparent wind velocity. Thus, G v = (12.0 ) cos 45°iˆ mph + (12.0 ) sin 45° ˆj mph + ( 8.0 ) iˆ mph
(
)
= ( 8.0 + 12.0cos45° ) iˆ mph + (12.0sin45° ) ˆj mph = 16.48iˆ + 8.48ˆj mph
θ = tan −1
8.48 mph = 27.2° 16.48 mph
The wind speed is v = 18.5 mph and the direction is from 27.2° south of west. G G (b) In this case, v′ = − (12.0 mph ) cos 45°iˆ − (12.0 mph ) sin 45° ˆj and V = ( 8.0 mph ) iˆ. So, G v = ⎡⎣ − (12.0 mph ) cos 45° + 8.0 mph ⎤⎦ iˆ − (12.0 mph ) sin 45° ˆj = −0.485iˆ − 8.485 ˆj
θ = tan −1
8.485 mph = 86.7° 0.485 mph
The speed is 8.5 mph and the direction is from 86.7° north of east or from 3.3° east of north.
Model: Let the ground frame be S and the car frame be S′. S′ moves relative to S with a velocity V along the x-direction. G G G G G Solve: The Galilean transformation of velocity is v = v′ + V where v and v′ are the velocities of the raindrops G in frames S and S′. While driving north, V = ( 25 m/s ) iˆ and v = −vR cosθ ˆj − vR sin θ iˆ. Thus, G G G v′ = v − V = ( −vR sinθ − 25 m/s ) iˆ − vR cosθ ˆj
4.59.
Since the observer in the car finds the raindrops making an angle of 38° with the vertical, we have
vR sin θ + 25 m/s = tan 38° vR cosθ G G While driving south, V = − ( 25 m/s ) iˆ, and v = −vR cosθ ˆj − vR sin θ iˆ. Thus, G v′ = ( −vR sin θ + 25 m/s ) iˆ − vR cosθ ˆj
Since the observer in the car finds the raindrops falling vertically straight, we have −vR sinθ + 25 m/s = tan 0° = 0 ⇒ vR sinθ = 25 m/s vR cosθ Substituting this value of vR sinθ into the expression obtained for driving north yields: 25 m/s + 25 m/s 50 m/s = tan 38° ⇒ vR cosθ = = 64.0 m/s vR cosθ tan 38° Therefore, we have for the velocity of the raindrops:
( vR sinθ )
2
+ ( vR cosθ ) = ( 25 m/s ) + ( 64.0 m/s ) ⇒ vR2 = 4721( m/s ) ⇒ vR = 68.7 m/s 2
2
tan θ =
2
vR sin θ 25 m/s = ⇒ θ = 21.3° vR cosθ 64 m/s
The raindrops fall at 69 m/s while making an angle of 21° with the vertical.
2
4.60. Model: Let the ground be frame S and the wind be frame S′. S′ moves relative to S. The ground frame S has the x-axis along the direction of east and the y-axis along the direction of north. Visualize:
Solve:
G G G (a) The Galilean velocity transformation is v = v′ + V , where G V = ( 50 mph ) cos30°iˆ + ( 50 mph ) sin 30° ˆj G v′ = ( 200 mph ) cosθ iˆ − ( 200 mph ) sin θ ˆj
G G Thus, v = ⎡⎣( 50cos30° + 200cosθ ) iˆ + ( 50sin 30° − 200sinθ ) ˆj ⎤⎦ mph. Because v should have no j-component, 50sin 30° − 200sin θ = 0 ⇒ θ = 7.18°
G (b) The pilot must head 7.18° south of east. Substituting this value of θ in the v equation gives 600 mi G v = ( 242 ) iˆ mph, along the direction of east. At a speed of 242 mph, the trip takes t = = 2.48 h 242 mph
4.61. Solve:
Model: We will use the particle model for the test tube which is in nonuniform circular motion. (a) The radial acceleration is 2
rev 1 min 2π rad ⎞ ⎛ 4 2 × × ar = rω 2 = ( 0.1 m ) ⎜ 4000 ⎟ = 1.75 × 10 m/s min 60 s 1 rev ⎠ ⎝ (b) An object falling 1 meter has a speed calculated as follows:
v12 = v02 + 2a y ( y1 − y0 ) = 0 m + 2 ( −9.8 m/s 2 ) ( −1.0 m ) ⇒ v1 = 4.43 m/s When this object is stopped in 1 × 10−3 s upon hitting the floor, v2 = v1 + a y ( t2 − t1 ) ⇒ 0 m s = −4.43 m s + a y (1× 10−3 s ) ⇒ a y = 4.4 × 103 m/s 2 This result is one-fourth of the above radial acceleration. Assess: The radial acceleration of the centrifuge is large, but it is also true that falling objects are subjected to large accelerations when they are stopped by hard surfaces.
4.62. Solve:
Model: We will use the particle model for the astronaut undergoing nonuniform circular motion. (a) The initial conditions are ω 0 = 0 rad/s, θ 0 = 0 rad, t0 = 0 s, and r = 6.0 m. After 30 s,
ω1 =
1 rev 1 rev 2π rad = × × = 4.83 rad/s 1.3 s 1.3 s rev
Using these values at t1 = 30 s,
ω1 = ω 0 + ( at r )( t1 − t0 ) = 0 + ( at r ) t1 ⎛ 1 ⎞ 2 ⇒ at = ( 6.0 m )( 4.83 rad/s ) ⎜ ⎟ = 0.97 m/s 30 s ⎝ ⎠
(b) The radial acceleration is
ar = rω12 = ( 6.0 m )( 4.83 rad/s ) Assess:
2
g = 14.3g 9.8 ( m/s2 )
The above acceleration is typical of what astronauts experience during liftoff.
4.63.
Model: Visualize:
Model the car as a particle in nonuniform circular motion.
Note that the tangential acceleration stays the same at 1.0 m/s2. As the tangential velocity increases, the radial acceleration increases as well. After a time t1 , as the car goes through an angle θ1 − θ 0 , the total acceleration will increase to 2.0 m/s2. Our objective is to find this angle. Solve: Using v1 = v0 + at ( t1 − t0 ) , we get
v1 = 0 m/s + (1.0 m/s 2 ) ( t1 − 0 s ) = (1.0 m/s 2 ) t1 2 2 rv 2 (1.0 m/s ) t1 t2 ⇒ ar = 21 = = 1 ( m/s 4 ) r 120 m 120 2
⇒ atotal = 2.0 m/s = a + a = 2
2 t
2 r
(1.0 m/s )
2 2
2
⎡ t2 ⎤ + ⎢ 1 ( m/s 4 ) ⎥ ⇒ t1 = 14.4 s 120 ⎣ ⎦
We can now determine the angle θ1 using 2 ⎛ at ⎞ ⎟ ( t1 − t0 ) r ⎝ ⎠
θ1 = θ 0 + ω 0 ( t1 − t0 ) + 12 ⎜ = 0 rad + 0 rad +
2 1 (1.0 m/s ) 2 (14.4 s ) = 0.864 rad = 49.5° 2 (120 m )
The car will have traveled through an angle of 50°.
4.64. Model: The earth is a rigid, rotating, and spherical body. Visualize:
Solve:
At a latitude of θ degrees, the radius is r = Re cos θ with Re = 6400 km = 6.400 × 106 m.
(a) In Miami θ = 26°, and we have r = (6.400 × 106 m)(cos 26°) = 5.752 × 106 m . The angular velocity of the earth is
ω=
2π 2π = = 7.272 × 10−5 rad/s T 24 × 3600 s
Thus, vstudent = rω = (5.752 × 106 m)(7.272 × 10−5 rad/s) = 418 m/s. (b) In Fairbanks θ = 65°, so r = (6.400 × 106 m)cos 65° = 2.705 × 106 m and vstudent = rω = (2.705 × 106 m)(7.272 × 10−5 rad/s) = 197 m/s.
4.65. Model:
The satellite is a particle in uniform circular motion.
Visualize:
Solve: (a) The satellite makes one complete revolution in 24 h about the center of the earth. The radius of the motion of the satellite is r = 6.37 × 106 m + 3.58 × 107 m = 4.22 × 107 m The speed of the satellite is v =
( distance traveled ) = 2π r = 3.07 × 103 m/s. 24 h ( time taken )
(b) The acceleration of the satellite is centripetal, with magnitude 3 v 2 ( 3.07 × 10 m/s ) ar = = = 0.223 m/s 2 r 4.22 × 107 m Assess: The small centripetal acceleration makes sense when realized it is for an object traveling in a circle with radius ≈ 26,400 miles. 2
4.66. Model: The magnetic computer disk is a rigid rotating body. Visualize:
Solve:
Using the rotational kinematic equation ω f = ωi + αΔt , we get
ω1 = 0 rad + (600 rad/s 2 )(0.5 s − 0 s) = 300 rad/s ω 2 = (300 rad/s) + (0 rad/s 2 )(1.0 s − 0.5 s) = 300 rad/s The speed of the painted dot v2 = rω 2 = (0.04 m)(300 rad/s) = 12 m/s. The number of revolutions during the time interval t0 to t2 is 1 2 1 θ 2 = θ1 + ω1 (t2 − t1 ) + α1 (t2 − t1 ) 2 2
1 2
θ1 = θ 0 + ω0 (t1 − t0 ) + α 0 (t1 − t0 ) 2 = 0 rad + 0 rad + (600 rad/s 2 )(0.5 s − 0 s) 2 = 75 rad
⎛ 1 rev ⎞ = 75 rad + (300 rad/s)(1.0 s − 0.5 s) + 0 rad = 225 rad = (225 rad) ⎜ ⎟ = 35.8 rev ⎝ 2π rad ⎠
4.67. Model: The drill is a rigid rotating body. Visualize:
The figure shows the drill’s motion from the top. Solve: (a) The kinematic equation
ωf = ωi + α (tf − ti )
becomes,
ωi = 2400 rpm = (2400)(2π )/60 = 251.3 rad/s, tf − ti = 2.5 s − 0 s = 2.5 s, and wf = 0 rad/s, 0 rad = 251.3 rad/s + α (2.5 s) ⇒ α = −100 rad/s 2
(b) Applying the kinematic equation for angular position yields:
1 2
θ f = θ i + ωi (tf − ti ) + α (tf − ti ) 2 1 = 0 rad + (251.3 rad/s)(2.5 s − 0 s) + (−100 rad/s2 )(2.5 s − 0 s)2 2 = 3.2 × 102 rad = 50 rev
after
using
4.68. Model: The turbine is a rigid rotating body. The known values are ωi = 3600 rpm = (3600)(2π )/60 = 120π rad/s, ti = 0 s, tf = 10 min = 600 s, ω f = 0 rad/s, and θ i = 0 rad Using the rotational kinematic equation ω f = ωi + α (tf − ti ), we get
Solve:
0 rad = (120π rad/s) + α (600 s − 0 s). Thus, α = −0.628 rad/s 2 . Now,
1 2
θ f = θ i + ωi (tf − ti ) + α (tf − ti ) 2 1 = 0 rad + (120π rad/s)(600 s − 0 s) + (−0.628 rad/s 2 )(600 s − 0 s) 2 2 = 113,100 rad = 1.80 × 104 rev Assess:
18,000 revolutions during 10 minutes when the starting angular velocity is 3600 rpm is reasonable.
4.69. Visualize: Please refer to Figure P4.69. Solve: Since ω f = ωi + (area under α vs t curve), at t = 3 s, the angular velocity is ω f = 60 rpm + ( 4.0 rad/s 2 ) ( 2 s − 1 s ) ⎛ 20 rev 60 s ⎞ = 60 rpm + ( 4 rad/s ) ⎜ × ⎟ ⎝ 2π rad 1 min ⎠ = 60 rpm + 38 rpm = 98 rpm
4.69. Visualize: Please refer to Figure P4.69. Solve: Since ω f = ωi + (area under α vs t curve), at t = 3 s, the angular velocity is ω f = 60 rpm + ( 4.0 rad/s 2 ) ( 2 s − 1 s ) ⎛ 20 rev 60 s ⎞ = 60 rpm + ( 4 rad/s ) ⎜ × ⎟ ⎝ 2π rad 1 min ⎠ = 60 rpm + 38 rpm = 98 rpm
4.70. Model: The car’s wheels are in nonuniform circular motion with constant angular acceleration. Solve: The problem must be solved in two steps: first, the angular acceleration α is found for the case of stopping in one revolution, then α is applied to the case of stopping with twice the initial angular velocity ωi . To find α for the case of stopping in 1.0 revolutions, set Δθ = 1.0 rev = 2π rad in ω f2 = ω i2 + 2αΔθ ⇒ ⎛ ω i2 ⎞ 0 m = ωi2 + 2α (2π rad) ⇒ α = − ⎜ ⎟ ⎝ 4π rad ⎠ Now use the same relationship, with that value of α , but this time Δθ is unknown and the initial angular velocity has been doubled.
⎛ −ωi2 ⎞ ωi2 2 2 Δθ ⇒ Δθ = 8π rad = 4.0 rev 0 m = ( 2ωi ) + 2 ⎜ ⎟ Δθ ⇒ 0 m = 4ω i − 2π rad ⎝ 4π rad ⎠ Assess: Doubling the initial velocity quadruples the number of turns the wheels make before locking. This is within reason.
4.71. Model: The bicycle wheel undergoes nonuniform circular motion with constant angular acceleration. Visualize:
Solve:
First find the angular acceleration α , then use it to find θ f . Using kinematics,
ω f = ωi + αΔt ⇒ 0 rpm = 100 rpm + α (1.0 min) ⇒ α = −100 (rpm)/min = −100 rev/min 2 The minus sign indicates the wheel is slowing down. The total number of revolutions the wheel makes while stopping is
θ f = 0 rev + (100 rpm)(1.0 min) + 12 (−100 rev/min 2 )(1.0 min) 2 = 50 rev Assess: A total of 50 revolutions in 60 s is on average less than one revolution per second, which is quite reasonable.
4.72. Model: Treat the rock as a particle in nonuniform circular motion with constant angular acceleration. Visualize:
Solve:
(a) The angular acceleration α is
α=
at −1.00 m/s 2 = = −1.667 rad/s 2 r 0.600 m
The minus sign indicates the angular acceleration is clockwise, as shown in the figure above, opposite to the 3.00 m/s angular velocity. The initial angular velocity is ωi = = 5.00 rad/s. Using angular kinematics, the 0.600 m angular velocity at some time t after braking starts is
ωf ( t ) = ωi + 0αΔt = 5.00 rad/s − 1.667 rad/s ( t − 0 s ) At t = 1.5 s, ω f = 2.50 rad/s while α = −1.667 rad/s. (b) The total acceleration of the rock is a = at2 + ar2 . The value of at = 1.00 m/s 2 is given. The radial
acceleration at a certain time t is ar = multiplying through by r = 0.600 m:
vt2 . Since vt = ω r , we can use the expression for ω f (t ) to find vt (t ) by r
ω f ( t ) r = ( 5.00 rad/s ) r − (1.667 rad/s ) rt ⇒ vt ( t ) = 3.00 m/s − (1.00m/s 2 ) t ⇒ t = The value of vt for a = g is needed. Setting a = g and substituting ar =
3.00 m/s − vt 1.00 m/s 2
vt2 , r
2
⎛ v2 ⎞ v4 g 2 = at2 + ⎜ t ⎟ ⇒ g 2 − at2 = t2 = ( 9.80 m/s 2 ) − (1.00 m/s 2 ) = 95.0 m 2 /s 4 r ⎝ r ⎠ ⇒ vt2 = ( 0.600 m ) 95.0 m 2 /s 4 ⇒ vt = 2.42 m/s 3.00 m/s − ( 2.42 m/s ) = 0.58 s. 1.00 m/s 2 Assess: The time of 0.58 s occurs early during the braking. This is reasonable since an acceleration equal to that supplied by gravity is fairly strong. Finally, t =
4.73. Model: The string is wrapped around the spool in such a way that it does not pile up on itself, and unwinds without slipping. Visualize:
Solve:
Since the string unwinds without slipping, the angular distance the spool turns as the string is pulled 1.0 Δx 1.0 m m is Δθ = = = 33 radians. r 3.0 × 10−2 m The angular acceleration of the spool due to the pull on the string is
α=
at 1.5 m/s 2 = = 50 rad/s 2 r 3.0 × 10−2 m
The angular velocity of the spool after pulling the string is found with kinematics.
ω f2 = ωi2 + 2αΔθ ⇒ ω f2 = 0 rad 2 /s 2 + 2 ( 50 rad/s 2 ) ( 33 rad ) ⇒ ω f2 = 57 rad/s
Converting to revolutions per minute,
( 57 rad/s ) ⎛⎜
rev ⎞⎛ 60 s ⎞ 2 ⎟⎜ ⎟ = 5.5 × 10 rpm π 2 rad min ⎝ ⎠⎝ ⎠
Assess: The angular speed of 57 rad/s ≈ 9 rev/s is reasonable for a medium-sized spool.
4.74.
Solve: (a) A golfer hits an iron shot with a new club as she approaches the green. She is pretty sure, based on past experience, that she hit the ball with a speed of 50 m/s, but she is not sure at what angle the golf ball took flight. She observed that the ball traveled 100 m before hitting the ground. What angle did she hit the ball?
(b) From the second equation, (4.9 m/s 2 )t12 − (50sinθ m/s) t1 = 0 ⇒ t1 = 0 s and t1 =
(50 m/s)sinθ 4.9 m/s 2
Using the above value for t1 in the first equation yields: 100 m =
(50cosθ )(50sinθ ) m 2 /s 2 4.9 m/s 2
⇒ 2cosθ sinθ = sin 2θ =
Assess:
9.8 = 0.392 ⇒ 2θ = 23.1 ⇒ θ = 11.5° 25
Although the original speed is reasonably high (50 m/s = 112 mph), the ball travels a distance of only
100 m, implying either a small launch angle around 10° or an angle closer to 80°. The calculated angle of 11.5° is thus pretty reasonable.
4.75.
Solve: (a) A submarine moving east at 3.0 m/s sees an enemy ship 100 m north of its path. The submarine’s torpedo tube happens to be stuck in a position pointing 45° west of north. The tube fires a torpedo with a speed of 6.0 m/s relative to the submarine. How far east or west of the ship should the sub be when it fires? (b) Relative to the water, the torpedo will have velocity components
vx = −6.0cos 45° m/s + 3.0 m/s = −4.24 m/s + 3 m/s = −1.24 m/s
v y = +6.0cos 45° m/s = +4.2 m/s The time to travel north to the ship is
100 m = ( 4.2 m/s ) t1 ⇒ t1 = 24 s Thus, x = (1.24 m/s )( 24 s ) = −30 m. That is, the ship should be 30 m west of the submarine.
4.76.
Solve: (a) A 1000 kg race car enters a 50 m radius curve and accelerates around the curve for 10.0 s. The forward force provided by the car’s wheels is 1500 N. After 10.0 s the car has moved 125 m around the track. Find the initial and final angular velocities. (b) From Newton’s second law,
Ft = mat ⇒ 1500 N = (1000 kg ) at ⇒ at = 1.5 m/s 2
θ f = θ i + ω it +
Δθ =
Δs 125 m = = 2.5 rad r 50 m
at 2 1.5 m/s 2 2 t ⇒ 2.5 rad = 0 rad + ωi (10 s ) + (10 s ) ⇒ ωi = 0.10 rad/s 2r 2 ( 50 m )
ωf = ωi +
at 1.5 m/s 2 t = 0.1 rad/s + (10 s ) = 0.40 rad/s r 50 m
4.77.
Solve: You decide to test fly your model airplane off of a 125 m tall building. The model’s engine starts fine and gets the airplane moving at 4.0 m/s but quits just as it gets to the edge of the building. The model proceeds to fall “like a rock.” How far from the edge of the building will it crash into the ground? (Assume g = 10 m/s 2 for easier calculation.) Visualize:
Using the equation y1 = y0 + ν 0 yt + 12 a y ( t0 − t1 ) , we get 2
y1 = −(5 m/s 2 ) t12 = −125 m ⇒ t1 =
The distance x1 = ( 4 m/s )( 5 s ) = 20 m.
125 m = 25 s 2 = 5 s 5 m/s 2
4.78.
Model: The ions are particles that move in a plane. They have vertical acceleration while between the acceleration plates, and they move with constant velocity from the plates to the tumor. The flight time will be so small, because of the large speeds, that we’ll ignore any deflection due to gravity. Visualize:
Solve: There’s never a horizontal acceleration, so the horizontal motion is constant velocity motion at vx = 5.0 × 106 m/s. The times to pass between the 5.0-cm-long acceleration plates and from the plates to the tumor are
0.050 m = 1.00 × 10−8 s 5.0 × 106 m/s 1.50 m t2 − t1 = = 3.00 × 10−7 s 5.0 × 106 m/s t1 − t0 = t1 =
Upon leaving the acceleration plates, the ion has been deflected sideways to position y1 and has velocity v1 y . These are y1 = y0 + v0 y t1 + 12 a y t12 = 12 a y t12 v1 y = v0 y + a yt1 = a yt1 In traveling from the plates to the tumor, with no vertical acceleration, the ion reaches position
y2 = y1 + v1 y (t2 − t1 ) = 12 a yt12 + (a yt1 )(t2 − t1 ) = ( 12 t12 + t1 (t2 − t1 ) ) a y We know y2 = 2.0 cm = 0.020 m, so we can solve for the acceleration a y that the ion had while between the plates: ay =
y2 0.020 m =1 = 6.6 × 1012 m/s2 2 −8 t + t1 (t2 − t1 ) 2 (1.00 × 10 s) + (1.00 × 10−8 s)(3.00 × 10−7 s)
1 2 2 1
Assess: This acceleration is roughly 1012 times larger than the acceleration due to gravity. This justifies our assumption that the acceleration due to gravity can be neglected.
4.79.
Model: We will use the particle model for the ball’s motion under constant-acceleration kinematic equations. Note that the ball’s motion on the smooth, flat board is a y = − g sin 20° = −3.352 m/s 2 .
Visualize:
Solve:
The ball’s initial velocity is v0 x = v0 cosθ = ( 3.0 m/s ) cosθ
v0 y = v0 sinθ = ( 3.0 m/s ) sinθ
Using x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) , 2
2.5 m = 0 m + ( 3.0 m/s ) cosθ ( t1 − 0 s ) + 0 m ⇒ t1 =
( 2.5 m ) ( 3.0 m/s ) cosθ
=
0.833 s cosθ
Using y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) and the above equation for t1 , 2
⎛ 0.833 s ⎞ 1 2 ( 0.833 s ) 0 m = 0 m + ( 3.0 m/s ) sinθ ⎜ ⎟ − 2 ( 3.352 m/s ) cos 2 θ ⎝ cosθ ⎠ ⇒ ( 2.5 m )
2
sinθ 1.164 = ⇒ 2.5sin θ cosθ = 1.164 ⇒ 2θ = 68.6° ⇒ θ = 34.3° cosθ cos 2 θ
4.80.
Model: Visualize:
Solve:
Use the particle model for the arrow and the constant-acceleration kinematic equations.
Using v1 y = v0 y + a y ( t1 − t0 ) , we get
v1 y = 0 m/s − gt1 ⇒ v1 y = − gt1 Also using x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) , 2
60 m = 0 m + v0 xt1 + 0 m ⇒ v0 x =
60 m = v1x t1
Since v1 y / v1x = − tan 3° = − 0.0524, using the components of v0 gives − gt1
( 60 m/ t1 )
= − 0.0524 ⇒ t1 =
( 0.0524 )( 60 m ) = 0.566 s
( 9.8 m/s ) 2
Having found t1 , we can go back to the x-equation to obtain v0 x = 60 m/0.566 s = 106 m/s. Assess: In view of the fact that the arrow took only 0.566 s to cover a horizontal distance of 60 m, a speed of 106 m/s or 237 mph for the arrow is understandable.
4.81.
Model: Use the particle model for the arrow and the constant-acceleration kinematic equations. We will assume that the archer shoots from 1.75 m above the slope (about 5′ 9′′). Visualize:
Solve:
For the y-motion:
y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) ⇒ y1 = 1.75 m + ( v0 sin 20° ) t1 − 12 gt12 2
⇒ y1 = 1.75 m + ( 50 m/s ) sin 20°t1 − 12 gt12
For the x-motion: x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) = 0 m + ( v0 cos 20° ) t1 + 0 m = ( 50 m/s ) (cos 20°)t1 2
Because y1 x1 = − tan15° = − 0.268,
1.75 m + ( 50 m/s ) (sin 20°)t1 − 12 gt12 = − 0.268 ⇒ t1 = 6.12 s and −0.058 s (unphysical) ( 50 m/s ) (cos 20°)t1 Using t1 = 6.12 s in the x- and y-equations above, we get y1 = −77.0 m and x1 = 287 m. This means the distance down the slope is
x12 + y12 =
( 287 m )
2
+ ( −77.0 m ) = 297 m. 2
Assess: With an initial speed of 112 mph (50 m/s) for the arrow, which is shot from a 15° slope at an angle of 20° above the horizontal, a horizontal distance of 287 m and a vertical distance of 77.0 m are reasonable numbers.
4.82.
Model: Visualize:
Treat the ball as a particle and apply the constant-acceleration equations of kinematics.
Solve: After the first bounce, the ball leaves the surface at 40° relative to the vertical or 50° relative to the horizontal. We first calculate the time t1 between the second bounce and the first bounce as follows:
x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) ⇒ 3.0 m = 0 m + ( v0 cos50° ) t1 + 0 m ⇒ t1 = 2
3.0 m v0 cos50°
In this time, the ball undergoes a vertical displacement of y1 − y0 = −(3.0 m) tan 20° = −1.092 m. Substituting these values in the equation for the vertical displacement yields: y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 )
2
⎛ 3.0 m ⎞ 1 3.0 m ⎞ 2 ⎛ −1.092 m = 0 m + ( v0 sin 50° ) t1 − 12 gt12 = ( v0 sin 50° ) ⎜ ⎟ − 2 ( 9.8 m/s ) ⎜ ⎟ ⎝ v0 cos50° ⎠ ⎝ v0 cos50° ⎠ ⇒ −1.092 m − 3.575 m = Assess:
−106.73 m3 /s 2 ⇒ v0 = 4.78 m/s, or v0 = 4.8 m/s v02
A speed of 4.8 m/s or 10.7 mph on the first bounce is reasonable.
2
4.83.
Model: Treat the skateboarder as a particle. Visualize: This is a two-part problem. Use an s-axis parallel to the slope for the first part, regular xycoordinates for the second. The skateboarder’s final velocity at the top of the ramp is her initial velocity as she becomes airborne.
Solve: Without friction, the skateboarder’s acceleration on the ramp is a0 = − g sin 30° = −4.90 m/s 2 . The length
of the ramp is s1 = (1.0 m) / sin 30° = 2.0 m. We can use kinematics to find her speed at the top of the ramp:
v12 = v02 + 2a0 ( s1 − s0 ) = v02 + 2a0 s1 ⇒ v1 = (7.0 m/s)2 + 2(−4.90 m/s 2) (2.0 m) = 5.4 m/s This is the skateboarder’s initial speed into the air, giving her velocity components v1x = v1 cos30° = 4.7 m/s and v1 y = v1 cos30° = 2.7 m/s. We can use the y-equation of projectile motion to find her time in the air: y2 = 0 m = y1 + v1 yt2 + 12 a1 yt22 = 1.0 m + (2.7 m/s) t2 − (4.90 m/s 2) t22 This quadratic equation has roots t2 = −0.253 s (unphysical) and t2 = 0.805 s. The x-equation of motion is thus x2 = x1 + v1xt2 = 0 m + (4.7 m/s) t2 = 3.8 m She touches down 3.8 m from the end of the ramp.
4.84.
Model: motion. Visualize:
Solve:
Use the particle model for the motorcycle daredevil and apply the kinematic equations of
We need to find the coordinates of the landing ramp ( x1 , y1 ). We have
x1 = x0 + v0 x ( t1 − t0 ) = 0 m + ( 40 m/s ) t1 y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) = 0 m + 0 m + 12 ( −9.8 m/s 2 ) t12 = − ( 4.9 m/s 2 ) t12 2
G This means we must find t1. Since v1 makes an angle of 20° below the horizontal we can find v1y as follows:
v1 y v1x
= tan 20° ⇒ v1 y = −v1x tan 20° = − ( 40 m/s ) tan 20° = −14.56 m/s
We now use this value of v1y , a y = −9.8 m/s 2 , and v0y = 0 m/s in the following equation to obtain t1: v1 y = v0 y + a y ( t1 − t0 ) ⇒ −14.56 m/s = 0 m/s − ( 9.8 m/s 2 ) t1 ⇒ t1 = 1.486 s Now, we are able to obtain x1 and y1 using the above x- and y-equations:
x1 = ( 40 m/s )(1.486 s ) = 59.4 m
y1 = − ( 4.9 m/s 2 ) (1.486 s ) = −10.82 m 2
That is, the landing ramp should be placed 10.8 m lower and 59 m away from the edge of the horizontal platform.
4.85. Model: The train and projectile are treated in the particle model. The height of the cannon above the tracks is ignored. Visualize:
Solve:
In the ground reference frame, the projectile is launched with velocity components
( v0 x )P = v0 cosθ + vtrain
(v )
0y P
= v0 sinθ
While the projectile is in free fall, vfy = viy − g Δt. The time for the projectile to rise to the highest point is
(with vfy = 0) Δt =
viy g
. So the time to vertically rise and fall is t1 = 2Δt = 2
(v )
0y P
g
=
2v0 sin θ g
During this time the projectile travels a horizontal distance
( x1 )P = ( v0 x )P t1 =
2v0 2 2v v sinθ cosθ + 0 train sinθ g g
During the same time, the train travels a horizontal distance
( x1 )T = ( v0 x )T t1 +
1 2 2v0vtrain 2v 2 a at1 = sin θ + 02 sin 2 θ 2 g g
The range R is the difference between the two horizontal distances:
R = ( x1 )P − ( x1 )T =
2v0 2 ⎛ a 2 ⎞ ⎜ sin θ cosθ − sin θ ⎟ g ⎝ g ⎠
Note that the range is independent of vtrain the train’s steady motion. This makes sense, since the train and projectile share that motion when the projectile is launched. 2v 2 dR = 0. Thus (ignoring the constant 0 ) Maximizing the range R requires g dθ
dR a a = cos 2 θ − sin 2 θ − ( 2sinθ cosθ ) = cos 2θ − sin 2θ = 0 dθ g g Solving for θ ,
sin 2θ g 1 ⎛g⎞ = tan 2θ = ⇒ θ = tan −1 ⎜ ⎟ a cos 2θ 2 ⎝a⎠ Note that
a > 0 (train speeding up) gives θ < 45D and
a < 0 (train slowing down) gives θ > 45D ⎛g⎞ since tan −1 ⎜ ⎟ will be in the 2nd quadrant. ⎝a⎠ Assess: As a check, see what the angle θ is for the limiting case in which the train does not accelerate: 1 1 a = 0 ⇒ θ = tan −1 ( ∞ ) = × 90° = 45° 2 2 This is the expected answer.
4.86.
Model: Let the earth be frame S and the river be frame S′. Assume the river flows toward the east, which is the x and x′-axis. Visualize:
Solve:
In frame S′, the river’s frame, the child is at rest. The boat can go directly to the child at angle
θ = tan (200 /1500) = 7.595°. The boat’s speed is 8.0 m/s, so the components of the boat’s velocity in S′ are −1
v′x = −(8.0 m/s)cos7.595° = −7.93 m/s
v′y = (8.0 m/s)sin 7.595° = 1.06 m/s →
The river flows with velocity V = 2.0iˆ m/s relative to the earth. In the earth’s frame, which is also the frame of the riverbank and the boat dock, the boat’s velocity is vx = v′x = Vx = −5.93 m/s and v y = v′y + Vy = 1.06 m/s Thus the boat’s angle with respect to the riverbank is θ = tan −1 (5.93/1.06) = 10.1°. Assess: The boat, like the child, is being swept downstream. This moves the boat’s angle away from the shore.
4.87. Model: The airplanes are modeled as particles, and are undergoing relative motion according to the Galilean transformations of position and velocity. We designate Uri’s plane as frame S′ and the earth as frame S. G Frame S′ moves relative to frame S with velocity V . Visualize:
G G G G According to the Galilean transformation of velocity v = v′ + V , where v is the velocity of Val’s plane G G relative to the earth, v′ is the velocity of Val’s plane relative to Uri’s plane, and V is the velocity of Uri’s plane relative to the earth. We have G v = − ( 500 mph ) cos30°iˆ + ( 500 mph ) sin 30° ˆj
Solve:
G V = − ( 500 mph ) cos 20°iˆ − ( 500 mph ) sin 20° ˆj G G G v′ = v − V = ( 36.8 mph ) iˆ + ( 421 mph ) ˆj ⎛ 421 ⎞ ⎟ = 85° ⎝ 36.8 ⎠
θ = tan −1 ⎜
The fuselage of Val’s plane points 30° north of west. Val sees her plane moving in a direction 85° north of east. Thus the angle between the fuselage and the direction of motion is
φ = 180° − 30° − 85° = 65°
4.87.
Model: The airplanes are modeled as particles, and are undergoing relative motion according to the Galilean transformations of position and velocity. We designate Uri’s plane as frame S′ and the earth as frame S. G Frame S′ moves relative to frame S with velocity V . Visualize:
G G G G According to the Galilean transformation of velocity v = v′ + V , where v is the velocity of Val’s plane G G relative to the earth, v′ is the velocity of Val’s plane relative to Uri’s plane, and V is the velocity of Uri’s plane relative to the earth. We have G v = − ( 500 mph ) cos30°iˆ + ( 500 mph ) sin 30° ˆj Solve:
G V = − ( 500 mph ) cos 20°iˆ − ( 500 mph ) sin 20° ˆj G G G v′ = v − V = ( 36.8 mph ) iˆ + ( 421 mph ) ˆj
⎛ 421 ⎞ ⎟ = 85° ⎝ 36.8 ⎠
θ = tan −1 ⎜
The fuselage of Val’s plane points 30° north of west. Val sees her plane moving in a direction 85° north of east. Thus the angle between the fuselage and the direction of motion is
4.88. Model: The ball is a particle launched into projectile motion by the wheel. Visualize:
Solve: The initial velocity of the projectile is the tangential velocity at the point of release, and the direction is tangential to the wheel. The strategy is to find the initial velocity vi = ω f r , then use ω f to find the required angular acceleration α .
Since the release point is Δθ =
11 1 of a complete circle, ϕ = ( 2π rad ) = 30° remains. In the coordinate system of the 12 12
figure,
xi = −20sin30°cm = 10.0 cm
yi = 20cos30°cm = 17.3 cm
xf = 100 cm
yf = 0
ay = − g
vi x = vi cosθ
viy = vi sinθ
ax = 0
The launch angle ϕ is identified by calculating the angles in the right triangle highlighted in the figure and realizing the initial velocity is tangential to the circle. Thus the launch angle is also ϕ . We will use Equations 4.17 for the position of a projectile as a function of time, and use the fact that the time to travel the horizontal distance to the cup is the same for the vertical motion. First, find the flight time required for the ball to hit the cup from the horizontal motion: 100 cm = −10.0 cm + ( vi cos30° ) t ⇒ t =
110 cm 127 cm = vi cos30° vi
Substitute this time in the equations for the vertical motion: ⎛ 127 cm ⎞ 1 ⎛ 127 cm ⎞ 0 cm = 17.3 cm + vi sin 30° ⎜ ⎟− g⎜ ⎟ ⎝ vi ⎠ 2 ⎝ vi ⎠ Solving for the required initial velocity, v =
(127 cm ) g . 2 ( 80.8 cm )
2
The centimeter units cancel, and vi = 2.78 m/s.
Now it is time to consider the angular motion. The required final angular velocity
ωf =
vi 2.78 m/s = = 13.9 rad/s 0.20 m r
Using ω f2 = ωi2 + 2αΔθ ,
ω f2 − ωi2 (13.9 rad/s ) = 16.8 rad/s 2 = ⎛ 11 ⎞ 2 Δθ 2 ⎜ ( 2π ) rad ⎟ 2
α=
⎝ 12
Assess:
⎠
An angular acceleration of 16.8 rad/s 2 ≈ 2.7 rev/s 2 seems reasonable for a spring-loaded wheel.
4-1
Kinematics in Two Dimensions
4-2
5.1. Visualize:
Assess: walls.
Note that the climber does not touch the sides of the crevasse so there are no forces from the crevasse
5.2. Visualize:
5.3. Visualize:
5.4. Model: Assume friction is negligible compared to other forces. Visualize:
5.5. Visualize:
Assess: The bow and archer are no longer touching the arrow, so do not apply any forces after the arrow is released.
5.6. Model: An object’s acceleration is linearly proportional to the net force.
Solve: (a) One rubber band produces a force F, two rubber bands produce a force 2F, and so on. Because F ∝ a and two rubber bands (force 2F) produce an acceleration of 1.2 m/s2, four rubber bands will produce an acceleration of 2.4 m/s 2 . (b) Now, we have two rubber bands (force 2F) pulling two glued objects (mass 2m). Using F = ma,
2 F = (2m) a ⇒ a = F/m = 0.6 m/s 2
5.7. Solve: Let the object have mass m and each rubber band exert a force F. For two rubber bands to accelerate the object with acceleration a, we must have a =
acceleration 3a to a mass
2F . We will need N rubber bands to give m
1 m. Find N: 2
3a = Three rubber bands are required.
NF ⎛ 2 F ⎞ 2 NF ⇒ 3⎜ ⇒ N = 3. ⎟= m m ⎝ m ⎠ 2
1
5.8. Visualize: Please refer to Figure EX5.8. Solve:
Mass is defined to be
m=
1 slope of the acceleration-versus-force graph
A larger slope implies a smaller mass. We know m2 = 0.20 kg, and we can find the other masses relative to m2 by comparing their slopes. Thus m1 1/slope 1 slope 2 1 2 = = = = = 0.40 m2 1/slope 2 slope 1 5 2 5 ⇒ m1 = 0.40m2 = 0.40 × 0.20 kg = 0.08 kg Similarly,
m3 1/slope 3 slope 2 1 5 = = = = = 2.50 m2 1/slope 2 slope 3 2 5 2 ⇒ m3 = 2.50m2 = 2.50 × 0.20 kg = 0.50 kg Assess: From the initial analysis of the slopes we had expected m3 > m2 and m1 < m2 . This is consistent with our numerical answers.
5.9. Visualize: Solve:
Please refer to Figure EX5.9. Mass is defined to be
m=
1 slope of the acceleration-versus-force graph
Thus −1
m1 ( slope of line 1) = m2 ( slope of line 2 )−1 −1
⎛ 5a1 ⎞ 3 ⎜ ⎟ 9 3N ⎠ ⎝ = =5= −1 5 25 ⎛ 3a1 ⎞ ⎜ ⎟ 3 ⎝ 5N ⎠
The ratio of masses is
Assess:
m1 9 = m2 25 More rubber bands produce a smaller acceleration on object 2, so it should be more massive.
5.10
Solve:
Use proportional reasoning. Given that distance traveled is proportional to the square of the d time, d ∝ t , so 2 should be constant. We have t 2.0 furlongs x = 2 2 ( 2.0 s ) ( 4.0 s ) Thus the distance traveled by the object in 4.0 s is x = 8.0 furlongs. Assess: A longer time should result in a longer distance traveled. 2
5.11 Solve: Use proportional reasoning. Let T = period of the pendulum, L = length of pendulum. We are T should be constant. We have L 3.0 s x = 2.0 m 3.0 m Solving, the period of the 3.0 m long pendulum is x = 3.7 s. Assess: Increasing the length increases the period, as expected.
told T ∝ L , so
5.12. Force is not necessary for motion. Constant velocity motion occurs in the absence of forces, that is, when the net force on an object is zero. Thus, it is incorrect to say that “force causes motion.” Instead, force causes acceleration. That is, force causes a change in the motion of an object, and acceleration is the kinematic quantity G that measures a change of motion. Newton’s second law quantifies this idea by stating that the net force Fnet on an object of mass m causes the object to undergo an acceleration: G G F a = net m The acceleration vector and the net force vector must point in the same direction.
5.13. Visualize:
Solve:
(a) Newton’s second law is F = ma. When F = 2 N, we have 2 N = (0.5 kg) a, hence a = 4 m/s 2 .
(b) When F = 1 N, we have 1 N = (0.5 kg) a, hence a = 2 m/s 2 . After repeating this procedure at various points, the above graph is obtained.
5.14 Solve: Newton’s second law tells us that F = ma. Compute F for each case: (a) F = (0.200 kg)(5 m/s 2 ) = 1N . (b) F = (0.200 kg)(10 m/s 2 ) = 2 N . Assess: To double the acceleration we must double the force, as expected.
5.15. Visualize: Please refer to Figure EX5.15.
Solve: Newton’s second law is F = ma. We can read a force and an acceleration from the graph, and hence find the mass. Choosing the force F = 1 N gives us a = 4 m/s 2 . Newton’s second law yields m = 0.25 kg.
5.16. Solve: (a) This problem calls for an estimate so we are looking for an approximate answer. Table 5.3 gives us no information on laptops, but does give the weight of a one-pound object. Place a pound weight in one hand and the laptop on the other. The sensation on your hand is the weight of the object. The sensation from the laptop is about five times the sensation from the pound weight. So we conclude the weight of the laptop is about five times the weight of the one-pound object or about 25 N. (b) According to Table 5.3, the propulsion force on a car is 5000 N. A bicycle (including the rider) is about 100 kg. This is about one-tenth of the mass of a car, which is about 1000 kg for a compact model. The acceleration of a bicycle is somewhat less than that of a car, let’s guess about one-fifth. We can write Newton’s second law as follows:
F ( bicycle ) =
1 1 5000 N = 100 N ( mass of car ) × ( acceleration of car ) = 10 5 50
So we would roughly estimate the propulsion force of a bicycle to be 100 N.
5.17. Solve: (a) This problem calls for an estimate so we are looking for an approximate answer. Table 5.3 gives us no information on pencils, but does give us the weight of the U.S. quarter. Put the quarter on one hand and a pencil on the other hand. The sensation on your hand is the weight of the object. The sensation from the quarter is about the same as the sensation from the pencil. So they both have about the same weight. We can estimate the weight of the pencil to be 0.05 N. (b) According to Table 5.3, the propulsion force on a car is 5000 N. The mass of a sprinter is about 100 kg. This is about one-tenth of the mass of a car, which is about 1000 kg for a compact model. The acceleration of a sprinter is somewhat less than that of a car, let’s guess about one-fifth. We can write Newton’s second law as follows: F ( sprinter ) =
1 1 5000 N = 100 N ( mass of car ) × ( acceleration of car ) = 10 5 50
So, we would roughly estimate the propulsion force of a sprinter to be 100 N. Assess: This is the same estimated number as we obtained in Exercise 5.16. This is reasonable since in both the cases the propulsion force comes from a human and it probably does not matter how the human is providing that force.
5.18. Visualize:
G G G Solve: The object will be in equilibrium if F3 has the same magnitude as F1 + F2 but is in the opposite direction so that the sum of all the three forces is zero.
5.19. Visualize:
G G G Solve: The object will be in equilibrium if F3 has the same magnitude as F1 + F2 but is in the opposite direction so that the sum of all three forces is zero.
5.20. Visualize:
G G G Solve: The object will be in equilibrium if F3 has the same magnitude as F1 + F2 but is in the opposite direction so that the sum of all the three forces is zero.
5.21. Visualize:
Solve: The free-body diagram shows two equal and opposite forces such that the net force is zero. The force directed down is labeled as a gravitational force, and the force directed up is labeled as a tension. With zero net force the acceleration is zero. So, a possible description is: “An object hangs from a rope and is at rest.” Or, “An object hanging from a rope is moving up or down with a constant speed.”
5.22. Visualize:
Solve: The free-body diagram shows three forces with a net force (and therefore net acceleration) upward. G G G There is a force labeled FG directed down, a force Fthrust directed up, and a force D directed down. So a possible description is: “A rocket accelerates upward.”
5.23. Visualize:
G The free-body diagram shows three forces. There is a gravitational force FG , which is down. There is a G G G normal force labeled n, which is up. The forces FG and n are shown with vectors of the same length so they are equal in magnitude and the net G vertical force is zero. So we have an object on the ground which is not moving vertically. There is also a force f k to the left. This must be a frictional force and we need to decide whether it is G static or kinetic friction. The frictional force is the only horizontal force so the net horizontal force must be f k . This means there is a net force to the left producing an acceleration to the left. This all implies motion and therefore the frictional force is kinetic. A possible description is: “A baseball player is sliding into second base.” Solve:
5.24. Visualize:
Assess: you.
G G There is a gravitational force FG . You are touching the park bench, so it exerts a contact force n on
5.25. Visualize:
Assess: The problem says that there is no friction and it tells you nothing about any drag; so we do not include either of these forces. The only remaining forces are the weight and the normal force.
5.26. Visualize:
Assess:
Since the velocity is constant, the acceleration is zero, and the net force is zero.
5.27. Visualize:
Assess: The problem uses the word “sliding.” Any real situation involves friction with the surface. Since we are not told to neglect it, we show that force.
5.28. Visualize:
Figure (a) shows velocity as downward, so the object is moving down. The length of the vector increases with each step showing that the speed is increasing (like a dropped ball). Thus, the acceleration is directed down. G G Since F = ma the force is in the same direction as the acceleration and must be directed down. Figure (b), however, shows the velocity as upward, so the object is moving upward. But the length of the vector decreases with each step showing that the speed is decreasing (like a ball thrown up). Thus, the acceleration is also directed down. As in part (a) the net force must be directed down.
5.29. Visualize:
The velocity vector in figure (a) is shown downward and to the left. So movement is downward and to the left. The velocity vectors get successively longer, which means the speed is increasing. Therefore the acceleration is G G downward and to the left. By Newton’s second law F = ma , the net force must be in the same direction as the acceleration. Thus, the net force is downward and to the left. The velocity vector in (b) is shown to be upward and to the right. So movement is upward and to the right. The velocity vector gets successively shorter, which means the speed is decreasing. Therefore the acceleration is downward and to the left. From Newton’s second law, the net force must be in the direction of the acceleration and so it is directed downward and to the left.
5.30. Visualize:
Solve: According to Newton’s second law F = ma, the force at any time is found simply by multiplying the value of the acceleration by the mass of the object.
5.31. Visualize:
Solve: According to Newton’s second law F = ma, the force at any time is found simply by multiplying the value of the acceleration by the mass of the object.
5.32. Visualize:
Solve: According to Newton’s second law F = ma, the acceleration at any time is found simply by dividing the value of the force by the mass of the object.
5.33. Visualize:
Solve: According to Newton’s second law F = ma, the acceleration at any time is found simply by dividing the value of the force by the mass of the object.
5.34. Model: Use the particle model for the object. Solve:
(a) We are told that for an unknown force (call it F0 ) acting on an unknown mass (call it m0 ) the
acceleration of the mass is 10 m/s 2 . According to Newton’s second law, F0 = m0 (10 m/s 2 ). The force then
becomes
1 2
F0 . Newton’s second law gives 1 1 F0 = m0 a = ⎡⎣ m0 (10 m s 2 ) ⎤⎦ 2 2
This means a is 5 m/s 2 . (b) The force is F0 and the mass is now
1 2
m0 . Newton’s second law gives
1 F0 = m0 a = m0 (10 m s 2 ) 2 This means a = 20 m/s 2 . (c) A similar procedure gives a = 10 m/s 2 . (d) A similar procedure gives a = 2.5 m/s 2 .
5.35. Model: Use the particle model for the object. Solve:
(a) We are told that for an unknown force (call it F0 ) acting on an unknown mass (call it m0 ) the
acceleration of the mass is 8 m/s 2 . According to Newton’s second law, F0 = m0 (8 m/s 2 ). The force then becomes
2 F0 . Newton’s second law gives 2 F0 = m0 a = 2 ⎡⎣ m0 ( 8 m s 2 ) ⎤⎦ This means a is 16 m/s 2 . (b) The force is F0 and the mass is now 2m0 . Newton’s second law gives
F0 = 2m0 a = m0 ( 8 m s 2 ) This means a = 4 m/s 2 . (c) A similar procedure gives a = 8 m/s 2 . (d) A similar procedure gives a = 32 m/s 2 .
5.36. Visualize:
Solve: (d) There are a normal force and a gravitational force which are equal and opposite, so this is an object on a horizontal surface. The description could be: “A tow truck pulls a stuck car out of the mud.”
5.37. Visualize:
Solve: (d) There is a normal force and a gravitational force which are equal and opposite, so this is an object on a horizontal surface, or at least balanced in the vertical direction. The description of this free-body diagram could be: “A jet plane is flying at constant speed.”
5.38. Visualize:
Solve: (d) This is an object on a surface because FG = n. It must be moving to the left because the kinetic friction is to the right. The description of the free-body diagram could be: “A compressed spring is shooting a plastic block to the left.”
5.39. Visualize:
Solve: (d) There is only a single force of weight. We are unable to tell the direction of motion. The description could be: “Galileo has dropped a ball from the Leaning Tower of Pisa.”
5.40. Visualize:
Solve: (d) There is an object on an inclined surface. The net force is down the plane so the acceleration is down the plane. The net force includes both the frictional force and the component of the gravitational force. The direction of the force of kinetic friction implies that the object is moving upward. The description could be: “A car is skidding up an embankment.”
5.41. Visualize:
Solve: (d) There is an object on an inclined surface with a tension force down the surface. There is a small frictional force up the surface implying that the object is sliding down the slope. A description could be: “A sled is being pulled down a slope with a rope that is parallel to the slope.”
5.42. Visualize:
Solve: (d) There is a thrust at an angle to the horizontal and a gravitational force. There is no normal force so the object is not on a surface. The description could be: “A rocket is fired at an angle to the horizontal and there is no drag force.”
5.43. Visualize:
Tension is the only contact force. The downward acceleration implies that FG > T .
5.44. Visualize:
5.45. Visualize:
The normal force is perpendicular to the ground. The thrust force is parallel to the ground and in the direction of acceleration. The drag force is opposite to the direction of motion.
5.46. Visualize:
The normal force is perpendicular to the hill. The frictional force is parallel to the hill.
5.47. Visualize:
The normal force is perpendicular to the hill. The kinetic frictional force is parallel to the hill and directed upward opposite to the direction of motion. The wind force is given as horizontal. Since the skier stays on the slope (that is, there is no acceleration away from the slope) the net force must be parallel to the slope.
5.48. Visualize:
As the rock slides there is kinetic friction between it and the rough concrete sidewalk. Since the rock stays on the level surface, the net force must be along that surface, and is equal to the kinetic friction.
5.49. Visualize:
The drag force due to air is opposite the motion.
5.50. Visualize:
The ball rests on the floor of the barrel because the gravitational force is equal to the normal force. There is a force of the spring to the right which causes an acceleration.
5.51. Visualize:
There are no contact forces on the rock. The gravitational force is the only force acting on the rock.
5.52. Visualize:
The gymnast experiences the long range force of gravity. There is also a contact force from the trampoline that is the normal force of the trampoline on the gymnast. The gymnast is moving downward and the trampoline is decreasing her speed, so the acceleration is upward and there is a net force upward. Thus the normal force must be larger than the gravitational force. The actual behavior of the normal force over time will be complicated as it involves the stretching of the trampoline and therefore tensions.
5.53. Visualize:
You can see from the motion diagram that the box accelerates to the right along with the truck. According to G G Newton’s second law, F = ma , there must be a force to the right acting on the box. This is friction, but not kinetic friction. The box is not sliding against the truck. Instead, it is static friction, the force that prevents slipping. Were it not for static friction, the box would slip off the back of the truck. Static friction acts in the direction needed to prevent slipping. In this case, friction must act in the forward (toward the right) direction.
5.54. Visualize:
You can see from the motion diagram that the bag accelerates to the left along with the car as the car slows G G down. According to Newton’s second law, F = ma , there must be a force to the left acting on the bag. This is friction, but not kinetic friction. The bag is not sliding across the seat. Instead, it is static friction, the force that prevents slipping. Were it not for static friction, the bag would slide off the seat as the car stops. Static friction acts in the direction needed to prevent slipping. In this case, friction must act in the backward (toward the left) direction.
5.55. Visualize: (a)
(b)
(c)
(d) The ball accelerates downward until the instant when it makes contact with the ground. Once it makes contact, it begins to compress and to slow down. The compression takes a short but nonzero distance, as shown in the motion diagram. The point of maximum compression is the turning point, where the ball has an instantaneous speed of v = 0 m/s and reverses direction. The ball then expands and speeds up until it loses G contact with the ground. The motion diagram shows that the acceleration vector a points upward the entireG time that the ball is in contact with the ground. An upward acceleration implies that there is a net upward force Fnet on the ball. The only two forces on the ball are the gravitational force downward and the normal force of the ground upward. To have a net force upward requires n > FG . So the ball bounces because the normal force of the ground exceeds the gravitational force, causing a net upward force during the entire time that the ball is in contact with the ground. This net upward force slows the ball, turns it, and accelerates it upward until it loses contact with the ground. Once contact with the ground is lost, the normal force vanishes and the ball is simply in free fall.
5.56. Visualize: (a)
You are sitting on a bench driving along to the right. Both you and the bench are moving with a constant speed. There is a force on you due to gravity, which is directed down. There is a contact force between you and the bench, which is directed up. Since you are not accelerating up or down the net vertical force on you is zero, which means the two vertical forces are equal in magnitude. The statement of the problem gives no indication of any other contact forces. Specifically, we are told that the bench is very slippery. We can take this to mean there is no frictional force. So our force diagram includes only the normal force up, the gravitational force down, and no horizontal force. (b) The above considerations lead to the free-body diagram that is shown. (c) The car (and therefore the bench) slows down. Does this create any new force on you? No. The forces remain the same. This means the pictorial representation and the free-body diagram are unchanged. (d) The car slows down because of some new contact force on the car (maybe the brakes lock the wheels and the road exerts a force on the tires). But there is no new contact force on you. So the force diagram for you remains unchanged. There are no horizontal forces on you. You do not slow down and you continue at an unchanged velocity until something in the picture changes for you (for example, you fall off the bench or hit the windshield). (e) The net force on you has remained zero because the net vertical force is zero and there are no horizontal forces at all. According to Newton’s first law if the net force on you is zero, then you continue to move in a straight line with a constant velocity. That is what happens to you when the car slows down. You continue to move forward with a constant velocity. The statement that you are “thrown forward” is misleading and incorrect. To be “thrown” there would need to be a net force on you and there is none. It might be correct to say that the car has been “thrown backward” leaving you to continue onward (until you part company with the bench). (f) We are now asked to consider what happens if the bench is NOT slippery. That implies there is a frictional force between the bench and you. This force is certainly horizontal (parallel to the surface of the bench). Is the frictional force directed forward (in the direction of motion) or backward? The car is slowing down and you are staying on the bench. That means you are slowing down with the bench. Your velocity to the right is decreasing (you are moving right and slowing down) so you are accelerating to the left. By Newton’s second law that means the force producing the acceleration must be to the left. That force is the force of static friction and it is shown on the free-body diagram below. Of course, when the car accelerates (increases in speed to the right) and you accelerate with it, then your acceleration is to the right and the frictional force must be to the right.
5-1
Force and Motion
5-2
6.1.
Model: Visualize:
Solve:
We can assume that the ring is a single massless particle in static equilibrium.
Written in component form, Newton’s first law is
( Fnet ) x = ΣFx = T1x + T2 x + T3 x = 0 N ( Fnet ) y = ΣFy = T1 y + T2 y + T3 y = 0 N Evaluating the components of the force vectors from the free-body diagram:
T1x = −T1 T2 x = 0 N T3 x = T3 cos30° T1 y = 0 N T2 y = T2 T3 y = −T3 sin 30° Using Newton’s first law: −T1 + T3 cos30° = 0 N T2 − T3 sin 30° = 0 N Rearranging:
T1 = T3 cos30° = (100 N )( 0.8666 ) = 86.7 N T2 = T3 sin 30° = (100 N )( 0.5 ) = 50.0 N G Assess: Since T3 acts closer to the x-axis than to the y-axis, it makes sense that T1 > T2 .
6.2.
Model: Visualize:
We can assume that the ring is a particle.
This is a static equilibrium problem. We will ignore the weight of the ring, because it is “very light,” so the only three forces are the tension forces shown in the free-body diagram. Note that the diagram defines the angle θ . G Solve: Because the ring is in equilibrium it must obey Fnet = 0 N. This is a vector equation, so it has both xand y-components:
( Fnet ) x = T3 cosθ − T2 = 0 N ⇒ T3 cosθ = T2 ( Fnet ) y = T1 − T3 sinθ = 0 N ⇒ T3 sinθ = T1 We have two equations in the two unknowns T3 and θ . Divide the y-equation by the x-equation:
T3 sin θ T 80 N = tanθ = 1 = = 1.6 ⇒ θ = tan −1 (1.6 ) = 58° T3 cosθ T2 50 N Now we can use the x-equation to find T3 =
T2 50 N = = 94 N cosθ cos58°
The tension in the third rope is 94 N directed 58° below the horizontal.
6.3.
Model: We assume the speaker is a particle in static equilibrium under the influence of three forces: gravity and the tensions in the two cables. Visualize:
Solve:
From the lengths of the cables and the distance below the ceiling we can calculate θ as follows:
sinθ =
2m = 0.677 ⇒ θ = sin −1 0.667 = 41.8° 3m
Newton’s first law for this situation is
( Fnet ) x = ΣFx = T1x + T2 x = 0 N ⇒ −T1 cosθ + T2 cosθ = 0 N ( Fnet ) y = ΣFy = T1 y + T2 y + wy = 0 N ⇒ T1 sinθ + T2 sinθ − w = 0 N The x-component equation means T1 = T2 . From the y-component equation:
( 20 kg ) ( 9.8 m/s w mg 2T1 sin θ = w ⇒ T1 = = = 2sin θ 2sinθ 2sin 41.8°
2
) = 196 N = 147 N 1.333
Assess: It’s to be expected that the two tensions are equal, since the speaker is suspended symmetrically from the two cables. That the two tensions add to considerably more than the weight of the speaker reflects the relatively large angle of suspension.
6.4.
Model: We can assume that the coach and his sled are a particle being towed at a constant velocity by the two ropes, with friction providing the force that resists the pullers. Visualize:
Solve:
Since the sled is not accelerating, it is in dynamic equilibrium and Newton’s first law applies:
( Fnet ) x = ΣFx = T1x + T2 x + f kx = 0 N
( Fnet ) y = ΣFy = T1 y + T2 y +
f ky = 0 N
From the free-body diagram: ⎛1 ⎞ ⎛1 ⎞ T1 cos ⎜ θ ⎟ + T2 cos ⎜ θ ⎟ − f k = 0 N ⎝2 ⎠ ⎝2 ⎠
⎛1 ⎞ ⎛1 ⎞ T1 sin ⎜ θ ⎟ − T2 sin ⎜ θ ⎟ + 0 N = 0 N ⎝2 ⎠ ⎝2 ⎠
From the second of these equations T1 = T2 . Then from the first:
2T1 cos10° = 1000 N ⇒ T1 =
1000 N 1000 N = = 508 N 2cos10° 1.970
Assess: The two tensions are equal, as expected, since the two players are pulling at the same angle. The two add up to only slightly more than 1000 N, which makes sense because the angle at which the two players are pulling is small.
6.5. Solve:
Visualize: Please refer to the Figure EX6.5. Applying Newton’s second law to the diagram on the left,
ax =
( Fnet ) x m
=
4 N−2N = 1.0 m/s 2 2 kg
ay =
( Fnet ) y m
=
3 N−3 N = 0 m/s 2 2 kg
For the diagram on the right: ax =
( Fnet ) x m
=
4 N−2 N = 1.0 m/s 2 2 kg
ay =
( Fnet ) y m
=
3 N −1 N − 2 N = 0 m/s 2 2 kg
6.6. Visualize: Please refer to Figure EX6.6. Solve: For the diagram on the left, three of the vectors lie along the axes of the tilted coordinate system. Notice that the angle between the 3 N force and the –y-axis is the same 20° by which the coordinates are tilted. Applying Newton’s second law, ax =
( Fnet ) x m
ay =
=
( Fnet ) y m
5 N − 1 N − ( 3sin 20° ) N = 1.49 m/s 2 2 kg
=
2.82 N − ( 3cos 20° ) N = 0 m/s 2 2 kg
For the diagram on the right, the 2-newton force in the first quadrant makes an angle of 15° with the positive xaxis. The other 2-newton force makes an angle of 15° with the negative y-axis. The accelerations are ax = ay =
( Fnet ) x ( 2cos15° ) N + ( 2sin15° ) N − 3 N = = −0.28 m/s 2 2 kg
m
( Fnet ) y m
=
1.414 N + ( 2sin15° ) N − ( 2cos15° ) N = 0 m/s 2 2 kg
6.7. Solve:
Visualize: Please refer to Figure EX6.7. (a) Apply Newton’s second law in both the x and y directions.
( Fnet ) x = ( 5.0 N ) cos37° − 2.0 N = ( 5.0 kg ) ax ⇒ ax = 0.40 m/s 2
( Fnet ) y = 2.0 N + ( 5.0 N ) sin 37° − 5.0 N = ( 5.0 kg ) a y ⇒ a y = 0.0 m/s 2
(b) The angle that the 5.0 N force makes with the –y-axis is 37°. Apply Newton’s second law for both the x and y direction.
( Fnet ) x = 3.0 N + ( 5.0 N ) sin 37° − 2.0 N = ( 5.0 kg ) ax ⇒ ax = 0.80 m/s 2
( Fnet ) y = 4.0 N − ( 5.0 N ) cos37° = ( 5.0 kg ) a y ⇒ a y = 0.0 m/s 2 Assess: The orientation of the coordinate axes is chosen for convenience, and does not always need to conform to the horizontal and vertical.
6.8. Visualize: Please refer to Figure EX6.8. Solve: We can use the constant slopes of the three segments of the graph to calculate the three accelerations. For t between 0 s and 3 s, ax =
Δvx 12 m/s − 0 s = = 4 m/s 2 3s Δt
For t between 3 s and 6 s, Δvx = 0 m/s, so ax = 0 m/s 2 . For t between 6 s and 8 s,
Δvx 0 m/s − 12 m/s = = −6 m/s 2 Δt 2s From Newton’s second law, at t = 1 s we have ax =
Fnet = max = (2.0 kg)(4 m/s 2 ) = 8 N At t = 4 s, ax = 0 m/s 2 , so Fnet = 0 N. At t = 7 s,
Fnet = max = (2.0 kg)(−6.0 m/s 2 ) = −12 N Assess: The magnitudes of the forces look reasonable, given the small mass of the object. The positive and negative signs are appropriate for an object first speeding up, then slowing down.
6.9. Visualize: Please refer to Figure EX6.9. Positive forces result in the object gaining speed and negative forces result in the object slowing down. The final segment of zero force is a period of constant speed. Solve: We have the mass and net force for all the three segments. This means we can use Newton’s second law to calculate the accelerations. The acceleration from t = 0 s to t = 3 s is ax =
Fx 4N = = 2 m/s 2 m 2.0 kg
ax =
−2 N Fx = = −1 m/s 2 m 2.0 kg
The acceleration from t = 3 s to t = 5 s is
The acceleration from t = 5 s to 8 s is ax = 0 m/s 2 . In particular, ax ( at t = 6 s ) = 0 m/s 2 . We can now use one-dimensional kinematics to calculate v at t = 6 s as follows: v = v0 + a1 ( t1 − t0 ) + a2 ( t2 − t0 )
= 0 + ( 2 m/s 2 ) ( 3 s ) + ( −1 m/s 2 ) ( 2 s ) = 6 m/s − 2 m/s = 4 m/s
Assess: The positive final velocity makes sense, given the greater magnitude and longer duration of the G positive F1. A velocity of 4 m/s also seems reasonable, given the magnitudes and directions of the forces and the mass involved.
6.10.
Model: We assume that the box is a particle being pulled in a straight line. Since the ice is frictionless, the tension in the rope is the only horizontal force. Visualize:
Solve: (a) Since the box is at rest, ax = 0 m/s 2 , and the net force on the box must be zero. Therefore, according to Newton’s first law, the tension in the rope must be zero. (b) For this situation again, ax = 0 m/s 2 , so Fnet = T = 0 N. (c) Here, the velocity of the box is irrelevant, since only a change in velocity requires a nonzero net force. Since ax = 5.0 m/s 2 ,
Fnet = T = max = ( 50 kg ) ( 5.0 m/s 2 ) = 250 N Assess: For parts (a) and (b), the zero acceleration immediately implies that the rope is exerting no horizontal force on the box. For part (c), the 250 N force (the equivalent of about half the weight of a small person) seems reasonable to accelerate a box of this mass at 5.0 m/s 2 .
6.11.
Model: We assume that the box is a point particle that is acted on only by the tension in the rope and the pull of gravity. Both the forces act along the same vertical line. Visualize:
Solve:
(a) Since the box is at rest, a y = 0 m/s 2 and the net force on it must be zero:
Fnet = T − FG = 0 N ⇒ T = FG = mg = ( 50 kg ) ( 9.8 m/s 2 ) = 490 N (b) Since the box is rising at a constant speed, again a y = 0 m/s 2 , Fnet = 0 N, and T = FG = 490 N. (c) The velocity of the box is irrelevant, since only a change in velocity requires a nonzero net force. Since a y = 5.0 m/s 2 ,
Fnet = T − FG = ma y = ( 50 kg ) ( 5.0 m/s 2 ) = 250 N ⇒ T = 250 N + w = 250 N + 490 N = 740 N (d) The situation is the same as in part (c), except that the rising box is slowing down. Thus a y = −5.0 m/s 2 and
we have instead Fnet = T − FG = ma y = ( 50 kg ) ( −5.0 m/s 2 ) = −250 N ⇒ T = −250 N + FG = −250 N + 490 N = 240 N Assess: For parts (a) and (b) the zero accelerations immediately imply that the gravitational force on the box must be exactly balanced by the upward tension in the rope. For part (c) the tension not only has to support the gravitational force on the box but must also accelerate it upward, hence, T must be greater than FG . When the
box accelerates downward, the rope need not support the entire gravitational force, hence, T is less than FG .
6.12.
Model: We assume the rocket is a particle moving in a vertical straight line under the influence of only two forces: gravity and its own thrust. Visualize:
Solve:
(a) Using Newton’s second law and reading the forces from the free-body diagram,
Fthrust − FG = ma ⇒ Fthrust = ma + mg Earth = ( 0.200 kg ) (10 m/s 2 + 9.80 m/s 2 ) = 3.96 N (b) Likewise, the thrust on the moon is (0.200 kg)(10 m/s 2 + 1.62 m/s 2 ) = 2.32 N. Assess: The thrust required is smaller on the moon, as it should be, given the moon’s weaker gravitational pull. The magnitude of a few newtons seems reasonable for a small model rocket.
6.13. Solve:
Model: The astronaut is treated as a particle. The mass of the astronaut is
m=
wearth 800 N = = 81.6 kg g earth 9.80 m/s 2
Therefore, the weight of the astronaut on Mars is wMars = mg Mars = ( 81.6 kg ) ( 3.76 m/s 2 ) = 307 N Assess: The smaller acceleration of gravity on Mars reveals that objects are less strongly attracted to Mars than to the earth, so the smaller weight on Mars makes sense.
6.14. Solve:
Model: Use the particle model for the woman. (a) The woman’s weight on the earth is
wearth = mg earth = ( 55 kg ) ( 9.80 m/s 2 ) = 540 N (b) Since mass is a measure of the amount of matter, the woman’s mass is the same on the moon as on the earth. Her weight on the moon is
wmoon = mg moon = ( 55 kg ) (1.62 m/s 2 ) = 89 N Assess: The smaller acceleration due to gravity on the moon reveals that objects are less strongly attracted to the moon than to the earth. Thus the woman’s smaller weight on the moon makes sense.
6.15.
Model: We assume that the passenger is a particle subject to two vertical forces: the downward pull of gravity and the upward push of the elevator floor. We can use one-dimensional kinematics and Equation 6.10. Visualize:
Solve:
(a) The weight is
⎛ a ⎞ ⎛ 0⎞ w = mg ⎜1 + y ⎟ = mg ⎜1 + ⎟ = mg = ( 60 kg ) ( 9.80 m/s 2 ) = 590 N g⎠ ⎝ g⎠ ⎝ (b) The elevator speeds up from v0 y = 0 m/s to its cruising speed at v y = 10 m/s. We need its acceleration before
we can find the apparent weight: ay =
Δv 10 m/s − 0 m/s = = 2.5 m/s 2 4.0 s Δt
The passenger’s weight is
⎛ ⎛ a ⎞ 2.5 m/s 2 ⎞ w = mg ⎜ 1 + y ⎟ = ( 590 N ) ⎜1 + = (590 N)(1.26) = 740 N 2 ⎟ g ⎠ ⎝ ⎝ 9.80 m/s ⎠ (c) The passenger is no longer accelerating since the elevator has reached its cruising speed. Thus, w = mg = 590 N as in part (a). Assess: The passenger’s weight is the gravitational force on the passenger in parts (a) and (c), since there is no acceleration. In part (b), the elevator must not only support the gravitational force but must also accelerate him upward, so it’s reasonable that the floor will have to push up harder on him, increasing his weight.
6.16.
Model: We assume that the passenger is a particle acted on by only two vertical forces: the downward pull of gravity and the upward force of the elevator floor. Visualize: Please refer to Figure EX6.16. The graph has three segments corresponding to different conditions: (1) increasing velocity, meaning an upward acceleration; (2) a period of constant upward velocity; and (3) decreasing velocity, indicating a period of deceleration (negative acceleration). Solve: Given the assumptions of our model, we can calculate the acceleration for each segment of the graph and then apply Equation 6.10. The acceleration for the first segment is
ay =
v1 − v0 8 m/s − 0 m/s = = 4 m/s 2 t1 − t0 2 s−0 s
⎛ ⎛ a ⎞ 4 m/s 2 ⎞ 4 ⎞ ⎛ ⇒ w = mg ⎜1 + y ⎟ = mg ⎜1 + = ( 75 kg ) ( 9.80 m/s 2 ) ⎜1 + ⎟ = 1035 N 2 ⎟ g 9.80 m/s 9.80 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ For the second segment, a y = 0 m/s 2 and the weight is ⎛ 0 m/s 2 ⎞ 2 w = mg ⎜1 + ⎟ = mg = ( 75 kg ) ( 9.80 m/s ) = 740 N g ⎠ ⎝ For the third segment,
ay =
v3 − v2 0 m/s − 8 m/s = = −2 m/s 2 t3 − t 2 10 s − 6 s
⎛ −2 m/s 2 ⎞ ⇒ w = mg ⎜1 + = ( 75 kg ) ( 9.80 m/s 2 ) (1 − 0.2 ) = 590 N 2 ⎟ 9.80 m/s ⎝ ⎠ Assess: As expected, the weight is greater than the gravitational force on the passenger when the elevator is accelerating upward and lower than normal when the acceleration is downward. When there is no acceleration the weight is the gravitational force. In all three cases the magnitudes are reasonable, given the mass of the passenger and the accelerations of the elevator.
6.17.
Model: We assume that the safe is a particle moving only in the x-direction. Since it is sliding during the entire problem, we can use the model of kinetic friction. Visualize:
Solve: The safe is in equilibrium, since it’s not accelerating. Thus we can apply Newton’s first law in the vertical and horizontal directions:
( Fnet ) x = ΣFx = FB + FC − f k = 0 N
⇒ f k = FB + FC = 350 N + 385 N = 735 N
( Fnet ) y = ΣFy = n − FG = 0 N ⇒ n = FG = mg = ( 300 kg ) ( 9.80 m/s2 ) = 2.94 × 103 N Then, for kinetic friction: fk = μk n ⇒ μk =
fk 735 N = = 0.250 n 2.94 × 103 N
Assess: The value of μ k = 0.250 is hard to evaluate without knowing the material the floor is made of, but it seems reasonable.
6.18.
Model: We assume that the mule is a particle acted on by two opposing forces in a single line: the farmer’s pull and friction. The mule will be subject to static friction until (and if!) it begins to move; after that it will be subject to kinetic friction. Visualize:
Solve: Since the mule does not accelerate in the vertical direction, the free-body diagram shows that n = FG = mg . The maximum friction force is
f smax = μs mg = ( 0.8 )(120 kg ) ( 9.80 m/s 2 ) = 940 N
The maximum static friction force is greater than the farmer’s maximum pull of 800 N; thus, the farmer will not be able to budge the mule. Assess: The farmer should have known better.
6.19.
Model: Visualize:
We will represent the crate as a particle.
G G (a) When the belt runs at constant speed, the crate has an acceleration a = 0 m/s 2 and is in dynamic G G equilibrium. Thus Fnet = 0. It is tempting to think that the belt exerts a friction force on the crate. But if it did, there would be a net force because there are no other possible horizontal forces to balance a friction force. Because there is no net force, there cannot be a friction force. The only forces are the upward normal force and the gravitational force on the crate. (A friction force would have been needed to get the crate moving initially, but no horizontal force is needed to keep it moving once it is moving with the same constant speed as the belt.) (b) If the belt accelerates gently, the crate speeds up without slipping on the belt. Because it is accelerating, the crate must have a net horizontal force. So now there is a friction force, and the force points in the direction of the crate’s motion. Is it static friction or kinetic friction? Although the crate is moving, there is no motion of the crate relative to the belt. Thus, it is a static friction force that accelerates the crate so that it moves without slipping on the belt. (c) The static friction force has a maximum possible value ( f s ) max = μ s n. The maximum possible acceleration of the crate is Solve:
amax =
( fs )max μs n =
m m If the belt accelerates more rapidly than this, the crate will not be able to keep up and will slip. It is clear from the free-body diagram that n = FG = mg . Thus, amax = μs g = (0.5)(9.80 m/s 2 ) = 4.9 m/s 2
6.20.
Model: Visualize:
We assume that the truck is a particle in equilibrium, and use the model of static friction.
Solve: The truck is not accelerating, so it is in equilibrium, and we can apply Newton’s first law. The normal force has no component in the x-direction, so we can ignore it here. For the other two forces:
( Fnet ) x = ΣFx =
fs − ( FG ) x = 0 N ⇒ fs = ( FG ) x = mg sin θ = ( 4000 kg ) ( 9.80 m/s 2 ) ( sin15° ) = 10,145 N
Assess: The truck’s weight (mg) is roughly 40,000 N. A friction force that is ≈ 25% of the truck’s weight seems reasonable.
6.21. Model: The car is a particle subject to Newton’s laws and kinematics. Visualize:
Solve: Kinetic friction provides a horizontal acceleration which stops the car. From the figure, applying Newton’s first and second laws gives
∑F
x
∑F
y
= − f k = max
= n − FG = 0 ⇒ n = FG = mg
Combining these two equations with f k = μ k n yields
ax = − μk g = − ( 0.50 ) ( 9.80 m/s 2 ) = −4.9 m/s Kinematics can be used to determine the initial velocity.
vf2 = vi2 + 2aΔx ⇒ vi2 = −2ax Δx Thus
vi = −2 ( −4.9 m/s 2 ) ( 65 m − 0 m ) = 25 m/s 2
Assess: The initial speed of 25 m/s 2 ≈ 56 mph is a reasonable speed to have initially for a vehicle to leave 65meter-long skid marks.
6.22.
Model: We assume that the plane is a particle accelerating in a straight line under the influence of two forces: the thrust of its engines and the rolling friction of the wheels on the runway. We can use one-dimensional kinematics. Visualize:
Solve:
We can use the definition of acceleration to find a, and then apply Newton’s second law. We obtain: a=
Δv 82 m/s − 0 m/s = = 2.34 m/s 2 Δt 35 s
( Fnet ) = ΣFx = Fthrust − f r = ma ⇒ Fthrust =
f r + ma
For rubber rolling on concrete, μ r = 0.02 (Table 6.1), and since the runway is horizontal, n = FG = mg. Thus:
Fthrust = μ r FG + ma = μ r mg + ma = m ( μ r g + a ) = ( 75,000 kg ) ⎡⎣( 0.02 ) ( 9.8 m/s 2 ) + 2.34 m/s 2 ⎤⎦ = 190,000 N Assess: It’s hard to evaluate such an enormous thrust, but comparison with the plane’s mass suggests that 190,000 N is enough to produce the required acceleration.
6.23.
Model: We treat the train as a particle subject to rolling friction but not to drag (because of its slow speed and large mass). We can use the one-dimensional kinematic equations. Visualize:
Solve: The locomotive is not accelerating in the vertical direction, so the free-body diagram shows us that n = FG = mg . Thus,
f r = μ r mg = ( 0.002 )( 50,000 kg ) ( 9.80 m/s 2 ) = 980 N
From Newton’s second law for the decelerating locomotive, −f −980 N = −0.01960 m/s 2 ax = r = m 50,000 kg
Since we’re looking for the distance the train rolls, but we don’t have the time:
v12 − v02 = 2ax ( Δx ) ⇒ Δx =
v12 − v02 ( 0 m/s ) − (10 m/s ) = = 2.55 × 103 m 2a x 2 ( −0.01960 m/s 2 ) 2
2
Assess: The locomotive’s enormous inertia (mass) and the small coefficient of rolling friction make this long stopping distance seem reasonable.
6.24.
Model: Visualize:
We can treat the sliding player as a particle experiencing kinetic friction.
Solve: We can assume a mass of 80 kg (which corresponds to a gravitational force of about 175 pounds). We have no value for μk of cloth sliding on loose dirt. It is probably greater than μk for wood on wood, due to the roughness of both surfaces. Let’s guess 0.40. From the free-body diagram, n = FG = mg , since there’s no vertical acceleration. Thus,
f k = μ k mg = ( 0.40 )( 80 kg ) ( 9.80 m/s 2 ) = 314 N
Assess: A frictional force of 314 N would produce an acceleration of
a=
Fnet −314 N = = − 4.0 m/s 2 m 80 kg
A player running initially at 8 m/s would thus be brought to a stop in about 2 seconds, which seems somewhat too long. Our estimate of μk is probably a bit low but 8 m/s is a large speed as well.
6.25.
Model: Visualize:
We assume that the skydiver is shaped like a box and is a particle.
The skydiver falls straight down toward the earth’s surface, that is, the direction of fall is vertical. Since the skydiver falls feet first, the surface perpendicular to the drag has the cross-sectional area A = 20 cm × 40 cm. The physical conditions needed to use Equation 6.16 for the drag force are satisfied. The terminal speed corresponds to the situation when the net force acting on the skydiver becomes zero. Solve: The expression for the magnitude of the drag with v in m/s is D≈
1 2 Av = 0.25 ( 0.20 × 0.40 ) v 2 N = 0.020v 2 N 4
The gravitational force on the skydiver is FG = mg = ( 75 kg ) ( 9.8 m/s 2 ) = 735 N. The mathematical form of the
condition defining dynamical equilibrium for the skydiver and the terminal speed is G G G Fnet = FG + D = 0 N 2 ⇒ 0.02vterm N − 735 N = 0 N ⇒ vterm =
735 ≈ 192 m/s 0.02
Assess: The result of the above simplified physical modeling approach and subsequent calculation, even if approximate, shows that the terminal velocity is very high. This result implies that the skydiver will be very badly hurt at landing if the parachute does not open in time.
6.26.
Model: Visualize:
We will represent the tennis ball as a particle.
The tennis ball falls straight down toward the earth’s surface. The ball is subject to a net force that is the resultant of the gravitational and drag force vectors acting vertically, in the downward and upward directions, respectively. Once the net force acting on the ball becomes zero, the terminal velocity is reached and remains constant for the rest of the motion. Solve: The mathematical equation defining the dynamical equilibrium situation for the falling ball is G G G G Fnet = FG + D = 0 N Since only the vertical direction matters, one can write: ΣFy = 0 N ⇒ Fnet = D − FG = 0 N When this condition is satisfied, the speed of the ball becomes the constant terminal speed v = vterm . The magnitudes of the gravitational and drag forces acting on the ball are:
FG = mg = m ( 9.80 m/s 2 )
1 2 2 2 2 = ( 0.25π )( 0.0325 m ) ( 26 m/s ) = 0.56 N ( Avterm ) = 0.25 (π R 2 ) vterm 4 The condition for dynamic equilibrium becomes: 0.56 N = 57 g ( 9.80 m/s2 ) m − 0.56 N = 0 N ⇒ m = 9.80 m/s 2 D≈
Assess: The value of the mass of the tennis ball obtained above seems reasonable.
6.27. Visualize:
We used the force-versus-time graph to draw the acceleration-versus-time graph. The peak acceleration was calculated as follows:
amax =
Fmax 10 N = = 2 m/s 2 m 5 kg
Solve: The acceleration is not constant, so we cannot use constant acceleration kinematics. Instead, we use the more general result that
v(t ) = v0 + area under the acceleration curve from 0 s to t The object starts from rest, so v0 = 0 m/s. The area under the acceleration curve between 0 s and 6 s is
1 2
(4 s)
(2 m/s ) = 4.0 m/s. We’ve used the fact that the area between 4 s and 6 s is zero. Thus, at t = 6 s, vx = 4.0 m/s. 2
6.28. Visualize:
The acceleration is ax = Fx m , so the acceleration-versus-time graph has exactly the same shape as the forceversus-time graph. The maximum acceleration is amax = Fmax m = ( 6 N ) ( 2 kg ) = 3 m/s 2 . Solve: The acceleration is not constant, so we cannot use constant-acceleration kinematics. Instead, we use the more general result that
v(t ) = v0 + area under the acceleration curve from 0 s to t The object starts from rest, so v0 = 0 m/s. The area under the acceleration curve between 0 s and 4 s is a rectangle (3 m/s 2 × 2 s = 6 m/s) plus a triangle
(
1 2
× 3 m/s 2 × 2 s = 3 m/s ). Thus vx = 9 m/s at t = 4 s.
6.29.
Model: Visualize:
Solve:
You can model the beam as a particle in static equilibrium.
Using Newton’s first law, the equilibrium equations in vector and component form are: G G G G G Fnet = T1 + T2 + FG = 0 N
( Fnet ) x = T1x + T2 x + FGx = 0 N ( Fnet ) y = T1 y + T2 y + FGy = 0 N
Using the free-body diagram yields: −T1 sinθ1 + T2 sin θ 2 = 0 N
T1 cosθ1 + T2 cosθ 2 − FG = 0 N
The mathematical model is reduced to a simple algebraic system of two equations with two unknowns, T1 and T2 . Substituting θ1 = 20°, θ 2 = 30°, and FG = mg = 9800 N, the simultaneous equations become −T1 sin 20° + T2 sin 30° = 0 N
T1 cos 20° + T2 cos30° = 9800 N
You can solve this system of equations by simple substitution. The result is T1 = 6397 N and T2 = 4376 N. Assess: The above approach and result seem reasonable. Intuition indicates there is more tension in the left rope than in the right rope.
6.30.
Model: Visualize:
The plastic ball is represented as a particle in static equilibrium.
Solve: The electric force, like the weight, is a long-range force. So the ball experiences the contact force of the string’s tension plus two long-range forces. The equilibrium condition is
( Fnet ) x = Tx + ( Felec ) x = T sin θ − Felec = 0 N ( Fnet ) y = Ty + ( FG ) y = T cosθ − mg = 0 N We can solve the y-equation to get T=
( 0.001 kg ) ( 9.8 m/s mg = cosθ cos 20°
2
) = 0.0104 N
Substituting this value into the x-equation,
Felec = T sin θ = (1.04 × 10−2 N ) sin 20° = 0.0036 N (b) The tension in the string is 0.0104 N.
6.31.
Model: Visualize:
Solve:
The piano is in static equilibrium and is to be treated as a particle.
(a) Based on the free-body diagram, Newton’s second law is
( Fnet ) x = 0 N = T1x + T2 x = T2 cosθ 2 − T1 cosθ1
( Fnet ) y = 0 N = T1 y + T2 y + T3 y + FGy = T3 − T1 sinθ1 − T2 sinθ 2 − mg Notice how the force components all appear in the second law with plus signs because we are adding forces. The negative signs appear only when we evaluate the various components. These are two simultaneous equations in the two unknowns T2 and T3 . From the x-equation we find T2 =
T1 cosθ1 ( 500 N ) cos15° = = 533 N cosθ 2 cos 25°
(b) Now we can use the y-equation to find
T3 = T1 sin θ1 + T2 sin θ 2 + mg = 5.25 × 103 N
6.32.
Model: We will represent Henry as a particle. His motion is governed by constant-acceleration kinematic equations. Visualize: Please refer to the Figure EX6.32. Solve: (a) Henry undergoes an acceleration from 0 s to 2.0 s, constant velocity motion from 2.0 s to 10.0 s, and another acceleration as the elevator brakes from 10.0 s to 12.0 s. The weight is the same as the gravitational force during constant velocity motion, so Henry’s weight w = FG = mg is 750 N. His weight is less than the gravitational force on him during the initial acceleration, so the acceleration is in a downward direction (negative a). Thus, the elevator’s initial motion is down. (b) Because the gravitational force on Henry is 750 N, his mass is m = FG /g = 76.5 kg. (c) The apparent weight (see Equation 6.10) during vertical motion is given by
⎛ w ⎞ ⎛ a⎞ w = mg ⎜1 + ⎟ ⇒ a = g ⎜ − 1⎟ g F ⎝ ⎠ ⎝ G ⎠ During the interval 0 s ≤ t ≤ 2 s, the elevator’s acceleration is ⎛ 600 N ⎞ a = g⎜ − 1⎟ = −1.96 m/s 2 ⎝ 750 N ⎠
At t = 2 s, Henry’s position is 1 1 2 2 y1 = y0 + v0 Δt0 + a ( Δt0 ) = a ( Δt0 ) = −3.92 m 2 2 and his velocity is v1 = v0 + aΔt0 = aΔt0 = −3.92 m/s During the interval 2 s ≤ t ≤ 10 s, a = 0 m/s 2 . This means Henry travels with a constant velocity v1 = −3.92 m/s. At t = 10 s he is at position y2 = y1 + v1Δt1 = −35.3 m and he has a velocity v2 = v1 = −3.92 m/s. During the interval 10 s ≤ t ≤ 12.0 s, the elevator’s acceleration is
⎛ 900 N ⎞ a = g⎜ − 1⎟ = +1.96 m/s 2 ⎝ 750 N ⎠ The upward acceleration vector slows the elevator and Henry feels heavier than normal. At t = 12.0 s Henry is at position 1 y3 = y2 + v2 (Δt2 ) + a (Δt2 ) 2 = −39.2 m 2 Thus Henry has traveled distance 39.2 m.
6.33.
Model: We’ll assume Zach is a particle moving under the effect of two forces acting in a single vertical line: gravity and the supporting force of the elevator. Visualize:
Solve:
(a) Before the elevator starts braking, Zach is not accelerating. His weight (see Equation 6.10) is
⎛ 0 m/s 2 ⎞ ⎛ a⎞ 2 w = mg ⎜1 + ⎟ = mg ⎜1 + ⎟ = mg = ( 80 kg ) ( 9.80 m/s ) = 784 N g g ⎝ ⎠ ⎝ ⎠ Zach’s weight is 7.8 × 102 N. (b) Using the definition of acceleration, a=
Δv v1 − v0 0 − ( −10 ) m/s = = = 3.33 m/s 2 Δt t1 − t0 3.0 s
⎛ 3.33 m/s 2 ⎞ ⎛ a⎞ ⇒ w = mg ⎜1 + ⎟ = ( 80 kg ) ( 9.80 m/s 2 ) ⎜1 + = (784 N)(1 + 0.340) = 1050 N 2 ⎟ ⎝ g⎠ ⎝ 9.80 m/s ⎠ Now Zach’s weight is 1.05 × 103 N. Assess: While the elevator is braking, it not only must support the gravitational force on Zach but must also push upward on him to decelerate him, so his weight is greater than the gravitational force.
6.34.
Model: We can assume your body is a particle moving in a straight line under the influence of two forces: gravity and the support force of the scale. Visualize:
Solve:
The weight (see Equation 6.10) of an object moving in an elevator is
⎛ ⎛ w ⎞ a⎞ w = mg ⎜1 + ⎟ ⇒ a = ⎜ − 1⎟ g g mg ⎝ ⎠ ⎝ ⎠ When accelerating upward, the acceleration is ⎛ 170 lb ⎞ a =⎜ − 1⎟ ( 9.80 m/s 2 ) = 1.3 m/s 2 ⎝ 150 lb ⎠ When braking, the acceleration is ⎛ 120 lb ⎞ a =⎜ − 1⎟ ( 9.80 m/s 2 ) = −2.0 m/s 2 ⎝ 150 lb ⎠
Assess: A 10-20% change in apparent weight seems reasonable for a fast elevator, as the ones in the Empire State Building must be. Also note that we did not have to convert the units of the weights from pounds to newtons because the weights appear as a ratio.
6.35.
Model: We can assume the foot is a single particle in equilibrium under the combined effects of gravity, the tensions in the upper and lower sections of the traction rope, and the opposing traction force of the leg itself. We can also treat the hanging mass as a particle in equilibrium. Since the pulleys are frictionless, the tension is the same everywhere in the rope. Because all pulleys are in equilibrium, their net force is zero. So they do not contribute to T. Visualize:
Solve:
(a) From the free-body diagram for the mass, the tension in the rope is
T = FG = mg = ( 6 kg ) ( 9.80 m/s 2 ) = 58.8 N (b) Using Newton’s first law for the vertical direction on the pulley attached to the foot,
( Fnet ) y = ΣFy = T sin θ − T sin15° − ( FG )foot = 0 N ⇒ sin θ =
T sin15° + ( FG )foot T
= sin15° +
( 4 kg ) ( 9.80 m/s mfoot g = 0.259 + T 58.8 N
2
) = 0.259 + 0.667 = 0.926
⇒ θ = sin −1 0.926 = 67.8° (c) Using Newton’s first law for the horizontal direction,
( Fnet ) x = ΣFx = T cosθ + T cos15° − Ftraction = 0 N ⇒ Ftraction = T cosθ + T cos15° = T ( cos67.8° + cos15° ) = (58.8 N)(0.3778 + 0.9659) = (58.8 N)(1.344) = 79.0 N
Assess: Since the tension in the upper segment of the rope must support the foot and counteract the downward pull of the lower segment of the rope, it makes sense that its angle is larger (a more direct upward pull). The magnitude of the traction force, roughly one-tenth of the gravitational force on a human body, seems reasonable.
6.36.
Model: We can assume the person is a particle moving in a straight line under the influence of the combined decelerating forces of the air bag and seat belt or, in the absence of restraints, the dashboard or windshield. Visualize:
Solve: (a) In order to use Newton’s second law for the passenger, we’ll need the acceleration. Since we don’t have the stopping time:
v12 = v02 + 2a ( x1 − x0 ) ⇒ a =
0 m 2 /s 2 − (15 m/s ) v12 − v02 = = −112.5 m/s 2 2 ( x1 − x0 ) 2 (1 m − 0 m ) 2
⇒ Fnet = F = ma = ( 60 kg ) ( −112.5 m/s 2 ) = −6750 N The net force is 6750 N to the left. (b) Using the same approach as in part (a), 0 m 2 /s 2 − (15 m/s ) v12 − v02 = ( 60 kg ) = −1,350,000 N 2 ( x1 − x0 ) 2 ( 0.005 m ) 2
F = ma = m
The net force is 1,350,000 N to the left. (c) The passenger’s weight is mg = (60 kg)(9.8 m/s 2 ) = 588 N. The force in part (a) is 11.5 times the passenger’s weight. The force in part (b) is 2300 times the passenger’s weight. Assess: An acceleration of 11.5g is well within the capability of the human body to withstand. A force of 2300 times the passenger’s weight, on the other hand, would surely be catastrophic.
6.37.
Model: Visualize:
The ball is represented as a particle that obeys constant-acceleration kinematic equations.
Solve: This is a two-part problem. During part 1 the ball accelerates upward in the tube. During part 2 the ball undergoes free fall (a = − g ). The initial velocity for part 2 is the final velocity of part 1, as the ball emerges from the tube. The free-body diagram for part 1 shows two forces: the air pressure force and the gravitational force. We need only the y-component of Newton’s second law:
ay = a =
( Fnet ) y m
=
Fair − FG Fair 2N = −g= − 9.80 m/s 2 = 30.2 m/s 2 m m 0.05 kg
We can use kinematics to find the velocity v1 as the ball leaves the tube:
v12 = v02 + 2a ( y1 − y0 ) ⇒ v1 = 2ay1 = 2 ( 30.2 m/s 2 ) (1 m ) = 7.77 m/s For part 2, free-fall kinematics v22 = v12 − 2 g ( y2 − y1 ) gives y2 − y1 =
v12 = 3.1 m 2g
6.38.
Model: Visualize:
We will represent the bullet as a particle.
Solve: (a) We have enough information to use kinematics to find the acceleration of the bullet as it stops. Then we can relate the acceleration to the force with Newton’s second law. (Note that the barrel length is not relevant to the problem.) The kinematic equation is
v02 ( 400 m/s ) = −6.67 × 105 m/s 2 =− 2Δx 2 ( 0.12 m ) G Notice that a is negative, in agreement with the vector a in the motion diagram. Turning to forces, the wood exerts two forces on the bullet. First, an upward normal force that keeps the bullet from “falling” through the G G wood. Second, a retarding frictional force f k that stops the bullet. The only horizontal force is f k , which points to the left and thus has a negative x-component. The x-component of Newton’s second law is 2
v12 = v02 + 2aΔx ⇒ a = −
( Fnet ) x = − f k = ma ⇒
f k = − ma = − ( 0.01 kg ) ( −6.67 × 105 m/s 2 ) = 6670 N
Notice how the signs worked together to give a positive value of the magnitude of the force. (b) The time to stop is found from v1 = v0 + aΔt as follows: Δt = −
v0 = 6.00 × 10−4 s = 600 μs a
(c)
Using the above kinematic equation, we can find the velocity as a function of t. For example at t = 60 μs,
vx = 400 m/s + (−6.667 × 105 m/s 2 )(60 × 10−6 s) = 360 m/s
6.39.
Model: Visualize:
Solve:
Represent the rocket as a particle that follows Newton’s second law.
(a) The y-component of Newton’s second law is
ay = a =
( Fnet ) y m
=
Fthrust − mg 3.0 × 105 N = − 9.80 m/s 2 = 5.2 m/s 2 m 20,000 kg
(b) At 5000 m the acceleration has increased because the rocket mass has decreased. Solving the equation of part (a) for m gives
m5000 m =
Fthrust 3.0 × 105 N = = 1.9 × 104 kg a5000 m + g 6.0 m/s 2 + 9.80 m/s 2
The mass of fuel burned is mfuel = minitial − m5000 m = 1.0 × 103 kg.
6.40.
Model: The steel block will be represented by a particle. Steel-on-steel has a static coefficient of friction μ s = 0.80 and a kinetic coefficient of friction μ k = 0.60. Visualize:
Solve:
(a) While the block is at rest, Newton’s second law is
( Fnet ) x = T − fs = 0 N ⇒ T = The static friction force has a maximum value
fs
( Fnet ) y = n − FG ⇒ n = FG = mg
( fs )max = μs mg. The string tension that will cause the block to slip
and start to move is T = μs mg = ( 0.80 )( 2.0 kg ) ( 9.80 m/s 2 ) = 15.7 N Any tension less than this will not be sufficient to cause the block to move, so this is the minimum tension for motion. (b) As the block is moving with a tension of 20 N in the string, we can find its acceleration from the xcomponent of Newton’s second law as follows:
( Fnet ) x = T − f k = max ⇒ ax =
T − fk m
The kinetic friction force f k = μk mg . The acceleration of the block is ax =
20 N − ( 0.60 )( 2 kg ) ( 9.8 m/s 2 ) 2 kg
= 4.12 m/s 2
Using kinematics, the block’s speed after moving 1.0 m will be v12 = 0 m 2 /s 2 + 2 ( 4.12 m/s 2 ) (1.0 m ) ⇒ v1 = 2.9 m/s (c) The only difference in this case is the coefficient of kinetic friction whose value is 0.050 instead of 0.60. The acceleration of the block is
ax =
20 N − ( 0.050 )( 2.0 kg ) ( 9.80 m/s 2 ) 2.0 kg
= 9.51 m/s 2
The block’s speed after moving 1.0 m will be v12 = 0 m 2 /s 2 + 2 ( 9.51 m/s 2 ) (1.0 m ) ⇒ v1 = 4.4 m/s
6.41.
Model: We assume that Sam is a particle moving in a straight horizontal line under the influence of two forces: the thrust of his jet skis and the resisting force of friction on the skis. We can use one-dimensional kinematics. Visualize:
Solve: (a) The friction force of the snow can be found from the free-body diagram and Newton’s first law, since there’s no acceleration in the vertical direction:
n = FG = mg = ( 75 kg ) ( 9.80 m/s 2 ) = 735 N ⇒ f k = μ k n = ( 0.10 )( 735 N ) = 73.5 N Then, from Newton’s second law:
( Fnet ) x = Fthrust − f k = ma0 ⇒ a0 =
Fthrust − f k 200 N − 73.5 N = = 1.687 m/s 2 m 75 kg
From kinematics:
v1 = v0 + a0t1 = 0 m/s + (1.687 m/s 2 ) (10 s ) = 16.9 m/s (b) During the acceleration, Sam travels to
1 1 2 x1 = x0 + v0t1 + a0t12 = (1.687 m/s 2 ) (10 s ) = 84 m 2 2 After the skis run out of fuel, Sam’s acceleration can again be found from Newton’s second law: F −73.5 N = −0.98 m/s 2 ( Fnet ) x = − f k = −73.5 N ⇒ a1 = net = m 75 kg
Since we don’t know how much time it takes Sam to stop: v22 = v12 + 2a1 ( x2 − x1 ) ⇒ x2 − x1 =
2 2 v22 − v12 0 m /s − (16.9 m/s ) = = 145 m 2a1 2 ( −0.98 m/s 2 ) 2
The total distance traveled is ( x2 − x1 ) + x1 = 145 m + 84 m = 229 m. Assess: A top speed of 16.9 m/s (roughly 40 mph) seems quite reasonable for this acceleration, and a coasting distance of nearly 150 m also seems possible, starting from a high speed, given that we’re neglecting air resistance.
6.42.
Model: We assume Sam is a particle moving in a straight line down the slope under the influence of gravity, the thrust of his jet skis, and the resisting force of friction on the snow. Visualize:
Solve:
From the height of the slope and its angle, we can calculate its length: h h 50 m = sin θ ⇒ x1 − x0 = = = 288 m x1 − x0 sin θ sin10°
Since Sam is not accelerating in the y-direction, we can use Newton’s first law to calculate the normal force:
( Fnet ) y = ΣFy = n − FG cosθ = 0 N
⇒ n = FG cosθ = mg cosθ = (75 kg)(9.80 m/s 2 )(cos10°) = 724 N
One-dimensional kinematics gives us Sam’s acceleration: v12 = v02 + 2ax ( x − x0 ) ⇒ ax =
v12 − v02 ( 40 m/s ) − 0 m 2 /s 2 = 2.78 m/s 2 = 2 ( x1 − x2 ) 2 ( 288 m ) 2
Then, from Newton’s second law and the equation f k = μ k n :
( Fnet ) x = ΣFx = FG sinθ + Fthrust − f k = max ⇒ μk = =
mg sin θ + Fthrust − ma n
( 75 kg ) ( 9.80 m/s 2 ) ( sin10° ) + 200 N − ( 75 kg ) ( 2.78 m/s 2 )
724 N Assess: This coefficient seems a bit high for skis on snow, but not impossible.
= 0.165
6.43.
Model: only. Visualize:
We assume the suitcase is a particle accelerating horizontally under the influence of friction
Solve: Because the conveyor belt is already moving, friction drags your suitcase to the right. It will accelerate until it matches the speed of the belt. We need to know the horizontal acceleration. Since there’s no acceleration in the vertical direction, we can apply Newton’s first law to find the normal force:
n = FG = mg = (10 kg ) ( 9.80 m/s 2 ) = 98.0 N The suitcase is accelerating, so we use μ k to find the friction force
f k = μ k mg = ( 0.3)( 98.0 N ) = 29.4 N We can find the horizontal acceleration from Newton’s second law:
( Fnet ) x = ΣFx =
f k = ma ⇒ a =
f k 29.4 N = = 2.94 m/s 2 m 10 kg
From one of the kinematic equations: v12 = v02 + 2a ( x1 − x0 ) ⇒ x1 − x0 =
v12 − v02 ( 2.0 m/s ) − ( 0 m/s ) = = 0.68 m 2a 2 ( 2.94 m/s 2 ) 2
2
The suitcase travels 0.68 m before catching up with the belt and riding smoothly. Assess: If we imagine throwing a suitcase at a speed of 2.0 m/s onto a motionless surface, 0.68 m seems a reasonable distance for it to slide before stopping.
6.44. Model: The box of shingles is a particle subject to Newton’s laws and kinematics. Visualize:
Solve: Newton’s laws can be used in the coordinate system in which the direction of motion of the box of JG shingles defines the +x-axis. The angle that FG makes with the –y-axis is 25°.
(∑ F )
(∑ F )
y
x
= FG sin25° − f k = ma
= n − FG cos25° = 0 ⇒ n = FG cos25°
We have used the observation that the shingles do not leap off the roof, so the acceleration in the y-direction is zero. Combining these equations with f k = μ k n and FG = mg yields mg sin 25° − μ k mg cos 25° = ma ⇒ a = ( sin25° − μ k cos25° ) g = −0.743 m/s 2 where the minus sign indicates the acceleration is directed up the incline. The required initial speed to have the box come to rest after 5.0 m is found from kinematics.
vf2 = vi2 + 2aΔx ⇒ vi2 = −2 ( −0.743 m/s 2 ) ( 5.0 m ) ⇒ vi = 2.7 m/s Assess: To give the shingles an initial speed of 2.7 m/s requires a strong, determined push, but is not beyond reasonable.
6.45.
Model: Visualize:
We will model the box as a particle, and use the models of kinetic and static friction.
The pushing force is along the +x-axis, but the force of friction acts along the –x-axis. A component of the gravitational force on the box acts along the –x-axis as well. The box will move up if the pushing force is at least equal to the sum of the friction force and the component of the gravitational force in the x-direction. Solve: Let’s determine how much pushing force you would need to keep the box moving up the ramp at steady speed. Newton’s second law for the box in dynamic equilibrium is
( Fnet ) x = ΣFx = nx + ( FG ) x + ( f k ) x + ( Fpush ) x = 0 N − mg sinθ − f k + Fpush = 0 N ( Fnet ) y = ΣFy = n y + ( FG ) y + ( f k ) y + ( Fpush ) y = n − mg cosθ + 0 N + 0 N = 0 N The x-component equation and the model of kinetic friction yield: Fpush = mg sinθ + f k = mg sinθ + μ k n Let us obtain n from the y-component equation as n = mg cosθ , and substitute it in the above equation to get Fpush = mg sinθ + μ k mg cosθ = mg (sinθ + μ k cosθ )
= (100 kg)(9.80 m/s 2 )(sin 20° + 0.60 cos 20°) = 888 N The force is less than your maximum pushing force of 1000 N. That is, once in motion, the box could be kept moving up the ramp. However, if you stop on the ramp and want to start the box from rest, the model of static friction applies. The analysis is the same except that the coefficient of static friction is used and we use the maximum value of the force of static friction. Therefore, we have Fpush = mg (sinθ + μ s cosθ ) = (100 kg)(9.80 m/s 2 )(sin 20° + 0.90 cos 20°) = 1160 N Since you can push with a force of only 1000 N, you can’t get the box started. The big static friction force and the weight are too much to overcome.
6.46.
Model: We will represent the wood block as a particle, and use the model of kinetic friction and kinematics. Assume w sin θ > fs , so it does not hang up at the top. Visualize:
The block ends where it starts, so x2 = x0 = 0 m. We expect v2 to be negative, because the block will be moving in the –x-direction, so we’ll want to take v2 as the final speed. Because of friction, we expect to find v2 < v0 . G → Solve: (a) The friction force is opposite to v, so f k points down the slope during the first half of the motion G G and up the slope during the second half. FG and n are the only other forces. Newton’s second law for the upward motion is
ax = a0 =
( Fnet ) x m
a y = 0 m/s 2 =
=
− FG sinθ − f k − mg sinθ − f k = m m
( Fnet ) y m
=
n − FG cosθ n − mg cosθ = m m
The friction model is f k = μ k n. First solve the y-equation to give n = mg cosθ . Use this in the friction model to get f k = μ k mg cosθ . Now substitute this result for f k into the x-equation: a0 =
−mg sin θ − μ k mg cosθ = − g ( sinθ + μ k cosθ ) = − ( 9.8 m/s 2 ) ( sin 30° + 0.20cos30° ) = −6.60 m/s 2 m
Kinematics now gives
v12 = v02 + 2a0 ( x1 − x0 ) ⇒ x1 =
2 2 v12 − v02 0 m /s − (10 m/s ) = = 7.6 m 2a0 2 ( −6.60 m/s 2 ) 2
The block’s height is then h = x1 sinθ = (7.6 m)sin 30° = 3.8 m. G (b) For the return trip, f k points up the slope, so the x-component of the second law is
ax = a1 =
( Fnet ) x m
=
− FG sin θ + f k −mg sin θ + f k = m m
Note the sign change. The y-equation and the friction model are unchanged, so we have a1 = − g ( sin θ − μ k cosθ ) = −3.20 m/s 2
The kinematics for the return trip are
v22 = v12 + 2a1 ( x2 − x1 ) ⇒ v2 = −2a1 x1 = 2 ( −3.20 m/s 2 ) ( −7.6 m ) = −7.0 m/s Notice that we used the negative square root because v2 is a velocity with the vector pointing in the –x-direction. The final speed is v2 = 7.0 m/s.
6.47.
Model: We will model the sled and friend as a particle, and use the model of kinetic friction because the sled is in motion. Visualize:
The net force on the sled is zero (note the constant speed of the sled). That means the component of the pulling force along the +x-direction is equal to the magnitude of the kinetic force of friction in the –x-direction. Also note that ( Fnet ) y = 0 N, since the sled is not moving along the y-axis. Solve:
Newton’s second law is ( Fnet ) x = ΣFx = nx + ( FG ) x + ( f k ) x + ( Fpull ) x = 0 N + 0 N − f k + Fpull cosθ = 0 N ( Fnet ) y = ΣFy = n y + ( FG ) y + ( f k ) y + ( Fpull ) y = n − mg + 0 N + Fpull sin θ = 0 N
The x-component equation using the kinetic friction model f k = μ k n reduces to
μ k n = Fpull cosθ The y-component equation gives
n = mg − Fpull sin θ
We see that the normal force is smaller than the gravitational force because Fpull has a component in a direction opposite to the direction of the gravitational force. In other words, Fpull is partly lifting the sled. From the xcomponent equation, μ k can now be obtained as
μk =
Fpull cosθ mg − Fpull sinθ
=
( 75 N )( cos30° ) = 0.12 ( 60 kg ) ( 9.80 m/s 2 ) − ( 75 N )( sin 30° )
Assess: A quick glance at the various μ k values in Table 6.1 suggests that a value of 0.12 for μ k is reasonable.
6.48.
Model: As long as the static force of friction between the box and the sled is sufficiently large for the box not to slip, the acceleration a of the box is the same as the acceleration of the sled. We will therefore model the box as a particle and use the model of static friction. Visualize:
The force on the box that is responsible for its acceleration is the force of static friction provided by the sled because the box would slide backward without this force. Solve: Newton’s second law for the box is
( Fnet ) x = ΣFx = nx + ( ( FG )box ) + ( fs ) x = 0 N + 0 N + f s max = max = mamax x
( Fnet ) y = ΣFy = n y + ( ( FG )box ) + ( fs ) y = n − mg + 0 N = ma y = 0 N y
The above equations can be combined along with the friction model to get amax =
fs max μs n μ s mg = = = μs g m m m
To find Tmax , we write Newton’s second law for the box-sled system, with total mass m + M , as
( Fnet ) x = ΣFx = N x + ( ( FG )sled+box ) + ( fs ) x + (Tmax ) x = 0 N + 0 N + 0 N + Tmax = (m + M )amax x
( Fnet ) y = ΣFy = N y + ( ( FG )sled+box ) + ( fs ) y + (Tmax ) y = N − (m + M ) g + 0 N + 0 N = 0 N y
Referring to Table 6.1 for the coefficient of friction, the x-component equation for the box-sled system yields
Tmax = (m + M )amax = ( m + M ) μs g = (0.50)(9.80 m/s 2 )(5.0 kg + 10 kg) = 74 N That is, the largest tension force for which the box does not slip is 74 N. Assess: If the box were glued to the sled, there would be no resistance to motion because the ice is frictionless, and the tension force could be as large as one wants. On the other hand, if the friction between the box and the sled were zero, one could not pull at all without causing the box to slide.
6.49.
Model: We will model the steel cabinet as a particle. It touches the truck’s steel bed, so only the steel bed can exert contact forces on the cabinet. As long as the cabinet does not slide, the acceleration a of the cabinet is equal to the acceleration of the truck. Visualize:
Solve: The shortest stopping distance is the distance for which the static friction force has its maximum value ( fs )max . Newton’s second law for the box and the model of static friction are
( Fnet ) x = ΣFx = nx + ( FG ) x + ( fs ) x = 0 N + 0 N − ( fs )max = max = ma ⇒ − ( f s )max = ma ( Fnet ) y = ΣFy = n y + ( FG ) y + ( f s ) y = n − mg + 0 N = 0 N ⇒ n = mg − ( f s )max = − μs n = − μs mg = ma These three equations can be combined together to get a = − μ s g . Because constant-acceleration kinematics
gives v12 = v02 + 2a ( x1 − x0 ) and μ s = 0.80 (Table 6.1), we find
( x1 − x0 ) =
v12 − v02 −v02 (15 m/s ) = = = 14.3 m g 2a 2 − μ 2 0.80 ( s ) ( )( ) ( 9.80 m/s 2 ) 2
Assess: The truck was moving at a speed of 15 m/s or at approximately 34 mph. A stopping distance without making the contents slide of about 14.3 m or approximately 47 feet looks reasonable.
Model: The antiques (mass = m) in the back of your pickup (mass = M ) will be treated as a particle. The antiques touch the truck’s steel bed, so only the steel bed can exert contact forces on the antiques. The pickup-antiques system will also be treated as a particle, and the contact force on this particle will be due to the road. Visualize:
6.50.
Solve: (a) We will find the smallest coefficient of friction that allows the truck to stop in 55 m, then compare that to the known coefficients for rubber on concrete. For the pickup-antiques system, with mass m + M , Newton’s second law is
( Fnet ) x = ΣFx = N x + ( ( FG )PA ) + ( f ) x = 0 N + 0 N − f = ( m + M ) ax = (m + M )a x
( Fnet ) y = ΣFy = N y + ( ( FG )PA ) + ( f ) y = N − (m + M ) g + 0 N = 0 N y
The model of static friction is f = μ N , where μ is the coefficient of friction between the tires and the road. These equations can be combined to yield a = − μ g . Since constant-acceleration kinematics gives
v12 = v02 + 2a( x1 + x0 ), we find v12 − v02 v02 ( 25 m/s ) ⇒ μ min = = = 0.58 2 ( x1 − x0 ) 2 g ( x1 − x0 ) ( 2 ) ( 9.8 m/s 2 ) ( 55 m ) 2
a=
The truck cannot stop if μ is smaller than this. But both the static and kinetic coefficients of friction, 1.00 and 0.80 respectively (see Table 6.1), are larger. So the truck can stop. (b) The analysis of the pickup-antiques system applies to the antiques, and it gives the same value of 0.58 for μ min . This value is smaller than the given coefficient of static friction ( μs = 0.60) between the antiques and the truck bed. Therefore, the antiques will not slide as the truck is stopped over a distance of 55 m.
Assess: The analysis of parts (a) and (b) are the same because mass cancels out of the calculations. According to the California Highway Patrol Web site, the stopping distance (with zero reaction time) for a passenger vehicle traveling at 25 m/s or 82 ft/s is approximately 43 m. This is smaller than the 55 m over which you are asked to stop the truck.
6.51.
Model: The box will be treated as a particle. Because the box slides down a vertical wood wall, we will also use the model of kinetic friction. Visualize:
Solve: The normal force due to the wall, Gwhich is perpendicular to the wall, is here to the right. JG The G box slides G down the wall at constant speed, so a = 0 and the box is in dynamic equilibrium. Thus, Fnet = 0. Newton’s second law for this equilibrium situation is ( Fnet ) x = 0 N = n − Fpush cos 45°
( Fnet ) y = 0 N = f k + Fpush sin 45° − FG = f k + Fpush sin 45° − mg The friction force is f k = μ k n. Using the x-equation to get an expression for n, we see that f k = μ k Fpush cos 45°. Substituting this into the y-equation and using Table 6.1 to find μ k = 0.20 gives,
μ k Fpush cos 45° + Fpush sin 45° − mg = 0 N ⇒ Fpush =
( 2.0 kg ) ( 9.80 m/s 2 ) mg = = 23 N μ k cos 45° + sin 45° 0.20cos 45° + sin 45°
6.52.
Model: Visualize:
Use the particle model for the block and the model of static friction.
Solve: The block is initially at rest, so initially the friction force is static friction. If the 12 N push is too strong, the box will begin to move up the wall. If it is too weak, the box will begin to slide down the wall. And if the pushing force is within the proper range, the box will remain stuck in place. First, let’s evaluate the sum of all the forces except friction:
ΣFx = n − Fpush cos30° = 0 N ⇒ n = Fpush cos30° ΣFy = Fpush sin 30° − FG = Fpush sin 30° − mg = (12 N)sin 30° − (1 kg)(9.8 m/s 2 ) = −3.8 N In the first equation we utilized the fact that any motion is parallel to the wall, so ax = 0 m/s 2 . These three forces add up to −3.8 ˆj N. This means the static friction force will be able to prevent the box from moving if
fs = +3.8 ˆj N. Using the x-equation and the friction model we get
( fs )max = μs n = μs Fpush cos30° = 5.2 N
G where we used μ s = 0.5 for wood on wood. The static friction force fs needed to keep the box from moving is less than ( fs )max . Thus the box will stay at rest.
6.53.
Model: We will model the skier along with the wooden skis as a particle of mass m. The snow exerts a contact force and the wind exerts a drag force on the skier. We will therefore use the models of kinetic friction and drag. Visualize:
We choose a coordinate system such that the skier’s motion is along the +x-direction. While the forces of kinetic G G friction f k and drag D act along the –x-direction opposing the motion of the skier, the gravitational force on the skier has a component in the +x-direction. At the terminal speed, the net force on the skier is zero as the forces along the +x-direction cancel out the forces along the –x-direction. Solve: Newton’s second law and the models of kinetic friction and drag are
( Fnet ) x = ΣFx = nx + ( FG ) x + ( f k ) x + ( D) x = 0 N + mg sin θ − f k
1 2 Av = max = 0 N 4
( Fnet ) y = ΣFy = n y + ( FG ) y + ( f k ) y + ( D) y = n − mg cosθ + 0 N + 0 N = 0 N
fk = μk n These three equations can be combined together as follows:
(1/4) Av 2 = mg sin θ − f k = mg sin θ − μ k n = mg sinθ − μ k mg cosθ 12
⎛ sinθ − μ k cosθ ⎞ ⇒ vterm = ⎜ mg ⎟ 1 A 4 ⎝ ⎠ Using μ k = 0.06 and A = 1.8 m × 0.40 m = 0.72 m 2 , we find
vterm
⎡ ⎛ sin 40° − 0.06cos 40° ⎞ ⎤ ⎟⎥ = ⎢( 80 kg ) ( 9.8 m/s 2 ) ⎜ 1 ⎜ ( 4 kg m3 )( 0.72 m 2 ) ⎟ ⎥ ⎢ ⎝ ⎠⎦ ⎣
12
= 51 m/s
Assess: A terminal speed of 51 m/s corresponds to a speed of ≈100 mph. This speed is reasonable but high due to the steep slope angle of 40° and a small coefficient of friction.
6.54. Model: The ball is a particle experiencing a drag force and traveling at twice its terminal velocity. Visualize:
Solve: (a) An object falling at greater than its terminal velocity will slow down to its terminal velocity. Thus the drag force is greater than the force of gravity, as shown in the free-body diagrams. When the ball is shot straight up,
(∑ F )
y
1 1 2 ⎛ ⎞ ⎛ 4mg ⎞ = ma = − ( FG + D ) = − ⎜ mg + Av 2 ⎟ = − mg − A ( 2vterm ) = − mg − A ⎜ ⎟ = −5mg 4 4 ⎝ ⎠ ⎝ A ⎠
Thus a = −5 g , where the minus sign indicates the downward direction. We have used Equations 6.16 for the drag force and 6.19 for the terminal velocity. (b) When the ball is shot straight down,
(∑ F )
y
= ma = D − FG =
1 2 A ( 2vterm ) − mg = 3mg 4
Thus a = 3 g , this time directed upward. (c)
The ball will slow down to its terminal velocity, slowing quickly at first, and more slowly as it gets closer to the terminal velocity because the drag force decreases as the ball slows.
6.55.
Model: We will model the sculpture as a particle of mass m. The ropes that support the sculpture will be assumed to have zero mass. Visualize:
Solve:
Newton’s first law in component form is
( Fnet ) x = ΣFx = T1x + T2 x + FGx = −T1 sin 30° + T2 sin 60° + 0 N = 0 N
( Fnet ) y = ΣFy = T1 y + T2 y + FGy = −T1 cos30° + T2 cos60° − FG = 0 N Using the x-component equation to obtain an expression for T1 and substituting into the y-component equation yields: T2 =
500 lbs FG = 250 lbs = ( sin 60°)( cos30°) + cos60° 2 sin 30°
Substituting this value of T2 back into the x-component equation, T1 = T2
sin 60° sin 60° = 250 lbs = 433 lbs sin 30° sin 30°
We will now find a rope size for a tension force of 433 lbs, that is, the diameter of a rope with a safety rating of 433 lbs. Since the cross-sectional area of the rope is 14 π d 2 , we have 12
⎡ ⎤ 4 ( 433 lbs ) ⎥ d =⎢ 2 ⎢⎣ π ( 4000 lbs/inch ) ⎥⎦
= 0.371 inch
Any diameter larger than 0.371 inch will ensure a safety rating of at least 433 lbs. The rope size corresponding to a diameter of 3/8 of an inch will therefore be appropriate. Assess: If only a single rope were used to hang the sculpture, the rope would have to support a gravitational force of 500 lbs. The diameter of the rope for a safety rating of 500 lbs is 0.399 inches, and the rope size jumps from a diameter of 3/8 to 4/8 of an inch. Also note that the gravitational force on the sculpture is distributed in the two ropes. It is the sum of the y-components of the tensions in the ropes that will equal the gravitational force on the sculpture.
6.56.
Model: We will model the container as a particle of mass m. The steel cable of the crane will be assumed to have zero mass. Visualize:
Solve: As long as the container is stationary or it is moving with a constant speed (zero acceleration), the net force on the container is zero. In these cases, the tension in the cable is equal to the gravitational force on the container:
T = mg = 44,000 N The cable should safely lift the load. More tension is required to accelerate the load. Newton’s second law is ( Fnet ) y = ΣFy = ( FG ) y + (T ) y = − mg + T = ma y
The crane’s maximum acceleration is amax = 1.0 m/s 2 . So the maximum cable tension is
Tmax = mg + mamax = 48,600 N This is less than the cable’s rating, so the cable must have been defective.
6.57.
Model: Visualize:
Solve:
We will model the skier as a particle, and use the model of kinetic friction.
G G G G Your best strategy, if it’s possible, is to travel at a very slow constant speed a = 0 so Fnet = 0 .
(
)
Alternatively, you want the smallest positive ax . A negative ax would cause you to slow and stop. Let’s find the G G value of μ k that gives Fnet = 0. Newton’s second law for the skier and the model of kinetic friction are
( Fnet ) x = ΣFx = nx + ( FG ) x + ( f k ) x + ( D) x = 0 + mg sin θ − f k − D cosθ = 0 N ( Fnet ) y = ΣFy = n y + ( FG ) y + ( f k ) y + ( D ) y = n − mg cosθ + 0 N − D sin θ = 0 N
fk = μk n The x- and y-component equations are f k = + mg sin θ − D cosθ n = mg cosθ + D sin θ From the model of kinetic friction,
μk =
2 f k mg sin θ − D cosθ 75 kg ( 9.8 m/s ) sin15° − ( 50 N ) cos15° = = = 0.196 n mg cosθ + D sin θ 75 kg ( 9.8 m/s 2 ) cos15° + ( 50 N ) sin15°
Yellow wax with μ k = 0.20 applied to skis will make the skis stick and hence cause the skier to stop. The skier’s next choice is to use the green wax with μ k = 0.15.
6.58.
Model: Visualize:
The ball hanging from the ceiling of the truck by a string is represented as a particle.
Solve: (a) You cannot tell from within the truck. Newton’s first law says that there is no distinction between “at rest” and “constant velocity.” In both cases, the net force acting on the ball is zero and the ball hangs straight down. (b) Now you can tell. If the truck is accelerating, then the ball is tilted back at an angle. (c) The ball moves with the truck, so its acceleration is 5 m/s 2 in the forward direction. G (d) The free-body diagram shows that the horizontal component of T provides a net force in the forward direction. This is the net force that causes the ball to accelerate in the forward direction along with the truck. (e) Newton’s second law for the ball is
ax =
( Fnet ) x m
=
Tx T sin10° = m m
a y = 0 m 2 /s 2 =
( Fnet ) y m
=
Ty + ( FG ) y m
We can solve the second equation for the magnitude of the tension:
T=
(1.0 kg ) ( 9.8 m/s 2 ) mg = = 9.95 N cos10° cos10°
Then the first equation gives the acceleration of the ball and truck: ax =
T sin10° ( 9.95 N ) sin10° = =1.73 m/s 2 m m
The truck’s velocity cannot be determined.
=
T cos10° − mg m
6.59.
Model: You will be treated as a particle. You will experience your weight force and the force (P) of the scale pulling up. Visualize:
G Solve: The weight of an object is the magnitude of the contact force supporting it. Here, the contact force is P, so the weight is w = P. Newton’s second law is
⎛ a⎞ Fnet = ΣFy = Py + ( FG ) y = ma y ⇒ Fnet = P − FG = ma ⇒ w = P = mg + ma = mg ⎜1 + ⎟ ⎝ g⎠
6.60. Solve: Using ax =
dvx , we express Newton’s second law as a differential equation, which we then use dt
to solve for vx . Fx = m
dvx F ct ⇒ dvx = x dt = dt dt m m
Integrating from the initial to final conditions for each variable of integration, vx
∫
v0 x
Thus
t
dvx =
c ct 2 t dt ⇒ v − v = x x 0 2m m ∫0 vx = v0 x +
ct 2 2m
6.61. Model: The astronaut is a particle oscillating on a spring. Solve:
(a) The position versus time function x(t) can be used to find the velocity versus time function v(t ) =
We have v(t ) =
d ( 0.30 m ) sin ( (π rad/s ) t ) = ( 0.30π m/s ) cos ( (π rad/s ) t ) dt
{
}
This can then be used to find the acceleration a (t ) =
a (t ) =
dv . dt
dv = − ( 0.30π 2 m/s 2 ) sin ( (π rad/s ) t ) dt
Newton’s second law yields a general expression for the force on the astronaut.
Fnet (t ) = ma(t ) = −(75 kg) ( 0.30π 2 m/s 2 ) sin ( (π rad/s ) t ) Evaluating this at t = 1.0 s gives Fnet (1.0 s ) = 0 N, since sin(π ) = 0. (b) Evaluating at t = 1.5 s, ⎛ 3π Fnet = −22.5π 2 N sin ⎜ ⎝ 2
Assess:
⎞ 2 ⎟ = 2.2 × 10 N ⎠
The force of 220 N is only one third of the astronaut’s weight on earth, so is easy for her to withstand.
dx . dt
6.62. Solve: (a) The terminal velocity for a falling object is reached when the downward gravitational force is balanced by the upward drag force.
FG = D mg = bvterm = 6πη Rvterm
⇒ vterm =
mg 6πη R
⎛4 ⎞ (b) The mass of the spherical sand grain of density p = 2400 kg/m3 is m = ρ ⎜ π R 3 ⎟ . ⎝3 ⎠ −4 3 2 2 ρ gR 2 2 ( 2400 kg/m )( 9.80 m/s )( 5.0 × 10 m ) vterm = = = 1.3 m/s Thus 9η 9 ⎛ −3 Ns ⎞ 1.0 × 10 ⎜ ⎟ m2 ⎠ ⎝ 50 m The time required for the sand grain to fall 50 m at this speed is t = = 38 s. 1.3 m/s Assess: The speed of 1.3 m/s for a sand grain falling through water seems about right. 2
6.63.
Solve: (a) A 1.0 kg block is pulled across a level surface by a string, starting from rest. The string has a tension of 20 N, and the block’s coefficient of kinetic friction is 0.50. How long does it take the block to move 1.0 m? (b) Newton’s second law for the block is
ax = a =
( Fnet ) x m
=
T − fk T − μk n = m m
a y = 0 m/s 2 =
( Fnet ) y m
=
n − FG n − mg = m m
where we have incorporated the friction model into the first equation. The second equation gives n = mg . Substituting this into the first equation gives
a=
T − μ k mg 20 N − 4.9 N = = 15.1 m/s 2 m 1.0 kg
Constant acceleration kinematics gives 2 (1.0 m ) 1 1 2 x1 2 2 x1 = x0 + v0 Δt + a ( Δt ) = a ( Δt ) ⇒ Δt = = = 0.36 s 2 2 a 15.1 m/s 2
6.64.
Solve: (a) A 15,000 N truck starts from rest and moves down a 15° hill with the engine providing a 12,000 N force in the direction of the motion. Assume the frictional force between the truck and the road is very small. If the hill is 50 m long, what will be the speed of the truck at the bottom of the hill? (b) Newton’s second law is
ΣFy = ny + FGy + f y + E y = ma y = 0 ΣFx = nx + FGx + f x + Ex = max ⇒ 0 N + FG sin θ + 0 N + 12,000 N = ma ⇒a=
mg sinθ + 12,000 N (15,000 N ) sin15° + 12,000 N = = 10.4 m/s 2 m (15,000 N/9.8 m/s2 )
where we have calculated the mass of the truck from the gravitational force on it. Using the constant-acceleration kinematic equation vx2 − v02 = 2ax,
vx2 = 2ax x = 2 (10.4 m/s 2 ) ( 50 m ) ⇒ vx = 32 m/s
6.65.
Solve: (a) A 1.0 kg box is pushed along an assembly line by a mechanical arm. The arm exerts a 20 N force at a downward angle of 30°. The box has a coefficient of kinetic friction on the assembly line of 0.25. What is the speed of the box after traveling 50 cm, starting from rest? (b) Newton’s second law for the box is
ax = a =
( Fnet ) x m
a y = 0 m/s 2 =
=
Fpush cos30° − f k
( Fnet ) y m
=
Fpush cos30° − μ k n
m m n − FG − Fpush sin 30° n − mg − Fpush sin 30° = = m m
The second equation is solved to give an expression for n. Substituting into the first equation:
a=
Fpush cos30° − μ k ( Fpush sin 30° + mg ) m
=
( 20 N ) cos30° − 4.9 N = 12.4 m/s 2 1.0 kg
Using kinematics,
v12 = v02 + 2aΔx = 2aΔx ⇒ v1 = 2aΔx = 2 (12.4 m/s 2 ) ( 0.50 m ) = 3.5 m/s
6.66.
Solve: (a) A driver traveling at 40 m/s in her 1500 kg auto slams on the brakes and skids to rest. How far does the auto slide before coming to rest? (b)
(c) Newton’s second law is ΣFy = n y + ( FG ) y = n − mg = ma y = 0 N
ΣFx = −0.80n = max
The y-component equation gives n = mg = (1500 kg ) ( 9.8 m/s 2 ) . Substituting this into the x-component equation
yields
(1500 kg ) ax = −0.80 (1500 kg ) ( 9.8 m/s 2 ) ⇒ ax = ( −0.80 ) ( 9.8 m/s2 ) = −7.8 m/s 2 Using the constant-acceleration kinematic equation v12 = v02 + 2aΔx, we find
( 40 m/s ) = 102 m v02 =− 2a 2 ( −7.8 m/s 2 ) 2
Δx = −
6.67. Solve: (a) A 20.0 kg wooden crate is being pulled up a 20° wooden incline by a rope that is connected to an electric motor. The crate’s acceleration is measured to be 2.0 m/s 2 . The coefficient of kinetic friction between the crate and the incline is 0.20. Find the tension T in the rope. (b)
(c) Newton’s second law for this problem in the component form is
( Fnet ) x = ΣFx = T − 0.20n − (20 kg)(9.80 m/s 2 )sin 20° = (20 kg)(2.0 m/s 2 ) ( Fnet ) y = ΣFy = n − (20 kg)(9.80 m/s 2 )cos 20° = 0 N Solving the y-component equation, n = 184.18 N. Substituting this value for n in the x-component equation yields T = 144 N.
6.68.
Solve: (a) You wish to pull a 20 kg wooden crate across a wood floor ( μ k = 0.20) by pulling on a rope
attached to the crate. Your pull is 100 N at an angle of 30° above the horizontal. What will be the acceleration of the crate? (b)
(c) Newton’s equations and the model of kinetic friction are
ΣFx = nx + Px + ( FG ) x + f x = 0 N + (100 N ) cos30° + 0 N − f k = (100 N ) cos30° − f k = max ΣFy = n y + Py + ( FG ) y + f y = n + (100 N ) sin 30° − mg − 0 N = ma y = 0 N fk = μk n From the y-component equation, n = 150 N. From the x-component equation and using the model of kinetic friction with μ k = 0.20,
(100 N)cos30° − (0.20)(150 N) = (20 kg) ax ⇒ ax = 2.8 m/s 2
6.69.
Model: Take the coin to be a particle held against the palm of your hand (which is like a vertical wall) by friction. The friction needs to be vertical and equal to the weight of the coin to hold the coin in place. Visualize:
The hand pushes on the coin giving a normal force to the coin and causing a friction force. Solve: (a) We need a force (push) to create a normal force. Since the force accelerates the coin, a minimum acceleration amin is needed. (b) Newton’s second law and the model of friction are
ΣFy = ( f min ) y − ( FG ) y = f min − mg = 0 N
ΣFx = nmin = mamin f min = μ nmin
Since you do not want the coin to slip down the hand you need μs . Combining the above three equations yields
f min = mg ⇒ μ s nmin = mg ⇒ μ s mamin = mg ⇒ amin =
g
μs
=
9.8 m/s 2 = 12.3 m/s 2 0.80
6.70.
Model: We will model the shuttle as a particle and assume the elastic cord to be massless. We will also use the model of kinetic friction for the motion of the shuttle along the square steel rail. Visualize:
Solve:
The upward tension component Ty = T sin 45° = 14.1 N is larger than the gravitational force on the
shuttle. Consequently, the elastic cord pulls the shuttle up against the rail and the rail’s normal force pushes downward. Newton’s second law in component form is ( Fnet ) x = ΣFx = Tx + ( f k ) x + (n) x + ( FG ) x = T cos 45° − f k + 0 N + 0 N = max = max ( Fnet ) y = ΣFy = Ty + ( f k ) y + ( n) y + ( FG ) y = T sin 45° + 0 N − n − mg = ma y = 0 N
The model of kinetic friction is f k = μ k n. We use the y-component equation to get an expression for n and hence f k . Substituting into the x-component equation and using the value of μ k in Table 6.1 gives us
T cos 45° − μ k (T sin 45° − mg ) m ( 20 N ) cos 45° − ( 0.60 ) ⎡⎣ + ( 20 N ) sin 45° − ( 0.800 kg ) ( 9.80 m/s 2 )⎤⎦ = = 13.0 m/s 2 0.800 kg
ax =
Assess: The x-component of the tension force is 14.1 N. On the other hand, the net force on the shuttle in the xdirection is max = (0.800 kg)(13.0 m/s 2 ) = 10.4 N. This value for ma is reasonable since a part of the 14.1 N tension force is used up to overcome the force of kinetic friction.
6.71.
Model: Assume the ball is a particle on a slope, and that the slope increases as the x-displacement increases. Assume that there is no friction and that the ball is being accelerated to the right so that it remains at rest on the slope. Visualize: Although the ball is on a slope, it is accelerating to the right. Thus we’ll use a coordinate system with horizontal and vertical axes.
Solve:
Newton’s second law is
ΣFx = n sin θ = max
ΣFy = n cosθ − FG = ma y = 0 N
Combining the two equations, we get max =
FG sin θ = mg tan θ ⇒ ax = g tan θ cosθ
The curve is described by y = x 2 . Its slope a position x is tanθ, which is also the derivative of the curve. Hence, dy = tan θ = 2 x ⇒ ax = (2 x) g dx
(b) The acceleration at x = 0.20 m is ax = (2)(0.20)(9.8 m/s 2 ) = 3.9 m/s 2 .
6.72.
Model: Visualize:
We will represent the widget as a particle.
Please refer to Figure CP6.72. Solve: (a) There are only two forces on the widget: the normal force of the table and the gravitational force. (b) Newton’s second law along the y-axis is
( Fnet ) y = ny + ( FG ) y = ny − mg = ma y ⇒ ny = m ( a y + g ) We know what the a y -vs-t graph looks like. To get the ny -vs-t graph, we need to add g y to the graph, which amounts to shifting the whole graph up by 9.8 m/s 2 , and multiplying by m = 5 kg. (c) The normal force is negative for t > 0.75 s. Physically, this means that the normal force is pointed in the downward direction. In other words, the table is pulling down on the widget rather than pushing up on the widget. It can do this because the widget is glued to the table rather than simply sitting on the table. (d) The weight is a maximum at t = 0 s, when the upward acceleration is maximum. (e) The weight is zero at t = 0.75 s when a y = −9.8 m/s 2 = − g . (f) If not glued down, the widget will fly off the table at t = 0.75 s, the instant at which its weight becomes zero. The table is accelerating in the negative direction so quickly after t = 0.75 s that the widget can stay on only if the table pulls downward on it. That is the significance of the negative value for ny . If the widget is not glued down, the largest downward acceleration it can achieve is the free-fall acceleration.
6.73. Visualize:
Solve: (a) The horizontal velocity as a function of time is determined by the horizontal net force. Newton’s second law as the x-direction gives ( Fnet ) x = max = − D cosθ = −bv cosθ = −bvx G JJG Note that D points opposite to v, so the angle θ with the x-axis is the same for both vectors, and the x components of both vectors have the same cosθ term. As the particle changes direction as it falls, the evolution of the horizontal motion depends only on the horizontal component of the velocity. Thus m vx ( t )
Separating and integrating,
∫
v0
dvx = −bvx dt
t
dvx b = − ∫ dt vx m0 ⎛ v (t ) ⎞ b ⇒ ln ⎜ x ⎟ = − t v m ⎝ 0 ⎠
Solving, vx ( t ) = v0e
−
bt m
= v0e
−
6πη Rt m
1 (b) The time to reach v ( t ) = v0 is found by solving for the time when 2 6πη Rt − 1 v0 = v0e m 2
Hence t=
m ln ( 2 ) 6πη R
With η = 1.0 × 10−3 Ns/m 2 , R = 2.0 × 10−2 m, and m = 0.033 kg, we get t = 61 s. 6πη R v = 1.1× 10−2 s-1 vx . This is a small fraction of the Assess: The magnitude of the acceleration is ax = m velocity, so a time of about one minute to slow to half the initial speed is reasonable.
(
)
6.74. Visualize:
dvx ⎛ dvx ⎞⎛ dx ⎞ dvx . =⎜ ⎟⎜ ⎟ = vx dt ⎝ dx ⎠⎝ dt ⎠ dx (b) The horizontal motion is determined by using Newton’s second law in the horizontal direction. Using the free-body diagram at a later time t, ( Fnet ) x = max = − D cosθ = −bv cosθ = −bvx G JJG Note that since D points opposite to v, the angle θ with the x-axis is the same for both vectors, and the xcomponents of both vectors have the same cosθ term. Thus
Solve:
(a) Using the chain rule, ax =
max = mvx
dvx = −bvx dx
vx ( x )
∫
Separating and integrating,
dvx = −
v0
b m
x (t )
∫ dx
x0
⇒ vx ( x ) − v0 = −
b ( x ( t ) − x0 ) m
Solving with x0 = 0, vx ( x ) = v0 −
b 6πη R x = v0 − x m m
(c) The marble stops after traveling a distance d when vx ( d ) = 0.
v0 =
Hence
⇒d =
6πη R d m
mv0 6πη R
Using v0 = 10 cm/s, R = 0.50 cm, m = 1.0 × 10−3 kg, and using η = 1.0 × 10−3 Ns/m, d= Assess:
(1.0 ×10
−3
kg ) ( 0.10 m/s )
6π (1.0 × 10 Ns/m 2 )( 5.0 × 10−3 m ) −3
= 1.1 m
The equation for d indicates that a marble with a faster initial velocity travels a further distance.
6.75.
Model: Visualize:
We will model the object as a particle, and use the model of drag.
Solve: (a) We cannot use the constant-acceleration kinematic equations since the drag force causes the acceleration to change with time. Instead, we must use ax = dvx /dt and integrate to find vx . Newton’s second law for the object is
( Fnet ) x = ΣFx = nx + wx + ( f k ) x + ( D) x = 0 N + 0 N + 0 N −
1 2 dv Avx = max = m x 4 dt
This can be written
dvx A = dt 2 vx 4m We can integrate this from the start (v0 x at t = 0) to the end (vx at t ):
∫
vx v0 x
dvx A t 1 1 A = dt ⇒ − + = t 2 ∫ 0 vx vx v0 x 4m 4m
Solving for vx gives vx =
v0 x 1 + Av0 xt 4m
(b) Using A = (1.6 m)(1.4 m) = 2.24 m 2 , v0 x = 20 m/s, and m = 1500 kg, we get
vx =
20 m/s 1 20 ⎛ ⎞ ⎛ 20 m/s ⎞ ⇒t =⎜ − 1⎟ = ⎟⎜ 2.24t ( 20 ) 1 + 0.007467t 0.007467 ⎝ ⎠ ⎝ vx ⎠ 1+ 4 × 1500
where t is in seconds. We can now obtain the time t for v = 10 m/s:
1 ⎛ ⎞ ⎛ 20 m/s ⎞ ⎛ 20 ⎞ t =⎜ − 1⎟ = 133.93⎜ − 1⎟ = 134 s ⎟⎜ ⎝ 0.007467 ⎠ ⎝ 10 m/s ⎠ ⎝ 10 ⎠ When vx = 5 m/s, then t = 402 s.
(c) If the only force acting on the object was kinetic friction with, say, μ k = 0.05, that force would be (0.05)(1500 kg) (9.8 m/s 2 ) = 735 N. The drag force at an average speed of 10 m/s is D = 14 ( 2.24 )(10 )2 N = 56 N. We conclude that it is not reasonable to neglect the kinetic friction force.
7.1.
Visualize:
Solve: (a) The weight lifter is holding the barbell in dynamic equilibrium as he stands up, so the net force on the barbell and on the weight lifter must be zero. The barbells have an upward contact force from the weight lifter and the gravitational force downward. The weight lifter has a downward contact force from the barbells and an upward one from the surface. Gravity also acts on the weight lifter.
(b) The system is the weight lifter and barbell, as indicated in the figure. (c)
7.2.
Visualize:
Solve: (a) Both the bowling ball and the soccer ball have a normal force from the surface and gravitational force on them. The interaction forces between the two are equal and opposite.
(b) The system consists of the soccer ball and bowling ball, as indicated in the figure. (c)
Assess: Even though the soccer ball bounces back more than the bowling ball, the forces that each exerts on the other are part of an action/reaction pair, and therefore have equal magnitudes. Each ball’s acceleration due to the forces on it is determined by Newton’s second law, a = Fnet / m, which depends on the mass. Since the masses of the balls are different, their accelerations are different.
7.3.
Visualize:
Solve: (a) Both the mountain climber and bag of supplies have a normal force from the surface on them, as well as a gravitational force vertically downward. The rope has gravity acting on it, along with pulls on each end from the mountain climber and supply bag. Both the mountain climber and supply bag also experience a frictional force with the surface of the mountain. In the case of the motionless mountain climber it is static friction, but the sliding supply bag experiences kinetic friction.
(b) The system consists of the mountain climber, rope, and bag of supplies, as indicated in the figure. (c)
Assess: Since the motion is along the surface, it is convenient to choose the x-coordinate axis along the surface. The free-body diagram of the rope shows pulls that are slightly off the x-axis since the rope is not massless.
7.4.
Visualize:
Solve: (a) The car and rabbit both experience a normal force and friction from the floor and a gravitational force from the Earth. The push each exerts on the other is a Newton’s third law force pair.
(b) The system consists of the car and stuffed rabbit, as indicated in the figure. (c)
7.5. Visualize: Please refer to Figure EX7.5. Solve: (a) Gravity acts on both blocks, and where Block A is in contact with the floor there is a normal force and friction. The string tension is the same on both blocks since the rope and pulley are massless, and the pulley is frictionless. There are two third law pairs of forces at the surface where the two blocks meet. Block B pushes against Block A with a normal force, while Block A has a reaction force that pushes back against Block B. There is also friction between the two blocks at the surface.
(b) A string that will not stretch constrains the two blocks to accelerate at the same rate but in opposite directions. Block A accelerates down the incline with the same acceleration that Block B has up the incline. The system consists of the two blocks, as indicated in the figure. (c)
Assess: The inclined coordinate systems allow the acceleration a to be purely along the x-axis. This is convenient since the one component of a is zero, simplifying the mathematical expression of Newton’s second law.
7.6. Visualize: Please refer to Figure EX7.6. Solve: (a) For each block, there is a gravitational force with the earth, a normal force and kinetic friction with the surface, and a tension force due to the rope.
(b) The tension in the massless ropes over the frictionless pulley is the same on both blocks. Block A accelerates down the incline with the same acceleration that Block B has up the incline. The system consists of the two blocks, as indicated in the figure. (c)
Assess: The inclined coordinate systems allow the acceleration a to be purely along the x-axis. This is convenient since then one component of a is zero, simplifying the mathematical expression of Newton’s second law.
7.7. Model: We will model the astronaut and the chair as particles. The astronaut and the chair will be denoted by A and C, respectively, and they are separate systems. The launch pad is a part of the environment. Visualize:
Solve:
(a) Newton’s second law for the astronaut is
∑( F )
on A y
= nC on A − ( FG )A = mA aA = 0 N ⇒ nC on A = ( FG )A = mA g
By Newton’s third law, the astronaut’s force on the chair is
nA on C = nC on A = mA g = ( 80 kg ) ( 9.8 m/s 2 ) = 7.8 × 102 N (b) Newton’s second law for the astronaut is:
∑( F )
on A y
= nC on A − ( FG )A = mA aA ⇒ nC on A = ( FG )A + mA aA = mA ( g + aA )
By Newton’s third law, the astronaut’s force on the chair is
nA on C = nC on A = mA ( g + aA ) = ( 80 kg ) ( 9.8 m/s 2 + 10 m/s 2 ) = 1.6 × 103 N Assess:
This is a reasonable value because the astronaut’s acceleration is more than g.
7.8. (a) Visualize: The upper magnet is labeled U, the lower magnet L. Each magnet exerts a long-range magnetic force on the other. Each magnet and the table exert a contact force (normal force) on each other. In addition, the table experiences a normal force due to the surface.
(b) Solve:
Each object is in static equilibrium with ( Fnet ) = 0. Start with the lower magnet. Because
FU on L = 3( FG ) L = 6.0 N, equilibrium requires nT on L = 4.0 N. For the upper magnet, FL on U = FU on L = 6.0 N because these are an action/ reaction pair. Equilibrium for the upper magnet requires nT on U = 8.0 N. For the table, action/reaction pairs are nL on T = nT on L = 4.0 N and nU on T = nT on U = 8.0. The table’s gravitational force is ( FG )T = 20 N, leaving nS on T = 24 N in order for the table to be in equilibrium. Summarizing, Upper magnet
Table
Lower magnet
( FG ) U = 2.0 N nT on U = 8.0 N FL on U = 6.0 N
( FG )T = 20 N nU on T = 8.0 N nL on T = 4.0 N nS on T = 24 N
( FG ) L = 2.0 N nT on L = 4.0 N FU on L = 6.0 N
Assess: The result nS on T = 24 N makes sense. The combined gravitational force on the table and two magnets is 24 N. Because the table is in equilibrium, the upward normal force of the surface has to exactly balance the total gravitational force on the table and magnets.
7.9. Model: The car and the truck will be modeled as particles and denoted by the symbols C and T, respectively. The surface of the ground will be denoted by the symbol S. Visualize:
Solve:
(a) The x-component of Newton’s second law for the car is
∑( F )
on C x
= FS on C − FT on C = mC aC
The x-component of Newton’s second law for the truck is
∑( F )
on T x
= FC on T = mT aT
Using aC = aT = a and FT on C = FC on T , we get
(F
C on S
⎛ 1 ⎞ ⎛ 1 ⎞ − FC on T ) ⎜ ⎟ = a ( FC on T ) ⎜ ⎟=a ⎝ mT ⎠ ⎝ mC ⎠
Combining these two equations,
(F
C on S
⎛ 1 ⎞ ⎛ 1 ⎛ 1 ⎞ ⎛ 1 ⎞ 1 ⎞ − FC on T ) ⎜ + ⎟ = ( FC on T ) ⎜ ⎟ = ( FC on S ) ⎜ ⎟ ⎟ ⇒ FC on T ⎜ m m m m ⎝ T⎠ T ⎠ ⎝ C⎠ ⎝ C ⎝ mC ⎠
⎛ mT ⎞ ⎛ ⎞ 2000 kg ⇒ FC on T = ( FC on S ) ⎜ ⎟ = ( 4500 N ) ⎜ ⎟ = 3000 N ⎝ 1000 kg + 2000 kg ⎠ ⎝ mC + mT ⎠ (b) Due to Newton’s third law, FT on C = 3000 N.
7.10. Model: The blocks are to be modeled as particles and denoted as 1, 2, and 3. The surface is frictionless and along with the earth it is a part of the environment. The three blocks are our three systems of interest. Visualize:
The force applied on block 1 is FA on 1 = 12 N. The acceleration for all the blocks is the same and is denoted by a. Solve: (a) Newton’s second law for the three blocks along the x-direction is
∑( F )
on 1 x
= FA on 1 − F2 on 1 = m1a
∑( F )
on 2 x
= F1 on 2 − F3 on 2 = m2 a
∑( F )
on 3 x
= F2 on 3 = m3a
Adding these three equations and using Newton’s third law ( F2 on 1 = F1 on 2 and F3 on 2 = F2 on 3 ), we get
FA on 1 = ( m1 + m2 + m3 ) a ⇒ (12 N ) = (1 kg + 2 kg + 3 kg ) a ⇒ a = 2 m/s 2 Using this value of a, the force equation on block 3 gives F2 on 3 = m3a = ( 3 kg ) ( 2 m/s 2 ) = 6 N (b) Substituting into the force equation on block 1,
12 N − F2 on 1 = (1 kg ) ( 2 m/s 2 ) ⇒ F2 on 1 = 10 N Assess: Because all three blocks are pushed forward by a force of 12 N, the value of 10 N for the force that the 2 kg block exerts on the 1 kg block is reasonable.
7.11. Model: The block (B) and the steel cable (C), the two objects of interest to us, are treated like particles. The motion of these objects is governed by the constant-acceleration kinematic equations. Visualize:
Solve:
Using v12x = v02x + 2ax ( x1 − x0 ) ,
( 4.0 m/s )
2
= 0 m 2 /s 2 + 2ax ( 2.0 m ) ⇒ ax = 4.0 m/s 2
From the free-body diagram on the block:
∑( F )
on B x
= FC on B = mBax ⇒ FC on B = ( 20 kg ) ( 4.0 m/s 2 ) = 80 N
Also, according to Newton’s third law FB on C = FC on B = 80 N. Newton’s second law on the cable is:
∑( F )
on C x
= Fext − FB on C = mC ax ⇒ 100 N − 80 N = mC ( 4.0 m/s 2 ) ⇒ mC = 5 kg
7.12. Model: The man (M) and the block (B) are interacting with each other through a rope. We will assume the pulley to be frictionless. This assumption implies that the tension in the rope is the same on both sides of the pulley. The system is the man and the block. Visualize:
Solve: Clearly the entire system remains in equilibrium since mB > mM . The block would move downward but it is already on the ground. From the free-body diagrams, we can write down Newton’s second law in the vertical direction as
∑( F )
on M y
= TR on M − ( FG )M = 0 N ⇒ TR on M = ( FG )M = ( 60 kg ) ( 9.8 m/s 2 ) = 588 N
Since the tension is the same on both sides, TB on R = TM on R = T = 588 N.
7.13. Model: Together the carp (C) and the trout (T) make up the system that will be represented through the particle model. The fishing rod line (R) is assumed to be massless. Visualize:
Solve: Jimmy’s pull T2 is larger than the total weight of the fish, so they accelerate upward. They are tied together, so each fish has the same acceleration a. Newton’s second law along the y-direction for the carp and the trout is
∑( F )
on C y
= T2 − T1 − ( FG )C = mC a
∑( F )
on T y
= T1 − ( FG )T = mT a
Adding these two equations gives a=
T2 − ( FG )C − ( FG )T
( mC + mT )
=
60 N − (1.5 kg ) ( 9.8 m/s 2 ) − ( 3 kg ) ( 9.8 m/s 2 ) 1.5 kg + 3.0 kg
= 3.533 m/s 2
Substituting this value of acceleration back into the force equation for the trout, we find that
T1 = mT ( a + g ) = ( 3 kg ) ( 3.533 m/s 2 + 9.8 m/s 2 ) = 40 N ( FG )T = mT g = ( 3 kg ) ( 9.8 m/s 2 ) = 29.4 N Thus, T2 > T1 > ( FG )T > ( FG )C .
( FG )C = mC g = (1.5 kg ) ( 9.8 m/s 2 ) = 14.7 N
7.14. Model: The block of ice (I) is a particle and so is the rope (R) because it is not massless. We must therefore consider both the block of ice and the rope as objects in the system. Visualize:
Solve:
G The force Fext acts only on the rope. Since the rope and the ice block move together, they have the same
acceleration. Also because the rope has mass, Fext on the front end of the rope is not the same as FI on R that acts on the rear end of the rope. Newton’s second law along the x-axis for the ice block and the rope is
∑( F )
= FR on I = mI a = (10 kg ) ( 2.0 m/s 2 ) = 20 N
∑( F )
= Fext − FI on R = mR a ⇒ Fext − FR on I = mR a
on I x
on R x
⇒ Fext = FR on I + mR a = 20 N + ( 0.500 kg ) ( 2.0 m/s 2 ) = 21 N
7.15. Model: The hanging block and the rail car are objects in the systems. Visualize:
Solve: The mass of the rope is very small in comparison to the 2000-kg block, so we will assume a massless G G rope. In that case, the forces T1 and T1′ act as if they are an action/reaction pair. The hanging block is in static G equilibrium, with Fnet = 0 N, so T1′ = mblock g = 19,600 N. The rail car with the pulley is also in static equilibrium:
T2 + T3 − T1 = 0 N Notice how the tension force in the cable pulls both the top and bottom of the pulley to the right. Now, T1 = T1′ = 19,600 N by Newton’s third law. Also, the cable tension is T2 = T3 = T . Thus, T = 12 T1′ = 9800 N.
7.16. Visualize:
Solve: The rope is treated as two 1.0-kg interacting objects. At the midpoint of the rope, the rope has a tension TB on T = TT on B ≡ T . Apply Newton’s first law to the bottom half of the rope to find T.
( Fnet ) y = 0 = T − ( FG ) B
⇒ T = mB g = (1.0 kg ) ( 9.80 m/s 2 ) = 9.8 N Assess: 9.8 N is half the gravitational force on the whole rope. This is reasonable since the top half is holding up the bottom half of the rope against gravity.
7.17. Model: The two hanging blocks, which can be modeled as particles, together with the two knots where rope 1 meets with rope 2 and rope 2 meets with rope 3 form a system. All the four objects in the system are in static equilibrium. The ropes are assumed to be massless. Visualize:
Solve: (a) We will consider both the two hanging blocks and the two knots. The blocks are G in static G equilibrium with Fnet = 0 N. Note that there are three action/reaction pairs. For Block 1 and Block 2, Fnet = 0 N and we have T4′ = ( FG )1 = m1 g T5′ = ( FG )2 = m2 g Then, by Newton’s third law:
T4 = T4′ = m1 g T5 = T5′ = m2 g
The knots are also in equilibrium. Newton’s law applied to the left knot is
( Fnet ) x = T2 − T1 cosθ1 = 0 N ( Fnet ) y = T1 sinθ1 − T4 = T1 sinθ1 − m1g = 0 N The y-equation gives T1 = m1 g sin θ1 . Substitute this into the x-equation to find
T2 =
m1 g cosθ1 m1 g = sin θ1 tanθ1
Newton’s law applied to the right knot is
( Fnet ) x = T3 cosθ 3 − T2′ = 0 N ( Fnet ) y = T3 sinθ 3 − T5 = T3 sinθ 3 − m2 g = 0 N These can be combined just like the equations for the left knot to give
T2′ =
m2 g cosθ 3 mg = 2 sin θ 3 tan θ 3
G G But the forces T2 and T2′ are an action/reaction pair, so T2 = T2′. Therefore,
m1 g mg m = 2 ⇒ tanθ 3 = 2 tanθ1 ⇒ θ 3 = tan −1 ( 2 tan 20° ) = 36° tan θ1 tanθ 3 m1 We can now use the y-equation for the right knot to find T3 = m2 g sinθ 3 = 67 N.
7.18. Visualize: Please refer to Figure P7.18.
Solve: Since the ropes are massless we can treat the tension force they transmit as a Newton’s third law force pair on the blocks. The connection shown in figure P7.18 has the same effect as a frictionless pulley on these massless ropes. The blocks are in equilibrium as the mass of A is increased until block B slides, which occurs when the static friction on B is at its maximum value. Applying Newton’s first law to the vertical forces on block B gives nB = ( FG ) B = mB g. The static friction force on B is thus
( fs )B = μs nB = μs mB g.
( fs )B = TA on B , and the same Since TA on B = TB on A , we have ( f s )B = mA g , so
Applying Newton’s first law to the horizontal forces on B gives vertical forces on A gives TB on A = ( FG )A = mA g .
μs mB g = mA g ⇒ mA = μs mB = ( 0.60 )( 20 kg ) = 12.0 kg
analysis of the
7.19. Model: The astronaut and the satellite, the two objects in our system, will be treated as particles. Visualize:
Solve: The astronaut and the satellite accelerate in opposite directions for 0.50 s. The force on the satellite and the force on the astronaut are an action/reaction pair, so both are 100 N. Newton’s second law for the satellite along the x-direction is
∑( F )
on S x
= FA on S = mSaS ⇒ aS =
FA on S mS
=
− (100 N ) = −0.156 m/s 2 640 kg
Newton’s second law for the astronaut along the x-direction is
∑( F )
on A x
= FS on A = mA aA ⇒ aA =
FS on A mA
=
FA on S mA
=
100 N = 1.25 m/s 2 80 kg
Let us first calculate the positions and velocities of the astronaut and the satellite at t1 = 0.50 s under the
accelerations aA and aS :
x1A = x0A + v0A ( t1 − t0 ) + 12 aA ( t1 − t0 ) = 0 m + 0 m + 12 (1.25 m/s 2 ) ( 0.50 s − 0 s ) = 0.156 m 2
2
x1S = x0S + v0S ( t1 − t0 ) + 12 aS ( t1 − t0 ) = 0 m + 0 m + 12 ( −0.156 m/s 2 ) ( 0.50 s − 0 s ) = −0.020 m 2
2
v1A = v0A + aA ( t1 − t0 ) = 0 m/s + (1.25 m/s 2 ) ( 0.50 s − 0 s ) = 0.625 m/s v1S = v0S + aS ( t1 − t0 ) = 0 m/s + ( −0.156 m/s 2 ) ( 0.5 s − 0 s ) = −0.078 m/s With x1A and x1S as initial positions, v1A and v1S as initial velocities, and zero accelerations, we can now obtain the new positions at ( t2 − t1 ) = 59.5 s :
x2A = x1A + v1A ( t2 − t1 ) = 0.156 m + ( 0.625 m/s )( 59.5 s ) = 37.34 m x2S = x1S + v1S ( t2 − t1 ) = − 0.02 m + ( −0.078 m/s )( 59.5 s ) = −4.66 m Thus the astronaut and the satellite are x2A − x2S = ( 37.34 m ) − ( −4.66 m ) = 42 m apart.
7.20. Model: The block (B) and the steel cable (C), the two objects in the system, are considered particles, and their motion is determined by the constant-acceleration kinematic equations. Visualize:
Solve:
Using v1x = v0 x + ax ( t1 − t0 ) ,
4.0 m/s = 0 m/s + ax ( 2.0 s − 0 s ) ⇒ ax = 2.0 m/s 2 Newton’s second law along the x-direction for the block is
∑( F )
on B x
= FC on B = mB ax = ( 20 kg ) ( 2.0 m/s 2 ) = 40 N
Fext acts on the right end of the cable and FB on C acts on the left end. According to Newton’s third law, FB on C = FC on B = 40 N. The difference in tension between the two ends of the cable is thus Fext − FB on C = 100 N − 40 N = 60 N
7.21. Model: The block (B) and the steel cable (C), the two objects in the system, are treated as particles, and their motion is determined by constant-acceleration kinematic equations. We ignore the hanging shape of the cable. Visualize:
Solve:
Using x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) , 2
4.0 m = 0 m + 0 m + 12 ax ( 2.0 s − 0 s ) ⇒ ax = 2.0 m/s 2 2
Newton’s second law along the x-direction for the block is
∑( F )
on B x
= FC on B = mBax = ( 20 kg ) ( 2.0 m/s 2 ) = 40 N
Therefore, the change in tension (T) in the cable from one Fext − FB on C = 100 N − 40 N = 60 N. The tension in the cable as a function of x is
( 60 N ) x
end
to
the
other
= ( 40 + 60 x ) N 1m with x in meters. x = 0 m is where the cable attaches to the box and x = 1.0 m is at the right end of the cable. T ( x ) = 40 N +
is
7.22. Visualize: Consider a segment of the rope of length y, starting from the bottom of the rope. The weight of this segment of rope is a downward force. It is balanced by the tension force at height y.
Solve: The mass m of this segment of rope is the same fraction of the total mass M = 2.2 kg as length y is a fraction of the total length L = 3.0 m. That is, m/M = y/L, from which we can write the mass of the rope segment
m=
M y L
This segment of rope is in static equilibrium, so the tension force pulling up on it is
T = FG = mg =
Mg (2.2 kg)(9.8 m /s 2 ) y= y = 7.19 y N 3.0 m L
where y is in m. The tension increases linearly from 0 N at the bottom ( y = 0 m) to 21.6 N at the top ( y = 3 m). This is shown in the graph.
7.23. Model: Sled A, sled B, and the dog (D) are treated like particles in the model of kinetic friction. Visualize:
Solve:
The acceleration constraint is ( aA ) x = ( aB ) x = ax . Newton’s second law on sled A is G G ∑ Fon A = nA − ( FG )A = 0 N ⇒ nA = ( FG )A = mA g ∑ Fon A = T1 on A − f A = mA ax
(
)
(
y
)
x
Using f A = μ k nA , the x-equation yields
T 1 on A − μ k nA = mA ax ⇒ 150 N − ( 0.1)(100 kg ) ( 9.8 m/s 2 ) = (100 kg ) ax ⇒ ax = 0.52 m/s 2 On sled B:
G
∑( F ) on B
y
= nB − ( FG )B = 0 N ⇒ nB = ( FG )B = mB g
G
∑( F ) on B
x
= T2 − T1 on B − f B = mB ax
T 1 on B and T 1 on A act as if they are an action/reaction pair, so T 1 on B = 150 N. Using f B = μ k nB = ( 0.10 )( 80 kg )
( 9.8 m/s ) = 78.4 N, 2
we get T2 − 150 N − 78.4 N = ( 80 kg ) ( 0.52 m/s 2 ) ⇒ T2 = 270 N
Thus the tension T2 = 2.7 × 102 N.
7.24. Model: The coffee mug (M) is the only object in the system, and it will be treated as a particle. The model of friction and the constant-acceleration kinematic equations will also be used. Visualize:
Solve: The mug and the car have the same velocity. If the mug does not slip, their accelerations will also be the same. Using v12x = v02x + 2ax ( x1 − x0 ) , we get
0 m 2 /s 2 = ( 20 m/s ) + 2ax ( 50 m ) ⇒ ax = −4.0 m/s 2 2
The static force needed to stop the mug is
( Fnet ) x = − fs = max = ( 0.5 kg ) ( −4.0 m/s2 ) = −2.0 N ⇒
f s = 2.0 N
The maximum force of static friction is ( f s ) max = μ s n = μ s FG = μ s mg = ( 0.50 )( 0.50 kg ) ( 9.8 m/s 2 ) = 2.45 N Since ( f s ) max < ( f s ) max , the mug does not slide.
7.25.
Visualize:
The car and the ground are denoted by C and S, respectively. Solve: (a) The car has an internal source of energy–fuel–that allows it to turn the wheels and exert the Gforce G FC on G . As the drive wheels turn they push backward against the ground. This is a static friction force FC on S because the wheels don’t slip against the ground. ByG Newton’s third law, the ground exerts a reaction force G FS on C . This reaction force is opposite in direction to FC on S , hence, is in the forward direction. This is the force that accelerates the car. Houses do not have an internal source of energy that allows them to push sideways against the ground. They also aren’t on wheels, which let the car slide across the ground with minimal friction. (b) The car presses down against the ground at both the drive wheels (assumed to be the front wheels F, although that is not critical) and the nondrive wheels. For this car, two-thirds of the gravitational force rests on the front G wheels. Physically, force FS on C is a static friction force. The maximum acceleration of the car on the ground (or concrete surface) occurs when the static friction reaches its maximum possible value.
F S on C = ( fs )max = μs nF = μs ( FG )F = (1.00 ) ( 23 ) (1500 kg ) ( 9.8 m/s 2 ) = 9800 N ⇒ amax =
FS on C m
=
9800 N = 6.533 m/s 2 1500 kg
7.26. Model: The starship and the shuttlecraft will be denoted as M and m, respectively, and both will be treated as particles. We will also use the constant-acceleration kinematic equations. Visualize:
G (a) The tractor beam is some kind of long-range force FM on m . Regardless of what kind of force it is, by G Newton’s third law there must be a reaction force Fm on M on the starship. As a result, both the shuttlecraft and the starship move toward each other (rather than the starship remaining at rest as it pulls the shuttlecraft in). However, the very different masses of the two crafts means that the distances they each move will also be very different. The pictorial representation shows that they meet at time t1 when xM1 = xm1. There’s only one force on each craft, so Newton’s second law is very simple. Furthermore, because the forces are an action/reaction pair, Solve:
FM on m = Fm on M = Ftractor beam = 4.0 × 104 N The accelerations of the two craft are aM =
Fm on M M
=
G FM on m −4.0 × 104 N 4.0 × 104 N 2 a = = = −2.0 m/s 2 = 0.020 m/s m 2.0 × 106 kg m 2.0 × 104 kg
Acceleration am is negative because the force and acceleration vectors point in the negative x-direction. The accelerations are very different even though the forces are the same. Now we have a constant-acceleration problem in kinematics. At a later time t1 the positions of the crafts are xM1 = xM0 + vM0 ( t1 − t0 ) + 12 aM ( t1 − t0 ) = 12 aM t12 2
xm1 = xm0 + vm0 ( t1 − t0 ) + 12 am ( t1 − t0 ) = xm0 + 12 amt12 2
The craft meet when xM1 = xm1 , so 1 2
aMt12 = xm0 + 12 amt12 ⇒ t1 =
2 (10,000 m ) 2 xm0 2 xm0 = = = 99.5 s aM − am aM + am 2.02 m/s 2
Knowing t1 , we can now find the starship’s position as it meets the shuttlecraft: xM1 = 12 aMt12 = 99 m
The starship moves 99 m as it pulls in the shuttlecraft from 10 km away.
7.27. Model: The rock (R) and Bob (B) are the two objects in our system, and will be treated as particles. We will also use the constant-acceleration kinematic equations. Visualize:
G Solve: (a) Bob exerts a forward force FB on R on the rock to accelerate it forward. The rock’s acceleration is calculated as follows: v2 ( 30 m/s ) = 450 m/s 2 v = v + 2a0R Δx ⇒ aR = 1R = 2Δx 2 (1.0 m ) 2
2 1R
2 0R
The force is calculated from Newton’s second law:
FB on R = mR aR = (.500 kg ) ( 450 m/s 2 ) =225 N Bob exerts a force of 2.3 × 102 N on the rock.
G G G (b) Because Bob pushes on the rock, the rock pushes back on Bob with a force FR on B . Forces FR on B and FB on R are an action/reaction pair, so FR on B = FB on R = 225 N. The force causes Bob to accelerate backward with an acceleration equal to
aB =
(F
)
net on B x
mB
=−
FR on B mB
=−
225 N = −3.0 m/s 2 75 kg
This is a rather large acceleration, but it lasts only until Bob releases the rock. We can determine the time interval by returning to the kinematics of the rock:
v1R = v0R + aR Δt = aR Δt ⇒ Δt =
v1R = 0.0667 s aR
At the end of this interval, Bob’s velocity is v1B = v0B + aBΔt = aBΔt = −0.20 m/s Thus his recoil speed is 0.20 m/s.
7.28. Model: The boy (B) and the crate (C) are the two objects in our system, and they will be treated in the particle model. We will also use the static and kinetic friction models. Visualize:
Solve: The fact that the boy’s feet occasionally slip means that the maximum force of static friction must exist between the boy’s feet and the sidewalk. That is, f sB = μ sB nB . Also f kC = μ kC nC . Newton’s second law for the crate is
∑( F )
on C y
= nC − ( FG )C = 0 N ⇒ nC = mC g
∑( F )
on C x
= FB on C − f kC = 0 N ⇒ FB on C = f kC = μ kC nC = μ kC mC g
∑( F )
= fsB − FC on B = 0 N ⇒ FC on B = f sB = μsB nB = μsB mB g
Newton’s second law for the boy is
∑( F )
on B y
= nB − ( FG )B = 0 N ⇒ nB = mB g
on B x
G G FC on B and FB on C are an action/reaction pair, so FC on B = FB on C ⇒ μsB mB g = μ kC mC g ⇒ mC =
μsB mB ( 0.8 )( 50 kg ) = = 2.0 × 102 kg μ kC ( 0.2 )
7.29. Model: Assume package A and package B are particles. Use the model of kinetic friction and the constant-acceleration kinematic equations. Visualize:
Solve: Package B has a smaller coefficient of friction. It will try to overtake package A and push against it. Package A will push back on B. The acceleration constraint is ( aA ) x = ( aB ) x = a.
Newton’s second law for each package is
∑( F )
on A x
= FB on A + ( FG )A sin θ − f kA = mA a
⇒ FB on A + mA g sin θ − μ kA ( mA g cosθ ) = mA a
∑( F )
on B x
= − FA on B − f kB + ( FG )B sinθ = mBa
⇒ − FA on B − μ kB ( mB g cosθ ) + mB g sinθ = mB a where we have used nA = mA cosθ g and nB = mB cosθ g . Adding the two force equations, and using
FA on B = FB on A because they are an action/reaction pair, we get a = g sin θ −
( μ kA mA + μ kBmB )( g cosθ ) mA + mB
= 1.82 m/s 2
Finally, using x1 = x0 + v0x ( t1 − t0 ) + 12 a ( t1 − t0 ) , 2
2.0 m = 0 m + 0 m + 12 (1.82 m/s 2 ) ( t1 − 0 s ) ⇒ t1 = 1.48 s 2
7.30. Model: The two blocks form a system of interacting objects. Visualize:
Please refer to Figure P7.30.
Solve: It is possible that the left-hand block (Block L) is accelerating down the slope faster than the right-hand block (Block R), causing the string to be slack (zero tension). If that were the case, we would get a zero or negative answer for the tension in the string. Newton’s first law applied to the y-direction on Block L yields
(∑ F )
L y
= 0 = nL − ( FG )L cos 20° ⇒ nL = mL g cos 20°
Therefore
( f k )L = ( μ k )L mL g cos 20° = ( 0.20 )(1.0 kg ) ( 9.80 m/s2 ) cos 20° = 1.84 N A similar analysis of the vertical forces on Block R gives ( f k )R = 1.84 N as well. Using Newton’s second law in the x-direction for Block L, ( ∑ FL ) x = mLa = TR on L − ( f k )L + ( FG )L sin 20° ⇒ mLa = TR on L − 1.84 N + mL g sin 20°. For Block R,
(∑ F )
R x
= mR a = ( FG )R sin 20° − 1.84 N − TL on R ⇒ mR a = mR g sin 20° − 1.84 N − TL on R
These are two equations in the two unknowns a and TL on R = TR on L ≡ T . Solving them, we obtain a = 2.12 m/s 2 and T = 0.61 N. Assess: The tension in the string is positive, and is about 1/3 of the kinetic friction force on each of the blocks, which is reasonable.
7.31. Model: The two ropes and the two blocks (A and B) will be treated as particles. Visualize:
Solve:
(a) The two blocks and two ropes form a combined system of total mass M = 2.5 kg. This combined
system is accelerating upward at a = 3.0 m/s 2 under the influence of a force F and the gravitational force − Mg ˆj. Newton’s second law applied to the combined system is
( Fnet ) y = F − Mg = Ma ⇒ F = M ( a + g ) = 32 N (b) The ropes are not massless. We must consider both the blocks and the ropes as systems. The force F acts only on block A because it does not contact the other objects. We can proceed to apply the y-component of Newton’s second law to each system, starting at the top. Each has an acceleration a = 3.0 m/s 2 . For block A:
(F
)
net on A y
= F − mA g − T1 on A = mA a ⇒ T1 on A = F − mA ( a + g ) = 19.2 N
(c) Applying Newton’s second law to rope 1: ( Fnet on 1 ) y = TA on 1 − m1g − TB on 1 = m1a G G G TA on 1 and T1 on A are an action/reaction pair. But, because the rope has mass, the two tension forces TA on 1 and G TB on 1 are not the same. The tension at the lower end of rope 1, where it connects to B, is
TB on 1 = TA on 1 − m1 ( a + g ) = 16.0 N (d) We can continue to repeat this procedure, noting from Newton’s third law that
T1 on B = TB on 1 and T2 on B = TB on 2 Newton’s second law applied to block B is
(F
)
net on B y
= T1 on B − mB g − T2 on B = mB a ⇒ T2 on B = T1 on B − mB ( a + g ) = 3.2 N
7.32. Model: The two blocks (1 and 2) are the systems of interest and will be treated as particles. The ropes are assumed to be massless, and the model of kinetic friction will be used. Visualize:
Solve: (a) The separate free-body diagrams for the two blocks show that there are two action/reaction pairs. G G Notice how block 1 both pushes down on block 2 (force n1′ ) and exerts a retarding friction force f 2 top on the top surface of block 2. Block 1 is in static equilibrium (a1 = 0 m/s 2 ) but block 2 is accelerating. Newton’s second law for block 1 is
(F
)
net on 1 x
= f1 − Trope = 0 N ⇒ Trope = f1
(F
)
net on 1 y
= n1 − m1 g = 0 N ⇒ n1 = m1 g
Although block 1 is stationary, there is a kinetic force of friction because there is motion between block 1 and block 2. The friction model means f1 = μ k n1 = μ k m1 g . Substitute this result into the x-equation to get the tension in the rope: Trope = f1 = μ k m1 g = 3.92 N (b) Newton’s second law for block 2 is ax = a =
(F
)
net on 2 x
m2
=
Tpull − f 2 top − f 2 bot m2
a y = 0 m/s 2 =
(F
)
net on 2 y
m2
=
n2 − n1′ − m2 g m2
G G Forces n1 and n1′ are an action/reaction pair, so n1′ = n1 = m1 g . Substituting into the y-equation gives
n2 = ( m1 + m2 ) g . This is not surprising because the combined weight of both objects presses down on the
surface. The kinetic friction on the bottom surface of block 2 is then f 2 bot = μ k n2 = μ k ( m1 + m2 ) g G G The forces f1 and f 2 top are an action/reaction pair, so f 2 bot = f1 = μ k m1 g . Inserting these friction results into
the x-equation gives
a=
(F
)
net on 2 x
m2
=
Tpull − μ k m1 g − μ k ( m1 + m2 ) g m2
= 2.16 m/s 2
7.33. Model: The 3-kg and 4-kg blocks are to be treated as particles. The models of kinetic and static friction and the constant-acceleration kinematic equations will be used. Visualize:
Solve: Minimum time will be achieved when static friction is at its maximum possible value. Newton’s second law for the 4-kg block is
∑( F )
on 4 y
= n3 on 4 − ( FG )4 = 0 N ⇒ n3 on 4 = ( FG )4 = m4 g = ( 4.0 kg ) ( 9.8 m/s 2 ) = 39.2 N
⇒ f s4 = ( f s ) max = μs n3 on 4 = ( 0.60 )( 39.2 N ) = 23.52 N Newton’s second law for the 3-kg block is
∑( F )
on 3 y
= n3 − n4 on 3 − ( FG )3 = 0 N ⇒ n3 = n4 on 3 + ( FG )3 = 39.2 N + ( 3.0 kg ) ( 9.8 m/s 2 ) = 68.6 N
Friction forces f and fs4 are an action/reaction pair. Thus
∑( F )
on 3 x
= fs3 − f k3 = m3a3 ⇒ f s4 − μ k n3 = m3a3 ⇒ 23.52 N − ( 0.20 )( 68.6 N ) = ( 3.0 kg ) a3 ⇒ a3 = 3.267 m/s 2
Since block 3 does not slip, this is also the acceleration of block 4. The time is calculated as follows: x1 − x0 + v0 x ( t1 − t0 ) + 12 a ( t1 − t0 ) ⇒ 5.0 m = 0 m + 0 m + 12 ( 3.267 m/s 2 ) ( t1 − 0 s ) ⇒ t1 = 1.75 s 2
2
7.34. Model: Blocks 1 and 2 make up the system of interest and will be treated as particles. Assume a massless rope and frictionless pulley. Visualize:
Solve: The blocks accelerate with the same magnitude but in opposite directions. Thus the acceleration constraint is a2 = a = −a1 , where a will have a positive value. There are two real action/reaction pairs. The two tension forces will act as if they are action/reaction pairs because we are assuming a massless rope and a frictionless pulley. Make sure you understand why the friction forces point in the directions shown in the freeG body diagrams, especially force f1′ exerted on block 2 by block 1. We have quite a few pieces of information to
include. First, Newton’s second law for blocks 1 and 2: G Fnet on 1 = f1 − T1 = μ k n1 − T1 = m1a1 = − m1a
(
)
x
(F
(F
)
net on 1 y
= n1 − m1 g = 0 N ⇒ n1 = m1 g
) = T − f ′− f − T = T − f ′− μ n − T = m a ( F ) = n − n′ − m g = 0 N ⇒ n = n′ + m g
net on 2 x
pull
net on 2 y
1
2
2
2
1
pull
2
1
k 2
2
2
1
2 2
= m2 a
2
We’ve already used the kinetic friction model in both x-equations. Next, Newton’s third law:
n1′ = n1 = m1 g f1′ = f1 = μ k n1 = μ k m1 g T1 = T2 = T Knowing n1′, we can now use the y-equation of block 2 to find n2 . Substitute all these pieces into the two xequations, and we end up with two equations in two unknowns:
μ k m1 g − T = −m1a Tpull − T − μ k m1 g − μ k ( m1 + m2 ) g = m2 a Subtract the first equation from the second to get
Tpull − μ k ( 3m1 + m2 ) g = ( m1 + m2 ) a ⇒ a =
Tpull − μ k ( 3m1 + m2 ) g m1 + m2
= 1.77 m/s 2
7.35. Model: The sled (S) and the box (B) will be treated in the particle model, and the model of friction will be used. Visualize:
In the sled’s free-body diagram nS is the normal (contact) force on the sled due to the snow. Similarly f kS is the force of kinetic friction on the sled due to snow. Solve: Newton’s second law on the box in the y-direction is
nS on B − ( FG )B cos 20° = 0 N ⇒ nS on B = (10 kg ) ( 9.8 m/s 2 ) cos 20° = 92.09 N G The static friction force fS on B accelerates the box. The maximum acceleration occurs when static friction reaches its maximum possible value.
( fs ) max = μSnS on B = ( 0.50 )( 92.09 N ) = 46.05 N Newton’s second law along the x-direction thus gives the maximum acceleration fS on B − ( FG )B sin 20° = mB a ⇒ 46.05 N − (10 kg ) ( 9.8 m/s 2 ) sin 20° = (10 kg ) a ⇒ a = 1.25 m/s 2 Newton’s second law for the sled along the y-direction is
nS − nB on S − ( FG )S cos 20° = 0 N ⇒ nS = nB on S + mS g cos 20° = ( 92.09 N ) + ( 20 kg ) ( 9.8 m/s 2 ) cos 20° = 276.27 N Therefore, the force of friction on the sled by the snow is f kS = ( μ k )nS = (0.06)(276.27 N) = 16.58 N Newton’s second law along the x-direction is
Tpull − wS sin 20° − f kS − f B on S = mSa The friction force f B on S = fS on B because these are an action/reaction pair. We’re using the maximum acceleration, so the maximum tension is
Tmax − ( 20 kg ) ( 9.8 m/s 2 ) sin 20° − 16.58 N − 46.05 N = ( 20 kg ) (1.25 m/s 2 ) ⇒ Tmax = 155 N
7.36. Model: The masses m and M are to be treated in the particle model. We will also assume a massless rope and frictionless pulley, and use the constant-acceleration kinematic equations for m and M. Visualize:
Solve:
Using y1 = y0 + v0 y ( t1 − t0 ) + 12 aM ( t1 − t0 ) , 2
( −1 m ) = 0 m + 0 m + 12 aM ( 6.0 s − 0 s )
2
⇒ aM = −0.0556 m/s 2
Newton’s second law for m and M is:
∑( F )
on m y
= TR on m − ( FG )m = mam
∑( F )
on M y
= TR on M − ( FG )M = MaM
The acceleration constraint is am = − aM . Also, the tensions are an action/reaction pair, thus TR on m = TR on M . With these, the second law equations are TR on M − Mg = MaM TR on M − mg = −maM
Subtracting the second from the first gives
⎡ g + aM ⎤ − Mg + mg = MaM + maM ⇒ m = M ⎢ ⎥ ⎣ g − aM ⎦ ⎡ 9.8 − 0.556 ⎤ = (100 kg ) ⎢ ⎥ = 99 kg ⎣ 9.8 + 0.556 ⎦ Assess: Note that am = − aM = 0.0556 m/s 2 . For such a small acceleration, a mass of 99 kg for m compared to M = 100 kg is understandable.
7.37. Model: Use the particle model for the block of mass M and the two massless pulleys. Additionally, the
G rope is massless and the pulleys are frictionless. The block is kept in place by an applied force F . Visualize:
Solve:
Since there is no friction on the pulleys, T2 = T3 and T2 = T5 . Newton’s second law for mass M is
T1 − FG = 0 N ⇒ T1 = Mg = (10.2 kg ) ( 9.8 m/s 2 ) = 100 N Newton’s second law for the small pulley is T2 + T3 − T1 = 0 N ⇒ T2 = T3 =
T1 = 50 N = T5 = F 2
Newton’s second law for the large pulley is T4 − T2 − T3 − T5 = 0 N ⇒ T4 = T2 + T3 + T5 = 150 N
7.38. Model: Assume the particle model for m1 , m2 , and m3 , and the model of kinetic friction. Assume the ropes to be massless, and the pulleys to be frictionless and massless. Visualize:
Solve:
Newton’s second law for m1 is T1 − ( FG )1 = m1a1. Newton’s second law for m2 is
∑( F ) on m2
∑ (F
y
= n2 − ( FG )2 = 0 N ⇒ n2 = ( 2.0 kg ) ( 9.8 m/s 2 ) = 19.6 N ) = T2 − f k2 − T = m2 a2 ⇒ T2 − μ k n2 − T1 = (2.0 kg)a2
on m2 x
Newton’s second law for m3 is
∑( F ) on m3
y
= T2 − ( FG )3 = m3a3
Since m1 , m2 , and m3 move together, a1 = a2 = −a3 = a. The equations for the three masses thus become
T1 − ( FG )1 = m1a = (1.0 kg ) a T2 − μ k n2 − T1 = m2 a = ( 2.0 kg ) a T2 − ( FG )3 = −m3a = − ( 3.0 kg ) a Subtracting the third equation from the sum of the first two equations yields: − ( FG )1 − μ k n2 + ( FG )3 = ( 6.0 kg ) a ⇒ − (1.0 kg ) ( 9.8 m/s 2 ) − ( 0.30 )(19.6 N ) + ( 3.0 kg ) ( 9.8 m/s 2 ) = ( 6.0 kg ) a ⇒ a = 2.3 m/s 2
7.39. Model: Assume the particle model for the two blocks, and the model of kinetic and static friction. Visualize:
Solve: (a) If the mass m is too small, the hanging 2.0 kg mass will pull it up the slope. We want to find the smallest mass that will stick because of friction. The smallest mass will be the one for which the force of static friction is at its maximum possible value: f s = ( f s )max = μs n. As long as the mass m is stuck, both blocks are at G rest with Fnet = 0 N. Newton’s second law for the hanging mass M is
( Fnet ) y = TM − Mg = 0 N ⇒ TM = Mg = 19.6 N For the smaller mass m,
( Fnet ) x = Tm − fs − mg sinθ = 0 N ( Fnet ) y = n − mg cosθ ⇒ n = mg cosθ
G G For a massless string and frictionless pulley, forces Tm and TM act as if they are an action/reaction pair. Thus Tm = TM . Mass m is a minimum when
( fs )max = μs n = μs mg cosθ .
Substituting these expressions into the x-
equation, TM − μ s mg cosθ − mg sinθ = 0 N ⇒ m =
TM
( μs cosθ + sinθ ) g
= 1.83 kg
(b) Because μ k < μS the 1.83 kg block will begin to slide up the ramp, and the 2.0 kg mass will begin to fall, if the block is nudged ever so slightly. Now the net force and the acceleration are not zero. Notice how, in the pictorial representation, we chose different coordinate systems for the two masses. This gives block M an acceleration with only a y-component and block m an acceleration with only an x-component. The magnitudes of the accelerations are the same because the blocks are tied together. But block M has a negative acceleration G component a y (vector a points down) whereas block m has a positive ax . Thus the acceleration constraint is
(am ) x = −(aM ) y = a, where a will have a positive value. Newton’s second law for block M is
( Fnet ) y = T − Mg = M ( aM ) y = − Ma For block m we have
( Fnet ) x = T − f k − mg sinθ = T − μ k mg cosθ − mg sinθ = m ( am ) x = ma In writing these equations, we used Newton’s third law to write Tm = TM = T . Also, we noticed that the yequation and the friction model for block m don’t change, except for μ s becoming μ k , so we already know f k from part (a). Notice that the tension in the string is not the gravitational force Mg. We have two equations in the two unknowns T and a: T − Mg = − Ma T − ( μ k cosθ + sin θ ) mg = ma
Subtracting the second equation from the first to eliminate T, − Mg + ( μ k cosθ + sinθ ) mg = − Ma − ma = −( M + m)a
⇒a=
M − ( μ k cosθ + sinθ ) m g = 1.32 m/s 2 M +m
7.40. Model: Assume the particle model for the two blocks. Visualize:
Solve: (a) The slope is frictionless, so the blocks stay in place only if held. Once m is released, the blocks will G move one way or the other. As long as m is held, the blocks are in static equilibrium with Fnet = 0 N. Newton’s second law for the hanging block M is ( Fnet on M ) = TM − Mg = 0 N ⇒ TM = Mg = 19.6 N y
By Newton’s third law, TM = Tm = T = 19.6 N is the tension in the string. (b) The free-body diagram shows box m after it is released. Whether it moves up or down the slope depends on whether the acceleration a is positive or negative. The acceleration constraint is (am ) x = a = −(aM ) y . Newton’s
second law for each system gives ( Fnet on m ) = T − mg sin 35° = m ( am )x = ma x
(F
)
net on M y
= T − Mg = M ( aM ) y = − Ma
We have two equations in two unknowns. Subtract the second from the first to eliminate T: M − m sin 35° − mg sin 35° + Mg = ( m + M ) a ⇒ a = g = −0.481 m/s 2 M +m Since a < 0 m/s 2 , the box accelerates down the slope. (c) It is now straightforward to compute T = Mg − Ma = 21 N. Notice how the tension is larger than when the blocks were motionless.
7.41. Model: Assume the particle model for the book (B) and the coffee cup (C), the models of kinetic and static friction, and the constant-acceleration kinematic equations. Visualize:
Solve:
(a) Using v12x = v02x + 2a ( x1 − x0 ) ,
0 m 2 /s 2 = ( 3.0 m/s ) + 2a ( x1 ) ⇒ ax1 = −4.5 m 2 /s 2 2
To find x1 , we must first find a. Newton’s second law for the book and the coffee cup is
∑( F )
on B y
= nB − ( FG )B cos 20° = 0 N ⇒ nB = (1.0 kg ) ( 9.8 m/s 2 ) cos 20° = 9.21 N
∑( F )
on B x
= −T − f k − ( FG )B sin 20° = mB aB
∑( F )
on C y
= T − ( FG )C = mC aC
The last two equations can be rewritten, using aC = aB = a, as −T − μ k nB − mB g sin 20° = mB a T − mC g = mC a Adding the two equations,
a ( mC + mB ) = − g ( mC + mB sin 20° ) − μ k ( 9.21 N )
⇒ (1.5 kg ) a = − ( 9.8 m/s 2 ) ⎣⎡0.500 kg + (1.0 kg ) sin 20° ⎦⎤ − ( 0.20 )( 9.21 N ) ⇒ a = −6.73 m/s 2 Using this value for a, we can now find x1 as follows: x1 =
−4.5 m 2 /s 2 −4.5 m 2 /s 2 = = 0.67 m −6.73 m/s 2 a
(b) The maximum static friction force is ( f s ) max = μs nB = ( 0.50 )( 9.21 N ) = 4.60 N. We’ll see if the force f s
needed to keep the book in place is larger or smaller than ( f s ) max . When the cup is at rest, the string tension is T = mC g . Newton’s first law for the book is
∑( F )
on B x
= fs − T − wB sin 20° = fs − mC g − mB g sin 20° = 0
⇒ fs = ( M C + M B sin 20° ) g = 8.25 N Because fs > ( f s ) max , the book slides back down.
7.42. Model: Use the particle model for the cable car and the counterweight. Assume a massless cable. Visualize:
Solve: (a) Notice the separate coordinate systems for the cable car (object 1) and the counterweight (object 2). G G G G Forces T1 and T2 act as if they are an action/reaction pair. The braking force FB works with the cable tension T1 G to allow the cable car to descend at a constant speed. Constant speed means dynamic equilibrium, so Fnet = 0 N for both systems. Newton’s second law for the cable car is
(F
)
net on 1 x
Newton’s second law for the counterweight is
(F
(F
= T1 + FB − m1 g sin θ1 = 0 N
)
net on 2 x
= m2 g sinθ 2 − T2 = 0 N
)
net on 1 y
(F
)
net on 2 y
= n1 − m1 g cosθ1 = 0 N
= n2 − m2 g cosθ 2 = 0 N
From the x-equation for the counterweight, T2 = m2 g sin θ 2 . By Newton’s third law, T1 = T2 . Thus the x-equation for the cable car becomes FB = m1 g sinθ1 − T1 = m1 g sin θ1 − m2 g sin θ 2 = 3770 N (b) If the brakes fail, then FB = 0 N. The car will accelerate down the hill on one side while the counterweight accelerates up the hill on the other side. Both will have negative accelerations because of the direction of the acceleration vectors. The constraint is a1x = a2 x = a, where a will have a negative value. Using T1 = T2 = T , the two x-equations are
( Fnet on 1 ) x = T − m1g sinθ1 = m1a1x = m1a
(F
)
net on 2 x
= m2 g sinθ 2 − T = m2 a2x = m2 a
Note that the y-equations aren’t needed in this problem. Add the two equations to eliminate T: m sin θ1 − m2 sinθ 2 g = −0.991 m/s 2 − m1 g sin θ1 + m2 g sin θ = ( m1 + m2 ) a ⇒ a = − 1 m1 + m2
Now we have a problem in kinematics. The speed at the bottom is calculated as follows: v12 = v02 + 2a ( x1 − x0 ) = 2ax1 ⇒ v1 = 2ax1 = 2 ( −0.991 m/s 2 ) ( −400 m ) = 28.2 m/s Assess: A speed of approximately 60 mph as the cable car travels a distance of 2000 m along a frictionless slope of 30° is reasonable.
7.43. Model: Assume the cable mass is negligible compared to the car mass and that the pulley is frictionless. Use the particle model for the two cars. Visualize: Please refer to Figure P7.43.
Solve: (a) The cars are moving at constant speed, so they are in dynamic equilibrium. Consider the descending car D. We can find the rolling friction force on car D, and then find the cable tension by applying Newton’s first law. In the y-direction for car D, ( Fnet ) y = 0 = nD − ( FG )D cos35°
⇒ nD = mD g cos35° So the rolling friction force on car D is
( f R )D = μ R nD = μ R mD g cos35°
Applying Newton’s first law to car D in the x-direction, ( Fnet ) x = TA on D + ( f R )D − ( FG )D sin 35° = 0 Thus
TA on D = mD g sin 35° − μ R mD g cos35°
= (1500 kg ) ( 9.80 m/s 2 ) ( sin 35° − 0.020cos35° )
= 8.2 × 103 N (b) Similarly, we find that for car A, ( f R )A = μ R mA g cos35°. In the x-direction for car A,
( Fnet ) x = Tmotor + TD on A − ( f R )A − ( FG )A sin 35° = 0 ⇒ Tmotor = mA g sin 35° + μ R mA g cos35° − ( mD g sin 35° − μ R mD g cos35° )
Here, we have used TA on D = TD on A . If we also use m A = m D , then Tmotor = 2μ R mA g cos35° = 4.8 × 102 N. Assess: Careful examination of the free-body diagrams for cars D and A yields the observation that Tmotor = 2( FR ) A in order for the cars to be in dynamic equilibrium. It is a tribute to the design that the motor must only provide such a small force compared to the tension in the cable connecting the two cars.
7.44. Model: The painter and the chair are treated as a single object and represented as a particle. We assume that the rope is massless and that the pulley is massless and frictionless. Visualize:
Solve: If the painter pulls down on the rope with force F, Newton’s third law requires the rope to pull up on the painter with force F. This is just the tension in the rope. With our model of the rope and pulley, the same tension force F also pulls up on the painter’s chair. Newton’s second law for (painter + chair) is
2F − FG = ( mP + mC ) a ⇒F =(
1 2
) ⎡⎣( mP + mC ) a + ( mP + mC ) g ⎤⎦ = 12 ( mP + mC )( a + g )
= ( 12 ) ( 70 kg + 10 kg ) ( 0.20 m/s 2 + 9.8 m/s 2 ) = 4.0 × 102 N Assess: A force of 400 N, which is approximately one-half the total gravitational force, is reasonable since the upward acceleration is small.
7.45. Model: Use the particle model for the tightrope walker and the rope. The rope is assumed to be massless, so the tension in the rope is uniform. Visualize:
Solve:
Newton’s second law for the tightrope walker is
FR on W − FG = ma ⇒ FR on W = m ( a + g ) = ( 70 kg ) ( 8.0 m/s 2 + 9.8 m/s 2 ) = 1.25 × 103 N Now, Newton’s second law for the rope gives
∑( F ) on R
y
= T sin θ + T sinθ − FW on R = 0 N ⇒ T =
FW on R 2sin10°
We used FW on R = FR on W because they are an action/reaction pair.
=
FR on W 2sin10°
=
1.25 × 103 N = 3.6 × 103 N 2sin10°
7.46. Model: Use the particle model for the wedge and the block. Visualize:
The block will not slip relative to the wedge if they both have the same horizontal acceleration a. Note: n1 on 2 = n2 on 1. Solve:
The y-component of Newton’s second law for block m2 is
∑( F )
on 2 y
= n1 on 2 cosθ − ( FG )2 = 0 N ⇒ n1 on 2 =
m2 g cosθ
Combining this equation with the x-component of Newton’s second law yields:
∑( F )
on 2 x
= n1 on 2 sin θ = m2 a ⇒ a =
n1 on 2 sin θ = g tan θ m2
Now, Newton’s second law for the wedge is
∑( F )
on 1 x
= F − n2 on 1 sinθ = m1a
⇒ F = m1a + n2 on 1 sin θ = m1a + m2 a = ( m1 + m2 )a = (m1 + m2 ) g tan θ
7.47. Model: Treat the basketball player (P) as a particle, and use the constant-acceleration kinematic equations.
Visualize:
Solve: (a) While in the process of jumping, the basketball player is pressing down on the floor as he straightens G his legs. He exerts a force FP on F on the floor. The player experiences a gravitational force FG as well as a P G G normal force by the floor nF on P . The only force that the floor experiences is the one exerted by the player FP on F . G G G (b) The player standing at rest exerts a force FP on F on the floor. The normal force nF on P is the reaction force to FP on F . G But nF on P = FP on F , so Fnet = 0 N. When the basketball player accelerates upward by straightening his legs, his
( )
speed has to increase from zero to v1y with which he leaves the floor. Thus, according to Newton’s second law, there must be a net upward force on him during this time. This can be true only if nF on P > ( FG )P . In other words, the player presses on the floor with a force FP on F larger than the gravitational force on him, which is equal to his G weight. The reaction force nF on P then exceeds his weight and accelerates him upward until his feet leave the floor. (c) The height of 80 cm = 0.80 m is sufficient to determine the speed v1y with which he leaves the floor. Once his feet are off the floor, he is simply in free fall, with a1 = − g. From kinematics, v22 y = v12y + 2a1 ( y2 − y1 ) ⇒ 0 m 2 /s 2 = v12y + 2 ( − g )( 0.80 m )
⇒ v1 y = 2g ( 0.80 m ) = 3.96 m/s The basketball player reached v1 y = 4.0 m/s by accelerating from rest through a distance of 0.60 m. (d) Assuming a0 to be constant during the jump, we find
v12y = v02y + 2a0 ( y1 − y0 ) = 0 m 2 /s 2 + 2a0 ( y1 − 0 m ) ⇒ a0 =
v12y 2 y1
( 3.96 m/s ) 2 ( 0.60 m )
2
=
= 13.1 m/s 2
(e) The scale reads the value of nF on P , the force exerted by the scale on the player. Before jumping,
nF on P − ( FG )P = 0 N ⇒ nF on P = ( FG )P = mg = (100 kg ) ( 9.8 m/s 2 ) = 980 N While accelerating up,
⎛ a ⎞ ⎛ 13.1 ⎞ nF on P − mg = ma0 ⇒ nF on P = ma0 + mg = mg ⎜ 1 + 0 ⎟ = ( 980 N ) ⎜1 + ⎟ = 2290 N 9.8 ⎠ g⎠ ⎝ ⎝ After leaving the scale, nF on P = 0 N because there is no contact with the scale.
7.48.
A 1.0 kg wood block is placed on top of a 2.0 kg wood block. A horizontal rope pulls the 2.0 kg block across a frictionless floor with a force of 21.0 N. Does the 1.0 kg block on top slide? Visualize:
Solve:
The 1.0 block is accelerated by static friction. It moves smoothly with the lower block if f s < ( fs ) max . It
slides if the force that would be needed to keep it in place exceeds ( f s ) max . Begin by assuming the blocks move together with common acceleration a. Newton’s second is
∑ ( Fon 1 ) x = fs = m1a Bottom block: ∑ ( Fon 2 ) x = Tpull − fs = m2 a
Top block:
Adding these two equations gives Tpull = (m1 + m2 )a, or a = (21.0 N)/(1.0 kg + 2.0 kg) = 7.0 m/s 2 . The static
friction force needed to accelerate the top block at 7.0 m/s2 is
fs m1a = (1.0 kg)(7.0 m/s 2 ) = 7.0 N To find the maximum possible static friction force ( f s ) max = μ s n1 , the y-equation of Newton’s second law for the top block shows that n1 = m1 g . Thus ( f s ) max = μs m1 g = (0.50)(1.0 kg)(9.80 m/s 2 ) = 4.9 N Because 7.0 N > 4.9 N, static friction is not sufficient to accelerate the top block. It slides.
7.49.
A 1.0 kg wood block is placed behind a 2.0 kg wood block on a horizontal table. The coefficients of kinetic friction with the table are 0.3 for the 1.0 kg block and 0.5 for the 2.0 kg block. The 1.0 kg block is pushed forward, against the 2.0 block, and released with a speed of 2.0 m/s. How far do the blocks travel before stopping? Visualize:
Solve: The 2.0 kg block in front has a larger coefficient of friction. Thus the 1.0 kg block pushes against the rear of the 2.0 kg block and, in reaction, the 2.0 kg block pushes backward against the 1.0 kg block. There’s no vertical acceleration, so n1 = m1 g and n2 = m2 g , leading to f1 = μ1m1 g and f 2 = μ 2 m2 g . Newton’s second law along the x-axis is
1 kg block: 2 kg block:
∑ (F
) = − F2 on 1 − f1 = − F2 on 1 − μ1m1 g = m1a
on 1 x
∑ ( Fon 2 ) x = F1 on 2 − f 2 = F2 on 1 − μ 2m2 g = m2a
where we used a1 = a2 = a. Also, F1 on 2 = F2 on 1 because they are an action/reaction pair. Adding these two equations gives − ( μ1m1 + μ 2 m2 ) g = ( m1 + m2 )a a=−
μ1m1 + μ 2 m2 m1 + m2
g =−
(0.3)(1.0 kg) + (0.5)(2.0 kg) × 9.80 m/s2 = −4.25 m/s2 1.0 kg + 2.0 kg
We can now use constant-acceleration kinematics to find v12x = 0 = v02x + 2a ( x1 − x0 ) ⇒ x1 = −
v02x (2.0 m/s) 2 =− = 0.47 m 2a 2( −4.25 m/s2)
7.50. Model: Visualize:
Treat the ball of clay and the block as particles.
G G Solve: (a) Forces FC on B and FB on C are an action/reaction pair, so FB on C = FC on B . Note that aB ≠ aC because the clay is decelerating while the block is accelerating. Newton’s second law in the x-direction is
∑ ( Fon C ) x = − FB on C = mC aC Block: ∑ ( Fon B ) x = FC on B = FB on C = mB aB
Clay:
Equating the two expressions for FB on C gives
aC = −
mB aB mC
Turning to kinematics, the velocity of each after Δt is (vC )1 = (vC )0 + aC Δt (vB )1 = (vB )0 + aB Δt = aBΔt But (vC )1 = (vB )1 because the clay and the block are moving together after Δt has elapsed. Equating these two expressions gives (vC )0 + aC Δt = aBΔt , from which we find aC = aB −
(vC )0 Δt
We can now equate the two expressions for aC : −
mB (v ) (vC )0 / Δt (10 m/s)(0.01 s) aB = aB − C 0 ⇒ aB = = = 100 m/s2 1 + mB / mC 1 + (900 g)(100 g) mC Δt
Then aC = −9aB = −900 m/s 2 . With the acceleration now known, we can use either kinematic equation to find
(vC )1 = (vB )1 = (100 m/s 2 )(0.010 s) = 1.0 m/s (b) FC on B = mB aB = (0.90 kg)(100 m/s 2 ) = 90 N. (c) FB on C = mC aC = (0.10 kg)(900 m/s 2 ) = 90 N. Assess: The two forces are of equal magnitude, as expected from Newton’s third law.
7.51. Model: Use the particle model for the two blocks. Assume a massless rope, and massless, frictionless pulleys. Visualize:
Note that for every meter block 1 moves forward, one meter is provided to block 2. So each rope on m2 has to be lengthened by one-half meter. Thus the acceleration constraint is a2 = − 12 a1. Solve: Newton’s second law for block 1 is T = m1a1. Newton’s second law for block 2 is 2T − ( FG ) 2 = m2 a2 . Combining these two equations gives
2 ( m1a1 ) − m2 g = m2 ( − 12 a1 ) ⇒ a1 [ 4m1 + m2 ] = 2m2 g ⇒ a1 =
2m2 g 4m1 + m2
where we have used a2 = − 12 a1. Assess:
If m1 = 0 kg, then a2 = − g . This is what is expected for a freely falling object.
7.52. Model: Use the particle model for the two blocks. Assume a massless rope and massless, frictionless pulleys. Visualize:
For every one meter that the 1.0-kg block goes down, each rope on the 2.0-kg block will be shortened by onehalf meter. Thus the acceleration constraint is a1 = −2a2 . Solve: Newton’s second law for the two blocks is
2T = m2 a2 T − ( FG )1 = m1a1 Since a1 = −2a2 , the above equations become 2T = m2 a2 T − m1 g = m1 ( −2a2 ) ⇒ m2
Assess:
2 (1.0 kg ) ( 9.8 m/s 2 ) a2 2m1 g + m1 ( 2a2 ) = m1 g ⇒ a2 = = = 3.3 m/s 2 2 m2 + 4m1 ( 2.0 kg + 4.0 kg )
If m1 = 0 kg, then a2 = 0 m/s 2 , which is expected.
7.53. Model: The hamster of mass m and the wedge with mass M will be treated as objects 1 and 2, respectively. They will be treated as particles. Visualize:
The scale is denoted by the letter s. G Solve: (a) The reading of the scale is the magnitude of the force n2 that the scale exerts upward. There are two G action/reaction pairs. Initially the hamster of mass m is stuck in place and is in static equilibrium with Fnet = 0 N. G Because of the shape of the blocks, it is not clear whether the scale has to exert a horizontal friction force fs on 2 to prevent horizontal motion. We’ve included one just in case. Newton’s second law for the hamster is
(F
)
net on 1 x
= mg sinθ − f 2 on 1 = 0 N ⇒ f 2 on 1 = mg sinθ
(F
)
net on 1 y
= n1 − mg cosθ = 0 N ⇒ n1 = mg cosθ
For the wedge, we see from Newton’s third law that n1′ = n1 = mg cosθ and that f 2 on 1 = f1 on 2 = mg sin θ . Using
these equations, Newton’s second law for the wedge is
( Fnet on 2 ) x =
(F
f1 on 2 cosθ + fs on 2 − n1′ sin θ = mg sinθ cosθ + fs on 2 − mg cosθ sinθ = 0 N ⇒ f s on 2 = 0 N
)
net on 2 y
= n2 − n1′ cosθ − f1 on 2 sin θ − Mg = n2 − mg cos 2 θ − mg sin 2 θ − Mg = 0 N
⇒ n2 = mg ( cos 2 θ + sin 2 θ ) + Mg = ( M + m ) g = ( 0.800 kg + 0.200 kg ) ( 9.8 m/s 2 ) = 9.80 N First we find that f s on 2 = 0 N, so no horizontal static friction is needed to prevent motion. More interesting, the scale reading is ( M + m ) g which is the total gravitational force resting on the scale. This is the expected result.
(b) Now suppose that the hamster is accelerating down the wedge. The total mass is still M + m, but is the reading still ( M + m) g ? The frictional forces between the systems 1 and 2 have now vanished, and system 1 now
has an acceleration. However, the acceleration is along the hamster’s x-axis, so a1 y = 0 m/s 2 . The hamster’s yequation is still
( Fnet on 1 ) y = n1 − mg cosθ = 0 N ⇒ n1 = mg cosθ We still have n1′ = n1 = mg cosθ , so the y-equation for block 2 (with a2 y = 0 m/s 2 ) is
(F
)
net on 2 y
= n2 − n1′ cosθ − Mg = n2 − mg cos 2 θ − Mg = 0 N
⇒ n2 = mg cos 2 θ + Mg = ( M + m cos 2 θ ) g = 8.99 N
Assess: The scale reads less than it did when the hamster was at rest. This makes sense if you consider the limit θ → 90°, in which case cosθ → 0. If the face of the wedge is vertical, then the hamster is simply in free fall and can have no effect on the scale (at least until impact!). So for θ = 90° we expect the scale to record Mg only, and that is indeed what the expression for n2 gives.
7.54. Model: The hanging masses m1 , m2 , and m3 are modeled as particles. Pulleys A and B are massless and frictionless. The strings are massless. Visualize:
Solve:
(a) The length of the string over pulley B is constant. Therefore,
( yB − y3 ) + ( yB − yA ) = LB ⇒ yA = 2 yB − y3 − LB The length of the string over pulley A is constant. Thus,
( yA − y2 ) + ( yA − y1 ) = LA = 2 yA − y1 − y2 ⇒ 2 ( 2 yB − y3 − LB ) − y1 − y2 = LA ⇒ 2 y3 + y2 + y1 = constant This constraint implies that 2
dy3 dy2 dy1 + + = 0 m/s = 2v3 y + v2 y + v1 y dt dt dt
Also by differentiation, 2a3 y + a2 y + a1 y = 0 m/s 2 . (b) Newton’s second law for the masses m3 , m2 , m1 , and pulley A is
TB − m3 g = m3a3 y TA − m2 g = m2 a2 y TA − m1 g = m1a1 y TB − 2TA = 0 N
The pulley equation is zero because the pulley is massless. These four equations plus the acceleration constraint are five equations for the five unknowns (two tensions and three accelerations). To solve for TA , multiply the m3 equation by 2, substitute 2TB = 4TA , then divide each of the mass equations by the mass. This gives the three equations 4TA /m3 − 2 g = 2a3 y TA /m 2 − g = a2 y TA /m 1 − g = a1 y If these three equations are added, the right side adds to zero because of the acceleration constraint. Thus
( 4/m3 + 1/m2 + 1/m2 )TA − 4 g = 0 ⇒ TA =
4g ( 4/m3 + 1/m2 + 1/m2 )
(c) Using numerical values, we find TA = 18.97 N. Then
a1 y = TA /m1 − g = −2.2 m/s 2 a2 y = TA /m 2 − g = 2.9 m/s 2 a3 y = 2TA /m 3 − g = −0.32 m/s 2 (d) m3 = m1 + m2 , so it looks at first like m3 should hang in equilibrium. For it to do so, tension TB would need
to equal m3 g . However, TB is not (m1 + m2 ) g because masses m1 and m2 are accelerating rather than hanging at rest. Consequently, tension TB is not able to balance the weight of m3 .
8.1.
Model: The model rocket and the target will be treated as particles. The kinematics equations in two dimensions apply. Visualize:
Solve:
For the rocket, Newton’s second law along the y-direction is
( Fnet ) y = FR − mg = maR ⇒ aR =
1 1 ⎡15 N − ( 0.8 kg ) ( 9.8 m/s 2 ) ⎤ = 8.95 m/s 2 ( FR − mg ) = ⎦ 0.8 kg ⎣ m
Using the kinematic equation y1R = y0R + (v0R ) y ( t1R − t0R ) + 12 aR ( t1R − t0R ) , 2
30 m = 0 m + 0 m + 12 ( 8.95 m/s 2 ) ( t1R − 0 s ) ⇒ t1R = 2.589 s 2
For the target (noting t1T = t1R ), x1T = x0T + (v0T ) x ( t1T − t0T ) + 12 aT ( t1T − t0T ) = 0 m + (15 m/s )( 2.589 s − 0 s ) + 0 m = 39 m 2
You should launch when the target is 39 m away. Assess: The rocket is to be fired when the target is at x0T . For a net acceleration of approximately 9 m/s 2 in the vertical direction and a time of 2.6 s to cover a vertical distance of 30 m, a horizontal distance of 39 m is reasonable.
8.2.
Model: The model rocket will be treated as a particle. Kinematic equations in two dimensions apply. Air resistance is neglected. Visualize:
The horizontal velocity of the rocket is equal to the speed of the car, which is 3.0 m/s. Solve: For the rocket, Newton’s second law along the y-direction is: 1 ⎡( 8.0 N ) − ( 0.5 kg ) ( 9.8 m/s 2 ) ⎤ = 6.2 m/s 2 ( Fnet ) y = FR − mg = maR ⇒ a y = ⎦ 0.5 kg ⎣
Thus using y1 = y0 + (v0 ) y ( t1 − t0 ) + 12 a y ( t1 − t0 ) , 2
( 20 m ) = 0 m + 0 m + 12 ( 6.2 m/s2 ) ( t1R − 0 s )
2
⇒ ( 20 m ) = ( 3.1 m/s 2 ) t12 ⇒ t1 = 2.54 s
Since t1 is also the time for the rocket to move horizontally up to the hoop, x1 = x0 + (v0 ) x ( t1 − t0 ) + 12 ax ( t1 − t0 ) = 0 m + ( 3.0 m/s )( 2.54 s − 0 s ) + 0 m = 7.6 m 2
Assess: In view of the rocket’s horizontal speed of 3.0 m/s and its vertical thrust of 8.0 N, the above-obtained value for the horizontal distance is reasonable.
8.3.
Model: The asteroid and the giant rocket will be treated as particles undergoing motion according to the constant-acceleration equations of kinematics. Visualize:
Solve:
(a) The time it will take the asteroid to reach the earth is
displacement 4.0 × 106 km = = 2.0 × 105 s = 56 h velocity 20 km/s (b) The angle of a line that just misses the earth is
tanθ =
⎛R⎞ R ⎛ 6400 km ⎞ ⇒ θ = tan −1 ⎜ ⎟ = tan −1 ⎜ ⎟ = 0.092° 6 y0 ⎝ 4.0 × 10 km ⎠ ⎝ y0 ⎠
(c) When the rocket is fired, the horizontal acceleration of the asteroid is
ax =
5.0 × 109 N = 0.125 m/s 2 4.0 × 1010 kg
(Note that the mass of the rocket is much smaller than the mass of the asteroid and can therefore be ignored completely.) The velocity of the asteroid after the rocket has been fired for 300 s is vx = v0 x + ax ( t − t0 ) = 0 m/s + ( 0.125 m/s 2 ) ( 300 s − 0 s ) = 37.5 m/s After 300 s, the vertical velocity is v y = 2 × 104 m/s and the horizontal velocity is vx = 37.5 m/s. The deflection due to this horizontal velocity is tan θ = That is, the earth is saved.
⎛ 37.5 m/s ⎞ vx ⇒ θ = tan −1 ⎜ ⎟ = 0.107° 4 vy ⎝ 2 × 10 m/s ⎠
8.4.
Model: We are using the particle model for the car in uniform circular motion on a flat circular track. There must be friction between the tires and the road for the car to move in a circle. Visualize:
Solve:
The centripetal acceleration is v 2 ( 25 m/s ) = = 6.25 m/s 2 100 m r 2
ar =
The acceleration points to the center of the circle, so the net force is G G Fr = ma = (1500 kg ) ( 6.25 m/s 2 , toward center ) = ( 9380 N, toward center )
This force is provided by static friction
f s = Fr = 9.4 kN
8.5.
Model: Visualize:
Solve:
We will use the particle model for the car which is in uniform circular motion.
The centripetal acceleration of the car is
v 2 (15 m/s ) = = 4.5 m/s 2 50 m r The acceleration is due to the force of static friction. f s = mar = (1500 kg ) ( 4.5m s 2 ) = 6750 N = 6.8 kN. 2
ar =
Assess:
The model of static friction is
The
( fs )max = nμs = mg μs ≈ mg ≈ 15,000 N
surface. We see that f s < ( fs )max , which is reasonable.
force
of
friction
since μ s ≈ 1 for a dry road
is
8.6.
Model: Treat the block as a particle attached to a massless string that is swinging in a circle on a frictionless table. Visualize:
Solve:
(a) The angular velocity and speed are
ω = 75
rev 2π rad 1min × = 471.2 rad/min vt = rω = ( 0.50 m )( 471.2 rad min ) × = 3.93 m/s min 1 rev 60 s
The tangential velocity is 3.9 m/s. (b) The radial component of Newton’s second law is
∑F
r
=T =
mv 2 r
Thus
T = ( 0.20 kg )
( 3.93 m/s ) 0.50 m
2
= 6.2 N
8.7.
Solve:
Newton’s second law is Fr = mar = mrω 2 . Substituting into this equation yields:
ω=
Fr = mr
8.2 × 10−8 N (9.1×10 kg )(5.3 × 10−11 m ) −31
= 4.37 × 1016 rad/s = 4.37 × 1016
Assess:
rad 1 rev × = 6.6 × 1015 rev/s s 2π rad
This is a very high number of revolutions per second.
8.8.
Model: Visualize:
The vehicle is to be treated as a particle in uniform circular motion.
On a banked road, the normal force on a vehicle has a horizontal component that provides the necessary centripetal acceleration. The vertical component of the normal force balances the gravitational force. Solve: From the physical representation of the forces in the r-z plane, Newton’s second law can be written
∑F
r
= n sin θ =
mv 2 r
∑F
z
= n cosθ − mg = 0 ⇒ n cosθ = mg
Dividing the two equations and making the conversion 90 km h = 25 m/s yields:
( 25 m/s ) v2 = = 0.128 ⇒ θ = 7.3° rg ( 9.8 m/s 2 ) 500 m 2
tanθ =
Assess: Such a banking angle for a speed of approximately 55 mph is clearly reasonable and within our experience as well.
8.9.
Model: The motion of the moon around the earth will be treated through the particle model. The circular motion is uniform. Visualize:
Solve:
The tension in the cable provides the centripetal acceleration. Newton’s second law is
⎛ 2π ⎞ ∑ Fr = T = mrω 2 = mr ⎜ T ⎟ ⎝ moon ⎠
2
2
⎡ 2π 1 day 1h ⎤ 20 = ( 7.36 × 1022 kg )( 3.84 × 108 m ) ⎢ × × ⎥ = 2.01 × 10 N 27.3 days 24 h 3600 s ⎣ ⎦ Assess: This is a tremendous tension, but clearly understandable in view of the moon’s large mass and the large radius of circular motion around the earth.
8.10.
Model: Visualize:
Solve:
Model the ball as a particle in uniform circular motion. Rolling friction is ignored.
The track exerts both an upward normal force and an inward normal force. From Newton’s second law, 2
⎡ 60 rev 2π rad 1min ⎤ Fnet = n2 = mrω 2 = ( 0.030 kg )( 0.20 m ) ⎢ × × = 0.24 N 1 rev 60 s ⎥⎦ ⎣ min
8.11.
Model: Visualize:
The satellite is considered to be a particle in uniform circular motion around the moon.
Solve: The radius of the moon is 1.738 × 106 m and the satellite’s distance from the center of the moon is the same quantity. The angular velocity of the satellite is
ω=
2π 2π rad 1min = × = 9.52 × 10−4 rad/s T 110 min 60 s
and the centripetal acceleration is
ar = rω 2 = (1.738 × 106 m )( 9.52 × 10−4 rad/s ) = 1.58 m/s 2 2
The acceleration of a body in orbit is the local “g” experienced by that body.
8.12.
Model: The earth is considered to be a particle in uniform circular motion around the sun. Solve: The earth orbits the sun in 365 days and is 1.5 × 1011 m from the sun. The angular velocity and centripetal acceleration are
ω=
2π rad 1 day 1h × × = 2.0 × 10−7 rad/s 365 days 24 h 3600 s
ar = g = rω 2 = (1.5 × 1011 m )( 2.0 × 10−7 rad/s ) = 6.0 × 10−3 m/s 2 2
Assess:
The smallness of this acceleration due to gravity is essentially due to the large earth-sun distance.
8.13.
Model: Visualize:
Use the particle model for the car which is undergoing circular motion.
Solve: The car is in circular motion with the center of the circle below the car. Newton’s second law at the top of the hill is
∑F = (F ) r
G r
− nr = mg − n = mar =
mv 2 n⎞ ⎛ ⇒ v2 = r ⎜ g − ⎟ m r ⎝ ⎠
Maximum speed is reached when n = 0 and the car is beginning to lose contact with the road.
vmax = rg = Assess:
( 50 m ) ( 9.8 m/s 2 ) = 22 m/s
A speed of 22 m/s is equivalent to 49 mph, which seems like a reasonable value.
8.14.
Model: Visualize:
The passengers are particles in circular motion.
Solve: The center of the circle of motion of the passengers is directly above them. There must be a net force pointing up that provides the needed centripetal acceleration. The normal force on the passengers is their weight. Ordinarily their weight is FG , so if their weight increases by 50%, n = 1.5 FG . Newton’s second law at the bottom of the dip is mv 2 ∑ Fr = n − FG = (1.5 − 1) FG = 0.5mg = r
⇒ v = 0.5 gr = 0.5 ( 9.8 m/s 2 ) ( 30 m ) = 12.1 m/s Assess:
A speed of 12.1 m/s is 27 mph, which seems very reasonable.
8.15.
Model: Model the roller coaster car as a particle at the top of a circular loop-the-loop undergoing uniform circular motion. Visualize:
Notice that the r-axis points downward, toward the center of the circle. G K Solve: The critical speed occurs when n goes to zero and FG provides all the centripetal force pulling the car in the vertical circle. At the critical speed mg = mvc2 r , therefore vc = rg . Since the car’s speed is twice the
critical speed, vt = 2vc and the centripetal force is
∑ Fr = n + FG =
2 mv 2 m ( 4vc ) m ( 4rg ) = = = 4mg r r r
Thus the normal force is n = 3 mg . Consequently, n FG = 3.
8.16.
Model: Visualize:
Model the roller coaster car as a particle undergoing uniform circular motion along a loop.
Notice that the r-axis points downward, toward the center of the circle. Solve: In this problem the normal force is equal to the gravitational force: n = FG = mg . We have
∑F
r
= n + FG =
mv 2 = mg + mg ⇒ v = 2rg = 2 ( 20 m ) ( 9.8 m/s 2 ) = 19.8 m/s r
8.17.
Model: Visualize:
Model the bucket of water as a particle in uniform circular motion.
Solve: Let us say the distance from the bucket handle to the top of the water in the bucket is 35 cm. This makes the shoulder to water distance 65 cm + 35 cm = 1.00 m. The minimum angular velocity for swinging a bucket of water in a vertical circle without spilling any water corresponds to the case when the speed of the bucket is critical. In this case, n = 0 N when the bucket is in the top position of the circular motion. We get
∑F
r
⇒ ωc = g / r =
= n + FG = 0 N + mg =
mvc 2 = mrω c2 r
9.8 m/s 2 1 rev 60 s = 3.13 rad/s = 3.13 rad s × × = 30 rpm 1.00 m 2π rad 1 min
8.18.
Model: Visualize:
Solve:
Use the particle model for the car, which is undergoing nonuniform circular motion.
The car is in circular motion with radius r =
d = 100 m. We require 2
1.5 m/s 2 1.5 m/s 2 = = 0.122 s −1 100 m r The definition of the angular velocity can be used to determine the time Δt using the angular acceleration a 1.5 m/s 2 α= t = = 1.5 × 10−2 s −2 . r 100 m ω = ωi + αΔt ar = ω 2 r = 1.5 m/s 2 ⇒ ω =
⇒ Δt =
ω − ωi 0.122 s −1 − 0 s −1 = = 8.2 s α 0.015 s −2
8.19.
Model: Visualize:
Solve:
The train is a particle undergoing nonuniform circular motion.
(a) Newton’s second law in the vertical direction is ( Fnet ) y = n − FG = 0
from which n = mg . The rolling friction is f R = μ R n = μ R mg . This force provides the tangential acceleration at = −
The angular acceleration is
fR = −μR g m
2 at − μ R g − ( 0.10 ) ( 9.8 m/s ) = = = −1.96 rad/s 2 r r 0.50 m ⎛ rev ⎞⎛ 1 min ⎞⎛ 2π rad ⎞ (b) The initial angular velocity is 30 ⎜ ⎟⎜ ⎟⎜ ⎟ = 3.14 rad/s. The time to come to a stop due to ⎝ min ⎠⎝ 60 sec ⎠⎝ rev ⎠ the rolling friction is ω − ωi 0 − 3.14 rad/s Δt = f = = 1.60 s −1.96 rad/s 2 α Assess: The original angular speed of π rad/s means the train goes around the track one time every 2 seconds, so a stopping time of less than 2 s is reasonable.
α=
8.20.
Model: The object is treated as a particle in the model of kinetic friction with its motion governed by constant-acceleration kinematics. Visualize:
The velocity v1x as the object sails off the edge is related to the initial velocity v0 x by
Solve:
v = v + 2ax ( x1 − x0 ) . Using Newton’s second law to determine ax while sliding gives 2 1x
2 0x
∑ Fx = − f k = max ⇒ ∑ Fy = n − mg = 0 N ⇒ n = mg
Using this result and the model of kinetic friction
( fk = μk n ) ,
the x-component equation can be written as
− μ k mg = max . This implies ax = − μ k g = − ( 0.50 ) ( 9.8 m/s 2 ) = −4.9 m/s 2 Kinematic equations for the object’s free fall can be used to determine v1x : y2 = y1 + v1 y ( t2 − t1 ) + 12 ( − g )( t2 − t1 ) ⇒ 0 m = 1.0 m + 0 m − 2
g 2 ( t2 − t1 ) ⇒ ( t2 − t1 ) = 0.4518 s 2
x2 = x1 + v1x ( t2 − t1 ) = 2.30 m = 2.0 m + v1x ( 0.4518 s ) ⇒ v1x = 0.664 m/s Having determined v1x and ax , we can go back to the velocity equation v12x = v02x + 2ax ( x1 − x0 ) :
( 0.664 m/s ) Assess:
2
= v02x + 2 ( −4.9 m/s 2 ) ( 2.0 m ) ⇒ v0 x = 4.5 m/s
v0 x = 4.5 m/s is about 10 mph and is a reasonable speed.
8.21.
Model: The rocket and puck together make a particle moving on frictionless ice. The thrust of the rocket motor is assumed to be constant. Visualize:
Solve The kinematics equations can be used to examine the motion for each coordinate axis independently as a function of time t, then combined to eliminate t. In the x-direction, x f = xi + v0 xt = 0 m + ( 2.0 m/s ) t
So t =
xf 2.0 m/s
. The acceleration of the rocket and puck in the y-direction is
ay =
( Fnet ) y mrocket + mpuck
=
8.0 N = 13.3 m/s 2 0.600 kg
In the y–direction, 1 1 yf = yi + v0 yt + a yt 2 = 0 m + ( 0 m/s ) t + (13.3 m/s 2 ) t 2 2 2 = ( 6.67 m/s 2 ) t 2 Substituting t from the equation for the x-direction, 2
⎛ x ⎞ 2 yf = ( 6.67 m/s 2 ) ⎜ ⎟ = 1.67 xf 2.0 m/s ⎝ ⎠ A graph of this equation is shown in the figure. Assess: When the puck has traveled 9 m in the x-direction it has traveled 15 m in the y-direction.
8.22.
Model: Treat Sam as a particle. Visualize: This is a two-part problem. Use an s-axis parallel to the slope for the first part, regular xycoordinates for the second. Sam’s final velocity at the top of the slope is his initial velocity as he becomes airborne.
Solve:
Sam’s acceleration up the slope is given by Newton’s second law:
( Fnet ) s = F − mg sin10° = ma0 a0 =
F 200 N − g sin10° = − (9.8 m/s 2 ) sin10° = 0.965 m/s 2 m 75 kg
The length of the slope is s1 = (50 m)/ sin10° = 288 m. His velocity at the top of the slope is v12 = v02 + 2a0 ( s1 − s0 ) = 2a0 s1 ⇒ v1 = 2(0.965 m/s 2) (288 m) = 23.6 m/s This is Sam’s initial speed into the air, giving him velocity components v1x = v1 cos10° = 23.2 m/s and
v1 y = v1 sin10° = 410 m/s. This is not projectile motion because Sam experiences both the force of gravity and the thrust of his skis. Newton’s second law for Sam’s acceleration is a1x = a1 y =
( Fnet ) y m
=
( Fnet ) x (200 N)cos10° = = 2.63 m/s 2 m 75 kg
(200 N)sin10° − (75 kg)(9.80 m/s 2 ) = −9.34 m/s 2 75 kg
The y-equation of motion allows us to find out how long it takes Sam to reach the ground: y2 = 0 m = y1 + v1 yt2 + 12 a1 yt22 = 50 m + (4.10 m/s) t2 − (4.67 m/s 2) t22
This quadratic equation has roots t2 = −2.86 s (unphysical) and t2 = 3.74 s. The x-equation of motion—this time with an acceleration—is x2 = x1 + v1xt2 + 12 a1xt22 = 0 m + (23.2 m/s) t2 − 12 (2.63 m/s 2) t22 = 105 m Sam lands 105 m from the base of the cliff.
8.23. Model: Treat the motorcycle and rider as a particle. Visualize: This is a two-part problem. Use an s-axis parallel to the slope for the first part, regular xycoordinates for the second. The motorcycle’s final velocity at the top of the ramp is its initial velocity as it becomes airborne.
Solve:
The motorcycle’s acceleration on the ramp is given by Newton’s second law:
( Fnet ) s = − f r − mg sin 20° = − μ r n − mg sin 20° = − μ r mg cos 20° − mg sin 20° = ma0 a0 = − g ( μ r cos 20° + sin 20°) = −(9.8 m/s 2 )((0.02)cos 20° + sin 20°) = −3.536 m/s 2 The length of the ramp is s1 = (2.0 m)/ sin 20° = 5.85 m. We can use kinematics to find its speed at the top of the ramp: v12 = v02 + 2a0 ( s1 − s0 ) = v02 + 2a0 s1 ⇒ v1 = (11.0 m/s) 2 + 2(−3.536 m/s 2) (5.85 m) = 8.92 m/s This is the motorcycle’s initial speed into the air, with velocity components v1x = v1 cos 20° = 8.38 m/s and
v1 y = v1 sin 20° = 3.05 m/s. We can use the y-equation of projectile motion to find the time in the air: y2 = 0 m = y1 + v1 yt2 + 12 a1 yt22 = 2.0 m + (3.05 m/s) t2 − (4.90 m/s 2) t22
This quadratic equation has roots t2 = −0.399 s (unphysical) and t2 = 1.021 s. The x-equation of motion is thus x2 = x1 + v1xt2 = 0 m + (8.38 m/s) t2 = 8.56 m
8.56 m < 10.0 m, so it looks like crocodile food.
8.24.
Model: Use the particle model and the constant-acceleration equations of kinematics for the rocket. Solve: (a) The acceleration of the rocket in the launch direction is obtained from Newton’s second law G G F = ma :
140,700 N = ( 5000 kg ) a ⇒ a = 28.14 m/s 2 Therefore, ax = a cos 44.7° = 20.0 m/s 2 and a y = a sin 44.7° = 19.8 m/s 2 . The net acceleration in the y-direction is thus
( anet ) y = a y − g = (19.8 − 9.8) m/s 2
= 10.0 m/s 2
With this acceleration, we can write the equations for the x- and y-motions of the rocket.
y = y0 + v0 y ( t − t0 ) + 12 ( anet ) y ( t − t0 ) = 0 m + 0 m + 12 (10.0 m/s 2 ) t 2 = ( 5.00 m/s 2 ) t 2 2
x = x0 + v0 x ( t − t0 ) + 12 ( anet ) x ( t − t0 ) = 0 m + 0 m + 12 ( 20.0 m/s 2 ) t 2 = (10.0 m/s 2 ) t 2 2
From these two equations, 2 2 x (10.0 m/s ) t = =2 y ( 5.00 m/s 2 ) t 2
The equation that describes the rocket’s trajectory is y = 12 x. (b) It is a straight line with a slope of 12 . (c) In general, v y = v0 y + ( anet ) y ( t1 − t0 ) = 0 + (10.0 m/s 2 ) t1
vx = v0 x + ( anet ) x ( t1 − t0 ) = 0 + ( 20.0 m/s 2 ) t1 v=
(10.0 m/s ) t + ( 20.0 m/s ) t = ( 22.36 m/s ) t 2 2 2 1
2 2 2 1
2
1
The time required to reach the speed of sound is calculated as follows: 330 m/s = ( 22.36 m/s 2 ) t1 ⇒ t1 = 14.76 s We can now obtain the elevation of the rocket. From the y-equation, y = ( 5.00 m/s 2 ) t12 = ( 5.00 m/s 2 ) (14.76 s ) = 1090 m 2
8.25.
Model: The hockey puck will be treated as a particle whose motion is determined by constantacceleration kinematic equations. We break this problem in two parts, the first pertaining to motion on the table and the second to free fall.
Visualize: Solve:
Newton’s second law is:
Fx = max ⇒ ax =
Fx 2.0 N = = 2.0 m/s 2 m 1.0 kg
The kinematic equation v12x = v02x + 2ax ( x1 − x0 ) yields:
v12x = 0 m 2 /s 2 + 2 ( 2.0 m/s 2 ) ( 4.0 m ) ⇒ v1x = 4.0 m/s Let us now find the time of free fall ( t2 − t1 ) :
y2 = y1 + v1 y ( t2 − t1 ) + 12 a1 y ( t2 − t1 )
2
⇒ 0 m = 2.0 m + 0 m + 12 ( −9.8 m/s 2 ) ( t2 − t1 ) ⇒ ( t2 − t1 ) = 0.639 s 2
Having obtained v1x and ( t2 − t1 ) , we can now find ( x2 − x1 ) as follows: x2 = x1 + v1x ( t2 − t1 ) + 12 ax ( t2 − t1 )
2
⇒ x2 − x1 = ( 4.0 m/s )( 0.639 s ) + 12 ( 2.0 m/s 2 ) ( 0.639 s ) = 3.0 m 2
Assess:
For a modest horizontal thrust of 2.0 N, a landing distance of 3.0 m is reasonable.
8.26.
Model: The model rocket is treated as a particle and its motion is determined by constant-acceleration kinematic equations. Visualize:
Solve:
As the rocket is accidentally bumped v0 x = 0.5 m/s and v0 y = 0 m/s. On the other hand, when the
engine is fired
Fx = max ⇒ ax =
Fx 20 N = = 40 m/s 2 m 0.500 kg
(a) Using y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) , 2
0 m = 40 m + 0 m + 12 ( −9.8 m/s 2 ) t12 ⇒ t1 = 2.857 s The distance from the base of the wall is x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) = 0 m + ( 0.5 m/s )( 2.857 s ) + 12 ( 40 m/s 2 ) ( 2.857 s ) = 165 m 2
2
(b) The x- and y-equations are y = y0 + v0 y ( t − t0 ) + 12 a y ( t − t0 ) = 40 − 4.9t 2 2
x = x0 + v0 x ( t − t0 ) + 12 ax ( t − t0 ) = 0.5t + 20t 2 2
Except for a brief interval near t = 0, 20t 2 0.5t. Thus x ≈ 20t 2 , or t 2 = x / 20. Substituting this into the yequation gives y = 40 − 0.245 x This is the equation of a straight line, so the rocket follows a linear trajectory to the ground.
8.27.
Model: Visualize:
Assume the particle model for the satellite in circular motion.
To be in a geosynchronous orbit means rotating at the same rate as the earth, which is 24 hours for one complete 3.58 × 107 m, r = 3.58 × 107 m, rotation. Because the altitude of the satellite is
re = 3.58 × 107 m + 6.37 × 106 m = 4.22 × 107 m. Solve: (a) The period (T ) of the satellite is 24.0 hours. (b) The acceleration due to gravity is 2
2
1 hr ⎞ ⎛ 2π ⎞ ⎛ 2π 7 2 × g = ar = rω 2 = r ⎜ ⎟ = ( 4.22 × 10 m ) ⎜ ⎟ = 0.223 m/s ⎝ T ⎠ ⎝ 24.0 hr 3600 s ⎠ (c) There is no normal force on a satellite, so the weight is zero. It is in free fall.
8.28.
Model: earth rotates. Visualize:
Treat the man as a particle. The man at the equator undergoes uniform circular motion as the
Solve: The scale reads the man’s weight FG = n, the force of the scale pushing up against his feet. At the north pole, where the man is in static equilibrium,
nP = FG = mg = 735 N At the equator, there must be a net force toward the center of the earth to keep the man moving in a circle. The raxis points toward the center, so
∑F = F r
G
− nE = mω 2 r ⇒ nE = mg − mω 2 r = nP − mω 2 r
The equator scale reads less than the north pole scale by the amount mω 2 r. The man’s angular velocity is that of the equator, or 2π 2π rad ω= = = 7.27 × 10−5 rad/s T 24 hours × (3600 s/1 h)
Thus the north pole scale reads more than the equator scale by
Δw = (75 kg)(7.27 × 10−5 rad/s) 2 (6.37 × 106 m) = 2.5 N Assess:
The man at the equator appears to have lost Δm = Δw/g ≈ 0.25 kg, or the equivalent of ≈
1 2
lb.
8.29.
Model: Visualize:
Solve:
Use the particle model for the (cart + child) system which is in uniform circular motion.
Newton’s second law along r and z directions can be written:
∑F
r
= T cos 20° − n sin 20° = mar
∑F
z
= T sin 20° − n cos 20° − mg = 0
The cart’s centripetal acceleration is 2
rev 1 min 2π rad ⎞ ⎛ 2 × × ar = rω 2 = ( 2.0cos20° m ) ⎜14 ⎟ = 4.04 m/s 1 rev ⎠ ⎝ min 60 s The above force equations can be rewritten as
0.94T − 0.342n = ( 25 kg ) ( 4.04 m/s 2 ) = 101 N 0.342T + 0.94n = ( 25 kg ) ( 9.8 m/s 2 ) = 245 N Solving these two equations yields T = 179 N for the tension in the rope. Assess: In view of the child + cart weight of 245 N, a tension of 179 N is reasonable.
8.30.
Model: Visualize:
Solve:
Model the ball as a particle which is in a vertical circular motion.
At the bottom of the circle,
∑ Fr = T − FG =
( 0.500 kg ) v ⇒ v = 5.5 m/s mv 2 ⇒ (15 N ) − ( 0.500 kg ) ( 9.8 m/s 2 ) = r (1.5 m ) 2
8.31.
Model: We will use the particle model for the car, which is undergoing uniform circular motion on a banked highway, and the model of static friction. Visualize:
Note that we need to use the coefficient of static friction μ s , which is 1.0 for rubber on concrete. Solve: Newton’s second law for the car is
mv 2 ∑ Fz = n cosθ − fs sinθ − FG = 0 N r Maximum speed is when the static friction force reaches its maximum value ( fs )max = μs n. Then
∑F
r
= f s cosθ + n sinθ =
mv 2 n ( cos15° − μs sin15° ) = mg r Dividing these two equations and simplifying, we get n ( μ s cos15° + sin15° ) =
μs + tan15° v 2 μ + tan15° = ⇒ v = gr s 1 − μs tan15° gr 1 − μ s tan15° = Assess:
( 9.80 m/s ) ( 70 m ) ( (1 − 0.268) ) = 34 m/s 2
1.0 + 0.268
The above value of 34 m/s ≈ 70 mph is reasonable.
8.32.
Model: Visualize:
Solve:
Use the particle model for the rock, which is undergoing uniform circular motion.
Newton’s second law is
mv 2 ∑ Fz = T sin10° − mg = 0 N r where the radius of the circular motion is r = (1.0 m ) cos10° = 0.985 m. Dividing these two equations, we get
∑F
r
tan10° = ⇒ω =
= T cos10° =
gr ⇒v= v2
gr = tan10°
( 9.8 m/s ) ( 0.985 m ) = 7.40 m/s 2
tan10°
v 7.40 m/s rad 1 rev 60 s = = 7.51 rad/s = 7.51 × × = 72 rpm r 0.985 m s 2π rad 1 min
8.33.
Model: motion. Visualize:
Use the particle model and static friction model for the coin, which is undergoing circular
Solve: The force of static friction is f s = μ s n = μs mg . This force is equivalent to the maximum centripetal force that can be applied without sliding. That is,
μs mg = m
vt 2 2 = m ( rω max ) ⇒ ωmax = r = 7.23
μs g r
=
( 0.80 ) ( 9.8 m/s 2 ) 0.15 m
= 7.23 rad/s
rad 1 rev 60 s × × = 69 rpm s 2π rad 1 min
So, the coin will stay still on the turntable. Assess: A rotational speed of approximately 1 rev per second for the coin to stay stationary seems reasonable.
8.34.
Model: Visualize:
Solve:
Use the particle model for the car, which is in uniform circular motion.
Newton’s second law is
∑ Fr = T sin 20° = mar =
mv 2 r
∑F
z
= T cos 20° − FG = 0 N
These equations can be written as T sin 20° =
mv 2 T cos 20° = mg r
Dividing these two equations gives
tan 20° = v 2 rg ⇒ v = rg tan 20° =
( 4.55 m ) ( 9.8 m/s 2 ) tan 20° = 4.03 m/s
8.35.
Model: Visualize:
Use the particle model for the ball in circular motion.
G Solve: (a) The mass moves in a horizontal circle of radius r = 20 cm. The acceleration a and the net force G vector point to the center of the circle, not along the string. The only two forces are the string tension T , which G does point along the string, and the gravitational force FG . These are shown in the free-body diagram. Newton’s second law for circular motion is mv 2 ∑ Fz = T cosθ − FG = T cosθ − mg = 0 N ∑ Fr = T sinθ = mar = r From the z-equation, T=
( 0.500 kg ) ( 9.8 m/s mg = cosθ cos11.54°
2
) = 5.00 N
(b) We can find the tangential speed from the r-equation:
v=
rT sin θ = 0.63 m/s m
The angular speed is v r
ω= =
0.63 ms = 3.15 rad/s = 30 rpm 0.20 m
8.36.
Model: Visualize:
Consider the passenger to be a particle and use the model of static friction.
Solve: The passengers stick to the wall if the static friction force is sufficient to support the gravitational force on them: fs = FG The minimum angular velocity occurs when static friction reaches its maximum possible value
( fs )max = μs n.
Although clothing has a range of coefficients of friction, it is the clothing with the smallest
coefficient ( μ s = 0.60) that will slip first, so this is the case we need to examine. Assuming that the person is stuck to the wall, Newton’s second law is
∑ F = n = mω r 2
r
The minimum frequency occurs when
∑ Fz = fs − w = 0 ⇒ fs = mg
2 f s = ( f s )max = μ s n = μ s mrω min
Using this expression for fs in the z-equation gives 2 fs = μ s mrω min = mg
⇒ ω min =
g 9.80 m/s 2 1 rev 60 s = = 2.56 rad/s = 2.56 rad/s × × = 24 rpm 0.60(2.5 m) 2π rad 1 min μs r
Assess: Note the velocity does not depend on the mass of the individual. Therefore, the minimum mass sign is not necessary.
8.37.
Model: Visualize:
Use the particle model for the marble in uniform circular motion.
Solve: The marble will roll in a horizontal circle if the static friction force is sufficient to support the gravitational on it: f s = FG If mg > ( fs )max then static friction is not sufficient and the marble will slip down the side as it rolls around the circumference. The r-equation of Newton’s second law is
∑
2
2π rad 1 min ⎞ ⎛ × Fr = n = mrω 2 = (0.010 kg)(0.060 m) ⎜150 rpm × ⎟ = 0.148 N 1 rev 60 s ⎠ ⎝
( fs )max = μs n = (0.80)(0.148 N) = 0.118 N. The friction force f s = mg = 0.098 N. We see that f s < ( fs )max , therefore friction is sufficient
Thus the maximum possible static friction is needed to support a 10 g marble is
and the marble spins in a horizontal circle. Assess: In reality, rolling friction will cause the marble to gradually slow down until point, it will begin to slip down the inside wall.
( fs )max < mg.
At that
8.38.
Model: Visualize:
Use the particle model for the car and the model of kinetic friction.
Solve: We will apply Newton’s second law to all three cars. Car A:
∑ F = n + ( f ) + ( F ) = 0 N − f + 0 N = ma ∑ F = n + ( f ) + y = n + 0 N − mg = 0 N x
k x
x
y
G x
k y
y
k
x
y
The y-component equation means n = mg . Since f k = μ k n, we have f k = μ k mg . From the x-component equation, −f − μ k mg ax = k = = − μ k g = −9.8 m/s 2 m m Car B: Car B is in circular motion with the center of the circle above the car. mv 2 ∑ Fr = nr + ( f k )r + ( FG )r = n + 0 N − mg = mar = r
∑F = n + ( f ) + (F ) t
k t
t
G t
= 0 N − f k + 0 N = + mat
From the r-equation
n = mg +
⎛ mv 2 v2 ⎞ ⇒ fk = μk n = μk m ⎜ g + ⎟ r r ⎠ ⎝
Substituting back into the t-equation,
at = −
2 ⎛ fk μ m⎛ v2 ⎞ ( 25 m/s ) ⎞⎟ = −12.9 m/s 2 = − k ⎜ g + ⎟ = − μ k ⎜ 9.8 m/s 2 + ⎜ 200 m ⎟⎠ m m ⎝ r ⎠ ⎝
Car C: Car C is in circular motion with the center of the circle below the car.
∑F
r
= n r + ( f k )r + ( FG )r = − n + 0 N + mg = mar =
∑F = n + ( f ) + (F ) t
t
k t
G t
mv 2 r
= 0 N − f k + 0 N = mat
From the r-equation n = m ( g − v 2 r ) . Substituting this into the t-equation yields at =
− fk −μk n = = − μ k ( g − v 2 r ) = −6.7 m/s 2 m m
8.39.
Model: Visualize:
Solve:
Model the ball as a particle that is moving in a vertical circle.
(a) The ball’s gravitational force FG = mg = ( 0.500 kg ) ( 9.8 m/s 2 ) = 4.9 N.
(b) Newton’s second law at the top is
∑F
r
= T1 + FG = mar = m
v2 r
⎡ ( 4.0 m/s )2 ⎤ ⎛ v2 ⎞ ⇒ T1 = m ⎜ − g ⎟ = ( 0.500 kg ) ⎢ − 9.8 m/s 2 ⎥ = 2.9 N ⎢⎣ 1.02 m ⎥⎦ ⎝ r ⎠
(c) Newton’s second law at the bottom is
∑ Fr = T2 − FG =
mv 2 r
2 ⎡ ⎛ ( 7.5 m/s ) ⎤ = 32 N v2 ⎞ ⇒ T2 = m ⎜ g + ⎟ = ( 0.500 kg ) ⎢9.8 m/s 2 + ⎥ 1.02 m ⎥⎦ r ⎠ ⎢⎣ ⎝
8.40.
Model: Visualize:
Solve:
Use the particle model for yourself while in uniform circular motion.
(a) The speed and acceleration are
2π r 2π (15 m ) v 2 ( 3.77 m/s ) = 0.95 m/s 2 = = 3.77 m/s ar = = 15 m r T 25 s 2
v=
So the speed is 3.8 m/s and the centripetal acceleration is 0.95 m/s 2 . (b) The weight w = m, the normal force. On the ground, your weight is the same as the gravitational force FG . Newton’s second law at the top is mv 2 ∑ Fr = FG − n = mar = r 2 ⎛ ⎛ ( 3.77 m/s ) ⎞⎟ = m 8.85 m/s 2 v2 ⎞ ⇒ n = w = m ⎜ g − ⎟ = m ⎜ 9.8 m/s 2 − ( ) ⎜ ⎟ 15 m r ⎠ ⎝ ⎝ ⎠
⇒
w 8.85 m s 2 = = 0.90 FG 9.8 m s 2
(c) Newton’s second law at the bottom is
∑ Fr = n − FG = mar =
mv 2 r
⎛ (3.77 m/s2 ) ⎟⎞ = 10.75 m/s2 m ⎛ v2 ⎞ ⇒ n = w = m ⎜ g + ⎟ = m ⎜ 9.8 m/s 2 + ( ) ⎜ ⎟ 15 m r ⎠ ⎝ ⎝ ⎠ ⇒
w 10.75 m s 2 = = 1.10 FG 9.8 m s 2
8.41.
Model: Visualize:
Solve:
Model a passenger as a particle rotating in a vertical circle.
(a) Newton’s second law at the top is
∑F
r
= nT + FG = mar =
mv 2 mv 2 ⇒ nT + mg = r r
The speed is
v=
2π r 2π ( 8.0 m ) = = 11.17 m/s T 4.5 s
⎡ (11.17 m/s )2 ⎤ ⎛ v2 ⎞ ⇒ nT = m ⎜ − g ⎟ = ( 55 kg ) ⎢ − 9.8 m/s 2 ⎥ = 319 N ⎢⎣ 8.0 m ⎥⎦ ⎝ r ⎠
That is, the ring pushes on the passenger with a force of 3.2 × 102 N at the top of the ride. Newton’s second law at the bottom: ⎛ v2 ⎞ mv 2 mv 2 ⇒ nB = + mg = m ⎜ + g ⎟ r r r ⎝ ⎠ ⎡ (11.17 m/s )2 ⎤ = ( 55 kg ) ⎢ + 9.8 m/s 2 ⎥ = 1397 N ⎢⎣ 8.0 m ⎥⎦ Thus the force with which the ring pushes on the rider when she is at the bottom of the ring is 1.4 kN. (b) To just stay on at the top, nT = 0 N in the r-equation at the top in part (a). Thus,
∑F
r
= nB − FG = mar =
2
⎛ 2π ⎞ mv 2 r 8.0 m = mrω 2 = mr ⎜ mg = = 2π = 5.7 s ⎟ ⇒ Tmax = 2π r g 9.8 m/s 2 ⎝ Tmax ⎠
8.42.
Model: Visualize:
Solve:
Model the chair and the rider as a particle in uniform circular motion.
Newton’s second law along the r-axis is
∑F
r
= Tr + ( FG )r = mar ⇒ T sin θ + 0 N = mrω 2
Since r = L sin θ , this equation becomes 2
⎡ 2π rad ⎤ T = mLω 2 = (150 kg )( 9.0 m ) ⎢ ⎥ = 3330 N ⎣ 4.0 s ⎦ Thus, the 3000 N chain is not strong enough for the ride.
8.43.
Model: Visualize:
Model the ball as a particle in motion in a vertical circle.
G Solve: If the ball moves in a complete circle, then there is a tension force T when the ball is at the top of the circle. The tension force adds to the gravitational force to cause the centripetal acceleration. The forces are along the r-axis, and the center of the circle is below the ball. Newton’s second law at the top is
( Fnet )r = T + FG = T + mg = ⇒ vtop = rg +
mv 2 r
rT m
The tension T can’t become negative, so T = 0 N gives the minimum speed vmin at which the ball moves in a
circle. If the speed is less than vmin , then the string will go slack and the ball will fall out of the circle before it reaches the top. Thus, vmin = rg ⇒ ω min =
rg vmin = = r r
g = r
( 9.8 m/s ) = 3.13 rad / s = 30 rpm 2
(1.0 m )
8.44.
Model: The ball is a particle on a massless rope in circular motion about the point where the rope is attached to the ceiling. Visualize:
Solve:
Newton’s second law in the radial direction is
(∑ F ) = T − F r
G
= T − mg =
mv 2 r
Solving for the tension in the rope and evaluating, 2 ⎛ ⎛ ( 5.5 m/s ) ⎞⎟ = 168 N v2 ⎞ T = m ⎜ g + ⎟ = (10.2 kg ) ⎜ 9.8 m/s 2 + ⎜ 4.5 m ⎟⎠ r ⎠ ⎝ ⎝ Assess: The tension in the rope is greater than the gravitational force on the ball in order to keep the ball moving in a circle.
8.45.
Model: Visualize:
Model the person as a particle in uniform circular motion.
Solve: The only force acting on the passengers is the normal force of the wall. Newton’s second law along the r-axis is:
∑F
r
= n = mrω 2
To create “normal” gravity, the normal force by the inside surface of the space station equals mg. Therefore,
mg = mrω 2 ⇒ ω = Assess:
2π = T
500 m g r ⇒ T = 2π = 2π = 45.0 s 9.8 m/s 2 r g
This is a fast rotation. The tangential speed is v=
2π r 2π ( 500 m ) = = 70 m/s ≈ 140 mph T 45 s
8.46. Model: Visualize:
Masses m1 and m2 are considered particles. The string is assumed to be massless.
Solve: The tension in the string causes the centripetal acceleration of the circular motion. If the hole is smooth, G G it acts like a pulley. Thus tension forces T1 and T2 act as if they were an action/reaction pair. Mass m1 is in circular motion of radius r, so Newton’s second law for m1 is
∑ Fr = T1 =
m1v 2 r
Mass m2 is at rest, so the y-equation of Newton’s second law is
∑F
y
= T2 − m2 g = 0 N ⇒ T2 = m2 g
Newton’s third law tells us that T1 = T2 . Equating the two expressions for these quantities:
m1v 2 m2 rg = m2 g ⇒ v = r m1
8.47.
Model: Visualize:
Model the ball as a particle swinging in a vertical circle, then as a projectile.
Solve: Initially, the ball is moving in a circle. Once the string is cut, it becomes a projectile. The final circularmotion velocity is the initial velocity for the projectile. The free-body diagram for circular motion is shown at the bottom of the circle. Since T > FG , there is a net force toward the center of the circle that causes the centripetal acceleration. The r-equation of Newton’s second law is
( Fnet )r = T − FG = T − mg =
mv 2 r
r 0.60 m ⎡5.0 N − ( 0.10 kg ) ( 9.8 m/s 2 ) ⎤ = 4.91 m/s (T − mg ) = ⎦ m 0.100 kg ⎣ G As a projectile the ball starts at y0 = 1.4 m with v0 = 4.91iˆ m/s. The equation for the y-motion is ⇒ vbottom =
y1 = 0 m = y0 + v0 y Δt − 12 g ( Δt ) = y0 − 12 gt12 2
This is easily solved to find that the ball hits the ground at time
t1 =
2 y0 = 0.535 s g
During this time interval it travels a horizontal distance x1 = x0 + v0 xt1 = ( 4.91 m/s )( 0.535 s ) = 2.63 m So the ball hits the floor 2.6 m to the right of the point where the string was cut.
8.48.
Model: Visualize:
Solve:
Use the particle model for a ball in motion in a vertical circle and then as a projectile.
For the circular motion, Newton’s second law along the r-direction is
∑F
r
= T + FG =
mvt2 r
Since the string goes slack as the particle makes it over the top, T = 0 N. That is,
FG = mg =
mvt2 ⇒ vt = gr = r
( 9.8 m/s ) ( 0.5 m ) = 2.21 m/s 2
The ball begins projectile motion as the string is released. The time it takes for the ball to hit the floor can be found as follows:
y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) ⇒ 0 m = 2.0 m + 0 m + 12 ( −9.8 m/s2 ) ( t1 − 0 s ) ⇒ t1 = 0.639 s 2
2
The place where the ball hits the ground is
x1 = x0 + v0 x ( t1 − t0 ) = 0 m + ( +2.21 m/s )( 0.639 s − 0 s ) = +1.41 m The ball hits the ground 1.41 m to the right of the point beneath the center of the circle.
8.49.
Model: Visualize:
Model the ball as a particle undergoing circular motion in a vertical circle.
Solve: Initially, the ball is moving in circular motion. Once the string breaks, it becomes a projectile. The final circular-motion velocity is the initial velocity for the projectile, which we can find by using the kinematic equation
v12 = v02 + 2a y ( y1 − y0 ) ⇒ 0 m 2 s 2 = ( v0 ) + 2 ( −9.8 m/s 2 ) ( 4.0 m − 0 m ) ⇒ v0 = 8.85 m/s 2
This is the speed of the ball as the string broke. The tension in the string at that instant can be found by using the r-component of the net force on the ball: ⎛ v2 ⎞ ( 8.85 m/s ) ∑ Fr = T = m ⎜⎜ r0 y ⎟⎟ ⇒ T = ( 0.100 kg ) 0.60 m = 13.1 N ⎝ ⎠ 2
8.50.
Model: Visualize:
Solve:
Model the car as a particle on a circular track.
(a) Newton’s second law along the t-axis is
∑ F = F = ma t
t
t
⇒ 1000 N = (1500 kg ) at ⇒ at = 2 3 m/s 2
With this tangential acceleration, the car’s tangential velocity after 10 s will be
v1t = v0t + at ( t1 − t0 ) = 0 m s + ( 2 3 m/s 2 ) (10 s − 0 s ) = 20 3 m/s The radial acceleration at this instant is
v12t ( 20 3 m/s ) 16 m/s 2 = = 25 m 9 r 2
ar = The car’s acceleration at 10 s has magnitude
a1 = at2 + ar2 =
( 2 3 m/s ) + (16 9 m/s ) 2 2
2 2
= 1.90 m/s 2 θ = tan −1
⎛ 23⎞ at = tan −1 ⎜ ⎟ = 21° ar ⎝ 16 9 ⎠
where the angle is measured from the r-axis. (b) The car will begin to slide out of the circle when the static friction reaches its maximum possible value ( fs )max = μs n. That is,
mv22t ⇒ v2t = rg = ( 25 m ) ( 9.8 m/s 2 ) = 15.7 m/s r In the above equation, n = mg follows from Newton’s second law along the z-axis. The time when the car begins to slide can now be obtained as follows:
∑F = ( f ) r
s max
= μ s n = μ s mg =
v2t = v0t + at ( t2 − t0 ) ⇒ 15.7 m/s = 0 m s + ( 2 3 m/s 2 ) ( t2 − 0 ) ⇒ t2 = 24 s
8.51.
Model: Visualize:
Solve:
Model the steel block as a particle and use the model of kinetic friction.
G (a) The components of thrust ( F ) along the r-, t-, and z-directions are
Fr = F sin 20° = ( 3.5 N ) sin 20° = 1.20 N Ft = F cos 20° = ( 3.5 N ) cos 20° = 3.29 N Fz = 0 N Newton’s second law is
( Fnet )r = T + Fr = mrω 2 ( Fnet )t = Ft − f k = mat
( Fnet ) z = n − mg = 0 N The z-component equation means n = mg . The force of friction is
f k = μ k n = μ k mg = ( 0.60 )( 0.500 kg ) ( 9.8 m/s 2 ) = 2.94 N Substituting into the t-component of Newton’s second law
( 3.29 N ) − ( 2.94 N ) = ( 0.500 kg ) at ⇒ at = 0.70 m/s 2 Having found at , we can now find the tangential velocity after 10 revolutions = 20π rad as follows: 1⎛ a ⎞
θ1 = ⎜ t ⎟ t12 ⇒ t1 = 2⎝ r ⎠
2rθ1 = 18.95 s at
⎛ at ⎞ ⎟ t1 = 6.63 rad/s ⎝r⎠
ω1 = ω 0 + ⎜
The block’s angular velocity after 10 s is 6.6 rad/s. (b) Substituting ω1 into the r-component of Newton’s second law yields:
T1 + Fr = mrω12 ⇒ T1 + (1.20 N ) = ( 0.500 kg )( 2.0 m )( 6.63 rad/s ) ⇒ T1 = 44 N 2
8.52.
Model: Visualize:
Solve:
Assume the particle model for a ball in vertical circular motion.
(a) Newton’s second law in the r- and t-directions is
( Fnet )r = T + mg cosθ = mar =
mvt2 r
( Fnet )t = −mg sinθ = mat
Substituting into the r-component,
( 20 N ) + ( 2.0 kg ) ( 9.8 m/s 2 ) cos30° = ( 2.0 kg )
vt2 ⇒ vt = 3.85 m/s ( 0.80 m )
The tangential velocity is 3.8 m/s. (b) Substituting into the t-component, − ( 9.8 m/s 2 ) sin 30° = at ⇒ at = −4.9 m/s 2 The radial acceleration is vt2 ( 3.85 m/s ) = = 18.5 m/s 2 0.80 m r 2
ar = Thus, the magnitude of the acceleration is a = ar2 + at2 =
(18.5 m/s ) + ( −4.9 m/s ) 2 2
2 2
The angle of the acceleration vector from the r-axis is
φ = tan −1 The angle is below the r-axis.
at 4.9 = tan −1 = 14.8° 18.5 ar
= 19.1 m/s 2
8.53.
Solve: (a) You are spinning a lead fishing weight in a horizontal 1.0 m diameter circle on the ice of a pond when the string breaks. You know that the test weight (breaking force) of the line is 60 N and that the lead weight has a mass of 0.30 kg. What was the weight’s angular velocity in rad/s and in rpm? 60 N rev 60 s ω2 = ⇒ ω = 20 rad/s × × = 191 rpm (b) 2π rad min ( 0.3 kg )( 0.5 m )
8.54.
Solve: (a) At what speed does a 1500 kg car going over a hill with a radius of 200 m have a weight of 11,760 N? (b) The weight is the normal force. 1500 kg v 2 2940 N = ⇒ v = 19.8 m/s 200 m
8.55.
Model: Visualize:
Assume the particle model and apply the constant-acceleration kinematic equations.
Solve:
(a) Newton’s second law for the projectile is G −F Fnet = − Fwind = max ⇒ ax = x m where Fwind is shortened to F. For the y-motion:
( )
y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) ⇒ 0 m = 0 m + ( v0 sin θ ) t1 − 12 gt12 ⇒ t1 = 0 s and t1 = 2
2v0 sin θ g
Using the above expression for t1 and defining the range as R we get from the x motion: x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 )
2
⎛ 2v sin θ ⎛ F⎞ ⇒ x1 − x0 = R = v0 xt1 + 12 ⎜ − ⎟ t12 = ( v0 cosθ ) ⎜ 0 g ⎝ m⎠ ⎝ =
⎞ F ⎛ 2v0 sin θ ⎞ ⎟− ⎜ ⎟ g ⎠ 2m ⎝ ⎠
2
2v02 2v 2 F cosθ sinθ − 0 2 sin 2 θ g mg
We will now maximize R as a function of θ by setting the derivative equal to 0: dR 2v02 2 Fv02 cos 2 θ − sin 2 θ ) − 2sinθ cosθ = 0 = ( dθ g mg 2
⎛ 2 Fv02 ⎞ ⎛ g ⎞ mg ⇒ cos 2 θ − sin 2 θ = cos 2θ = ⎜ sin 2θ ⇒ tan 2θ = 2 ⎟⎜ 2 ⎟ F ⎝ mg ⎠⎝ 2v0 ⎠ Thus the angle for maximum range is θ = 12 tan −1 ( mg / F ) (b) We have
2 mg ( 0.50 kg ) ( 9.8 m/s ) = = 8.167 ⇒ θ = 12 tan −1 ( 8.167 ) = 41.51° F 0.60 N The maximum range without air resistance is
R′ =
2v02 sin 45° cos 45° v02 = g g
Therefore, we can write the equation for the range R as
R = 2 R′ sin 41.51° cos 41.51° −
⇒
2F R′ sin 2 41.51° = R′ ( 0.9926 − 0.1076 ) = 0.885R′ mg
R R′ − R = 0.8850 ⇒ = 1 − 0.8850 = 0.115 R′ R′
Thus R is reduced from R′ by 11.5%. Assess: The condition for maximum range ( tan 2θ = mg / F ) means 2θ → 90° as F → 0. That is, θ = 45° when F = 0, as is to be expected.
8.56.
Visualize:
G ⎛1 ⎞ From Chapter 6 the drag on a projectile is D = ⎜ Av 2 , direction opposite to motion ⎟ , where A is the 4 ⎝ ⎠ cross- sectional area. Using the free-body diagram above, apply Newton’s second law to each direction. In the x-direction, 1 ( Fnet ) x D cosθ Av 2 cosθ ax = =− =−4 m n m 1 ( Fnet ) y Av 2 sin θ D sin θ − FG ay = =− =−4 −g m m m Solve:
Since v = vx2 + v y2 , vx = v cosθ , and v y = v sinθ , we can rewrite these as ax = −
Avx vx2 + v y2 A ( v cosθ ) v =− 4m 4m
ay = −
Av y vx2 + v y2 A ( v sin θ ) v − g = −g − 4m 4m
8.57.
Model: Visualize:
Solve:
Use the particle model and apply the constant-acceleration kinematic equations of motion.
(a) The thrust of the rocket burn must be such that vx decreases from 2.0 km/s to 0 during the same
interval that v y increases from 0 to 1.0 km/s. The forward distance in which to accomplish this is
x1 = (500 km)cos30° = 433 km. We can find the burn time by combining two kinematic equations: v1x = 0 = v0 x + axt1 ⇒ axt1 = −v0 x x1 = 0 + v0 xt1 + 12 axt12 = v0 xt1 + 12 (−v0 x )t1 = 12 v0 xt1 ⇒ t1 =
2 x1 2(433 km) = = 433 s 2.0 km/s v0 x
The required acceleration is ax = −v0 y /t1 = −(2000 m/s)/(433 s) = −4.62 m/s. During this interval, v y increases
from 0 to 1.0 km/s, thus v1 y = 0 + a yt1 ⇒ a y =
1000 m/s = 2.31 m/s 2 433 s
The thrust force is given by Newton’s second law: K G Fthrust = ma = (20,000 kg)(–4.62iˆ + 2.31 ˆj m/s 2 ) = (−92,400iˆ + 46,200 ˆj ) N The magnitude of the thrust force is 103,300 N and it is at angle θ = tan −1 (46,200/92,400) = 26.6° below the +xaxis (i.e., to the right and down). In order to point the thrusters in this direction, you must rotate the rocket counterclockwise 180° − θ = 153.4°. To make it home with one rocket burn, you need to rotate your rocket to 153.4° and fire with a thrust of 103,300 N for 433 s. (b) The x- and y-positions during the burn are x = (2000 m/s) t − 12 (4.62 m/s 2) t 2 y = 12 (2.31 m/s 2) t 2
The y-position at t1 = 433 s is 216,500 m = 216.5 km, 33.5 km away from the entrance. With the rocket off, you’re now coasting straight toward the entrance at 1.0 km/s and will cover this distance in 33.5 s. Thus you pass through the entrance at 433 s + 33.5 s = 466.5 s. The following table calculates your position at intervals of 50 s and, for accuracy, at t1 = 433 s when you end the burn. The trajectory is a parabola that intersects the x = 433 km line at y = 216.5 km, then a vertical line into the entrance.
8.58.
Model: Visualize:
Solve:
Use the particle model for the ball, which is in uniform circular motion.
From Newton’s second law along r and z directions,
∑F
r
= n cosθ =
mv 2 r
∑F
z
= n sin θ − mg = 0 ⇒ n sin θ = mg
Dividing the two force equations gives tan θ =
gr v2
From the geometry of the cone, tanθ = r y . Thus r gr = ⇒ v = gy y v2
8.59.
Model: Visualize:
Model the block as a particle and use the model of kinetic friction.
Solve: The only radial force is tension, so we can use Newton’s second law to find the angular velocity ω max at which the tube breaks:
∑ F = T = mω r ⇒ ω 2
r
max
=
Tmax 50 N = = 9.12 rad/s mr (0.50 kg)(1.2 m)
The compressed air and friction exert tangential forces, and the second law along the tangential direction is
∑F = F − f t
at =
t
k
= Ft − μ k n = Ft − μ k mg = mat
4.0 N Ft − μk g = − (0.60)(9.80 m/s 2 ) = 2.12 m/s 2 0.50 kg m
The time needed to accelerate to 9.12 rad/s is given by rω max (1.2)(9.12 rad/s) ⎛ at ⎞ = = 5.16 s ⎟ t1 ⇒ t1 = at 2.12 m/s 2 ⎝r⎠
ω1 = ω max = 0 + ⎜
During this interval, the block turns through angle
⎛ 2.12 m/s 2 ⎞ 1 rev ⎛a ⎞ 2 Δθ = θ1 − θ 0 = ω 0t1 + 12 ⎜ t ⎟ t12 = 0 + 12 ⎜ = 3.7 rev ⎟ (5.16 s) = 23.52 rad × π rad 1.2 m 2 ⎝r⎠ ⎝ ⎠
8.60.
Model: Visualize:
Solve:
Assume the particle model for a sphere in circular motion at constant speed.
(a) Newton’s second law along the r and z axes is:
mvt2 ∑ Fz = T1 cos30° + T 2 cos60° − FG = 0 N r Since we want T1 = T2 = T , these two equations become
∑ Fr = T1 sin 30° + T2 sin 60° =
T ( sin 30° + sin 60° ) =
mvt2 T ( cos30° + cos60° ) = mg r
Since sin 30° + sin 60° = cos30° + cos 60°, mvt2 ⇒ vt = rg r The triangle with sides L1 , L2 , and 1.0 m is isosceles, so L2 = 1.0 m and r = L2 cos30°. Thus mg =
L2 cos 30° g = (b) The tension is
T=
(1.0 m ) cos 30° g = ( 0.866 m ) ( 9.8 m/s2 ) = 2.9 m/s
( 2.0 kg ) ( 9.8 m/s mg = cos30° + cos60° 0.866 + 0.5
2
) = 14.3 N
8.61.
Model: Visualize:
Solve:
Use the particle model for a sphere revolving in a horizontal circle.
Newton’s second law in the r- and z-directions is
mvt2 ∑ ( F ) z = T1 sin 30° − T2 sin 30° − FG = 0 N r r Using r = (1.0 m ) cos30° = 0.886 m, these equations become
∑( F )
= T1 cos30° + T2 cos30° =
mvt2 ( 0.300 kg )( 7.5 m/s ) = 22.5 N = r cos30° ( 0.866 m )( 0.866 ) 2
T1 + T2 =
T1 − T2 =
( 0.300 kg ) ( 9.8 m/s 2 ) mg = = 5.88 N sin 30° ( 0.5)
Solving for T1 and T2 yields T1 = 14.2 N and T2 = 8.3 N.
8.62.
Model: Visualize:
Solve:
Use the particle model for the ball.
(a) Newton’s second law along the r- and z-directions is
∑F
r
= n cosθ = mrω 2
∑F
z
= n sin θ − FG = 0 N
Using FG = mg and dividing these equations yields:
tan θ =
g R− y = rω 2 r
where you can see from the figure that tanθ = ( R − y ) r . Thus ω =
g . R− y
(b) ω will be minimum when ( R − y) is maximum or when y = 0 m. Then ω min = g / R . (c) Substituting into the above expression,
ω=
g 9.8 m/s 2 rad 60 s 1 rev = = 9.9 × × = 95 rpm R−y 0.20 m − 0.10 m s 1 min 2π rad
8.63.
Model: Visualize:
Use the particle model for the airplane.
G Solve: In level flight, the lift force L balances the gravitational force. When turning, the plane banks so that the radial component of the lift force can create a centripetal acceleration. Newton’s second law along the r- and z-directions is
∑F
r
= L sin θ =
mvt2 r
∑F
z
= L cosθ − mg = 0 N
These can be written: sin θ =
mv 2 mg cosθ = L rL
Dividing the two equations gives: 2
miles 1 hr 1610 m ⎤ ⎡ ⎢⎣ 400 hour × 3600 s × 1 mile ⎥⎦ v2 v2 ⇒r= = = 18.5 km tanθ = ⎡⎣9.8 m s 2 ⎤⎦ tan10° gr g tan θ The diameter of the airplane’s path around the airport is 2 × 18.5 km = 37 km.
8.64.
Model: Visualize:
Use the particle model for a small volume of water on the surface.
Solve: Consider a particle of water of mass m at point C on the surface. Newton’s second law along the r- and zdirections is
( Fnet )r = n cosθ = mrω 2 ⇒ cosθ =
mrω 2 n
( Fnet ) z = n sinθ − mg = 0 N ⇒ sinθ =
mg n
Dividing both equations gives tanθ = g rω 2 . For a parabola z = ar 2 . This means dz 1 1 = 2ar = slope of the curve at C = tan φ = tan ( 90° − θ ) = ⇒ tan θ = dr tan θ 2ar Equating the two equations for tanθ , we get
1 g ω2 = 2 ⇒a= 2ar rω 2g Thus the surface is described by the equation z=
ω2 2g
r2
which is the equation of a parabola.
8-1
9.1. Model: Model the car and the baseball as particles. Solve:
(a) The momentum p = mv = (1500 kg )(10 m/s ) = 1.5 × 104 kg m/s.
(b) The momentum p = mv = ( 0.2 kg )( 40 m/s ) = 8.0 kg m/s.
9.2. Model: Model the bicycle and its rider as a particle. Also model the car as a particle. Solve:
From the definition of momentum,
pcar = pbicycle ⇒ mcar v car = mbicyclevbicycle ⇒ vbicycle =
⎛ 1500 kg ⎞ mcar v car = ⎜ ⎟ ( 5.0 m/s ) = 75 m/s mbicycle ⎝ 100 kg ⎠
Assess: This is a very high speed (≈ 168 mph). This problem shows the importance of mass in comparing two momenta.
9.3. Solve:
Visualize: Please refer to Figure EX9.3. The impulse J x is defined in Equation 9.6 as tf
J x = ∫ Fx ( t ) dt = area under the Fx (t ) curve between ti and tf ti
Jx =
1 ( 4 ms )(1000 N ) + ( 6 − 4 ms )(1000 N ) = 4 Ns 2
9.4. Model: The particle is subjected to an impulsive force. Visualize: Please refer to Figure EX9.4. Solve: Using Equation 9.6, the impulse is the area under the curve. From 0 s to 2 ms the impulse is
∫ Fdt = ( −500 N ) ( 2 × 10 s ) = −0.5 N s 1 2
−3
From 2 ms to 8 ms the impulse is
∫ Fdt = ( +2000 N )(8 ms − 2 ms ) = +6.0 N s 1 2
From 8 ms to 10 ms the impulse is
∫ Fdt = ( −500 N )(10 ms − 8 ms ) = −0.5 N s 1 2
Thus, from 0 s to 10 ms the impulse is ( −0.5 + 6.0 − 0.5 ) N s = 5.0 N s.
9.5. Visualize: Please refer to Figure EX9.5. Solve:
The impulse is defined in Equation 9.6 as tf
J x = ∫ Fx ( t ) dt = area under the Fx (t ) curve between ti and tf ti
⇒ 6.0 N s =
1 2
( Fmax )(8 ms ) ⇒ Fmax = 1.5 × 103 N
9.6. Model: Model the object as a particle and the interaction as a collision. Visualize: Please refer to Figure EX9.6. Solve: The object is initially moving to the right (positive momentum) and ends up moving to the left (negative momentum). Using the impulse-momentum theorem pfx = pix + J x ,
−2 kg m/s = +6 kg m/s + J x ⇒ J x = −8 kg m/s = −8 N s Since J x = Favg Δt , we have
Favg Δt = −8 N s ⇒ Favg = G The force is F = ( 8 × 102 N, left ) .
−8 N s = −8 × 102 N 10 ms
9.7. Model: Model the object as a particle and the interaction with the force as a collision. Visualize: Please refer to Figure EX9.7. Solve: Using the equations tf
pfx = pix + J x and J x = ∫ Fx (t ) dt = area under force curve ti
( 2.0 kg ) vfx = ( 2.0 kg )(1.0 m/s ) + (area under the force curve) ⇒ vfx = (1.0 m/s ) +
1 (1.0 s )( 2.0 N ) = 2.0 m/s 2.0 kg
Becaue vfx is positive, the object moves to the right at 2.0 m/s. Assess: For an object with positive velocity, a positive impulse increases the object’s speed. The opposite is true for an object with negative velocity.
9.8. Model: Model the object as a particle and the interaction with the force as a collision. Visualize: Please refer to Figure EX9.8. Solve: Using the equations tf
pfx = pix + J x and J x = ∫ Fx (t ) dt = area under force curve ti
( 2.0 kg ) vfx = ( 2.0 kg )(1.0 m/s ) + (area under the force curve) ⎛ 1 ⎞ ⇒ vfx = (1.0 m/s ) + ⎜ ⎟ ( −2.0 N )( 0.50 s ) = 0.50 m/s ⎝ 2.0 kg ⎠ Assess: For an object with positive velocity, a negative impulse slows the object. The opposite is true for an object with negative velocity.
9.9. Model: Use the particle model for the sled, the model of kinetic friction, and the impulse-momentum theorem. Visualize:
Note that the force of kinetic friction f k imparts a negative impulse to the sled. Solve:
Using Δpx = J x , we have tf
tf
ti
ti
pfx − pix = ∫ Fx (t ) dt = − f k ∫ dt = − f k Δt ⇒ mvfx − mvix = − μ k nΔt = − μ k mg Δt We have used the model of kinetic friction f k = μ k n, where μ k is the coefficient of kinetic friction and n is the normal (contact) force by the surface. The force of kinetic friction is independent of time and was therefore taken out of the impulse integral. Thus,
Δt =
1
μk g
( vix − vfx )
=
1
( 0.25) ( 9.8 m/s 2 )
( 8.0 m/s − 5.0 m/s ) = 1.22 s
9.10. Model: Use the particle model for the falling object and the impulse-momentum theorem. Visualize:
Note that the object is acted on by the gravitational force, whose magnitude is mg. Solve: Using the impulse-momentum theorem, tf
pfy − piy = J y = ∫ Fy (t ) dt ⇒ mvfy − mviy = −mg Δt ti
⇒ Δt = Assess:
viy − vfy g
=
−5.5 m/s − ( −10.4 m/s ) 9.8 m/s 2
= 0.50 s
Since Fy = − mg is independent of time, we have taken it out of the impulse integral.
9.11. Model: Model the tennis ball as a particle, and its interaction with the wall as a collision. Visualize:
The force increases to Fmax during the first two ms, stays at Fmax for two ms, and then decreases to zero during the last two ms. The graph shows that Fx is positive, so the force acts to the right. Solve: Using the impulse-momentum theorem pfx = pix + J x , 6 ms
( 0.06 kg )( 32 m/s ) = ( 0.06 kg )( −32 m/s ) + ∫
Fx ( t ) dt
0
The impulse is 6 ms
1 1 ∫ F ( t ) dx = area under force curve = 2 F ( 0.002 s ) + F ( 0.002 s ) + 2 F ( 0.002 s ) = ( 0.004 s ) F max
x
0
⇒ Fmax =
max
max
( 0.06 kg )( 32 m/s ) + ( 0.06 kg )( 32 m/s ) = 9.6 × 102 N 0.004 s
max
9.12. Model: Model the ball as a particle, and its interaction with the wall as a collision in the impulse approximation. Visualize: Please refer to Figure EX9.12. Solve: Using the equations tf
pfx = pix + J x and J x = ∫ Fx (t ) dt = area under force curve ti
( 0.250 kg ) vfx = ( 0.250 kg )( −10 m/s ) +
(500 N)(8.0 ms)
⎛ 4.0 N ⎞ ⇒ vfx = ( −10 m/s ) + ⎜ ⎟ = 6 m/s ⎝ 0.250 kg ⎠ Assess:
The ball’s final velocity is positive, indicating it has turned around.
9.13. Model: Model the glider cart as a particle, and its interaction with the spring as a collision. Visualize:
Solve:
Using the impulse-momentum theorem pfx − pix = ∫ Fdt ,
( 0.6 kg )( 3 m/s ) − ( 0.6 kg )( −3 m/s ) = area under force curve = 12 ( 36 N )( Δt ) ⇒ Δt = 0.20 s
9.14. Model: Visualize:
Solve:
Choose car + gravel to be the system. Ignore friction in the impulse approximation.
There are no external forces on the car + gravel system, so the horizontal momentum is conserved. This
means pfx = pix . Hence,
(10,000 kg + 4000 kg ) vfx = (10,000 kg )( 2.0 m/s ) + ( 4000 kg )( 0.0 m/s )
⇒ vfx = 1.43 m/s
9.15. Model: Visualize:
Choose car + rainwater to be the system.
There are no external horizontal forces on the car + water system, so the horizontal momentum is conserved. Solve: Conservation of momentum is pfx = pix . Hence,
( mcar + mwater )( 20 m/s ) = ( mcar )( 22 m/s ) + ( mwater )( 0 m/s ) ⇒ ( 5000 kg + mwater )( 20 m/s ) = ( 5000 kg )( 22 m/s ) ⇒ mwater = 5.0 × 102 kg
9.16. Model: Visualize:
Choose skydiver + glider to be the system in the impulse approximation.
Note that there are no external forces along the x-direction (ignoring friction in the impulse approximation), implying conservation of momentum along the x-direction. Solve: The momentum conservation equation pfx = pix is
( 680 kg − 60kg )( vG ) x + ( 60 kg )( vD ) x = ( 680 kg )( 30 m/s ) Immediately after release, the skydiver’s horizontal velocity is still ( vD ) x = 30 m/s. Thus
( 620 kg )( vG ) x + ( 60 kg )( 30 m/s ) = ( 680 kg )( 30 m/s ) ⇒ ( vG ) x = 30 m/s Assess:
The skydiver’s motion in the vertical direction has no influence on the glider’s horizontal motion.
9.17. Model: We will define our system to be bird + bug. This is the case of an inelastic collision because the bird and bug move together after the collision. Horizontal momentum is conserved because there are no external forces acting on the system during the collision in the impulse approximation. Visualize:
Solve:
The conservation of momentum equation pfx = pix is
( m1 + m2 ) vfx = m1 ( vix )1 + m2 ( vix )2
⇒ ( 300 g + 10 g ) vfx = ( 300 g )( 6.0 m/s ) + (10 g )( −30 m/s ) ⇒ vfx = 4.8 m/s
Assess: We left masses in grams, rather than convert to kilograms, because the mass units cancel out from both sides of the equation. Note that (vix ) 2 is negative.
9.18. Model: The two cars are not an isolated system because of external frictional forces. But during the collision friction is not going to be significant. Within the impulse approximation, the momentum of the Cadillac + Volkswagen system will be conserved in the collision. Visualize:
Solve:
The momentum conservation equation pfx = pix is
( mC + mVW ) vfx = mC ( vix )C + mVW ( vix )VW ⇒ 0 kg mph = ( 2000 kg )(1.0 mph ) + (1000 kg )( vix )VW ⇒ ( vix )VW = −2.0 mph You need a speed of 2.0 mph.
9.19. Model: Because of external friction and drag forces, the car and the blob of sticky clay are not exactly an isolated system. But during the collision, friction and drag are not going to be significant. The momentum of the system will be conserved in the collision, within the impulse approximation. Visualize:
Solve:
The conservation of momentum equation pfx = pix is
( mC + mB )( vf ) x = mB ( vix )B + mC ( vix )C ⇒ 0 kg m/s = (10 kg )( vix )B + (1500 kg )( −2.0 m/s ) ⇒ (vix ) B = 3.0 × 102 m/s
Assess: This speed of the blob is around 600 mph, which is very large. However, we must point out that a very large speed is expected in order to stop a car with only 10 kg of clay.
9.20. Model: We will define our system to be archer + arrow. The force of the archer (A) on the arrow (a) is equal to the force of the arrow on the archer. These are internal forces within the system. The archer G is Gstanding on frictionless ice, and the normal force by ice on the system balances the weight force. Thus Fext = 0 on the system, and momentum is conserved. Visualize:
The initial momentum pix of the system is zero, because the archer and the arrow are at rest. The final moment pfx must also be zero. Solve:
We have M A vA + ma va = 0 kg m/s. Therefore, vA =
−ma va − ( 0.100 kg )(100 m/s ) = = − 0.20 m/s 50 kg mA
The archer’s recoil speed is 0.20 m/s. Assess: It is the total final momentum that is zero, although the individual momenta are nonzero. Since the arrow has forward momentum, the archer will have backward momentum.
9.21. Model: We will define our system to be Bob + rock. Bob’s (B) force on the rock (R) is equal to the rock’s force on Bob. These are internal forces within the Bob is standing on frictionless ice, and the G G system. normal force by ice on the system balances the weight. Fext = 0 on the system, and thus momentum is conserved. Visualize:
The initial momentum pix of the system is zero because Bob and the rock are at rest. Thus pfx = 0 kg m/s. Solve: We have mBvB + mR vR = 0 kg m/s. Hence,
vB = −
⎛ 0.500 kg ⎞ mR vR = − ⎜ ⎟ ( 30 m/s ) = −0.20 m/s mB ⎝ 75 kg ⎠
Bob’s recoil speed is 0.20 m/s. Assess: Since the rock has forward momentum, Bob’s momentum is backward. This makes the total momentum zero.
9.22. Model: We will define our system to be Dan + skateboard, and their interaction as an explosion. While friction is present between the skateboard and the ground, it is negligible in the impulse approximation. Visualize:
The system has nonzero initial momentum pix . As Dan (D) jumps backward off the gliding skateboard (S), the skateboard will move forward in such a way that the final total momentum of the system pfx is equal to pix . Solve: We have mS ( vfx )S + mD ( vfx )D = ( mS + mD ) vix . Hence,
( 5.0 kg )( 8.0 m/s ) + ( 50 kg )( vfx )D = ( 5.0 kg + 50 kg )( 4.0 m/s ) ⇒ ( vfx )D = 3.6 m/s
9.23. Model: We assume that the momentum is conserved in the collision. Visualize: Please refer to Figure EX9.23. Solve: The conservation of momentum equation yields
( pfx )1 + ( pfx )2 = ( pix )1 + ( pix )2 ⇒ ( pfx )1 + 0 kg m/s = 2 kg m/s − 4 kg m/s ⇒ ( pfx )1 = −2 kg m/s
(p ) +(p ) =(p ) +(p ) fy 1
fy 2
iy 1
iy 2
⇒ ( pfy ) − 1 kg m/s = 2 kg m/s + 1 kg m/s ⇒ ( pfy ) = 4 kg m/s 1
Thus, the final momentum of particle 1 is ( −2iˆ + 4 ˆj ) kg m/s.
1
9.24. Model: This problem deals with the conservation of momentum in two dimensions in an inelastic collision. Visualize:
Solve:
G G The conservation of momentum equation pbefore = pafter is m1 ( vix )1 + m2 ( vix )2 = ( m1 + m2 ) vfx m1 ( viy ) + m2 ( viy ) = ( m1 + m2 ) vfy 1
2
Substituting in the given values,
(.02 kg )( 3.0 m/s ) + 0 kg m/s = (.02 kg + .03 kg ) vf cosθ 0 kg m/s + (.03 kg )( 2.0 m/s ) = (.02 kg + .03 kg ) vf sin θ
⇒ vf cosθ = 1.2 m/s vf sinθ = 1.2 m/s ⇒ vf =
(1.2 m/s )
2
+ (1.2 m/s ) = 1.7 m/s θ = tan −1
The ball of clay moves 45° north of east at 1.7 m/s.
2
vy vx
= tan −1 (1) = 45°
9.25. Model: Model the ball as a particle. We will also use constant-acceleration kinematic equations.
Solve:
(a) The momentum just after throwing is
p0 x = p0 cos30° = mv0 cos30° = ( 0.050 kg )( 25 m/s ) cos30° = 1.083 kg m/s p0 y = p0 sin 30° = mv0 sin 30° = ( 0.050 kg )( 25 m/s ) sin 30° = 0.625 kg m/s The momentum just after throwing is (1.08, 0.63) kg m/s. The momentum at the top is p1x = mv1x = mv0 x = 1.08 kg m/s p1 y = mv1 y = 0 kg m/s Just before hitting the ground, the momentum is p2 x = mv2 x = mv0 x = 1.08 kg m/s p2 y = mv2 y = − mv0 y = −0.63 kg m/s (b) px is constant because no forces act on the ball in the x-direction. Mathematically,
Fx =
dpx = 0 N ⇒ px = constant dt
(c) The change in the y-component of the momentum during the ball’s flight is
Δp y = −0.625 kg m/s − 0.625 kg m/s = −1.250 kg m/s From kinematics, the time to reach the top is obtained as follows:
v1 y = v0 y + a y ( t1 − t0 ) ⇒ t1 =
−v0 y ay
=
− (12.5 m/s ) −9.8 m/s 2
= 1.276 s
The time of flight is thus Δt = 2t1 = 2 (1.276 s ) = 2.552 s. Multiplying this time by –mg, the y-component of the weight, yields –1.25 kg m/s. This follows from the impulse-momentum theorem: Δp y = − mg Δt
9.26. Model: Model the rocket as a particle, and use the impulse-momentum theorem. The only force acting on the rocket is due to its own thrust. Visualize: Please refer to Figure P9.26. Solve: (a) The impulse is
J x = ∫ Fx dt = area of the Fx (t ) graph between t = 0 s and t = 30 s =
1 2
(1000 N )( 30 s ) = 1.5 × 104 N s
(b) From the impulse-momentum theorem, pfx = pix + 1.5 × 104 N s. That is, the momentum or velocity increases as
long as J x increases. When J x increases no more, the speed will be a maximum. This happens at t = 30 s. At this time, mvfx = mvix + 1.5 × 104 N s ⇒ ( 425 kg ) vfx = ( 425 kg )( 75 m/s ) + 1.5 × 10 4 N s ⇒ vfx = 110 m/s
9.27.
Solve: Using the equation
area under the force curve m 2.0 s 1 2π t ⎞ = 0 m/s + (10 N ) sin ⎛⎜ ⎟ dt 0.250 kg ∫0 ⎝ 4.0 s ⎠
v fx = vix +
2.0 s ⎞ ⎛ 4.0 s ⎞ ⎛ ⎛ 2π t ⎞ ⎜ ⎟ = ( 40 N/m ) ⎜ − cos ⎟⎜ ⎜ ⎟ ⎝ 2π ⎠ ⎝ ⎝ 4.0 s ⎠ 0 ⎟⎠
⎛ 80 ⎞ = −⎜ m/s ⎟ ( cos (π ) − cos ( 0 ) ) ⎝π ⎠ = 25 m/s ⎛ 2π t ⎞ The force is applied for half the period of 4.0 s. During that time, sin ⎜ ⎟ is positive, so an object ⎝ 4.0 s ⎠ initially at rest acquires a positive velocity. Assess:
9.28. Model: Let the system be ball + racket. During the collision of the ball and racket, momentum is conserved because all external interactions are insignificantly small. Visualize:
Solve:
(a) The conservation of momentum equation pfx = pix is
mR ( vfx )R + mB ( vfx )B = mR ( vix )R + mB ( vix )B
(1.000 kg )( vfx )R + ( 0.060 kg )( 40 m/s ) = (1.000 kg )(10 m/s ) + ( 0.060 kg )( −20 m/s ) ⇒ ( vfx )R = 6.4 m/s (b) The impulse on the ball is calculated from ( pfx ) B = ( pix ) B + J x as follows:
( 0.060 kg )( 40 m/s ) = ( 0.060 kg )( −20 m/s ) + J x ⇒ Favg =
⇒ J x = 3.6 N s = ∫ Fdt = Favg Δt
3.6 Ns = 3.6 × 102 N 10 ms
Let us now compare this force with the gravitational ball ( FG )B = mB g = ( 0.060 kg ) ( 9.8 m/s 2 ) = 0.588 N. Thus, Favg = 612( FG ) B .
force
on
the
Assess: This is a significant force and is reasonable because the impulse due to this force changes the direction as well as the speed of the ball from approximately 45 mph to 90 mph.
9.29. Model: Model the ball as a particle that is subjected to an impulse when it is in contact with the floor. We will also use constant-acceleration kinematic equations. Ignore any forces other than the interaction between the floor and the ball during the collision in the impulse approximation. Visualize:
Solve:
To find the ball’s velocity just before and after it hits the floor:
v12y = v02y + 2a y ( y1 − y0 ) = 0 m 2 /s 2 + 2 ( −9.8 m/s 2 ) ( 0 − 2.0 m ) ⇒ v1 y = −6.261 m/s v32y = v22 y + 2a y ( y3 − y2 ) ⇒ 0 m 2 /s 2 = v22 y + 2 ( −9.8 m/s 2 ) (1.5 m − 0 m ) ⇒ v2 y = 5.422 m/s The force exerted by the floor on the ball can be found from the impulse-momentum theorem:
mv2 y = mv1 y + ∫ Fdt = mv1 y + area under the force curve
⇒ ( 0.200 kg )( 5.422 m/s ) = − ( 0.200 kg )( 6.261 m/s ) + 12 Fmax ( 5.0 × 10−3 s )
⇒ Fmax = 9.3 × 102 N Assess:
A maximum force of 9.3 × 102 N exerted by the floor is reasonable.
9.30. Model: Model the rubber ball as a particle that is subjected to an impulsive force when it comes in contact with the floor. We will also use constant-acceleration kinematic equations and the impulse-momentum theorem. Visualize:
Solve: (a) To find the magnitude and direction of the impulse that the floor exerts on the ball, we use the impulse-momentum theorem: pfy = piy + J y ⇒ J y = p2 y − p1 y = m ( v2 y − v1 y ) Let us now find v1y and v2 y by using kinematics. For the falling and rebounding ball,
v12y = v02y + 2a y ( y1 − y0 ) = 0 m 2 /s 2 + 2 ( −9.8 m/s 2 ) ( 0 m − 1.8 m ) ⇒ v1 y = −5.940 m/s v32y = v22 y + 2a y ( y3 − y2 ) ⇒ 0 m 2 /s 2 = v22 y + 2 ( −9.8 m/s 2 ) (1.2 m − 0 m ) ⇒ v2 y = 4.850 m/s Going back to the impulse-momentum equation, we find J y = ( 0.040 kg ) ⎡⎣ 4.850 m/s − ( −5.940 m/s ) ⎤⎦ = 0.432 N s
The impulse is 0.43 N s upward. (b) As the ball compresses, the force of contact increases and the ball slows to v y = 0 m/s. Then in
decompression the ball is accelerated upward. To a good approximation, the force due to the floor as a function of time is shown in the figure. (c) For a rubber ball, Δt is likely in the range 5 to 10 ms. For 10 ms, J y = Favg Δt = 0.432 N s ⇒ Favg ≈ 40 N
The force is ≈80 N if Δt = 5 ms. Altogether, 40 to 80 N is a reasonable estimate.
9.31. Model: Model the cart as a particle rolling down a frictionless ramp. The cart is subjected to an impulsive force when it comes in contact with a rubber block at the bottom of the ramp. We will use the impulsemomentum theorem and the constant-acceleration kinematic equations. Visualize:
Solve:
From the free-body diagram on the cart, Newton’s second law “before the collision” is
∑( F )
x
= FG sin θ = max ⇒ ax =
mg sin θ g = g sin 30° = m 2
Using this acceleration, we can find the cart’s speed just before its contact with the rubber block: v12x = v02x + 2ax ( x1 − x0 ) = 0 m 2 /s 2 + 2 ( 12 g )(1.0 m − 0 m) ⇒ v1x = 3.13 m/s
Now we can use the impulse-momentum theorem to obtain the velocity just after the collision:
mv2 x = mv1x + ∫ Fx dt = mv1x + area under the force graph ⇒ ( 0.500 kg ) v2 x = ( 0.500 kg )( 3.13 m/s ) − 12 ( 200 N ) ( 26.7 × 10−3 s ) ⇒ v2 x = −2.21 m/s Note that the given force graph is positive, but in this coordinate system the impulse of the force is to the left up the slope. That is the reason to put a minus sign while evaluating the ∫ Fx dt integral. We can once again use a kinematic equation to find how far the cart will roll back up the ramp: v32x = v22x + 2ax ( x3 − x2 ) ⇒ ( 0 m/s ) = ( −2.21 m/s ) + 2 ( − 12 g ) ( x3 − x2 ) ⇒ ( x3 − x2 ) = 0.50 m 2
2
9.32.
Solve: Using Newton’s second law for the x–direction, Fx =
Fx =
dpx . Therefore, dt
d ( 6t 2 kg m/s ) = 12t N dt
Assess: The x-component of the net force on an object is equal to the time rate of change of the x-component of the object’s momentum.
9.33.
Visualize:
Solve:
Using Newton’s second law for the y-direction and the chain rule,
( Fnet ) y =
dp y dt
=
⎛ dv ⎞ d dm mv y ) = vy ) + m ⎜ y ⎟ ( ( dt dt ⎝ dt ⎠
= ( −0.50 kg/s )(120 m/s ) + ( 48 kg ) (18 m/s 2 ) = 8.0 × 102 N Assess:
Since the rocket is losing mass,
dm < 0. The time derivative of the velocity is the acceleration. dt
9.34. Model: Model the train cars as particles. Since the train cars stick together, we are dealing with perfectly inelastic collisions. Momentum is conserved in the collisions of this problem in the impulse approximation, in which we ignore external forces during the time of the collision. Visualize:
Solve:
In the collision between the three-car train and the single car:
mv1x + ( 3m ) v2 x = 4mv3 x ⇒ v1x + 3v2 x = 4v3 x ⇒ ( 4.0 m/s ) + 3( 2.0 m/s ) = 4v3 x ⇒ v3 x = 2.5 m/s In the collision between the four-car train and the stationary car:
( 4m ) v3 x + mv4 x = ( 5m ) v5 x
⇒ 4v3 x + 0 m/s = 5v5 x ⇒ v5 x =
4v3 x = ( 0.8 )( 2.5 m/s ) = 2.0 m/s 5
9.35.
Model: The dart and cork are particles in free fall until they have a perfectly inelastic head-on collision. Ignore air friction. Visualize:
Solve:
The positions of the dart and cork just before the collision are 1 ( y1 )D = ( y0 )D + ( v0 )D t1 − gt12 2 1 2 ( y1 )C = ( y0 )C − gt1 2 In the particle model, the dart and cork have no physical size, so ( y1 )D = ( y1 )C . Hence
1 1 1 9.8 m/s 2 ) t12 = 3 m − ( 9.8 m/s 2 ) t12 ⇒ t1 = s ( 2 2 3 At this time, the velocities of the dart and cork are 1 ( v1 )D = ( v0 )D − gt1 = 9.0 m/s − ( 9.8 m/s 2 ) ⎛⎜ s ⎞⎟ = 5.73 m/s ⎝3 ⎠ 1 ( v1 )C = ( v0 )C − gt1 = − ( 9.8 m/s 2 ) ⎜⎛ s ⎟⎞ = −3.27 m/s ⎝3 ⎠ These are the initial velocities to use with momentum conservation in a perfectly inelastic collision. pfy = piy 0 m + ( 9.0 m/s ) t1 −
( mD + mC ) v = mD ( v1 )D + mC ( v1 )C Thus v=
( 0.030 kg )( 5.73 m/s ) + ( 0.020 kg )( −3.27 m/s ) = 2.13 m/s, up ( 0.030 kg + 0.20 kg )
Assess: The heavier, faster upward-going dart has more momentum than the falling cork, so the total momentum is upwards.
9.36. Model: Model the earth (E) and the asteroid (A) as particles. Earth + asteroid is our system. Since the two stick together during the collision, this is a case of a perfectly inelastic collision. Momentum is conserved in the collision since no significant external force acts on the system. Visualize:
Solve:
(a) The conservation of momentum equation pfx = pix is
mA ( vix )A + mE ( vix )E = ( mA + mE ) vfx ⇒ (1.0 × 1013 kg )( 4.0 × 104 m s ) + 0 kg m/s = (1.0 × 1013 kg + 5.98 × 1024 kg ) vfx ⇒ vfx = 6.7 × 10−8 m/s (b) The speed of the earth going around the sun is
vE =
11 2π r 2π (1.50 × 10 m ) = = 3.0 × 104 m/s T 3.15 × 107 s
Hence, vfx vE = 2 × 10−12 = 2 × 10−10%. Assess: The earth’s recoil speed is insignificant compared to its orbital speed because of its large mass.
9.37. Model: Model the skaters as particles. The two skaters, one traveling north (N) and the other traveling west (W), are a system. Since the two skaters hold together after the “collision,” this is a case of a perfectly inelastic collision in two dimensions. Momentum is conserved since no significant external force in the x-y plane acts on the system in the “collision.” Visualize:
Solve:
(a) The x-component of the conservation of momentum is
( mN + mW ) vfx = mN ( vix ) N + mW ( vix )W
⇒ ( 75 kg + 60 kg ) vfx = 0 kg m/s + ( 60 kg )( −3.5 m/s ) ⇒ vfx = −1.556 m/s
The y-component of the conservation of momentum is
( mN + mW ) vfy = mN ( viy )N + mW ( viy )W
⇒ ( 75 kg + 60 kg ) vfy = ( 75 kg )( 2.5 m/s ) + 0 kg m/s
⇒ vfy = 1.389 m/s ⇒ vf =
( vfx )
2
+ ( vfy ) = 2.085 m/s 2
The time to glide to the edge of the rink is radius of the rink 25 m = = 12.0 s vf 2.085 m/s (b) The location is θ = tan −1 vfy vfx = 42° north of west. Assess:
A time of 12.0 s in covering a distance of 25 m at a speed of ≈ 2 m/s is reasonable.
9.38. Model: This problem deals with a case that is the opposite of a collision. The two ice skaters, heavier and lighter, will be modeled as particles. The skaters (or particles) move apart after pushing off against each other. During the “explosion,” the total momentum of the system is conserved. Visualize:
Solve:
The initial momentum is zero. Thus the conservation of momentum equation pfx = pix is
mH ( vfx )H + mL ( vfx )L = 0 kg m/s ⇒ ( 75 kg )( vfx )H + ( 50 kg )( vfx )L = 0 kg m/s Using the observation that the heavier skater takes 20 s to cover a distance of 30 m, we find ( vfx )H = 30 m 20 s = 1.5 m/s. Thus,
( 75 kg )(1.5 m/s ) + ( 50 kg )( vfx )L = 0 kg m/s
⇒ ( vfx )L = −2.25 m/s
Thus, the time for the lighter skater to reach the edge is 30 m 30 m = = 13.3 s v 2.25 m/s ( fx )L Assess: Conservation of momentum leads to a higher speed for the lighter skater, and hence a shorter time to reach the edge of the ice rink.
9.39. Model: This problem deals with a case that is the opposite of a collision. Our system is comprised of three coconut pieces that are modeled as particles. During the blow up or “explosion,” the total momentum of the system is conserved in the x-direction and the y-direction. Visualize:
Solve:
The initial momentum is zero. From pfx = pix , we get
+ m1 ( vfx )1 + m3 ( vf )3 cosθ = 0 kg m/s ⇒ ( vf )3 cosθ =
−m1 ( vfx )1 m3
=
−m ( −20 m/s ) = 10 m/s 2m
=
− m ( −20 m/s ) = 10 m/s 2m
From pfx = pix , we get
+ m2 ( vfy ) + m3 ( vf )3 sin θ = 0 kg m/s ⇒ ( vf )3 sin θ = 2
⇒ ( vf )3 =
(10 m/s )
The velocity is 14.1 m/s at 45° east of north.
2
− m2 ( vfy ) m3
2
+ (10 m/s ) = 14.1 m/s θ = tan −1 (1) = 45° 2
9.40. Model: The billiard balls will be modeled as particles. The two balls, m1 (moving east) and m2 (moving west), together are our system. This is an isolated system because any frictional force during the brief collision period is going to be insignificant. Within the impulse approximation, the momentum of our system will be conserved in the collision. Visualize:
Note that m1 = m2 = m. Solve:
The equation pfx = pix yields:
m1 ( vfx )1 + m2 ( vfx )2 = m1 ( vix )1 + m2 ( vix )2 ⇒ m1 ( vf )1 cosθ + 0 kg m/s = m1 ( vix )1 + m2 ( vix )2 ⇒ ( vf )1 cosθ = ( vix )1 + ( vix )2 = 2.0 m/s − 1.0 m/s = 1.0 m/s The equation pfy = piy yields:
+ m1 ( vfy ) sin θ + m2 ( vfy ) = 0 kg m/s ⇒ ( vf )1 sinθ = − ( vfy ) = −1.41 m/s 1
2
⇒ ( vf )1 =
2
(1.0 m/s )
2
+ ( −1.41 m/s ) = 1.73 m/s 2
⎛ 1.41 m/s ⎞ ⎟ = 55° ⎝ 1.0 m/s ⎠
θ = tan −1 ⎜ The angle is below +x axis, or south of east.
9.41. Model: This is a two-part problem. First, we have an inelastic collision between the wood block and the bullet. The bullet and the wood block are an isolated system. Since any external force acting during the collision is not going to be significant (the impulse approximation), the momentum of the system will be conserved. The second part involves the dynamics of the block + bullet sliding on the wood table. We treat the block and the bullet as particles. Visualize:
Solve:
The equation pfx = pix gives
( mB + mW ) vfx = mB ( vix )B + mW ( vix )W ⇒ ( 0.010 kg + 10 kg ) vfx = ( 0.010 kg )( vix )B + (10 kg )( 0 m/s ) ⇒ vfx =
1 ( vix )B 1001
From the model of kinetic friction, f k = − μ k n = − μ k ( mB + mW ) g = ( mB + mW ) ax ⇒ ax = − μ k g Using the kinematic equation v12x = v22x + 2ax ( x1 − x0 ) , 2
2 ⎛ 1 ⎞ v12x = v22x − 2 μ k g ( x1 − x0 ) ⇒ 0 m 2 s 2 = vf2x − 2μ k g x1 ⇒ ⎜ ⎟ ( vix )B = 2 μ k g x1 ⎝ 1001 ⎠
⇒ ( vix )B = 1001 2μ k g x1 = 1001 2 ( 0.20 ) ( 9.8 m/s 2 ) ( 0.050 m ) = 443 m/s The bullet’s speed is 4.4 × 102 m/s. Assess: The bullet’s speed is reasonable (≈ 900 mph).
9.42. Model: This is a two-part problem. First, we have an inelastic collision between Fred (F) and Brutus (B). Fred and Brutus are an isolated system. The momentum of the system during collision is conserved since no significant external force acts on the system. The second part involves the dynamics of the Fred + Brutus system sliding on the ground. Visualize:
Note that the collision is head-on and therefore one-dimensional. Solve: The equation pfx = pix is
( mF + mB ) vfx = mF ( vix )F + mB ( vix )B ⇒ ( 60 kg + 120 kg ) vfx = ( 60 kg )( −6.0 m/s ) + (120 kg )( 4.0 m/s ) ⇒ vfx = 0.667 m/s The positive value indicates that the motion is in the direction of Brutus. The model of kinetic friction yields: f k = − μ k n = − μ k ( mF + mB ) g = ( mF + mB ) ax ⇒ ax = − μ k g Using the kinematic equation v12x = v02x + 2ax ( x1 − x0 ) , we get v12x = v02x − 2 μ k g x1 ⇒ 0 m 2 s 2 = vf2x − 2 ( 0.30 ) ( 9.8 m/s 2 ) x1 ⇒ 0 m 2 s 2 = ( 0.667 m/s ) − ( 5.9 m/s 2 ) x1 ⇒ x1 = 7.6 cm 2
They slide 7.6 cm in the direction Brutus was running. Assess: After the collision, Fred and Brutus slide with a small speed but with a good amount of kinetic friction. A stopping distance of 7.6 cm is reasonable.
9.43. Model: Model the package and the rocket as particles. This is a two-part problem. First we have an inelastic collision between the rocket (R) and the package (P). During the collision, momentum is conserved since no significant external force acts on the rocket and the package. However, as soon as the package + rocket system leaves the cliff they become a projectile motion problem. Visualize:
Solve: The minimum velocity after collision that the package + rocket must have to reach the explorer is v0 x , which can be found as follows:
y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) ⇒ −200 m = 0 m + 0 m + 12 ( −9.8 m/s 2 ) t12 ⇒ t1 = 6.389 s 2
With this time, we can now find v0 x using x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) . We obtain 2
30 m = 0 m + v0 x ( 6.389 s ) + 0 m ⇒ v0 x = 4.696 m/s = vfx We now use the momentum conservation equation pfx = pix which can be written
( mR + mP ) vfx = mR ( vix )R + mP ( vix )P ⇒ (1.0 kg + 5.0 kg )( 4.696 m/s ) = (1.0 kg )( vix )R + ( 5.0 kg )( 0 m/s ) ⇒ ( vix )R = 28 m/s
9.44.
Model: Visualize:
Solve:
The clay balls undergo a perfectly inelastic collision in the impulse approximation.
(a) The total momentum in reference frame S is P = ( px )1 + ( px )2 = m1 ( vx )1 + m2 ( vx )2
= ( 0.020 kg )(12 m/s ) + 0 kg m/s
= 0.24 kg m/s (b) In the S′ frame, the total momentum is zero. P′ = ( px )1′ + ( px )2′ = 0 ⇒ m1 ( vx )1′ = − m2 ( vx )2′ Equation 4.24 relates velocities in the two reference frames. ( vx )1′ = ( vx )1 − Vx
( vx )2′ = ( vx )2 − Vx Thus
m1 ( ( vx )1 − Vx ) = − m2 ( ( vx )2 − Vx ) ⇒ Vx =
m1 ( vx )1 + m2 ( vx )2 m1 + m2
= 4.0 m/s
(c) Since the total momentum is zero, the final velocity of the resulting 60 g clay ball must be ( vx )f ′ = 0 m/s. (d) Using Equation 4.24 again,
( vx )f = ( vx )f ′ + Vx = 0 m/s + 4.0 m/s = 4.0 m/s Assess: The final velocity of both balls in the S frame is less than the original velocity of the smaller ball, and is in the same direction. The S′ frame is called the “center of mass frame.” Some collisions are more easily studied in the center of mass frame, such as high energy particle collisions at particle accelerator facilities.
9.45.
Model: This is a two-part problem. First, we have an explosion that creates two particles. The momentum of the system, comprised of two fragments, is conserved in the explosion. Second, we will use kinematic equations and the model of kinetic friction to find the displacement of the lighter fragment. Visualize:
The initial momentum is zero. Using momentum conservation pfx = pix during the explosion,
Solve:
mH ( v1x )H + mL ( v1x )L = mH ( v0 x )H + mL ( v0 x )L ⇒ 7m ( v1x )H + m ( v1x )L = 0 kg m/s ⇒ ( v1x )H = − ( 17 ) ( v1x )L Because mH slides to x2H = −8.2 m before stopping, we have f k = μ k nH = μ k wH = μ k mH g = mH aH ⇒ aH = μ k g
Using kinematics,
( v2 x )H = ( v1x )H + 2aH ( x2H − x1H ) 2
2
⇒ 0 m 2 s 2 = ( 17 ) ( v1x )L + 2μ k g ( −8.2 m − 0 m ) ⇒ ( v1x )L = −88.74 μ k m/s 2
2
How far does mL slide? Using the information obtained above in the following kinematic equation,
( v2 x )L = ( v1x )L + 2aL ( x2L − x1L ) ⇒ 0 2
Assess:
2
m 2 s 2 = μ k ( 88.74 ) − 2μ k gx2L ⇒ x2L = 4.0 × 102 m 2
Note that aH is positive, but aL is negative, and both are equal in magnitude to μ k g . Also, x2H is
negative but x2L is positive.
9.46. Model: We will model the two fragments of the rocket after the explosion as particles. We assume the explosion separates the two parts in a vertical manner. This is a three-part problem. In the first part, we will use kinematic equations to find the vertical position where the rocket breaks into two pieces. In the second part, we will apply conservation of momentum to the system (that is, the two fragments) in the explosion. In the third part, we will again use kinematic equations to find the velocity of the heavier fragment just after the explosion. Visualize:
Solve:
The rocket accelerates for 2.0 s from rest, so
v1 y = v0 y + a y ( t1 − t0 ) = 0 m/s + (10 m/s 2 ) ( 2 s − 0 s ) = 20 m/s
y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) = 0 m + 0 m + 12 (10 m/s 2 ) ( 2 s ) = 20 m 2
2
At the explosion the equation pfy = piy is
mL ( v2 y ) + mH ( v2 y ) = ( mL + mH ) v1 y ⇒ ( 500 kg ) ( v2 y ) + (1000 kg ) ( v2 y ) = (1500 kg )( 20 m/s ) L
H
L
H
To find (v2 y ) H we must first find (v2 y ) L , the velocity after the explosion of the upper section. Using kinematics,
(v ) = (v ) 2
2
3y L
2y L
+ 2 ( −9.8 m/s 2 ) ( y3L − y2L ) ⇒ ( v2 y ) = 2 ( 9.8 m/s 2 ) ( 530 m − 20 m ) = 99.98 m/s L
Now, going back to the momentum conservation equation we get
( 500 kg )( 99.98 m/s ) + (1000 kg ) ( v2 y )H = (1500 kg )( 20 m/s ) ⇒ ( v2 y )H = −20 m/s The negative sign indicates downward motion.
9.47. Model: Let the system be bullet + target. No external horizontal forces act on this system, so the horizontal momentum is conserved. Model the bullet and the target as particles. Since the target is much more massive than the bullet, it is reasonable to assume that the target undergoes no significant motion during the brief interval in which the bullet passes through. Visualize:
Solve: (a) By assuming that the target has negligible motion during the interval in which the bullet passes through, the time is that needed to slow from 1200 m/s to 900 m/s in a distance of 30 cm. We’ll use kinematics to first find the acceleration, then the time.
(v1x ) B2 = (v0 x ) B2 + 2ax Δx ⇒ ax =
⇒ Δt =
(v1x ) B2 − (v0 x ) B2 (900 m/s)2 − (1200 m/s)2 = = −1.05 × 106 m/s 2 2Δx 2(0.30 m) (v1x ) B = (v0 x ) B + ax Δt (v1x ) B − (v0 x ) B 900 m/s − 1200 m/s = = 2.86 × 10−4 s = 286 μ s ax −1.05 × 106 m/s 2
The average force on the bullet is Favg = m ax = 26 kN. (b) Now we can use the conservation of momentum equation p1x = p0 x to find
mT (v1x )T + mB (v1x ) B = mT (v0 x )T + mB (v0 x ) B = 0 + mB (v0 x ) B ⇒ (v1x )T =
mB (v0 x ) B − mB (v1x ) B 0.025 kg = ( (1200 m/s) − (900 m/s) ) = 0.021 m/s mT 350 kg
9.48. Model: Model the two blocks (A and B) and the bullet (L) as particles. This is a two-part problem. First, we have a collision between the bullet and the first block (A). Momentum is conserved since no external force acts on the system (bullet + block A). The second part of the problem involves a perfectly inelastic collision between the bullet and block B. Momentum is again conserved for this system (bullet + block B). Visualize:
Solve:
For the first collision the equation pfx = pix is
mL ( v1x )L + mA ( v1x )A = mL ( v0 x )L + mA ( v0 x )A ⇒ ( 0.010 kg )( v1x )L + ( 0.500 kg )( 6.0 m/s ) = ( 0.010 kg )( 400 m/s ) + 0 kg m/s ⇒ ( v1x )L = 100 m/s The bullet emerges from the first block at 100 m/s. For the second collision the equation pfx = pix is
( mL + mB ) v2 x = mL ( v1x )L ⇒ ( 0.010 kg + 0.500 kg ) v2 x = ( 0.010 kg )(100 m/s ) ⇒ v2 x = 1.96 m/s
9.49. Model: Model Brian (B) along with his wooden skis as a particle. The “collision” between Brian and
Ashley lasts for a short time, and during this time no significant external forces act on the Brian + Ashley system. Within the impulse approximation, we can then assume momentum conservation for our system. After finding the velocity of the system immediately after the collision, we will apply constant-acceleration kinematic equations and the model of kinetic friction to find the final speed at the bottom of the slope. Visualize:
Solve:
Brian skiing down for 100 m:
( v1x )B = ( v0 x )B + 2ax ( x1B − x0B ) 2
2
= 0 m 2 s 2 + 2ax (100 m − 0 m ) ⇒ ( v1x )B =
( 200 m ) ax
To obtain ax , we apply Newton’s second law to Brian in the x and y directions as follows:
∑( F )
on B x
= wB sinθ − f k = mBax
∑( F )
on B y
= n − wB cosθ = 0 N ⇒ n = w cosθ
From the model of kinetic friction, f k = μ k n = μ k wB cosθ . The x-equation thus becomes wB sin θ − μ k wB cosθ = mB ax
⇒ ax = g ( sin θ − μ k cosθ ) = ( 9.8 m/s 2 ) ⎡⎣sin 20° − ( 0.060 ) cos 20°⎤⎦ = 2.80 m/s 2 Using this value of ax ,
( v1x )B = ( 200 m ) ( 2.80 m/s 2 ) = 23.7 m/s.
In the collision with Ashley the conservation
of momentum equation pfx = pix is
( mB + mA ) v2 x = mB ( v1x )B ⇒ v2 x =
mB 80 kg ( v1x )B = ( 23.66 m/s ) = 14.56 m/s mB + mA 80 kg + 50 kg
Brian + Ashley skiing down the slope:
v32x = v22x + 2ax ( x3 − x2 ) = (14.56 m/s ) + 2 ( 2.80 m/s 2 ) (100 m ) ⇒ v3 x = 27.8 m/s 2
That is, Brian + Ashley arrive at the bottom of the slope with a speed of 27.8 m/s. Note that we have used the same value of ax in the first and the last parts of this problem. This is because ax is independent of mass. Assess:
A speed of approximately 60 mph on a ski slope of 200 m length and 20° slope is reasonable.
9.50. Model: Model the spy plane (P) and the rocket (R) as particles. The plane and the rocket undergo a
perfectly inelastic collision. During the brief collision time, momentum for the plane + rocket system will be conserved because no significant external forces act on this system during the collision. After finding the velocity of the system immediately after the collision, we will apply projectile equations to find where the system will hit the ground. Visualize:
Solve:
For the collision we have pfx = pix and pfy = piy . This means the magnitude of the final momentum is
pf = ⇒ vf =
( pfx )
2
+ ( pfy ) = 2
1
( pix )
2
+ ( piy ) ⇒ ( mR + mP ) vf = ⎡⎣ mR ( vix )R ⎤⎦ + ⎡ mP ( viy )P ⎤ ⎣ ⎦ 2
2
⎡⎣(1280 kg )( 725 m/s ) ⎤⎦ + ⎡⎣( 575 kg )( 450 m/s ) ⎤⎦ = 519.4 m/s 2
(1280 kg + 575 kg )
2
2
⎛ ( 575 kg )( 450 m/s ) ⎞ ⎛ pfy ⎞ −1 ⎟⎟ = 15.58° north of east ⎟ = tan ⎜⎜ ⎝ pfx ⎠ ⎝ (1280 kg )( 725 m/s ) ⎠
θ = tan −1 ⎜
We call this direction the x′ axis in the x-y plane. Let us now look at the falling plane + rocket system in the x′-z plane where z is the vertical axis perpendicular to the x-y plane. We have z0 = 2700 m
z1 = 0 m
v0 z = 0 m/s
x0′ = 0 m
t0 = 0 s
v0 x′ = 519.4 m/s
Using the kinematic equation z1 = z0 + v0 z ( t1 − t0 ) + 12 az ( t1 − t0 ) , we can find the time to fall: 2
0 m = 2700 m + 0 m + 12 ( −9.8 m/s 2 ) t12 ⇒ t1 = 23.47 s In this time t1 , the wreckage travels horizontally to
x1′ = x0′ + v0 x′ ( t1 − t0 ) = 0 m + ( 519.4 m/s )( 23.47 s ) = 12,192 m = 12.19 km The enmeshed plane + rocket system lands 12.19 km from the collision point at an angle of 15.58° north of east.
9.51. Model: This is an isolated system, so momentum is conserved in the explosion. Momentum is a vector
G quantity, so the direction of the initial velocity vector v1 establishes the direction of the momentum vector. The
final momentum vector, after the explosion, must still point in the +x-direction. The two known pieces continue to move along this line and have no y-components of momentum. The missing third piece cannot have a ycomponent of momentum if momentum is to be conserved, so it must move along the x-axis—either straight forward or straight backward. We can use conservation laws to find out. Visualize:
Solve:
From the conservation of mass, the mass of piece 3 is
m3 = mtotal − m1 − m2 = 7.0 × 105 kg To conserve momentum along the x-axis, we require
[ pi = mtotalvi ] = [ pf
= p1f + p2f + p3f = m1v1f + m2v2f + p3f ]
⇒ p3f = mtotalvi − m1v1f − m2v2f = +1.02 × 1013 kg m/s Because p3f > 0, the third piece moves in the +x-direction, that is, straight forward. Because we know the mass m3 , we can find the velocity of the third piece as follows: v3f =
p3f 1.02 × 1013 kg m/s = = 1.46 × 107 m/s m3 7.0 × 105 kg
9.52.
Model: The two railcars make up a system. The impulse approximation is used while the spring is expanding, so friction can be ignored. The spring is massless. Visualize:
Solve: holds.
Since the cars are at rest initially, the total momentum of the system is zero. Conservation of momentum
0 = m1 ( vfx )1 + m2 ( vfx )2
We are only told that the relative velocity of the two cars after the spring expands is 4.0 m/s, so
( vfx )2 − ( vfx )1 = 4.0 m/s The positive-going car 2 velocity is written first to ensure that the relative velocity is positive. Substitute ( vfx )2 = ( vfx )1 + 4.0 m/s into the conservation of momentum equation, then solve for ( vfx )1 .
0 = m1 ( vfx )1 + m2 ( ( vfx )1 + 4.0 m/s ) ⇒ ( vfx )1 = −
Assess:
m2 ( 4.0 m/s ) ( 90 tons )( 4.0 m/s ) = −3.0 m/s =− ( m1 + m2 ) ( 30 tons + 90 tons )
The other more massive railcar has a velocity ( vfx )2 = ( vfx )1 + 4.0 m/s = 1.0 m/s. A slower speed for the
more massive car makes sense.
9.53. Model: This is a three-part problem. In the first part, the shell, treated as a particle, is launched as a projectile and reaches its highest point. We will use constant-acceleration kinematic equations for this part. The shell, which is our system, then explodes at the highest point. During this brief explosion time, momentum is conserved. In the third part, we will again use the kinematic equations to find the horizontal distance between the landing of the lighter fragment and the origin. Visualize:
Solve:
The initial velocity is
v0 x = v cosθ = (125 m/s ) cos55° = 71.7 m/s v0 y = v sinθ = (125 m/s ) sin 55° = 102.4 m/s
At the highest point, v1 y = 0 m/s and v1x = 71.7 m/s. The conservation of momentum equation pfx = pix is mL ( v1x )L + mH ( v1x )H = ( mL + mH ) v1x The heavier particle falls straight down, so (v1x ) H = 0 m/s. Thus,
(15 kg )( v1x )L + 0 kg m s = (15 kg + 60 kg )( 71.7 m/s ) ⇒ ( v1x )L = 358 m/s That is, the velocity of the smaller fragment immediately after the explosion is 358 m/s and this velocity is in the horizontal x-direction. Note that (v1 y ) L = 0 m/s. To find x2 , we will first find the displacement x1 − x0 and then x2 − x1. For x1 − x0 ,
v1 y = v0 y + a y ( t1 − t0 ) ⇒ 0 m s = (102.4 m/s ) + ( −9.8 m/s 2 ) ( t1 − 0 s ) ⇒ t1 = 10.45 s x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) ⇒ x1 − x0 = ( 71.7 m/s )(10.45 s ) + 0 m = 749 m 2
For x2 − x1 : x2 = x1 + ( v1x ) L ( t2 − t1 ) + 12 ax ( t2 − t1 ) ⇒ x2 − x1 = ( 358 m/s )(10.45 s ) + 0 m=3741 m 2
⇒ x2 = ( x2 − x1 ) + ( x1 − x0 ) = 3741 m + 749 m = 4490 m = 4.5 km Assess: Note that the time of ascent to the highest point is equal to the time of descent to the ground, that is, t1 − t0 = t2 − t1.
9.54. Model: Model the proton (P) and the gold atom (G) as particles. The two constitute our system, and momentum is conserved in the collision between the proton and the gold atom. Visualize:
Solve:
The conservation of momentum equation pfx = pix is
mG ( vfx )G + mP ( vfx )P = mP ( vix )P + mG ( vix )G ⇒ (197 u )( vfx )G + (1 u ) ( −0.90 × 5.0 × 107 m/s ) = (1 u ) ( 5.0 × 107 m/s ) + 0 u m/s ⇒ ( vfx )G = 4.8 × 105 m/s
9.55. Model: Model the proton (P) and the target nucleus (T) as particles. The proton and the target nucleus make our system and in the collision between them momentum is conserved. This is due to the impulse approximation because the collision lasts a very short time and the external forces acting on the system during this time are not significant. Visualize:
Solve:
The conservation of momentum equation pfx = pix is
mT ( vfx )T + mP ( v fx ) = mT ( vix )T + mP ( vix )P P
⇒ mT ( 3.12 × 10 m/s ) + (1 u ) ( −0.75 × 2.5 × 106 m/s ) = 0 u m/s + (1 u ) ( 2.5 × 106 m/s ) 5
⇒ mT = 14.0 u Assess:
This is the mass of the nucleus of a nitrogen atom.
9.56. Model: This problem deals with a case that is the opposite of a collision. It is a case of an “explosion” in which a 214 Po nucleus (P) decays into an alpha-particle (A) and a daughter nucleus (N). During the “explosion” or decay, the total momentum of the system is conserved. Visualize:
Solve: Conservation of mass requires the daughter nucleus to have mass mN = 214 u − 4 u = 210 u. The conservation of momentum equation pfx = pix is
mN ( vfx ) N + mA ( vfx )A = ( mN + mA ) vP ⇒ ( 210 u )( vfx ) N + ( 4 u ) ( −1.92 × 107 m / s ) = 0 u m/s ⇒ ( vfx ) N = 3.66 × 105 m/s
9.57. Model: The neutron’s decay is an “explosion” of the neutron into several pieces. The neutron is an isolated system, so its momentum should be conserved. The observed decay products, the electron and proton, move in opposite directions. Visualize:
Solve:
(a) The initial momentum is pix = 0 kg m/s. The final momentum pfx = meve + mpvp is
2.73 × 10−23 kg m/s − 1.67 × 10−22 kg m/s = −1.40 × 10−22 kg m/s No, momentum does not seem to be conserved. (b) and (c) If the neutrino is needed to conserve momentum, then pe + pP + pneutrino = 0 kg m/s. This requires pneutrino = − ( pe + pP ) = +1.40 × 10−22 kg m/s The neutrino must “carry away” 1.40 × 10−22 kg m/s of momentum in the same direction as the electron.
9.58. Model: Model the two balls of clay as particles. Our system comprises these two balls. Momentum is conserved in the perfectly inelastic collision. Visualize:
Solve:
The x-component of the final momentum is
pfx = pix = m1 ( vix )1 + m2 ( vix )2 = ( 0.020 kg )( 2.0 m/s ) − ( 0.030 kg )(1.0 m/s ) cos30° = 0.0140 kg m/s The y-component of the final momentum is
pfy = piy = m1 ( viy ) + m2 ( viy ) 1
2
= ( 0.02 kg )( 0 m/s ) − ( 0.03 kg )(1.0 m/s ) sin 30° = −0.0150 kg m/s
⇒ pf =
( 0.014 kg m/s )
2
+ ( −0.015 kg m/s ) = 0.0205 kg m/s 2
Since pf = ( m1 + m2 ) vf = 0.0205 kg m/s, the final speed is
vf =
0.0205 kg m/s = 0.41 m/s ( 0.02 + 0.03) kg
and the direction is
θ = tan −1
pfy pfx
= tan −1
0.015 = 47° south of east 0.014
9.59. Model: Model the three balls of clay as particle 1 (moving north), particle 2 (moving west), and particle 3 (moving southeast). The three stick together during their collision, which is perfectly inelastic. The momentum of the system is conserved. Visualize:
Solve:
The three initial momenta are G G pi1 = m1vi1 = ( 0.020 kg )( 2.0 m/s ) ˆj = 0.040ˆj kg m/s
(
)
G G pi2 = m2vi2 = ( 0.030 kg ) −3.0 m/s iˆ = −0.090iˆ kg m/s
(
)
G G pi3 = m3vi3 = ( 0.040 kg ) ⎡⎣( 4.0 m/s ) cos 45°iˆ − ( 4.0 m/s ) sin 45° ˆj ⎤⎦ = 0.113iˆ − 0.113 ˆj kg m/s G G G G Since pf = pi = pi1 + pi2 + pi3 , we have G
( m1 + m2 + m3 ) vf
(
)
(
)
G = 0.023iˆ − 0.073 ˆj kg m/s ⇒ vf = 0.256iˆ − 0.811 ˆj m/s
⇒ vf =
( 0.256 m/s )
θ = tan −1
vfy vfx
2
+ ( −0.811 m/s ) = 0.85 m/s
= tan −1
2
0.811 = 72° below +x 0.256
9.60. Model: Model the truck (T) and the two cars (C and C′) as particles. The three forming our system stick together during their collision, which is perfectly inelastic. Since no significant external forces act on the system during the brief collision time, the momentum of the system is conserved. Visualize:
Solve:
The three momenta are G G piT = mT viT = ( 2100 kg )( 2.0 m/s ) iˆ = 4200iˆ kg m/s G G piC = mCviC = (1200 kg )( 5.0 m/s ) ˆj = 6000ˆj kg m/s G G piC′ = mC′viC′ = (1500 kg )(10 m/s ) iˆ = 15,000iˆ kg m/s
(
)
G G G G G pf = pi = piT + piC + piC′ = 19,200iˆ + 6000ˆj kg m/s ⇒ pf = ( mT + mC + mC′ ) vf = ⇒ vf = 4.2 m/s θ = tan −1
(19,200 kg m/s ) py px
= tan −1
2
+ ( 6000 kg m/s )
2
6000 = 17.4° above +x 19,200
Assess: A speed of 4.2 m/s for the entangled three vehicles is reasonable since the individual speeds of the cars and the truck before entanglement were of the same order of magnitude.
9.61. Model: The 14 C atom undergoes an “explosion” and decays into a nucleus, an electron, and a neutrino. Momentum is conserved in the process of “explosion” or decay. Visualize:
Solve:
G G The conservation of momentum equation pf = pi = 0 kg m/s is G G G G G G G G pe + pn + pN = 0 N ⇒ pN = − ( pe + pn ) = − meve − mn vn
(
)
= − ( 9.11 × 10−31 kg )( 5.0 × 107 m/s ) iˆ − ( 8.0 × 10−24 kg m/s ) ˆj = − 45.55 × 10−24iˆ + 8.0 × 10−24 ˆj kg m/s ⇒ pN = mN vN =
( 45.55 × 10 ) + (8.0 × 10 ) −24 2
−24
2
kg m/s
⇒ ( 2.34 × 10−26 kg ) vN = 4.62 × 10−23 kg m/s ⇒ vN = 1.97 × 103 m/s
9.62. (a) A 100 g ball traveling to the left at 30 m/s is batted back to the right at 40 m/s. The force curve for the force of the bat on the ball can be modeled as a triangle with a maximum force of 1400 N. How long is the ball in contact with the bat? (b)
(c) The solution is Δt = 0.100 s = 10 ms.
9.63. (a) A 200 g ball of clay traveling to the right overtakes and collides with a 400 g ball of clay traveling to the right at 3.0 m/s. The balls stick and move forward at 4.0 m/s. What was the speed of the 200 g ball of clay? (b)
(c) The solution is ( vix )2 = 6.0 m/s.
9.64. (a) A 2000 kg auto traveling east at 5.0 m/s suffers a head-on collision with a small 1000 kg auto traveling west at 4.0 m/s. They lock bumpers and stick together after the collision. What will be the speed and direction of the combined wreckage after the collision? (b)
(c) The solution is vfx = 2.0 m/s along +x direction.
9.65. (a) A 150 g spring-loaded toy is sliding across a frictionless floor at 1.0 m/s. It suddenly explodes into two pieces. One piece, which has twice the mass of the second piece, continues to slide in the forward direction at 7.5 m/s. What is the speed and direction of the second piece? (b)
(c) The solution is (vfx )1 = −12 m/s. The minus sign tells us that the second piece moves backward at 12 m/s.
9.66. Model: Visualize:
Solve:
The cart + man (C + M) is our system. It is an isolated system, and momentum is conserved.
The conservation of momentum equation pfx = pix is
mM ( vfx )M + mC ( vfx )C = mM ( vix )M + mC ( vix )C Note that (vfx ) M and (vfx )C are the final velocities of the man and the cart relative to the ground. What is given in this problem is the velocity of the man relative to the moving cart. The man’s velocity relative to the ground is
( vfx )M = ( vfx )C − 10 m/s With this form for (vfx ) M, we rewrite the momentum conservation equation as
mM ⎡⎣( vfx )C − 10 m/s ⎤⎦ + mC ( vfx )C = mM ( 5.0 m/s ) + mC ( 5.0 m/s )
⇒ ( 70 kg ) ⎡⎣( vfx )C − 10 m/s ⎤⎦ + (1000 kg )( vfx )C = (1000 kg + 70 kg )( 5.0 m/s ) ⇒ ( vfx )C [1000 kg + 70 kg ] = (1070 kg )( 5.0 m/s ) + ( 70 kg )(10 m/s ) ⇒ ( vfx )C = 5.7 m/s
9.67. Model: Model Ann and cart as particles. The initial momentum is pi = 0 kg m/s in a coordinate system attached to the ground. As Ann begins running to the right, the cart will have to recoil to the left to conserve momentum. Visualize:
Solve: The difficulty with this problem is that we are given Ann’s velocity of 5.0 m/s relative to the cart. If the cart is also moving with velocity vcart then Ann’s velocity relative to the ground is not 5.0 m/s. Using the Galilean transformation equation for velocity, Ann’s velocity relative to the ground is
( vfx )Ann = ( vfx )cart + 5.0 m/s Now, the momentum conservation equation pix = pfx is
0 kg m/s = mAnn ( vfx )Ann + mcart ( vfx )cart ⇒ 0 kg m/s = ( 50 kg ) ⎡⎣( vfx )cart + 5.0 m/s ⎤⎦ + ( 500 kg )( vfx )cart ⇒ ( vfx )cart = −0.45 m/s Using the recoil velocity (vfx )cart relative to the ground, we find Ann’s velocity relative to the ground to be (vfx ) Ann = 5.00 m/s − 0.45 m/s = 4.55 m/s The distance Ann runs relative to the ground is Δx = (vfx ) Ann Δt , where Δt is the time it takes to reach the end of the cart. Relative to the cart, which is 15 m long, Ann’s velocity is 5 m/s. Thus Δt = (15 m ) ( 5 m/s ) = 3.00 s. Her
distance over the ground during this interval is
Δx = ( vfx )Ann Δt = ( 4.55 m/s )( 3.00 s ) = 13.6 m
9.68. Model: Visualize:
Solve:
The projectile + wood ball are our system. In the collision, momentum is conserved.
The momentum conservation equation pfx = pix is
( mP + mB ) vfx = mP ( vix )P + mB ( vix )B
⇒ (1.0 kg + 20 kg ) vfx = (1.0 kg )( vix )P + 0 kg m/s ⇒ ( vix )P = 21vfx
We therefore need to determine vfx . Newton’s second law for circular motion is
T − FG = T − ( mP + mB ) g =
( mP + mB ) vf2x r
Using Tmax = 400 N, this equation gives
400 N − (1.0 kg + 20 kg )( 9.8 m/s ) =
(1.0 kg + 20 kg ) vf2x 2.0 m
⇒ ( vfx )max = 4.3 m/s
Going back to the momentum conservation equation,
( vix )P = 21vfx = ( 21)( 4.3 m/s ) = 90 m/s That is, the largest speed this projectile can have without causing the cable to break is 90 m/s.
9.69. Model: This is an “explosion” problem and momentum is conserved. The two-stage rocket is our system. Visualize:
Solve:
Relative to the ground, the conservation of momentum equation pfx = pix is
m1 ( vfx )1 + m2 ( vfx )2 = ( m1 + m2 ) v1x ⇒ 3 m 2 ( vfx )1 + m2 ( vfx )2 = ( 4 m 2 )(1200 m/s ) ⇒ 3( vfx )1 + ( vfx )2 = 4800 kg m/s The fact that the first stage is pushed backward at 35 m/s relative to the second can be written
( vfx )1 = −35 m/s + ( vfx )2 Substituting this form of (vfx )1 in the conservation of momentum equation, 3 ⎡⎣ −35 m/s + ( vfx )2 ⎤⎦ + ( vfx )2 = 4800 kg m/s ⇒ ( vfx )2 = 1226 m/s
9.70. Model: Model the bullet and the vehicle as particles, and use the impulse-momentum theorem to find the impulse provided to the vehicle by bullet(s). Because the final speed of the vehicle is small compared to the bullet speed, and the mass of the bullet is so much smaller than the mass of the vehicle, we will assume that each bullet exerts the same impulse on the vehicle. Visualize:
Solve: For one bullet, the impulse-momentum theorem Δpx = J x allows us to find that the impulse exerted on the bullet by the vehicle’s sail is
( J x ) B = mB (Δv) B = (0.020 kg) ( ( −200 m/s) − (400 m/s) ) = −12.0 kg m/s From Newton’s third law, the impulse that the bullet exerts on the vehicle is ( J x ) V = −( J x ) B = 12.0 kg m/s. A second use of the impulse momentum theorem allows us to find the vehicle’s increase in velocity due to one bullet: (Δv) V =
( J x ) V 12.0 kg m/s = = 0.120 m/s mV 100 kg
This increase in velocity is independent of the vehicle’s speed, so as long as the impulse per bullet stays essentially constant (which it does because the bullet speeds are so much larger than the vehicle speed), the number of impacts needed to increase the vehicle speed to 12 m/s is N bullets =
12 m/s = 100 bullets 0.12 m/s per bullet
To reach this speed in 20 s requires the bullets to be fired at a rate of 5 bullets per second.
9.71. Model: Let the system be rocket + bullet. This is an isolated system, so momentum is conserved. Visualize: The fact that the bullet’s velocity relative to the rocket is 139,000 can be written (vf ) B = (vf ) R + 139,000 m/s.
Solve:
Consider the firing of one bullet when the rocket has mass M and velocity vi . The conservation of
momentum equation pf = pi is
( M − 5kg)vf + (5 kg)(vf + 139,000 m/s) = Mvi ⇒ Δv = vf − vi = −
5 kg (139,000 m/s) M
The rocket starts with mass M = 2000 kg, which is much larger than 5 kg. If only a few bullets are needed, M will not change significantly as the rocket slows. If we assume that M remains constant at 2000 kg, the loss of speed per bullet is Δv = −347.5 m/s = −1250 km/h. Thus exactly 8 bullets will reduce the speed by 10,000 km/h, from 25,000 km/h to 15,000 km/h. If you’re not sure that treating M as a constant is valid, you can calculate Δv for each bullet and reduce M by 5 kg after each shot. The loss of mass causes Δv to increase slightly for each bullet. An eight-step calculation then finds that 8 bullets will slow the rocket to 14,900 km/h. Seven bullets wouldn’t be enough, and 9 would slow the rocket far too much.
9.72.
Visualize:
Solve: Ladies and gentlemen of the jury, how far would the chair slide if it was struck with a bullet from my client’s gun? We know the bullet’s velocity as it leaves the gun is 450 m/s. The bullet travels only a small distance to the chair, so we will neglect any speed loss due to air resistance. The bullet and chair can be considered an isolated system during the brief interval of the collision. The bullet embedded itself in the chair, so this was a perfectly inelastic collision. Momentum conservation allows us to calculate the velocity of the chair immediately after the collision as follows:
pix = pfx ⇒ mB ( vi )B = ( mB + mC ) vf ⇒ vf =
mB ( vi )B mB + mC
=
( 450 m/s )( 0.010 kg ) = 0.225 m/s 20.01 kg
This is the velocity immediately after the collision when the chair starts to slide but before it covers any distance. For the purpose of the problem in dynamics, call this the initial velocity v0 . The free-body diagram of the chair shows three forces. Newton’s second law applied to the chair (with the embedded bullet) is
ax = a =
( Fnet ) x mtot
=
( Fnet ) y n − mtot g − fk μn = = − k a y = 0 m/s 2 = mtot mtot mtot mtot
where we’ve used the friction model in the x-equation. The y-equation yields n = mtot g , and the x-equation yields
a = − μ k g = −1.96 m/s 2 . We know the coefficient of kinetic friction because it is a wood chair sliding on a wood floor. Finally, we have to determine the stopping distance of the chair. The motion of the chair ends with v1 = 0 m/s after sliding a distance Δx, so
( 0.225 m/s ) = 0.013 m = 1.3 cm v02 =− 2a 2 ( −1.96 m/s 2 ) 2
v12 = 0 m 2 /s 2 = v02 + 2aΔx ⇒ Δx = −
If the bullet lost any speed in the air before hitting the chair, the sliding distance would be even less. So you can see that the most the chair could slide if it had been struck by a bullet from my client’s gun, would be 1.3 cm. But in actuality, the chair slid 3 cm, more than twice as far. The murder weapon, ladies and gentlemen, was a much more powerful gun than the one possessed by my client. I rest my case.
9-1
10.1. Model: We will use the particle model for the bullet (B) and the running student (S). Visualize:
Solve:
For the bullet,
1 1 K B = mBvB2 = (0.010 kg)(500 m/s) 2 = 1250 J 2 2 For the running student, 1 1 KS = mSvS2 = (75 kg)(5.5 m/s) 2 = 206 J 2 2 Thus, the bullet has the larger kinetic energy. Assess: Kinetic energy depends not only on mass but also on the square of the velocity. The above calculation shows this dependence. Although the mass of the bullet is 7500 times smaller than the mass of the student, its speed is more than 90 times larger.
10.2. Model: Model the hiker as a particle. Visualize:
The origin of the coordinate system chosen for this problem is at sea level so that the hiker’s position in Death Valley is y0 = −8.5 m. Solve: The hiker’s change in potential energy from the bottom of Death Valley to the top of Mt. Whitney is
ΔU = U gf − U gi = mgyf − mgyi = mg ( yf − yi ) = (65 kg)(9.8 m/s 2 )[4420 m − (−85 m)] = 2.9 × 106 J Assess:
Note that ΔU is independent of the origin of the coordinate system.
10.3. Model: Model the compact car (C) and the truck (T) as particles. Visualize:
Solve:
For the kinetic energy of the compact car and the kinetic energy of the truck to be equal,
1 1 20,000 kg mT (25 km/hr) = 112 km/hr K C = K T ⇒ mCvC2 = mT vT2 ⇒ vC = vT = 2 2 1000 kg mC Assess:
A smaller mass needs a greater velocity for its kinetic energy to be the same as that of a larger mass.
10.4.
Model: Model the car (C) as a particle. This is an example of free fall, and therefore the sum of kinetic and potential energy does not change as the car falls. Visualize:
Solve:
(a) The kinetic energy of the car is
KC =
1 1 mCvC2 = (1500 kg)(30 m/s) 2 = 6.75 × 105 J 2 2
The car’s kinetic energy is 6.8 × 105 J. (b) Let us relabel K C as K f and place our coordinate system at yf = 0 m so that the car’s potential energy U gf
is zero, its velocity is vf , and its kinetic energy is K f . At position yi , vi = 0 m/s or K i = 0 J, and the only energy the car has is U gi = mgyi . Since the sum K + U g is unchanged by motion, K f + U gf = K i + U gi . This
means K f + mgyf = K i + mgyi ⇒ K f + 0 = K i + mgyi ⇒ yi =
( K f − Ki ) (6.75 × 105 J − 0 J) = = 46 m mg (1500 kg)(9.8 m / s 2 )
(c) From part (b),
1 2 1 2 2 2 ( K f − K i ) 2 mvf − 2 mvi ( vf − vi ) yi = = = mg mg 2g Free fall does not depend upon the mass.
10.5.
Model: This is a case of free fall, so the sum of the kinetic and gravitational potential energy does not change as the ball rises and falls. Visualize:
The figure shows a ball’s before-and-after pictorial representation for the three situations in parts (a), (b) and (c). Solve: The quantity K + U g is the same during free fall: K f + U gf = Ki + U gi . We have (a)
1 2 1 mv1 + mgy1 = mv02 + mgy0 2 2 ⇒ y1 = ( v02 − v12 ) 2 g = [(10 m/s) 2 − (0 m/s) 2 ]/(2 × 9.8 m / s 2 ) = 5.10 m
5.1 m is therefore the maximum height of the ball above the window. This is 25.1 m above the ground. 1 2 1 (b) mv2 + mgy2 = mv02 + mgy0 2 2 Since y2 = y0 = 0, we get for the magnitudes v2 = v0 = 10 m / s. (c)
1 2 1 mv3 + mgy3 = mv02 + mgy0 ⇒ v32 + 2 gy3 = v02 + 2 gy0 ⇒ v32 = v02 + 2 g ( y0 − y3 ) 2 2 ⇒ v32 = (10 m / s) 2 + 2(9.8 m / s 2 )[0 m − (−20 m)] = 492 m 2 / s 2
This means the magnitude of v3 is equal to 22 m/s. Assess: Note that the ball’s speed as it passes the window on its way down is the same as the speed with which it was tossed up, but in the opposite direction.
10.6. Model: This is a problem of free fall. The sum of the kinetic and gravitational potential energy for the ball, considered as a particle, does not change during its motion. Visualize:
The figure shows the ball’s before-and-after pictorial representation for the two situations described in parts (a) and (b). Solve: The quantity K + U g is the same during free fall. Thus, K f + U gf = Ki + U gi . 1 2 1 mv1 + mgy1 = mv02 + mgy0 ⇒ v02 = v12 + 2 g ( y1 − y0 ) (a) 2 2 ⇒ v02 = (0 m/s) 2 + 2(9.8 m/s 2 )(10 m − 1.5 m) = 166.6 m 2 /s 2 ⇒ v0 = 12.9 m/s (b)
1 2 1 mv2 + mgy2 = mv02 + mgy0 ⇒ v22 = v02 + 2 g ( y0 − y2 ) 2 2 2 2 ⇒ v2 = 166.6 m /s 2 + 2(9.8 m/s 2 )(1.5 m − 0 m) ⇒ v2 = 14.0 m/s
Assess: An increase in speed from 12.9 m/s to 14.0 m/s as the ball falls through a distance of 1.5 m is reasonable. Also, note that mass does not appear in the calculations that involve free fall.
10.7. Model: Model the oxygen and the helium atoms as particles. Visualize: We denote the oxygen and helium atoms by O and He, respectively. Note that the oxygen atom is four times heavier than the helium atom, so mO = 4 mHe . Solve:
The energy conservation equation K O = K He is
1 1 v 2 2 mOvO2 = mHevHe ⇒ (4 mHe )vO2 = mHevHe ⇒ He = 2.0 2 2 vO Assess: The result vHe = 2vO , combined with the fact that mHe = 14 mO , is a consequence of the way kinetic energy is defined: It is directly proportional to the mass and to the square of the speed.
10.8. Model: Model the ball as a particle undergoing rolling motion with zero rolling friction. The sum of the ball’s kinetic and gravitational potential energy, therefore, does not change during the rolling motion. Visualize:
Solve:
Since the quantity K + U g does not change during rolling motion, the energy conservation equations
apply. For the linear segment the energy conservation equation K 0 + U g0 = K1 + U g1 is 1 2 1 1 1 1 mv1 + mgy1 = mv02 + mgy0 ⇒ m v12 + mg (0 m) = m(0 m/s) 2 + mgy0 ⇒ mv12 = mgy0 2 2 2 2 2
For the parabolic part of the track, K1 + U g1 = K 2 + U g 2 is 1 2 1 1 1 1 mv1 + mgy1 = mv22 + mgy2 ⇒ mv12 + mg (0 m) = m(0 m / s) 2 + mgy2 ⇒ mv12 = mgy2 2 2 2 2 2
Since from the linear segment we have 12 m v12 = mgy0 , we get mgy0 = mgy2 or y2 = y0 = 1.0 m. Thus, the ball rolls up to exactly the same height as it started from. Assess: Note that this result is independent of the shape of the path followed by the ball, provided there is no rolling friction. This result is an important consequence of energy conservation.
10.9. Model: Model the skateboarder as a particle. Assuming that the track offers no rolling friction, the sum of the skateboarder’s kinetic and gravitational potential energy does not change during his rolling motion. Visualize:
The vertical displacement of the skateboarder is equal to the radius of the track. Solve: The quantity K + U g is the same at the upper edge of the quarter-pipe track as it was at the bottom. The energy conservation equation K f + U gf = K i + U gi is
1 2 1 mvf + mgyf = mvi2 + mgyi ⇒ vi2 = vf2 + 2 g ( yf − yi ) 2 2 vi2 = (0 m/s) 2 + 2(9.8 m/s 2 )(3.0 m − 0 m) = 58.8 m 2 /s 2 ⇒ vi = 7.7 m/s Assess:
Note that we did not need to know the skateboarder’s mass, as is the case with free-fall motion.
10.10. Model: Model the puck as a particle. Since the ramp is frictionless, the sum of the puck’s kinetic and gravitational potential energy does not change during its sliding motion. Visualize:
Solve:
The quantity K + U g is the same at the top of the ramp as it was at the bottom. The energy conservation
equation K f + U gf = K i + U gi is
1 2 1 mvf + mgyf = mvi2 + mgyi ⇒ vi2 = vf2 + 2 g ( yf − yi ) 2 2 ⇒ vi2 = (0 m/s) 2 + 2(9.8 m/s 2 )(1.03 m − 0 m) = 20.2 m 2 /s 2 ⇒ vi = 4.5 m/s Assess: An initial push with a speed of 4.5 m/s ≈ 10 mph to cover a distance of 3.0 m up a 20° ramp seems reasonable.
10.11.
Model: In the absence of frictional and air-drag effects, the sum of the kinetic and gravitational potential energy does not change as the pendulum swings from one side to the other. Visualize:
The figure shows the pendulum’s before-and-after pictorial representation for the two situations described in parts (a) and (b). Solve: (a) The quantity K + U g is the same at the lowest point of the trajectory as it was at the highest point. Thus, K1 + U g1 = K 0 + U g0 means
1 2 1 mv1 + mgy1 = mv02 + mgy0 ⇒ v12 + 2 gy1 = v02 + 2 gy0 2 2 ⇒ v12 + 2 g (0 m) = (0 m/s) 2 + 2 gy0 ⇒ v1 = 2 gy0 From the pictorial representation, we find that y0 = L − L cos30°. Thus,
v1 = 2 gL(1 − cos30°) = 2(9.8 m/s 2 )(0.75 m)(1 − cos30°) = 1.403 m/s The speed at the lowest point is 1.40 m/s. (b) Since the quantity K + U g does not change, K 2 + U g 2 = K1 + U g1. We have 1 2 1 mv2 + mgy2 = mv12 + mgy1 ⇒ y2 = ( v12 − v22 ) 2 g 2 2 ⇒ y2 = [(1.403 m/s) 2 − (0 m/s) 2 ]/(2 × 9.8 m/s 2 ) = 0.100 m Since y2 = L − L cosθ , we obtain
cosθ =
L − y2 (0.75 m) − (0.10 m) = = 0.8667 ⇒ θ = cos −1 (0.8667) = 30° L (0.75 m)
That is, the pendulum swings to the other side by 30°.
Assess: The swing angle is the same on either side of the rest position. This result is a consequence of the fact that the sum of the kinetic and gravitational potential energy does not change. This is shown as well in the energy bar chart in the figure.
10.12. Model: Model the child and swing as a particle, and assume the chain to be massless. In the absence of frictional and air-drag effects, the sum of the kinetic and gravitational potential energy does not change during the swing’s motion. Visualize:
Solve:
The quantity K + U g is the same at the highest point of the swing as it is at the lowest point. That is,
K 0 + U g0 = K1 + U g1. It is clear from this equation that maximum kinetic energy occurs where the gravitational potential energy is the least. This is the case at the lowest position of the swing. At this position, the speed of the swing and child will also be maximum. The above equation is 1 2 1 mv0 + mgy0 = mv12 + mgy1 ⇒ v12 = v02 + 2 g ( y0 − y1 ) 2 2 2 ⇒ v1 = (0 m/s) 2 + 2 g ( y0 − 0 m) ⇒ v1 = 2 gy0 We see from the pictorial representation that
y0 = L − L cos 45° = (3.0 m) − (3.0 m)cos 45° = 0.879 m ⇒ v1 = 2 gy0 = 2(9.8 m/s 2 )(0.879 m) = 4.2 m/s Assess: We did not need to know the swing’s or the child’s mass. Also, a maximum speed of 4.2 m/s is reasonable.
10.13. Model: Model the car as a particle with zero rolling friction. The sum of the kinetic and gravitational potential energy, therefore, does not change during the car’s motion. Visualize:
Solve:
The initial energy of the car is
1 1 K 0 + U g0 = mv02 + mgy0 = (1500 kg)(10.0 m/s) 2 + (1500 kg)(9.8 m/s 2 )(10 m) = 2.22 × 105 J 2 2 The car increases its height to 15 m at the gas station. The conservation of energy equation K 0 + U g0 = K1 + U g1 is
1 1 2.22 × 105 J = mv12 + mgy1 ⇒ 2.22 × 105 J = (1500 kg)v12 + (1500 kg)(9.8 m/s 2 )(15 m) 2 2 ⇒ v1 = 1.41 m/s Assess: A lower speed at the gas station is reasonable because the car has decreased its kinetic energy and increased its potential energy compared to its starting values.
10.14.
Model: Assume that the spring is ideal and obeys Hooke’s law. Also model the sled as a particle. G Visualize: The only horizontal force acting on the sled is Fsp .
Solve:
Applying Newton’s second law to the sled gives
∑ (F
) = Fsp = max ⇒ k Δy = max
on sled x
⇒ ax = k Δ x / m = (150 N/m)(0.20 m)/20 kg = 1.50 m/s 2
10.15.
Model: Assume that the spring is ideal and obeys Hooke’s law. Visualize: According to Hooke’s law, the spring force acting on a mass (m) attached to the end of a spring is given as Fsp = k Δ x, where Δx is the change in length of the spring. If the mass m is at rest, then Fsp is also equal
to the gravitational force FG = mg . Solve:
We have Fsp = k Δ x = mg . We want a 0.100 kg mass to give Δ x = 0.010 m. This means k = mg / Δx = (0.100 kg)(9.8 N/m)/(0.010 m) = 98 N/m
10.16. Model: Assume an ideal spring that obeys Hooke’s law. Visualize:
Solve:
(a) The spring force on the 2.0 kg mass is Fsp = − k Δy. Notice that Δy is negative, so Fsp is positive.
This force is equal to mg, because the 2.0 kg mass is at rest. We have − k Δy = mg . Solving for k:
k = −( mg / Δy ) = −(2.0 kg)(9.8 m/s 2 )/(−0.15 m − (−0.10 m)) = 392 N/m The spring constant is 3.9 × 102 N/m. (b) Again using − k Δy = mg: Δy = − mg / k = −(3.0 kg)(9.8 m/s 2 )/(392 N/m) y′ − ye = −0.075 m ⇒ y′ = ye − 0.075 m = −0.10 m − 0.075 m = −0.175 m = −17.5 cm The length of the spring is 17.5 cm when a mass of 3.0 kg is attached to the spring. The position of the end of the spring is negative because it is below the origin, but length must be a positive number.
10.17. Model: Assume that the spring is ideal and obeys Hooke’s law. We also model the 5.0 kg mass as a particle. Visualize:
Solve:
We will use the subscript s for the scale and sp for the spring.
(a) The scale reads the upward force Fs on m that it applies to the mass. Newton’s second law gives
∑ (F
) = Fs on m − FG = 0 ⇒ Fs on m = FG = mg = (5.0 kg)(9.8 m/s 2 ) = 49 N
on m y
(b) In this case, the force is
∑ (F
) = Fs on m + Fsp − FG = 0 ⇒ 20 N + k Δy − mg = 0
on m y
⇒ k = (mg − 20 N)/ Δy = (49 N − 20 N)/0.02 m = 1450 N/m The spring constant for the lower spring is 1.45 × 103 N/m. (c) In this case, the force is ∑ ( Fon m ) y = Fsp − FG = 0 ⇒ k Δy − mg = 0 ⇒ Δy = mg / k = (49 N)/(1450 N/m) = 0.0338 m = 3.4 cm
10.18. Model: Model the student (S) as a particle and the spring as obeying Hooke’s law. Visualize:
Solve:
According to Newton’s second law the force on the student is
∑ (F
) = Fspring on S − FG = ma y
on S y
⇒ Fspring on S = FG + ma y = mg + ma y = (60 kg)(9.8 m/s 2 + 3.0 m/s 2 ) = 768 N Since Fspring on S = FS on spring = k Δy, k Δy = 768 N. This means Δy = (768 N)/(2500 N/m) = 0.31 m.
10.19. Model: Assume an ideal spring that obeys Hooke’s law. Solve: The elastic potential energy of a spring is defined as U s = 12 k (Δs ) 2 , where Δs is the magnitude of the stretching or compression relative to the unstretched or uncompressed length. We have Δs = 20 cm = 0.20 m and k = 500 N/m. This means
1 1 U s = k (Δs ) 2 = (500 N/m)(0.20 m) = 10 J 2 2 Assess: Since Δs is squared, U s is positive for a spring that is either compressed or stretched. U s is zero when the spring is in its equilibrium position.
10.20. Model: Assume an ideal spring that obeys Hooke’s law. Solve:
The elastic potential energy of a spring is defined as U s = 12 k (Δs ) 2 , where Δs is the magnitude of the
stretching or compression relative to the unstretched or uncompressed length. ΔU s = 0 when the spring is at its equilibrium length and Δs = 0. We have U s = 200 J and k = 1000 N/m. Solving for Δs : Δs =
2U s / k = 2(200 J) / 1000 N/m = 0.632 m
10.21. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, so the mechanical energy K + U s is conserved. Also model the book as a particle. Visualize:
The figure shows a before-and-after pictorial representation. The compressed spring will push on the book until the spring has returned to its equilibrium length. We put the origin of our coordinate system at the equilibrium position of the free end of the spring. The energy bar chart shows that the potential energy of the compressed spring is entirely transformed into the kinetic energy of the book. Solve: The conservation of energy equation K 2 + U s2 = K1 + U s1 is
1 2 1 1 1 mv2 + k ( x2 − xe ) 2 = mv12 + k ( x1 − xe ) 2 2 2 2 2 Using x2 = xe = 0 m and v1 = 0 m/s, this simplifies to 1 2 1 (1250 N/m)(0.040 m) 2 kx12 mv2 = k ( x1 − 0 m) 2 ⇒ v2 = = = 2.0 m/s 2 2 (0.500 kg) m
Assess: This problem cannot be solved using constant-acceleration kinematic equations. The acceleration is not a constant in this problem, since the spring force, given as Fs = − k Δx, is directly proportional to Δx or | x − xe |.
10.22. Model: Assume an ideal spring that obeys Hooke’s law. Since there is no friction, the mechanical energy K + U s is conserved. Also, model the block as a particle. Visualize:
The figure shows a before-and-after pictorial representation. We have put the origin of our coordinate system at the equilibrium position of the free end of the spring. This gives us x1 = xe = 0 cm and x2 = 2.0 cm. Solve:
The conservation of energy equation K 2 + U s2 = K1 + U s1 is
1 2 1 1 1 mv2 + k ( x2 − xe ) 2 = mv12 + k ( x1 − xe ) 2 2 2 2 2 Using v2 = 0 m/s, x1 = xe = 0 m, and x2 − xe = 0.020 m, we get
m 1 1 k ( x2 − xe ) 2 = mv12 ⇒ Δx = ( x2 − xe ) = v1 k 2 2 That is, the compression is directly proportional to the velocity v1. When the block collides with the spring with twice the earlier velocity (2v1 ), the compression will also be doubled to 2( x2 − xe ) = 2(2.0 cm) = 4.0 cm. Assess: This problem shows the power of using energy conservation over using Newton’s laws in solving problems involving nonconstant acceleration.
10.23. Model: Model the grocery cart as a particle and the spring as an ideal that obeys Hooke’s law. We will also assume zero rolling friction during the compression of the spring, so that mechanical energy is conserved. Visualize:
The figure shows a before-and-after pictorial representation. The “before” situation is when the cart hits the spring in its equilibrium position. We put the origin of our coordinate system at this equilibrium position of the free end of the spring. This give x1 = xe = 0 and ( x2 − xe ) = 60 cm. Solve:
The conservation of energy equation K 2 + U s2 = K1 + U s1 is
1 2 1 1 1 mv2 + k ( x2 − xe ) 2 = mv12 + k ( x1 − xe ) 2 2 2 2 2 Using v2 = 0 m/s,( x2 − xe ) = 0.60 m, and x1 = xe = 0 m gives:
1 1 k 250 N/m k ( x2 − xe ) 2 = m v12 ⇒ v1 = ( x2 − xe ) = (0.60 m) = 3.0 m/s 2 2 m 10 kg
10.24. Model: Model the jet plane as a particle, and the spring as an ideal that obeys Hooke’s law. We will also assume zero rolling friction during the stretching of the spring, so that mechanical energy is conserved. Visualize:
The figure shows a before-and-after pictorial representation. The “before” situation occurs just as the jet plane lands on the aircraft carrier and the spring is in its equilibrium position. We put the origin of our coordinate system at the right free end of the spring. This gives x1 = xe = 0 m. Since the spring stretches 30 m to stop the
plane, x2 − xe = 30 m. Solve:
The conservation of energy equation K 2 + U s2 = K1 + U s1 for the spring-jet plane system is 1 2 1 1 1 mv2 + k ( x2 − xe ) 2 = mv12 + k ( x1 − xe ) 2 2 2 2 2
Using v2 = 0 m/s, x1 = xe = 0 m, and x2 − xe = 30 m yields
1 1 k 60,000 N/m k ( x2 − xe ) 2 = mv12 ⇒ v1 = ( x2 − x1 ) = (30 m) = 60 m/s 2 2 m 15,000 kg Assess:
A landing speed of 60 m/s or ≈120 mph is reasonable.
10.25. Model: We assume this is a one-dimensional collision that obeys the conservation laws of momentum and mechanical energy. Visualize:
Note that momentum conservation alone is not sufficient to solve this problem because the two final velocities (vfx )1 and (vfx ) 2 are unknowns and can not be determined from one equation. Solve:
Momentum conservation: m1 (vix )1 + m2 (vix ) 2 = m1 (vfx )1 + m2 (vfx ) 2 Energy conservation:
1 1 1 1 m1 (vix )12 + m2 (vix ) 22 = m1 (vfx )12 + m2 (vfx ) 22 2 2 2 2
These two equations can be solved for (vfx )1 and (vfx ) 2 , as shown by Equations 10.39 through 10.43, to give
(vfx )1 =
m1 − m2 50 g − 20 g (vix )1 = (2.0 m/s) = 0.86 m/s m1 + m2 50 g + 20 g
(vfx ) 2 =
2m1 2(50 g) (vix )1 = (2.0 m/s) = 2.9 m/s m1 + m2 50 g + 20 g
Assess: These velocities are of a reasonable magnitude. Since both these velocities are positive, both balls move along the +x-direction.
10.26. Model: This is a case of a perfectly elastic collision between a proton and a carbon atom. The collision obeys the momentum as well as the energy conservation law. Visualize:
Solve:
Momentum conservation: mP (vix ) P + mC (vix ) C = mP (vfx ) P + mC (vfx ) C
1 1 1 1 mP (vix ) 2P + mC (vix )C2 = mP (vfx ) P2 + mC (vfx ) C2 2 2 2 2 These two equations can be solved, as described in the text through Equations 10.39 to 10.43: Energy conservation:
⎛ m − 12mP ⎞ mP − mC 7 7 mP + mC (vix ) P = ⎜ P ⎟ (2.0 × 10 m/s) = −1.69 × 10 m/s + mP + mC m 12 m ⎝ P P ⎠ ⎛ 2mP ⎞ 2mP 7 6 (vfx )C = (vix ) P = ⎜ ⎟ (2.0 × 10 m/s) = 3.1 × 10 m/s mP + mC ⎝ mP + 12mP ⎠ (vfx ) P =
After the elastic collision the proton rebounds at 1.69 × 107 m/s and the carbon atom moves forward at
3.08 × 106 m/s.
10.27. Model: This is the case of a perfectly inelastic collision. Momentum is conserved because no external force acts on the system (clay + brick). We also represent our system as a particle. Visualize:
Solve:
(a) The conservation of momentum equation pfx = pix is
(m1 + m2 )vfx = m1 (vix )1 + m2 (vix ) 2 Using (vix )1 = v0 and (vix ) 2 = 0, we get
vfx =
m1 0.050 kg (vix )1 = (vix )1 = 0.0476(vix )1 = 0.0476 v0 m1 + m2 (1.0 kg + 0.050 kg)
The brick is moving with speed 0.048v0 . (b) The initial and final kinetic energies are given by 1 1 1 1 2 K i = m1 (vix )12 + m2 (vix ) 22 = (0.050 kg)v02 + (1.0 kg) ( 0 m/s ) = (0.025 kg)v02 2 2 2 2 1 1 K f = (m1 + m2 )vf2x = (1.0 kg + 0.050 kg)(0.0476) 2 v02 = 0.00119 v02 2 2 ⎛ K − Kf The percent of energy lost = ⎜ i ⎝ Ki
⎞ ⎛ 0.00119 ⎞ ⎟ × 100% = ⎜1 − ⎟ × 100% = 95% 0.025 ⎠ ⎝ ⎠
10.28. Model: In this case of a one-dimensional collision, the momentum conservation law is obeyed whether the collision is perfectly elastic or perfectly inelastic. Visualize:
Solve: In the case of a perfectly elastic collision, the two velocities (vfx )1 and (vfx ) 2 can be determined by combining the conservation equations of momentum and mechanical energy. By contrast, a perfectly inelastic collision involves only one final velocity vfx and can be determined from just the momentum conservation equation. (a)
Momentum conservation: m1 (vix )1 + m2 (vix ) 2 = m1 (vfx )1 + m2 (vfx ) 2
1 1 1 1 m1 (vix )12 + m2 (vix ) 22 = m1 (vfx )12 + m2 (vfx ) 22 2 2 2 2 These two equations can be solved as shown in Equations 10.39 through 10.43: m − m2 (100 g) − (300 g) (vfx )1 = 1 (vix )1 = (10 m/s) = −5.0 m/s m1 + m2 (100 g) + (300 g) Energy conservation:
(vfx ) 2 =
2m1 2(100 g) (vix )1 = (10 m/s) = +5.0 m/s m1 + m2 (100 g) + (300 g)
(b) For the inelastic collision, both balls travel with the same final speed vfx . The momentum conservation equation
pfx = pix is (m1 + m2 )vfx = m1 (vix )1 + m2 (vix ) 2 ⎛ ⎞ 100 g ⇒ vfx = ⎜ ⎟ (10 m/s) + 0 m/s = 2.5 m/s ⎝ 100 g + 300 g ⎠ Assess: In the case of the perfectly elastic collision, the two balls bounce off each other with a speed of 5.0 m/s. In the case of the perfectly inelastic collision, the balls stick together and move together at 2.5 m/s.
10.29. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant. Visualize:
The particle is released from rest at x = 1.0 m. That is, K = 0 at x = 1.0 m. Since the total energy is given by E = K + U , we can draw a horizontal total energy (TE) line through the point of intersection of the potential energy curve (PE) and the x = 1.0 m line. The distance from the PE curve to the TE line is the particle’s kinetic energy. These values are transformed as the position changes, causing the particle to speed up or slow down, but the sum K + U does not change. Solve: (a) We have E = 4.0 J and this energy is a constant. For x < 1.0, U > 4.0 J and, therefore, K must be negative to keep E the same (note that K = E − U or K = 4.0 J − U ). Since negative kinetic energy is unphysical, the particle can not move to the left. That is, the particle will move to the right of x = 1.0 m. (b) The expression for the kinetic energy is E − U . This means the particle has maximum speed or maximum kinetic energy when U is minimum. This happens at x = 4.0 m. Thus,
K max = E − U min = (4.0 J) − (1.0 J) = 3.0 J
1 2 2(3.0 J) 8.0 J mvmax = 3.0 J ⇒ vmax = = = 17.3 m/s 2 m 0.020 kg
The particle possesses this speed at x = 4.0 m. (c) The total energy (TE) line intersects the potential energy (PE) curve at x = 1.0 m and x = 6.0 m. These are the turning points of the motion.
10.30. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant. Visualize:
The particle with a mass of 500 g is released from rest at A. That is, K = 0 at A. Since E = K + U = 0 J + U , we can draw a horizontal TE line through U = 5.0 J. The distance from the PE curve to the TE line is the particle’s kinetic energy. These values are transformed as the position changes, causing the particle to speed up or slow down, but the sum K + U does not change. Solve: The kinetic energy is given by E − U , so we have
1 2 mv = E − U ⇒ v = 2( E − U )/ m 2 Using U B = 2.0 J, U C = 3.0 J, and U D = 0 J, we get vB = 2(5.0 J − 2.0 J)/0.500 kg = 3.5 m/s vD = 2(5.0 J − 0 J)/0.500 kg = 4.5 m/s
vC = 2(5.0 J − 3.0 J)/0.500 kg = 2.8 m/s
10.31. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant. Visualize:
Since the particle oscillates between x = 2.0 mm and x = 8.0 mm, the speed of the particle is zero at these points. That is, for these values of x, E = U = 5.0 J, which defines the total energy (TE) line. The distance from the potential energy (PE) curve to the TE line is the particle’s kinetic energy. These values are transformed as the position changes, but the sum K + U does not change. Solve: The equation for total energy E = U + K means K = E − U , so that K is maximum when U is minimum. We have
⇒ vmax
1 2 K max = mvmax = 5.0 J − U min 2 = 2(5.0 J − U min )/ m = 2(5.0 J − 1.0 J)/0.0020 kg = 63 m/s
10.32. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant. Visualize:
For the speed of the particle at A that is needed to reach B to be a minimum, the particle’s kinetic energy as it reaches the top must be zero. Similarly, the minimum speed at B for the particle to reach A obtains when the particle just makes it to the top with zero kinetic energy. Solve: (a) The energy equation K A + U A = K top + U top is
1 2 mvA + U A = 0 J + U top 2 ⇒ vA = 2(U top − U A )/ m = 2(5.0 J − 2.0 J)/0.100 kg = 7.7 m/s (b) To go from point B to point A, K B + U B = K top + U top is
1 2 mvB + U B = 0 J + U top 2 ⇒ vB = 2(U top − U B )/ m = 2(5.0 J − 0 J)/0.100 kg = 10.0 m/s Assess:
The particle requires a higher kinetic energy to reach A from B than to reach B from A.
10.33. Model: Model your vehicle as a particle. Assume zero rolling friction, so that the sum of your kinetic and gravitational potential energy does not change as the vehicle coasts down the hill. Visualize:
The figure shows a before-and-after pictorial representation. Note that neither the shape of the hill nor the angle of the downward slope is given, since these are not needed to solve the problem. All we need is the change in potential energy as you and your vehicle descend to the bottom of the hill. Also note that
35 km / hr = (35,000 m / 3600 s) = 9.722 m / s Solve:
Using yf = 0 and the equation K i + U gi = K f + U gf we get
1 2 1 mvi + mgyi = mvf2 + mgyf ⇒ vi2 + 2 gyi = vf2 2 2 ⇒ vf = vi2 + 2 gyi = (9.722 m/s) 2 + 2(9.8 m/s 2 )(15 m) = 19.7 m/s = 71 km/h You are driving over the speed limit. Yes, you will get a ticket. Assess: A speed of 19.7 m/s or 71 km/h at the bottom of the hill, when your speed at the top of the hill was 35 km/s, is reasonable. From the energy bar chart, we see that the initial potential energy is completely transformed into the final kinetic energy.
10.34. Model: This is case of free fall, so the sum of the kinetic and gravitational potential energy does not change as the cannon ball falls. Visualize:
The figure shows a before-and-after pictorial representation. To express the gravitational potential energy, we put the origin of our coordinate system on the ground below the fortress. Solve: Using yf = 0 and the equation K i + U gi = K f + U gf we get
1 2 1 mvi + mgyi = mvf2 + mgyf ⇒ vi2 + 2 gyi = vf2 2 2 vf = vi2 + 2 gyi = (80 m/s) 2 + 2(9.8 m/s 2 )(10 m) = 81 m/s Assess: Note that we did not need to use the tilt angle of the cannon, because kinetic energy is a scalar. Also note that using the energy conservation equation, we can find only the magnitude of the final velocity, not the direction of the velocity vector.
10.35. Model: This is a case of free fall, so the sum of the kinetic and gravitational potential energy does not change as the rock is thrown. Model the Frisbee and the rock as particles. Visualize:
The coordinate system is put on the ground for this system, so that yf = 16 m. The rock’s final velocity vf must be at least 5.0 m/s to dislodge the Frisbee. Solve: (a) The energy conservation equation for the rock K f + U gf = K i + U gi is
1 2 1 mvf + mgyf = mvi2 + mgyi 2 2 This equation involves only the velocity magnitudes and not the angle at which the rock is to be thrown to dislodge the Frisbee. This equation is true for all angles that will take the rock to the Frisbee. (b) Using the above equation we get
vf2 + 2 gyf = vi2 + 2 gyi
vi = vf2 + 2 g ( yf − yi ) = (5.0 m/s)2 + 2(9.8 m/s 2 )(16 m − 2.0 m) = 17.3 m/s
Assess: Kinetic energy is defined as K = 12 mv 2 and is a scalar quantity. Scalar quantities do not have directional properties.
10.36. Model: For the ice cube sliding around the inside of a smooth pipe, the sum of the kinetic and gravitational potential energy does not change. Visualize:
We use a coordinate system with the origin at the bottom of the pipe, that is, y1 = 0. The radius (R) of the pipe is 10 cm, and therefore ytop = y2 = 2 R = 0.20 m. At an arbitrary angle θ , measured counterclockwise from the bottom of the circle, y = R − R cosθ . Solve: (a) The energy conservation equation K 2 + U g2 = K1 + U g1 is
1 1 ⇒ mv22 + mgy2 = mv12 + mgy1 2 2 ⇒ v2 = v12 + 2 g ( y1 − y2 ) = (3.0 m/s) 2 + 2(9.8 m/s 2 )(0 m − 0.20 m) = 2.25 m/s (b) Expressing the energy conservation equation as a function of θ :
1 1 K (θ ) + U g (θ ) = K1 + U g1 ⇒ mv 2 (θ ) + mgy (θ ) = mv12 + 0 J 2 2 ⇒ v (θ ) = v12 − 2 gy (θ ) = v12 − 2 gR (1 − cosθ ) Using v1 = 3.0 m/s, g = 9.8 m/s 2 , and R = 0.10 m, we get v(θ ) = 9 − 1.96(1 − cosθ ) (m/s) (c) The accompanying figure shows a graph of v for a complete revolution (0° ≤ θ ≤ 360°).
Assess: Beginning with a speed of 3.0 m/s at the bottom, the marble’s potential energy increases and kinetic energy decreases as it gets toward the top of the circle. At the top, its speed is 2.25 m/s. This is reasonable since some of the kinetic energy has been transformed into the marble’s potential energy.
10.37. Model: Assume that the rubber band behaves similar to a spring. Also, model the rock as a particle. Visualize:
Please refer to Figure P10.37. Solve: (a) The rubber band is stretched to the left since a positive spring force on the rock due to the rubber band results from a negative displacement of the rock. That is, ( Fsp ) x = − kx, where x is the rock’s displacement
from the equilibrium position due to the spring force Fsp . (b) Since the Fsp versus x graph is linear with a negative slope and can be expressed as Fsp = − kx, the rubber
band obeys Hooke’s law. (c) From the graph, |ΔFsp | = 20 N for |Δx| = 10 cm. Thus, k=
|ΔFsp | |Δx|
=
20 N = 200 N/m = 2.0 × 102 N/m 0.10 m
(d) The conservation of energy equation K f + U sf = Ki + U si for the rock is
1 2 1 2 1 2 1 2 1 1 1 1 mvf + kxf = mvi + kxi ⇒ mvf2 + k (0 m) 2 = m(0 m/s) 2 + kxi2 2 2 2 2 2 2 2 2 k 200 N/m vf = xi = (0.30 m) = 19.0 m/s m 0.050 kg Assess: Note that xi is Δx, which is the displacement relative to the equilibrium position, and xf is the equilibrium position of the rubber band, which is equal to zero.
10.38. Model: Model the block as a particle and the springs as ideal springs obeying Hooke’s law. There is no friction, hence the mechanical energy K + U s is conserved. Visualize:
Note that xf = xe and xi − xe = Δx. The before-and-after pictorial representations show that we put the origin of the coordinate system at the equilibrium position of the free end of the springs. Solve: The conservation of energy equation K f + U sf = Ki + U si for the single spring is
1 2 1 1 1 mvf + k ( xf − xe ) 2 = mvi2 + k ( xi − xe ) 2 2 2 2 2 Using the value for vf given in the problem, we get 1 2 1 1 1 mv0 + 0 J = 0 J + k (Δx) 2 ⇒ mv02 = k (Δx ) 2 2 2 2 2
Conservation of energy for the two-spring case: 1 1 1 mVf2 + 0 J = 0 J + k ( xi − xe ) 2 + k ( xi − xe ) 2 2 2 2 Using the result of the single-spring case,
1 mVf2 = k (Δx ) 2 2
1 mVf2 = mv02 ⇒ Vf = 2v 0 2 Assess:
The block separates from the spring at the equilibrium position of the spring.
10.39. Model: Model the block as a particle and the springs as ideal springs obeying Hooke’s law. There is no friction, hence the mechanical energy K + U s is conserved. Visualize:
The springs in both cases have the same compression Δx. We put the origin of the coordinate system at the equilibrium position of the free end of the spring for the single-spring case (a), and at the free end of the two connected springs for the two-spring case (b). Solve: The conservation of energy for the single-spring case: 1 1 1 1 K f + U sf = K i + U si ⇒ mvf2 + k ( xf − xe ) 2 = mvi2 + k ( xi − xe ) 2 2 2 2 2 Using xf = xe = 0 m, vi = 0 m/s, and vf = v0, this equation simplifies to
1 2 1 mv0 = k (Δx) 2 2 2 Conservation of energy in the case of the two springs in series, where each spring compresses by Δx /2, is 1 1 1 1 K f + U sf = K i + U si ⇒ mVf2 + 0 = mvi2 + k (Δx /2) 2 + k (Δx /2) 2 2 2 2 2 Using xf = xe′ = 0 m and vi = 0 m/s, we get 1 1 ⎡1 ⎤ mVf2 = ⎢ k (Δx) 2 ⎥ 2 2 ⎣2 ⎦ Comparing the two results we see that Vf = v0 / 2. Assess: The block pushes on the spring until the spring has returned to its equilibrium length.
10.40. Model: Model the ball and earth as particles. An elastic collision between the ball and earth conserves both momentum and mechanical energy. Visualize:
The before-and-after pictorial representation of the collision is shown in the figure. The ball as it is dropped from a height of 10 m has zero velocity. Its speed just before it hits earth can be found using kinematics. We can then apply the conservation laws to find earth’s recoil velocity. Solve: (a) To get the speed of the ball at collision:
(v1 ) 2B = (v0 ) B2 − 2 g ( y1 − y0 ) B = (0 m/s) − 2(9.8 m/s 2 )(−10.0 m) ⇒ (v1 ) B = 2(9.8 m/s 2 )(10 m) = 14.0 m/s In an elastic collision, the velocity of the object that is struck is
(v2 ) E =
2mB 2(0.500 kg) (v1 ) B = (14.0 m / s) = 2.3 × 10−24 m/s mE + mB 5.98 × 1024 kg
(b) The time it would take earth to move 1.0 mm at this speed is given by speed =
distance 1.0 × 10−3 m = = 4.3 × 1020 s = 1.36 × 1013 years velocity 2.3 × 10−24 m/s
10.41. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, and therefore the mechanical energy K + U s + U g is conserved. Visualize:
The figure shows a before-and-after pictorial representation. We have chosen to place the origin of the coordinate system at the position where the ice cube has compressed the spring 10 cm. That is, y0 = 0. Solve:
The energy conservation equation K 2 + U s2 + U g2 = K 0 + U s0 + U g0 is 1 2 1 1 1 mv2 + k ( xe − xe ) 2 + mgy0 = mv02 + k ( x − xe ) 2 + mgy0 2 2 2 2
Using v2 = 0 m / s, y0 = 0 m, and v0 = 0 m / s,
k ( x − xe ) 2 1 (25 N/m)(0.10 m) 2 = = 25.5 cm mgy2 = k ( x − xe ) 2 ⇒ y2 = 2 2 mg 2(0.050 kg)(9.8 m/s 2 ) The distance traveled along the incline is y2 / sin 30° = 25.5 cm/ sin 30° = 51 cm. Assess: The net effect of the launch is to transform the potential energy stored in the spring into gravitational potential energy. The block has kinetic energy as it comes off the spring, but we did not need to know this energy to solve the problem.
10.42. Model: Model the two packages as particles. Momentum is conserved in both inelastic and elastic collisions. Kinetic energy is conserved only in a perfectly elastic collision. Visualize:
Solve:
For a package with mass m the conservation of energy equation is
1 1 K1 + U g1 = K 0 + U g0 ⇒ m(v1 ) 2m + mgy1 = m(v0 ) m2 + mgy0 2 2
Using (v0 ) m = 0 m / s and y1 = 0 m, 1 m(v1 ) 2m = mgy0 ⇒ (v1 ) m = 2 gy0 = 2(9.8 m/s 2 )(3.0 m) = 7.668 m/s 2 (a) For the perfectly inelastic collision the conservation of momentum equation is
pfx = pix ⇒ ( m + 2m)(v2 )3m = m(v1 ) m + (2m)(v1 ) 2m Using (v1 ) 2m = 0 m / s, we get (v2 )3m = (v1 ) m / 3 = 2.56 m/s The packages move off together at a speed of 2.6 m/s. (b) For the elastic collision, the mass m package rebounds with velocity m − 2m 1 (v1 ) m = − ( 7.668 m/s ) = −2.56 m/s m + 2m 3 The negative sign with (v3 ) m shows that the package with mass m rebounds and goes to the position y4 . We can (v3 ) m =
determine y4 by applying the conservation of energy equation as follows. For a package of mass m: 1 1 K f + U gf = K i + U gi ⇒ m(v4 ) 2m + mgy4 = m(v3 ) m2 + mgy3 2 2 Using (v3 )m = −2.55 m/s, y3 = 0 m, and (v4 ) m = 0 m/s, we get 1 mgy4 = m(−2.56 m/s) 2 ⇒ y4 = 33 cm 2
10.43. Model: Model the marble and the steel ball as particles. We will assume an elastic collision between the marble and the ball, and apply the conservation of momentum and the conservation of energy equations. We will also assume zero rolling friction between the marble and the incline. Visualize:
This is a two-part problem. In the first part, we will apply the conservation of energy equation to find the marble’s speed as it exits onto a horizontal surface. We have put the origin of our coordinate system on the horizontal surface just where the marble exits the incline. In the second part, we will consider the elastic collision between the marble and the steel ball. Solve: The conservation of energy equation K1 + U g1 = K 0 + U g0 gives us:
1 1 2 2 mM (v1 ) M + mM gy1 = mM (v0 ) M + mM gy0 2 2 1 2 (v1 ) M = gy0 ⇒ (v1 )M = 2 gy0 . When the marble collides with the 2 steel ball, the elastic collision gives the ball velocity Using (v0 ) M = 0 m / s and y1 = 0 m, we get
(v2 )S =
2mM 2mM (v1 ) M = 2 gy0 mM + mS mM + mS
Solving for y0 gives 2
⎤ 1 ⎡ mM + mS y0 = (v2 )S ⎥ = 0.258 m = 25.8 cm ⎢ 2 g ⎣ 2mM ⎦
10.44. Model: Assume an ideal spring that obeys Hooke’s law. Since this is a free-fall problem, the mechanical energy K + U g + U s is conserved. Also, model the safe as a particle. Visualize:
We have chosen to place the origin of our coordinate system at the free end of the spring, which is neither stretched nor compressed. The safe gains kinetic energy as it falls. The energy is then converted into elastic potential energy as the safe compresses the spring. The only two forces are gravity and the spring force, which are both conservative, so energy is conserved throughout the process. This means that the initial energy—as the safe is released—equals the final energy—when the safe is at rest and the spring is fully compressed. Solve: The conservation of energy equation K1 + U g1 + U s1 = K 0 + U g0 + U s0 is
1 2 1 1 1 mv1 + mg ( y1 − ye ) + k ( y1 − ye ) 2 = mv02 + mg ( y0 − ye ) + k ( ye − ye ) 2 2 2 2 2 Using v1 = v0 = 0 m/s and ye = 0 m, the above equation simplifies to 1 mgy1 + ky12 = mgy0 2
⇒k =
2mg ( y0 − y1 ) 2(1000 kg)(9.8 m/s 2 )(2.0 m − ( −0.50 m)) = = 1.96 × 105 N/m y12 ( −0.50 m) 2
Assess: By equating energy at these two points, we do not need to find how fast the safe was moving when it hit the spring.
10.45. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction and hence the mechanical energy K + U s + U g is conserved. Visualize:
Solve:
(a) When releasing the block suddenly, K 2 + U s2 + U g 2 = K1 + U s1 + U g1 1 2 1 1 1 mv2 + k ( y2 − ye ) 2 + mgy2 = mv12 + k ( y1 − ye ) 2 + mgy1 2 2 2 2
Using v2 = 0 m/s, v1 = 0 m/s, and y1 = ye , we get
1 0 J + (490 N/m)( y2 − y1 ) 2 + mgy2 = 0 J + 0 J + mgy1 ⇒ (245 N/m)( y2 − y1 ) 2 = mg ( y1 − y2 ) 2 ⇒ (245 N/m)( y1 − y2 ) 2 = (5.0 kg)(9.8 m/s 2 )( y1 − y2 ) ⇒ ( y1 − y2 ) = 0.20 m (b) When lowering the block gently until it rests on the spring, the block reaches a point of static equilibrium. Fnet = k Δy − mg = 0 ⇒ Δy =
mg (5.0 kg)(9.8 m/s 2 ) = = 0.10 m k 490 N/m
(c) In part (b), at a point 0.10 m down, the forces balance. But in part (a) the block has kinetic energy as it reaches 0.10 m. So the block continues on past the equilibrium point until all the gravitational potential energy is stored in the spring.
10.46. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, and thus the mechanical energy K + U s + U g is conserved. Visualize:
We place the origin of our coordinate system at the spring’s compressed position y1 = 0. The rock leaves the
spring with velocity v2 as the spring reaches its equilibrium position. Solve: (a) The conservation of mechanical energy equation is 1 2 1 1 1 K 2 + U s 2 + U g 2 = K1 + U s1 + U g1 mv2 + k ( y2 − ye ) 2 + mgy2 = mv12 + k ( y1 − ye ) 2 + mgy1 2 2 2 2 Using y2 = ye , y1 = 0 m, and v1 = 0 m/s, this simplifies to 1 2 1 mv2 + 0 J + mgy2 = 0 J + k ( y1 − ye ) 2 + 0 2 2 1 1 (0.400 kg)v22 + (0.400 kg)(9.8 m/s 2 )(0.30 m) = (1000 N/m)(0.30 m) 2 ⇒ v2 = 14.8 m/s 2 2 (b) Let us use the conservation of mechanical energy equation once again to find the highest position ( y3 ) of the rock where its speed (v3 ) is zero:
1 1 K 3 + U g3 = K 2 + U g 2 ⇒ mv32 + mgy3 = mv22 + mgy2 2 2 1 v 2 (14.8 m/s) 2 ⇒ 0 + g ( y3 − y2 ) = v22 ⇒ ( y3 − y2 ) = 2 = = 11.2 m 2 2 g 2(9.8 m/s 2 ) If we assume the spring’s length to be 0.5 m, then the distance between ground and fruit is 11.2 m + 0.5 m = 11.7 m. This is much smaller than the distance of 15 m between fruit and ground. So, the rock does not reach the fruit, and the contestants go hungry.
10.47. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, so the mechanical energy K + U g + U s is conserved. Visualize:
Place the origin of the coordinate system at the end of the unstretched spring, making ye = 0 m. Solve: The clay is in static equilibrium while resting in the pan. The net force on it is zero. We can start by using this to find the spring constant.
Fsp = FG ⇒ − k ( y1 − ye ) = − ky1 = mg ⇒ k = −
mg (0.10 kg)(9.8 m/s2 ) =− = 9.8 N/m y1 −0.10 m
Now apply conservation of energy. Initially, the spring is unstretched and the clay ball is at the ceiling. At the end, the spring has maximum stretch and the clay is instantaneously at rest. Thus K 2 + (U g ) 2 + (U s ) 2 = K 0 + (U g )0 + (U s )0 ⇒ 12 mv22 + mgy2 + 12 ky22 = 12 mv02 + mgy0 + 0 J
Since v0 = 0 m/s and v2 = 0 m/s, this equation becomes
2mg 2mgy0 y2 − =0 k k y22 + 0.20 y2 − 0.10 = 0
mgy2 + 12 ky22 = mgy0 ⇒ y22 +
The numerical values were found using known values of m, g, k, and y0 . The two solutions to this quadratic equation are y2 = 0.231 m and y2 = −0.432 m. The point we’re looking for is below the origin, so we need the negative root. The distance of the pan from the ceiling is
L = y2 + 50 cm = 93 cm
10.48. Model: Assume an ideal spring that obeys Hooke’s law. Also assume zero rolling friction between the roller coaster and the track, and a particle model for the roller coaster. Since no friction is involved, the mechanical energy K + U s + U g is conserved. Visualize:
We have chosen to place the origin of the coordinate system on the end of the spring that is compressed and touches the roller coaster car. Solve: (a) The energy conservation equation for the car going to the top of the hill is K 2 + U g 2 + U s2 = K 0 + U g0 + U s0
1 2 1 1 1 mv2 + mgy2 + k ( xe − xe ) 2 = mv02 + mgy0 + k ( x0 − xe ) 2 2 2 2 2
Noting that y0 = 0 m, v2 = 0 m/s, v1 = 0 m/s, and x0 − xe = 2.0 m, we obtain 1 0 J + mgy2 + 0 J = 0 J + 0 J + k (2.0 m) 2 2 2mgy2 2(400 kg)(9.8 m/s 2 )(10 m) ⇒k = = = 1.96 × 104 N/m (2.0 m) 2 (2.0 m) 2 We now increase this value for k by 10% for safety, giving a value of 2.156 × 104 N/m ≈ 2.2 × 104 N/m. (b) The energy conservation equation K 3 + U g3 + U s3 = K 0 + U g 0 + U s0 is 1 2 1 1 1 mv3 + mgy3 + k ( xe − xe ) 2 = mv02 + mgy0 + k ( x0 − xe ) 2 2 2 2 2
Using y3 = −5.0 m, v0 = 0 m / s, y0 = 0 m, and | x0 − xe | = 2.0 m, we get 1 2 1 mv3 + mg (−5.0 m) + 0 J = 0 J + 0 J + k ( x0 − xe ) 2 2 2 1 1 (400 kg)v32 − (400 kg)(9.8 m/s 2 )(5.0 m) = (2.156 × 104 N/m)(2.0 m) 2 2 2 ⇒ v3 = 17.7 m/s
10.49. Model: Since there is no friction, the sum of the kinetic and gravitational potential energy does not change. Model Julie as a particle. Visualize:
We place the coordinate system at the bottom of the ramp directly below Julie’s starting position. From geometry, Julie launches off the end of the ramp at a 30º angle. 1 1 Solve: Energy conservation: K1 + U g1 = K 0 + U g0 ⇒ mv12 + mgy1 = mv02 + mgy0 2 2 Using v0 = 0 m/s, y0 = 25 m, and y1 = 3 m, the above equation simplifies to
1 2 mv1 + mgy1 = mgy0 ⇒ v1 = 2 g ( y0 − y1 ) = 2(9.8 m/s 2 )(25 m − 3 m) = 20.77 m/s 2 We can now use kinematic equations to find the touchdown point from the base of the ramp. First we’ll consider the vertical motion: 1 1 y2 = y1 + v1 y (t2 − t1 ) + a y (t2 − t1 ) 2 0 m = 3 m + (v1 sin 30°)(t2 − t1 ) + (−9.8 m / s 2 )(t2 − t1 ) 2 2 2 (20.77 m/s)sin 30° (3 m) 2 ⇒ (t2 − t1 ) − =0 (t2 − t1 ) − (4.9 m/s 2 ) (4.9 m/s 2 ) (t2 − t1 ) 2 − (2.119 s)(t2 − t1 ) − (0.6122 s 2 ) = 0 ⇒ (t2 − t1 ) = 2.377 s For the horizontal motion: 1 x2 = x1 + v1x (t2 − t1 ) + ax (t2 − t1 ) 2 2 x2 − x1 = (v1 cos30°)(t2 − t1 ) + 0 m = (20.77 m/s)(cos30°)(2.377 s) = 43 m Assess: Note that we did not have to make use of the information about the circular arc at the bottom that carries Julie through a 90° turn.
10.50. Model: We assume the spring to be ideal and to obey Hooke’s law. We also treat the block (B) and the ball (b) as particles. In the case of an elastic collision, both the momentum and kinetic energy equations apply. On the other hand, for a perfectly inelastic collision only the equation of momentum conservation is valid. Visualize:
Place the origin of the coordinate system on the block that is attached to one end of the spring. The before-andafter pictorial representations of the elastic and perfectly inelastic collision are shown in figures (a) and (b), respectively. Solve: (a) For an elastic collision, the ball’s rebound velocity is
(vf ) b =
−80 g mb − mB (vi ) b = (5.0 m/s) = −3.33 m/s mb + mB 120 g
The ball’s speed is 3.3 m/s. (b) An elastic collision gives the block speed
(vf ) B =
2mB 40 g (vi ) b = (5.0 m/s) = 1.667 m/s mb + mB 120 g
To find the maximum compression of the spring, we use the conservation equation of mechanical energy for the block + spring system. That is K1 + U s1 = K 0 + U s0 : 1 1 1 1 mB (vf′ ) 2B + k ( x1 − x0 ) 2 = mB (vf ) B2 + k ( x0 − x0 ) 2 2 2 2 2
0 + k ( x1 − x0 ) 2 = mB (vf )B2 + 0
(x1 − x0 ) = (0.100 kg )(1.667 m/s )2 /(20 N/m ) = 11.8 cm (c) Momentum conservation pf = pi for the perfectly inelastic collision means (mB + mB )vf = mb (vi ) b + mB (vi ) B (0.100 kg + 0.020 kg)vf = (0.020 kg)(5.0 m/s) + 0 m/ v ⇒ vf = 0.833 m/s
The maximum compression in this case can now be obtained using the conservation of energy equation K1 + U s1 = K 0 + U s0 :
0 J + (1/ 2)k (Δx) 2 = (1/ 2)(mB + mb )vf 2 + 0 J ⇒ Δx =
mB + mb 0.120 kg vf = (0.833m/s) = 0.0645 m = 6.5 cm k 20 N/m
10.51.
Model: Assume an ideal spring that obeys Hooke’s law. We treat the bullet and the block in the particle model. For a perfectly inelastic collision, the momentum is conserved. Furthermore, since there is no friction, the mechanical energy of the system (bullet + block + spring) is conserved.
Visualize:
We place the origin of our coordinate system at the end of the spring that is not anchored to the wall. Solve: (a) Momentum conservation for perfectly inelastic collision states pf = pi . This means ⎛ m (m + M )vf = m(vi ) m + M (vi ) M ⇒ (m + M )vf = mvB + 0 kg m/s ⇒ vf = ⎜ ⎝m+M
⎞ ⎟ vB ⎠
where we have used vB for the initial speed of the bullet. The mechanical energy conservation equation K1 + U s1 = K e + U se as the bullet embedded block compresses the spring is: 1 1 1 1 m(v′f ) 2 + k ( x1 − xe ) 2 = (m + M )(vf ) 2 + k ( xe − xe ) 2 2 2 2 2 2
1 1 (m + M )kd 2 ⎛ m ⎞ 2 0 J + kd 2 = (m + M ) ⎜ 0 J v + ⇒ v = B ⎟ B 2 2 m2 ⎝m+M ⎠ (b) Using the above formula with m = 5.0 g, M = 2.0 kg, k = 50 N/m, and d = 10 cm,
vB = (0.0050 kg + 2.0 kg)(50 N/m)(0.10 m) 2 /(0.0050) 2 = 2.0 × 102 m/s (c) The fraction of energy lost is 1 2 1 2 2 mvB − (m + M )vf2 m+M ⎛ vf ⎞ m+M ⎛ m ⎞ 2 2 =1− = − 1 ⎜ ⎟ ⎜ ⎟ 1 2 m ⎝ vB ⎠ m ⎝m+M ⎠ mvB 2 m 0.0050 kg =1− =1− = 0.9975 m+M (0.0050 kg + 2.0 kg)
That is, during the perfectly inelastic collision 99.75% of the bullet’s energy is lost. The energy is dissipated inside the block. Although it is common to say, “The energy is lost to heat,” in the next chapter we’ll see that it is more accurate to say, “The energy is transformed to thermal energy.”
10.52. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, hence the mechanical energy K + U g + U s is conserved. Visualize:
We have chosen to place the origin of the coordinate system on the free end of the spring that is neither stretched nor compressed, that is, at the equilibrium position of the end of the unstretched spring. The bullet’s mass is m and the block’s mass is M. Solve: (a) The energy conservation equation K 2 + U s2 + U g2 = K1 + U s1 + U g1 becomes
1 1 1 1 (m + M )v22 + k ( y2 − ye )2 + (m + M ) g ( y2 − ye ) = (m + M )v12 + k ( y1 − ye )2 − (m + M ) g ( y1 − ye ) 2 2 2 2 Noting v2 = 0 m/s, we can rewrite the above equation as
k (Δy2 ) 2 + 2( m + M ) g ( Δy2 + Δy1 ) = (m + M )v12 + k (Δy1 ) 2
Let us express v1 in terms of the bullet’s initial speed vB by using the momentum conservation equation pf = pi which is (m + M )v1 = mvB + Mvblock . Since vblock = 0 m/s, we have ⎛ m ⎞ v1 = ⎜ ⎟ vB ⎝m+M ⎠
We can also find the magnitude of y1 from the equilibrium condition k ( y1 − ye ) = Mg . Δy1 =
Mg k
With these substitutions for v1 and Δy1 , the energy conservation equation simplifies to k (Δy2 ) 2 + 2(m + M ) g (Δy1 + Δy2 ) =
m 2vB2 ⎛ Mg ⎞ + k⎜ ⎟ (m + M ) ⎝ k ⎠
2
2
(m + M ) M 2 g 2 ⎛m+M ⎞ ⎛m+M ⎞ 2 ⇒ vB2 = 2 ⎜ Δ + Δ − + k⎜ g y y ( ) 1 2 ⎟ ⎟ ( Δy 2 ) 2 2 m k ⎝ m ⎠ ⎝ m ⎠ We still need to include the spring’s maximum compression (d) into this equation. We assume that d = Δy1 + Δy2 , that is, maximum compression is measured from the initial position ( y1 ) of the block. Thus, using Δy2 = d − Δy1 = (d − Mg / k ), we have ⎡ ⎛ m + M ⎞2 ⎛m+M vB = ⎢ 2 ⎜ ⎟ gd − ⎜ 2 ⎝ m ⎢⎣ ⎝ m ⎠
2 2 ⎞M g ⎛m+M + k⎜ ⎟ 2 ⎠ k ⎝ m
12
⎤ ⎞ 2 ⎟ ( d − Mg / k ) ⎥ ⎠ ⎥⎦
(b) Using m = 0.010 kg, M = 2.0 kg, k = 50 N/m, and d = 0.45 m, 2
2
⎛ 2.0 kg ⎞ ⎛ 2.010 kg ⎞ 2 2 2 vB2 = 2 ⎜ ⎟ (9.8 m/s ) /(50 N/m) ⎟ (9.8 m/s )(0.45 m) − (2.010 kg) ⎜ 0.010kg 0.010 kg ⎝ ⎠ ⎝ ⎠ 1 + (50 N/m)(2.010 kg) [0.45 m − (2.0 kg) × (9.8 m/s 2 )/50 N/m]2 (0.010 kg) 2 ⇒ vB = 453 m/s The bullet has a speed of 4.5 × 10 2 m/s. Assess: This is a reasonable speed for the bullet.
10.53. Model: Because the track is frictionless, the sum of the kinetic and gravitational potential energy does not change during the car’s motion. Visualize:
We place the origin of the coordinate system at the ground level directly below the car’s starting position. This is a two-part problem. If we first find the maximum speed at the top of the hill, we can use energy conservation to find the maximum initial height. Solve: Because its motion is circular, at the top of the hill the car has a downward-pointing centripetal accelG eration ac = −(mv 2 / r ) ˆj. Newton’s second law at the top of the hill is
( Fnet ) y = n y + ( FG ) y = n − mg = m( ac ) y = −
⎛ mv 2 mv 2 v2 ⎞ ⇒ n = mg − = m⎜ g − ⎟ R R R⎠ ⎝
If v = 0 m / s, n = mg as expected in static equilibrium. As v increases, n gets smaller—the weight of the car and passengers decreases as they go over the top. But n has to remain positive for the car to be on the track, so the maximum speed vmax occurs when n → 0. We see that vmax = gR . Now we can use energy conservation to relate the top of the hill to the starting height:
1 1 1 K f + U f = Ki + U i ⇒ mvf2 + mgyf = mv12 + mgy1 ⇒ mgR + mgR = 0 J + mghmax 2 2 2 where we used vf = vmax and yf = R. Solving for hmax gives hmax = 32 R. (b) If R = 10 m, then hmax = 15 m.
10.54. Model: This is a two-part problem. In the first part, we will find the critical velocity for the block to go over the top of the loop without falling off. Since there is no friction, the sum of the kinetic and gravitational potential energy is conserved during the block’s motion. We will use this conservation equation in the second part to find the minimum height the block must start from to make it around the loop. Visualize:
We place the origin of our coordinate system directly below the block’s starting position on the frictionless track. Solve: The free-body diagram on the block implies mv 2 FG + n = c R For the block to just stay on track, n = 0. Thus the critical velocity vc is
FG = mg =
mvc2 ⇒ vc2 = gR R
The block needs kinetic energy 12 mvc2 = 12 mgR to go over the top of the loop. We can now use the conservation of mechanical energy equation to find the minimum height h. 1 1 K f + U gf = K i + U gi ⇒ mvf2 + mgyf = mvi2 + mgyi 2 2
Using vf = vc = gR , yf = 2 R, vi = 0 m/s, and yi = h, we obtain
1 gR + g (2 R ) = 0 + gh ⇒ h = 2.5R 2
10.55. Model: Model Lisa (L) and the bobsled (B) as particles. We will assume the ramp to be frictionless,
so that the mechanical energy of the system (Lisa + bobsled + spring) is conserved. Furthermore, during the collision, as Lisa leaps onto the bobsled, the momentum of the Lisa + bobsled system is conserved. We will also assume the spring to be an ideal one that obeys Hooke’s law. Visualize:
We place the origin of our coordinate system directly below the bobsled’s initial position. Solve: (a) Momentum conservation in Lisa’s collision with bobsled states p1 = p0 , or (mL + mB )v1 = mL (v0 ) L + mB (v0 ) B ⇒ (mL + mB )v1 = mL (v0 ) L + 0 ⎛ mL ⎞ ⎛ ⎞ 40 kg ⇒ v1 = ⎜ ⎟ (v0 ) L = ⎜ ⎟ (12 m/s) = 8.0 m/s m + m 40 kg + 20 kg ⎝ ⎠ B ⎠ ⎝ L The energy conservation equation: K 2 + U s2 + U g 2 = K1 + U s1 + U g1 is 1 1 1 1 (mL + mB )v22 + k ( x2 − xe ) 2 + ( mL + mB ) gy2 = (mL + mB )v12 + k ( xe − xe ) 2 + (mL + mB ) gy1 2 2 2 2 Using v2 = 0 m/s, k = 2000 N/m, y2 = 0 m, y1 = (50 m)sin 20° = 17.1 m, v1 = 8.0 m/s, and (mL mB ) = 60 kg, we get 1 1 0 J + (2000 N / m)( x2 − xe ) 2 + 0 J = (60 kg)(8.0 m / s) 2 + 0 J + (60 kg)(9.8 m / s 2 )(17.1 m) 2 2 Solving this equation yields ( x2 − xe ) = 3.5 m. (b) As long as the ice is slippery enough to be considered frictionless, we know from conservation of mechanical energy that the speed at the bottom depends only on the vertical descent Δy. Only the ramp’s height h is important, not its shape or angle.
10.56. Model: We can divide this problem into two parts. First, we have an elastic collision between the 20 g ball (m) and the 100 g ball (M). Second, the 100 g ball swings up as a pendulum. Visualize:
The figure shows three distinct moments of time: the time before the collision, the time after the collision but before the two balls move, and the time the 100 g ball reaches its highest point. We place the origin of our coordinate system on the 100 g ball when it is hanging motionless. Solve: For a perfectly elastic collision, the ball moves forward with speed
(v1 ) M =
2mm 1 (v0 ) m = (v0 ) m mm + mM 3
In the second part, the sum of the kinetic and gravitational potential energy is conserved as the 100 g ball swings up after the collision. That is, K 2 + U g 2 = K1 + U g1. We have 1 1 M (v2 ) 2M + Mgy2 = M (v1 ) 2M + Mgy1 2 2 Using (v2 ) M = 0 J, (v1 ) M =
(v0 ) m , y1 = 0 m, and y2 = L − L cosθ , the energy equation simplifies to 3
g ( L − L cosθ ) =
1 (v0 ) 2m 2 9
⇒ (v0 ) m = 18 g L(1 − cosθ ) = 18(9.8 m/s 2 )(1.0 m)(1 − cos50°) = 7.9 m/s
10.57. Model: Model the balls as particles. We will use the Galilean transformation of velocities (Equation 10.44) to analyze the problem of elastic collisions. We will transform velocities from the lab frame S to a frame S′ in which one ball is at rest. This allows us to apply Equations 10.43 to the case of a perfectly elastic collision in S′, find the final velocities of the balls in S′, and then transform these velocities back to the lab frame S. Visualize: Let S′ be the frame of the 200 g ball. Denoting masses as m1 = 100 g and m2 = 200 g, the initial velocities in the S frame are (vix )1 = 4 m/s and (vix ) 2 = −3 m/s.
Figure (a) shows the before-collision situation as seen in frame S, and figure (b) shows the before-collision situation as seen in frame S′. The after-collision velocities in S′ are shown in figure (c), and figure (d) indicates velocities in S after they have been transformed to frame S from S′. Solve: (a) In the S frame, (vix )1 = 4 m/s and (vix ) 2 = −3 m/s. S′ is the reference frame of the 200 g ball, so V = −3 m/s. The velocities of the two balls in this frame can be obtained using the Galilean transformation of velocities v′ = v − V . So, (vix )1′ = (vix )1 − V = 4 m/s − ( − 3 m/s) = 7 m/s
(vix )′2 = (vix ) 2 − V = −3 m/s − ( − 3 m/s) = 0 m/s
Figure (b) shows the “before” situation, where ball 2 is at rest. Now we can use Equations 10.43 to find the after-collision velocities in frame S′: (vfx )1′ =
m1 − m2 100 g − 200 g 7 (vix )1′ = (7 m/s) = − m/s m1 + m2 100 g + 200 g 3
(vfx )′2 =
2m1 2(100) g 14 (vix )1′ = (7 m/s) = m/s m1 + m2 100 g + 200 g 3
Finally, we need to apply the reverse Galilean transformation v = v′ + V , with the same V, to transform the aftercollision velocities back to the lab frame S: 7 (vfx )1 = (vfx )1′ + V = − m/s − 3 m/s = −5.33 m/s 3 14 (vfx ) 2 = (vfx )′2 + V = m/s − 3 m/s = 1.67 m/s 3 Figure (d) shows the “after” situation in the lab frame. The 100 g ball is moving left at 5.3 m/s; the 200 g ball is moving right at 1.7 m/s.
(b) The momentum conservation equation pfx = pix for a perfectly inelastic collision is
(m1 + m2 )vfx = m1 (vix )1 + m2 (vix ) 2 (0.100 kg + 0.200 kg)vfx = (0.100 kg)(4.0 m/s) + (0.200 kg)(−3.0 m/s) ⇒ vfx = −0.667 m/s Both balls are moving left at 0.67 m/s.
10.58. Model: Model the balls as particles. We will use the Galilean transformation of velocities (Equation 10.44) to analyze the problem of elastic collisions. We will transform velocities from the lab frame S to a frame S′ in which one ball is at rest. This allows us to apply Equations 10.43 to a perfectly elastic collision in S′. After finding the final velocities of the balls in S′, we can then transform these velocities back to the lab frame S. Visualize: Let S′ be the frame of the 400 g ball. Denoting masses as m1 = 100 g and m2 = 400 g, the initial velocities in the S frame are (vix )1 = + 4.0 m/s and (vix ) 2 = +1.0 m/s.
Figures (a) and (b) show the before-collision situations in frames S and S′, respectively. The after-collision velocities in S′ are shown in figure (c). Figure (d) indicates velocities in S after they have been transformed to S from S′. Solve: In frame S, (vix )1 = 4.0 m/s and (vix ) 2 = 1.0 m/s. Because S′ is the reference frame of the 400 g ball, V = 1.0 m/s. The velocities of the two balls in this frame can be obtained using the Galilean transformation of velocities v′ = v − V . So, (vix )1′ = (vix )1 − V = 4.0 m/s − 1.0 m/s = 3.0 m/s (vix )′2 = (vix ) 2 − V = 1.0 m/s − 1.0 m/s = 0 m/s
Figure (b) shows the “before” situation in frame S′ where the ball 2 is at rest. Now we can use Equations 10.43 to find the after-collision velocities in frame S′: (vfx )1′ =
m1 − m2 100 g − 400 g (vix )1′ = (3.0 m/s) = −1.80 m/s m1 + m2 100 g + 400 g
(vfx )′2 =
2m1 2(100 g) (vix )1′ = (3.0 m/s) = 1.20 m/s m1 + m2 100 g + 400 g
Finally, we need to apply the reverse Galilean transformation v = v′ + V , with the same V, to transform the aftercollision velocities back to the lab frame S. (vfx )1 = (vfx )1′ + V = −1.80 m/s + 1.0 m/s = −0.80 m/s (vfx ) 2 = (vfx )′2 + V = 1.20 m/s + 1.0 m/s = 2.20 m/s
Figure (d) shows the “after” situation in frame S. The 100 g ball moves left at 0.80 m/s, the 400 g ball right at 2.2 m/s. Assess: The magnitudes of the after-collision velocities are similar to the magnitudes of the before-collision velocities.
10.59. Model: Use the model of the conservation of mechanical energy. Visualize:
Solve: (a) The turning points occur where the total energy line crosses the potential energy curve. For E = 12 J, this occurs at the points x = 1 m and x = 7 m. (b) The equation for kinetic energy K = E − U gives the distance between the potential energy curve and total energy line. U = 8 J at x = 2 m, so K = 12 J − 8 J = 4 J. The speed corresponding to this kinetic energy is
v=
2K 2(4 J) = = 4.0 m/s m 0.5 kg
(c) Maximum speed occurs for minimum U. This occurs at x = 1 m and x = 4 m, where U = 0 J and K = 12 J. The speed at these two points is
v=
2K 2(12 J) = = 6.9 m/s m 0.500 kg
(d) The particle leaves x = 1 m with v = 6.9 m/s. It gradually slows down, reaching x = 2 m with a speed of 4.0 m/s. After x = 2 m, it speeds up again, returning to a speed of 6.9 m/s as it crosses x = 4 m. Then it slows again, coming instantaneously to a halt (v = 0 m/s) at the x = 7 m turning point. Now it will reverse direction and move back to the left. (e) If the particle has E = 4 J it cannot cross the 8 J potential energy “mountain” in the center. It can either oscillate back and forth over the range 1.0 m ≤ x ≤ 1.5 m or over the range 3 m ≤ x ≤ 5 m.
10.60. Model: We will use the conservation of mechanical energy. Visualize:
The potential energy (U ) of the nitrogen atom as a function of z exhibits a double-minimum behavior; the two minima correspond to the nitrogen atom’s position on both sides of the plane containing the three hydrogens. Solve: (a) At room temperature, the total energy line is below the “hill” in the center of the potential energy curve. That is, the nitrogen atom does not have sufficient energy to pass from one side of the molecule to the other. There’s a stable equilibrium position on either side of the hydrogen-atom plane at the points where U = 0. Since E > 0, the nitrogen atom will be on one side of the plane and will make small vibrations back and forth along the z-axis—that is, toward and away from the hydrogen-atom plane. In the figure above, the atom oscillates between points A and B. (b) The total energy line is now well above the “hill,” and the turning points of the nitrogen atom’s motion (where the total energy line crosses the potential curve) are at points C and D. In other words, the atom oscillates from one side of the H3 plane to the other. It slows down a little as it passes through the plane of hydrogen atoms, but it has sufficient energy to get through.
dU = 0. dx dU 1 = 0 = 1 + 2cos ( 2 x ) ⇒ cos ( 2 x ) = − dx 2 1 ⎛ 1⎞ ⇒ x = cos −1 ⎜ − ⎟ 2 ⎝ 2⎠
10.61. Solve: (a) The equilibrium positions are located at points where
1 2π 4π ⎛ 1⎞ is in radians and x is in meters. The function cos −1 ⎜ − ⎟ may have values and . Thus 2 3 3 2 ⎝ ⎠ there are two values of x, π 2π x1 = and x2 = 3 3 within the interval 0 m ≤ x ≤ π m. (b) A point of stable equilibrium corresponds to a local minimum, while a point of unstable equilibrium corresponds to a local maximum. Compute the concavity of U(x) at the equilibrium positions to determine their stability. d 2U = −4sin ( 2 x ) dx 2 ⎛ 3⎞ d 2U π π d 2U π At x1 = , x < 0, x1 = is a local maximum, so x1 = is a point x = −4 ⎜⎜ ⎟⎟ = −2 3. Since 2 ( 1) 2 ( 1) 3 3 dx dx 3 2 ⎝ ⎠ of unstable equilibrium. ⎛ 2π d 2U 3⎞ d 2U 2π 2π − x2 ) = −4 ⎜⎜ − At x2 = > 0, x2 = is a local minimum, so x2 = is a , ⎟⎟ = +2 3. Since 2 ( 2 3 dx 2 dx 3 3 ⎝ ⎠ point of stable equilibrium.
Note that −
10.62. Model: The potential energy of two nucleons interacting via the strong force is U = U 0 [1 − e − x / x0 ] where x is the distance between the centers of the two nucleons, x0 = 2.0 × 10−15 m, and U 0 = 6.0 × 10−11 J. Visualize: Nucleons are protons and neutrons, and they are held together in the nucleus by a force called the strong force. This force exists between nucleons at very small separations. Solve: (a)
(b) For x = 5.0 × 10−15 m, U = 55.1 × 10−12 J. This energy is represented by a total energy line. (c) Due to conservation of total energy, the potential energy when x = 5.0 × 10−15 J is transformed into kinetic energy until x = twice the radius = 1.0 × 10−15 m. At this separation, u = 15.7 × 10−12 J. Thus,
1 2 1 2 mv + mv + 15.7 × 10−12 J = 55.1 × 10 −12 J ⇒ v = 1.53 × 108 m/s 2 2 Assess: A speed of 1.53 × 108 m/s is approximately 0.5 c where c is the speed of light. This speed is understandable for the present model.
10.63. A 2.5 kg ball is thrown upward at a speed of 4.0 m/s from a height of 82 cm above a vertical spring. When the ball comes down it lands on and compresses the spring. If the spring has a spring constant of k = 600 N/m, by how much is it compressed?
10.64. (a) A 1500 kg auto coasts up a 10.0 m high hill and reaches the top with a speed of 5.0 m/s. What initial speed must the auto have had at the bottom of the hill? (b)
We place the origin of our coordinate system at the bottom of the hill. (c) The solution of the equation is vi = 14.9 m/s. This is approximately 30 mph and is a reasonable value for the speed at the bottom of the hill.
10.65. (a) A spring gun is compressed 15 cm to launch a 200 g ball on a horizontal, frictionless surface. The ball has a speed of 2.0 m/s as it loses contact with the spring. Find the spring constant of the gun. (b)
We place the origin of our coordinate system on the free end of the spring in the equilibrium position. Because the surface is frictionless, the mechanical energy for the system (ball + spring) is conserved. (c) The conservation of energy equation is
K f + U sf = K i + U si 1 2 1 1 1 mv1 + k (0 m) 2 = m(0 m/s) 2 + k ( −0.15 m) 2 2 2 2 2 (0.200 kg)(2.0 m/s) 2 = k ( −0.15 m) 2 k = 36 N/m
10.66. (a) A 100 g lump of clay traveling at 3.0 m/s strikes and sticks to a 200 g lump of clay at rest on a
frictionless surface. The combined lumps smash into a horizontal spring with k = 3.0 N/m. The other end of the spring is firmly anchored to a fixed post on the surface. How far will the spring compress? (b)
(c) Solving the conservation of momentum equation we get v1x = 1.0 m/s. Substituting this value into the conservation of energy equation yields Δx2 = 32 cm.
10.67. (a) A spring with spring constant 400 N/m is anchored at the bottom of a frictionless 30° incline. A 500 g block is pressed against the spring, compressing the spring by 10 cm, then released. What is the speed with which the block is launched up the incline? (b) The origin is placed at the end of the uncompressed spring. This is the point from which the block is launched as the spring expands.
(c) Solving the energy conservation equation, we get vf = 2.6 m/s.
10.68. Model: Assume an ideal spring that obeys Hooke’s law. The mechanical energy K + U s + U g is conserved during the launch of the ball. Visualize:
This is a two-part problem. In the first part, we use projectile equations to find the ball’s velocity v2 as it leaves the spring. This will yield the ball’s kinetic energy as it leaves the spring. Solve: Using the equations of kinematics,
1 x3 = x2 + v2 x (t3 − t2 ) + ax (t3 − t2 ) 2 ⇒ 5.0 m = 0 m + (v2 cos30°)(t3 − 0 s) + 0 m 2 (v2 cos30°)t3 = 5.0 m ⇒ t3 = (5.0 m/ v2 cos30°) 1 y3 = y2 + v2 y (t3 − t2 ) + a y (t3 − t2 ) 2 2 1 −1.5 m = 0 + (v2 sin 30°)(t3 − 0 s) + (9.8 m / s 2 )(t3 − 0 s) 2 2 ⎛ 5.0 m ⎞ 5.0 m ⎞ 2 ⎛ Substituting the value for t3 , (−1.5 m) = (v2 sin 30°) ⎜ ⎟ − ( 4.9 m/s ) ⎜ ⎟ ⎝ v2 cos30° ⎠ ⎝ v2 cos30° ⎠
⇒ (−1.5 m) = + (2.887 m) −
2
163.33 ⇒ v2 = 6.102 m/s v22
The conservation of energy equation K 2 + U s2 + U g 2 = K1 + U s1 + U g1 is 1 2 1 1 1 mv2 + k (0 m) 2 + mgy2 = mv12 + k ( Δs ) 2 + mgy1 2 2 2 2
Using y2 = 0 m, v1 = 0 m/s, Δs = 0.20 m, and y1 = −(Δs )sin g 30°, we get 1 2 1 mv2 + 0 J + 0 J = 0 J + k ( Δs ) 2 − mg (Δs )sin 30° (Δs ) 2 k = mv22 + 2mg (Δs )sin 30° 2 2 (0.20 m) 2 k = (0.020 kg)(6.102 m/s)2 + 2(0.020 kg)(9.8 m/s 2 )(0.20)(0.5) ⇒ k = 19.6 N/m Assess: Note that y1 = −( Δs )sin 30° is with a minus sign and hence the gravitational potential energy mgy1 is − mg (Δs )sin 30°.
10.69. Model: This is a two-part problem. In the first part, we will find the critical velocity for the ball to go over the top of the peg without the string going slack. Using the energy conservation equation, we will then obtain the gravitational potential energy that gets transformed into the critical kinetic energy of the ball, thus determining the angle θ . Visualize:
We place the origin of our coordinate system on the peg. This choice will provide a reference to measure gravitational potential energy. For θ to be minimum, the ball will just go over the top of the peg. Solve: The two forces in the free-body force diagram provide the centripetal acceleration at the top of the circle. Newton’s second law at this point is mv 2 FG + T = r where T is the tension in the string. The critical speed vc at which the string goes slack is found when T → 0. In this case, mg =
mvC2 ⇒ vC2 = gr = gL 3 r
The ball should have kinetic energy at least equal to 1 2 1 ⎛ L⎞ mvC = mg ⎜ ⎟ 2 2 ⎝3⎠ for the ball to go over the top of the peg. We will now use the conservation of mechanical energy equation to get the minimum angle θ . The equation for the conservation of energy is 1 1 K f + U gf = K i + U gi ⇒ mvf2 + mgyf = mvi2 + mgyi 2 2
Using vf = vc , yf = 13 L, vi = 0, and the above value for vC2 , we get 1 L L L mg + mg = mgyi ⇒ yi = 2 3 3 2 That is, the ball is a vertical distance
1 2
L above the peg’s location or a distance of ⎛ 2L L ⎞ L − ⎟= ⎜ ⎝ 3 2⎠ 6
below the point of suspension of the pendulum, as shown in the figure on the right. Thus, cosθ =
L /6 1 = ⇒ θ = 80.4° L 6
10.70. Model: Choose yourself + spring + earth as the system. There are no forces from outside this system, so it is an isolated system. The interaction forces within the system are the spring force of the bungee cord and the gravitational force. These are both conservative forces, so mechanical energy is conserved. Visualize:
We can equate the system’s initial energy, as you step off the bridge, to its final energy when you reach the lowest point. We do not need to compute your speed at the point where the cord starts to stretch. We do, however, need to note that the end of the unstretched cord is at y0 = y1 − 30 m = 70 m, so U 2s = 12 k ( y2 − y0 ) 2 .
Also note that U1s = 0, since the cord is not stretched. The energy conservation equation is 1 K 2 + U 2g + U 2s = K1 + U1g + U1s ⇒ 0 J + mgy2 + k ( y2 − y0 ) 2 = 0 J + mgy1 + 0 J 2 Multiply out the square of the binomial and rearrange: 1 1 mgy2 + ky22 − ky0 y2 + ky02 = mgy1 2 2 ⎛ 2mg ⎞ ⎛ 2 2mgy1 ⎞ 2 2 ⇒ y2 + ⎜ − 2 y0 ⎟ y2 + ⎜ y0 − ⎟ = y2 − 100.8 y2 + 980 = 0 k k ⎝ ⎠ ⎝ ⎠ This is a quadratic equation with roots 89.9 m and 10.9 m. The first is not physically meaningful because it is a height above the point where the cord started to stretch. So we find that your distance from the water when the bungee cord stops stretching is 10.9 m.
10.71. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, hence the mechanical energy K + U g + U s is conserved. Visualize:
We have chosen to place the origin of the coordinate system at the point of maximum compression. We will use lengths along the ramp with the variable s rather than x. Solve: (a) The conservation of energy equation K 2 + U g2 + U s2 = K1 + U g1 + U s1 is
1 2 1 1 mv2 + mgy2 + k ( Δs ) 2 = mv12 + mgy1 + k (0 m) 2 2 2 2 1 1 1 m(0 m/s) 2 + mg (0 m) + k ( Δs ) 2 = m(0 m/s) 2 + mg (4.0 m + Δs )sin 30° + 0 J 2 2 2 1 ⎛1⎞ (250 N/m)( Δs ) 2 = (10 kg)(9.8 m/s 2 )(4.0 m + Δs ) ⎜ ⎟ 2 ⎝ 2⎠ This gives the quadratic equation: (125 N/m)(Δs ) 2 − (49 kg ⋅ m/s 2 ) Δs − 196 kg ⋅ m 2 /s 2 = 0 ⇒ Δs = 1.46 m and − 1.07 m (unphysical)
The maximum compression is 1.46 m. (b) We will now apply the conservation of mechanical energy to a point where the vertical position is y and the block’s velocity is v. We place the origin of our coordinate system on the free end of the spring when the spring is neither compressed nor stretched.
1 2 1 1 1 mv + mgy + k ( Δs ) 2 = mv12 + mgy1 + k (0 m) 2 2 2 2 2 1 2 1 mv + mg (−Δs sin 30°) + k ( Δs ) 2 = 0 J + mg (4.0 m sin 30°) + 0 J 2 2 1 1 2 k (Δs ) − ( mg sin 30°) Δs + mv 2 − mg sin 30°(4.0 m) = 0 2 2 To find the compression where v is maximum, take the derivative of this equation with respect to Δs : 1 1 dv k 2(Δs ) − (mg sin 30°) + m 2v −0=0 2 2 d Δs
Since
dv = 0 at the maximum, we have d Δs Δs = (mg sin 30°)/ k = (10 kg)(9.8 m/s 2 )(0.5)/(250 N/m) = 19.6 cm
10.72. Model: Assume an ideal, massless spring that obeys Hooke’s law. Let us also assume that the cannon (C) fires balls (B) horizontally and that the spring is directly behind the cannon to absorb all motion. Visualize:
The before-and-after pictorial representation is shown, with the origin of the coordinate system located at the spring’s free end when the spring is neither compressed nor stretched. This free end of the spring is just behind the cannon. Solve: The momentum conservation equation pfx = pix is
mB (vfx ) B + mC (vfx )C = mB (vix ) B + mC (vix )C Since the initial momentum is zero,
(vfx ) B = −
⎛ 200 kg ⎞ mC (vfx )C = − ⎜ ⎟ (vfx )C = −20(vfx )C mB ⎝ 10 kg ⎠
The mechanical energy conservation equation for the cannon + spring K f + U sf = Ki + U si is 1 1 1 1 1 m(vf )C2 + k (Δx) 2 = m(vi )C2 + 0 J ⇒ 0 J + k (Δx) 2 = m(vfx )C2 2 2 2 2 2 k (20,000 N/m) ⇒ (vfx )C = ± Δx = ± (0.50 m) = ±5.0 m/s m 200 kg To make this velocity physically correct, we retain the minus sign with (vfx )C . Substituting into the momentum conservation equation yields: (vfx ) B = −20(−5.0 m/s) = 100 m/s
10.73. Model: This is a collision between two objects, and momentum is conserved in the collision. In addition, because the interaction force is a spring force and the surface is frictionless, energy is also conserved. Visualize:
Let part 1 refer to the time before the collision starts, part 2 refer to the instant when the spring is at maximum compression, and part 3 refer to the time after the collision. Notice that just for an instant, when the spring is at maximum compression, the two blocks are moving side by side and have equal velocities: vA2 = vB2 = v2 . This is an important observation. Solve: First relate part 1 to part 2. Conservation of energy is 1 1 1 1 1 1 2 2 2 mA vA1 = mA vA2 + mBvB2 + k (Δxmax ) 2 = ( mA + mB )v22 + k (Δxmax ) 2 2 2 2 2 2 2 where Δxmax is the spring’s compression. U g is not in the equation because there are no elevation changes. Also note
that K 2 is the sum of the kinetic energies of all moving objects. Both v2 and Δxmax are unknowns. Now add the conservation of momentum: mA vA1 mA vA1 = mA vA 2 + mBvB2 = (mA + mB )v2 ⇒ v2 = = 2.667 m/s mA + mB
Substitute this result for v2 into the energy equation to find: 1 1 1 2 k (Δxmax ) 2 = mA vA1 − (mA + mB )v22 2 2 2 2 mAvA1 − (mA + mB )v22 = 0.046 m = 4.6 cm k Notice how both conservation laws were needed to solve this problem. (b) Again, both energy and momentum are conserved between “before” and “after.” Energy is 1 1 1 1 2 2 2 2 2 mA vA1 = mA vA3 + mBvB3 ⇒ 16 = vA3 + vB3 2 2 2 2 The spring is no longer compressed, so the energies are purely kinetic. Momentum is mA vA1 = mA vA3 + mBvB3 ⇒ 8 = 2vA3 + vB3
⇒ Δxmax =
We have two equations in two unknowns. From the momentum equation, we can write vB3 = 2(4 − vA3 ) and use this in the energy equation to obtain: 2 1 2 2 2 2 16 = vA3 + ⋅ 4 ( 4 − vA3 = 3vA3 − 16vA3 + 32 ⇒ 3vA3 − 16vA3 + 16 = 0 ) 2 This is a quadratic equation for vA3 with roots vA3 = (4 m/s, 1.33 m/s). Using vB3 = 2(4 − vA3 ), these two roots give vB3 = (0 m/s, 5.333 m/s). The first pair of roots corresponds to a “collision” in which A misses B, so each keeps its initial speed. That’s not the situation here. We want the second pair of roots, from which we learn that the blocks’ speeds after the collision are vA3 = 1.33 m/s and vB3 = 5.3 m/s.
10.74. Model: Mechanical energy and momentum are conserved during the expansion of the spring. Visualize: Please refer to Figure CP10.74. Solve: Example 10.8 is a very similar problem, except that the objects are initially at rest. We can use the solution from Example 10.8 for this problem in a reference frame S′ in which the two carts are initially at rest, then transform the answer to the frame S in which the carts are initially moving. Thus in the S′ frame,
( vfx )′2 =
k ( Δxi ) m m2 1 + 2
( vfx )1′ = −
(
2
m1
)
m2 ( vfx )2′ m1
Let the 100 g cart be Cart 1 and the 300 g cart be Cart 2. With k = 120 N/m and Δxi = 4.0cm,
( vfx )2′ = 0.40 m/s, ( vfx )1′ = −1.2 m/s An object at rest in the S′ frame is traveling to the right at 1.0 m/s in the S frame. The equation of transformation is therefore vx = vx′ + 1.0 m/s In the S frame, the velocities of the carts are ( vfx )1 = −1.2 m/s + 1.0 m/s = −0.2 m/s
( vfx )2 = 0.40 m/s + 1.0 m/s = 1.4 m/s
Assess:
Cart 1 is moving slowly to the left while the heavier Cart 2 is moving quickly to the right.
10.75. Model: Model the balls as particles, and assume a perfectly elastic collision. After the collision is over, the balls swing out as pendulums. The sum of the kinetic energy and gravitational potential energy does not change as the balls swing out. Visualize:
In the pictorial representation we have identified before-and-after quantities for both the collision and the pendulum swing. We have chosen to place the origin of the coordinate system at a point where the two balls at rest barely touch each other. Solve: As the ball with mass m1 , whose string makes an angle θ il with the vertical, swings through its equilibrium position, it lowers its gravitational energy from m1 gy0 = m1 g ( L − L cosθ i1 ) to zero. This change in potential energy transforms into a change in kinetic energy. That is, 1 m1 g ( L − L cosθ i1 ) = m1 (v1 )12 ⇒ (v1 )1 = 2 gL(1 − cosθ i1 ) 2
Similarly, (v1 )2 = 2 gL(1 − cosθ i2 ). Using θ i1 = θ i2 = 45°, we get (v1 )1 = 2.396 m/s = (v1 ) 2 . Both balls are moving at the point where they have an elastic collision. Since our analysis of elastic collisions was for a situation in which ball 2 is initially at rest, we need to use the Galilean transformation to change to a frame S′ in which ball 2 is at rest. Ball 2 is at rest in a frame that moves with ball 2, so choose S′ to have V = −2.396 m/s, with the minus sign because this frame (like ball 2) is moving to the left. In this frame, ball 1 has velocity (v1′ )1 = (v1 )1 − V = 2.396 m/s + 2.396 m/s = 4.792 m/s and ball 2 is at rest. The elastic collision causes the balls to move with velocities m − m2 1 (v2′ )1 = 1 (v1′ )1 = – (4.792 m/s)= − 1.597 m/s 3 m1 + m2 (v′2 ) 2 =
2m1 2 (v1′ )1 = (2.396 m/s)=3.194 m/s m1 + m2 3
We can now use v = v′ + V to transform these back into the laboratory frame: (v2 )1 = −1.597 m/s − 2.396 m/s = −3.99 m/s (v2 ) 2 = 3.195 m/s − 2.396 m/s = 0.799 m/s
Having determined the velocities of the two balls after collision, we will once again use the conservation equation Kf + U gf = Ki + U gi for each ball to solve for the θ f1 and θ f2 . 1 1 m1 (v3 )12 + m1 gL(1 − cosθ f 1 ) = m1 (v2 )12 + 0 J 2 2 Using (v3 )1 = 0, this equation simplifies to
1 1 gL(1 − cosθ f 1 ) = (−3.99 m/s) 2 ⇒ cosθ f 1 = 1 − (−3.99 m/s) 2 ⇒ θ f 1 = 79.3° 2 2 gL The 100 g ball rebounds to 79º. Similarly, for the other ball: 1 1 m2 (v3 ) 22 + m2 gL(1 − cosθ f 2 ) = m2 (v2 ) 22 + 0 J 2 2 Using (v3 ) 2 = 0, this equation becomes
⎛ 1 ⎞ 2 cosθ f 2 = 1 − ⎜ ⎟ (0.799) ⇒ θ f 2 = 14.7° ⎝ 2gL ⎠ The 200 g ball rebounds to 14.7°.
10.76. Model: Model the sled as a particle. Because there is no friction, the sum of the kinetic and gravitational potential energy is conserved during motion. Visualize:
Place the origin of the coordinate system at the center of the hemisphere. Then y0 = R and, from geometry,
y1 = R cos φ . Solve:
The energy conservation equation K1 + U1 = K 0 + U 0 is
1 2 1 1 mv1 + mgy1 = mv02 + mgy0 ⇒ mv12 + mgR cos φ = mgR ⇒ v1 = 2 gR (1 − cos φ ) 2 2 2 G (b) If the sled is on the hill, it is moving in a circle and the r-component of Fnet has to point to the center with magnitude Fnet = mv 2 / R. Eventually the speed gets so large that there is not enough force to keep it in a circular trajectory, and that is the point where it flies off the hill. Consider the sled at angle φ . Establish an r-axis pointing toward the center of the circle, as we usually do for circular motion problems. Newton’s second law along this axis requires:
mv 2 R ⎛ mv 2 v2 ⎞ ⇒ n = mg cos φ − = m ⎜ g cos φ − ⎟ R R⎠ ⎝
( Fnet ) r = FG cos φ − n = mg cos φ − n = mar =
The normal force decreases as v increases. But n can’t be negative, so the fastest speed at which the sled stays on the hill is the speed vmax that makes n → 0. We can see that vmax = gR cos φ . (c) We now know the sled’s speed at angle φ , and we know the maximum speed it can have while remaining on the hill. The angle at which v reaches vmax is the angle φmax at which the sled will fly off the hill. Combining the two expressions for v1 and vmax gives: 2 gR (1 − cos φ ) = gR cos φ ⇒ 2 R (1 − cos φmax ) = R cos φmax
⇒ cos φmax =
2 ⎛ 2⎞ ⇒ φmax = cos −1 ⎜ ⎟ = 48° 3 ⎝3⎠
11.1. Visualize: Please refer to Figure EX11.1. G G Solve: (a) A ⋅ B = AB cosα = (4)(5)cos 40° = 15.3. (b) (c)
G G C ⋅ D = CD cosα = (2)(4)cos120° = −4.0. G G E ⋅ F = EF cosα = (3)(4)cos90° = 0.
11.2. Visualize: Please refer to Figure EX11.2. G G Solve: (a) A ⋅ B = AB cosα = (3)(4)cos110° = −4.1. (b) (c)
G G C ⋅ D = CD cosα = (4)(5)cos180° = −20. G G E ⋅ F = EF cosα = (4)(3)cos30° = 10.4.
11.3. Solve: (a) (b)
G G A ⋅ B = Ax Bx + Ay By = (3)( −2) + ( −4)(6) = −30.
G G A ⋅ B = Ax Bx + Ay By = (2)(6) + (3)(−4) = 0.
11.4. Solve: (a) (b)
G G A ⋅ B = Ax Bx + Ay By = (4)(−3) + (−2)( −2) = −16.
G G A ⋅ B = Ax Bx + Ay By = (−4)(−1) + (2)(−2) = 0.
G
G
G G
G
11.5. Solve: (a) The length of A is A = A = A ⋅ A = ( 3) + ( −4 ) = 25 = 5. The length of B is B=
( −2 )
2
2
2
G G 2 + ( 6 ) = = 40 = 2 10. Using the answer A ⋅ B = −30 from Ex11.3(a),
G G A ⋅ B = AB cosα
(
)
⇒ −30 = ( 5 ) 2 10 cosα G G (b) From EX11.3 (b), A ⋅ B = 0. Thus
⇒ α = cos
−1
( −0.9487 ) = 162°
cosα = 0 α = 90°
Assess: In part (a) the vectors are nearly pointing in opposite directions, while in part (b) the vectors are perpendicular.
G
G G G G 2 2 A = A = A ⋅ A = ( 4 ) + ( 2 ) = 20. The length of B is G G 13. Using the answer A ⋅ B = −16 from EX11.4(a), G G A ⋅ B = AB cosα
11.6. Solve: (a) The length of A is B=
( −3)
2
+ ( −2 ) = 2
⇒ −16 =
(
⇒ α = cos
20 −1
)( 13 ) cosα
( −0.9923) = 173°
G G (b) From EX11.4(b), A ⋅ B = 0. Thus
cosα = 0 α = 90° Assess: In part (a) the vectors are nearly pointing in opposite directions, while in part (b) the vectors are perpendicular.
G
G G G W = F ⋅ Δr = (6.0iˆ − 3.0 ˆj ) ⋅ (2.0 ˆj ) N ⋅ m = (12.0iˆ ⋅ ˆj − 6.0 ˆj ⋅ ˆj ) J = −6.0 J.
11.7. Solve: (a) W = F ⋅ Δr = (6.0iˆ − 3.0 ˆj ) ⋅ (2.0iˆ) N ⋅ m = (12.0iˆ ⋅ iˆ − 3.0 ˆj ⋅ iˆ) J = 12.0 J. (b)
G
G
G G G G W = F ⋅ Δr = ( −4.0iˆ − 6.0 ˆj ) N ⋅ (−3.0iˆ + 2.0 ˆj ) m = (12.0 − 12.0) J=0 J.
11.8. Solve: (a) W = F ⋅ Δr = (−4.0iˆ − 6.0 ˆj ) N ⋅ (3.0iˆ) m = ( −12.0iˆ ⋅ iˆ + 12.0 ˆj ⋅ iˆ) J= − 12.0 J. (b)
11.9. Model: Use the work-kinetic energy theorem to find the net work done on the particle. Visualize:
Solve:
From the work-kinetic energy theorem, 1 1 1 1 W = ΔK = mv12 − mv02 = m ( v12 − v02 ) = (0.020 kg)[(30 m/s) 2 − (−30 m/s) 2 ] = 0 J 2 2 2 2
Assess: Negative work is done in slowing down the particle to rest, and an equal amount of positive work is done in bringing the particle to the original speed but in the opposite direction.
G
G
G
G
G
11.10. Model: Work done by a force F on a particle is defined as W = F ⋅ Δr , where Δr is the particle’s displacement. Visualize:
Solve:
(a) The work done by gravity is G G Wg = FG ⋅ Δr = (− mgjˆ) N ⋅ (2.25 − 0.75) ˆj m = −(2.0 kg)(9.8 m/s 2 )(1.50 m) J = −29 J
G G (b) The work done by hand is WH = Fhand on book ⋅ Δr . As long as the book does not accelerate, G G Fhand on book = − Fearth on book = −(− mgjˆ) = mgjˆ
⇒ WH = ( mgjˆ) ⋅ (2.25 − 0.75) ˆj m = (2.0 kg)(9.8 m/s 2 )(1.50 m)=29 J
G
G
G
11.11. Model: Model the piano as a particle and use W = F ⋅ Δr , where W is the work done by the force F G through the displacement Δr . Visualize:
G For the force FG : G G G G W = F ⋅ Δr = FG ⋅ Δr = ( Fg )(Δr )cos0° = (2500 N)(5.00 m)(1) = 1.250 × 104 J G For the tension T1 : G G W = T1 ⋅ Δr = (T1 )(Δr )cos(150°) = (1830 N)(5.00 m)(−0.8660) = −7.92 × 103 J G For the tension T2 : G G W = T2 ⋅ Δr = (T2 )(Δr )cos(135°) = (1295 N)(5.00 m)(−0.7071) = −4.58 × 103 J
Solve:
Assess:
G Note that the displacement Δr in all the above cases is directed downwards along − ˆj.
G
G
G
11.12. Model: Model the crate as a particle and use W = F ⋅ Δr , where W is the work done by a force F on G a particle and Δr is the particle’s displacement. Visualize:
Solve:
G For the force f k :
G For the tension T1 :
G For the tension T2 :
G G W = f k ⋅ Δr = f k (Δr )cos(180°) = (650 N)(3.0 m)(−1) = −1.95 kJ
G G W = T1 ⋅ Δr = (T1 )( Δr )cos 20° = (600 N)(3.0 m)(0.9397) =1.69 kJ
G G W = T2 ⋅ Δr = (T2 )(Δr )cos30° = (410 N)(3.0 m)(0.866) = 1.07 kJ G Assess: Negative work done by the force of kinetic friction ( f k ) means that 1.95 kJ of energy has been transferred out of the crate.
11.13. Model: Model the 2.0 kg object as a particle, and use the work-kinetic energy theorem. Visualize: Please refer to Figure EX11.13. For each of the five intervals the velocity-versus-time graph gives the initial and final velocities. The mass of the object is 2.0 kg. Solve: According to the work-kinetic energy theorem:
1 1 1 W = ΔK = mvf2 = mvi2 = m ( vf2 − vi2 ) 2 2 2 1 Interval AB: vi = 2 m/s, vf = −2 m/s ⇒ W = (2.0 kg)[(−2 m/s) 2 − (2 m/s) 2 ] = 0 J 2 1 Interval BC: vi = −2 m/s, vf = −2 m/s ⇒ W = (2.0 kg)[(−2 m/s) 2 − (−2 m/s) 2 ] = 0 J 2 1 Interval CD: vi = −2 m/s, vf = 0 m/s ⇒ W = (2.0 kg)[(0 m/s) 2 − (−2 m/s) 2 ] = −4.0 J 2 1 Interval DE: vi = 0 m/s, vf = 2 m/s ⇒ W = (2.0 kg)[(2 m/s) 2 − (0 m/s)2 ] = +4.0 J 2 1 Interval EF: vi = 2 m/s, vf = 1 m/s ⇒ W = (2.0 kg)[(1 m/s)2 − (2 m/s)2 ] = −3.0 J 2 Assess: The work done is zero in intervals AB and BC. In the interval CD + DE the total work done is zero. It is not whether v is positive or negative that counts because K ∝ v 2 . What is important is the magnitude of v and how v changes.
11.14. Model: Use the definition of work. Visualize: Please refer to Figure EX11.14. Solve: Work is defined as the area under the force-versus-position graph: sf
W = ∫ Fs ds = area under the force curve si
Interval 0–1 m: W = (4.0 N)(1.0 m − 0.0 m) = 4.0 J Interval 1–2 m: W = (4.0 N)(0.5 m) + (− 4.0 N)(0.5 m) = 0 J 1 Interval 2–3 m: W = ( −4.0 N)(1 m) = −2.0 J 2
11.15. Model: Use the work-kinetic energy theorem to find velocities. Visualize: Please refer to Figure EX11.15. Solve: The work-kinetic energy theorem is x
f 1 1 ΔK = mvf2 − mvi2 = W = ∫ Fx dx = area under the force curve from xi to xf 2 2 xi
x
f 1 1 1 ⇒ mvf2 − (0.500 kg)(2.0m/s) 2 = mvf2 − 1.0 J = ∫ Fx dx = area from 0 to x 2 2 2 0m
1 (0.500 kg)vf2 − 1.0 J = 12.5 J ⇒ vf = 7.35 m/s 2 1 At x = 2 m: (0.500 kg)vf2 − 1.0 J = 20 J ⇒ vf = 9.17 m/s 2 1 At x = 3 m: (0.500 kg)vf2 − 1.0 J = 22.5 J ⇒ vf = 9.70 m/s 2 At x =1 m:
11.16. Model: Use the work-kinetic energy theorem. Visualize: Please refer to Figure EX11.16. Solve: The work-kinetc energy theorem is x
f 1 1 ΔK = mvf2 − mvi2 = W = ∫ Fx dx = area under the force curve from xi to xf 2 2 xi
x
f 1 1 1 ⇒ mvf2 − (2.0 kg)(4.0 m/s) 2 = mvf2 − 16.0 J = ∫ Fx dx = area from 0 to x 2 2 2 0m
1 1 (2.0 kg)vf2 − 16.0 J = (10 N)(2 m) = 10 J ⇒ vf = 5.1 m/s 2 2 1 2 At x = 4 m : (2.0 kg)vf − 16.0 J = 0 J ⇒ vf = 4.0 m/s 2
At x = 2 m :
11.17. Model: Use the work-kinetic energy theorem. Visualize: Please refer to Figure EX11.17. Solve: The work-kinetic energy theorem is xf
ΔK = W = ∫ Fx dx = area of the Fx -versus- x graph between xi and xf xi
1 2 1 2 1 mvf − mvi = ( Fmax )(2 m) 2 2 2 Using m = 0.500 kg, vf = 6.0 m/s, and vi = 2.0 m/s, the above equation yields Fmax = 8.0 N. Assess: Problems in which the force is not a constant can not be solved using constant-acceleration kinematic equations.
11.18. Model: Use the definition Fs = − dU / ds. Visualize: Please refer to Figure EX11.18. Solve: Fx is the negative of the slope of the potential energy graph at position x. Between x = 0 cm and x = 10 cm the slope is
slope = (U f − U i ) / ( xf − xi ) = (0 J − 10 J) / (0.10 m − 0.0 m) = −100 N Thus, Fx = 100 N at x = 5 cm. The slope between x = 10 cm and x = 20 cm is zero, so Fx = 0 N at x = 15 cm. Between 20 cm and 40 cm, slope = (10 J − 0 J) / (0.40 m − 0.20 m) = 50 N At x = 25 cm and x = 35 cm, therefore, Fx = −50 N.
11.19. Model: Use the definition Fs = − dU ds . Visualize: Please refer to Figure EX11.19. Solve: Fx is the negative of the slope of the potential energy graph at position x.
⎛ dU ⎞ Fx = − ⎜ ⎟ ⎝ dx ⎠
Between x = 0 m and x = 3 m, the slope is slope = (U f − U i ) / ( xf − xi ) = (60 J − 0 J) / (3 m − 0 m) = 20 N Thus, Fx = −20 N at x = 1 m. Between x = 3 m and x = 5 m, the slope is
slope = (U f − U i ) / ( xf − xi ) = ( 0 J − 60 J ) / ( 5 m − 3 m ) = −30 N
Thus, Fx = 30 N at x = 4 m.
11.20. Model: Use the negative derivative of the potential energy to determine the force acting on a particle. Solve:
(a)
The graph of the potential energy U = 4 y 3 is shown. (b) The y-component of the force is
Fy = −
dU d = − (4 y 3 ) = −12 y 2 dy dy
At y = 0 m, Fy = 0 N; at y = 1 m, Fy = −12 N; and at y = 2 m, Fy = −48 N.
11.21. Model: Use the negative derivative of the potential energy to determine the force acting on a particle. Solve:
(a)
The graph of the potential energy U = 10 / x is shown. (b) The x-component of the force is
dU d ⎛ 10 ⎞ 10 =− ⎜ ⎟= dx dx ⎝ x ⎠ x 2 10 10 = 2 = 2.5 N Fx = 5 m = 2 x x=2 m x
Fx = − Fx = 2 m
= 0.40 N x =5 m
Fx =8 m =
10 x2
= 0.156 N x =8 m
11.22. Model: Assume the carbon-carbon bond acts like an ideal spring that obeys Hooke’s law. Visualize:
The quantity ( x − xe ) is the stretching relative to the spring’s equilibrium length. In the present case, bond stretching is analogous to spring stretching. Solve: (a) The kinetic energy of the carbon atom is
1 1 K = mv 2 = (2.0 × 10−26 kg)(500 m/s) 2 = 2.5 × 10−21 J 2 2 (b) The energy of the spring is given by
1 U s = k ( x − xe ) 2 = K 2 2K 2(2.5 × 10−21 J) N ⇒k = = = 2.0 2 ( x − xe ) (0.050 × 10−9 m) 2 m
11.23. Visualize: One mole of helium atoms in the gas phase contains N A = 6.02 × 1023 atoms. Solve:
If each atom moves with the same speed v, the microscopic total kinetic energy will be
2 K micro 2(3700 J) ⎛1 ⎞ = = 1360 m/s K micro = N A ⎜ mv 2 ⎟ = 3700 J ⇒ v = −27 × 2 (6.68 10 kg)(6.02 × 1023 ) mN ⎝ ⎠ A
11.24. Solve: (a) The car has an initial kinetic energy K i . That energy is transformed into thermal energy of the car and the road surface. The gravitational potential energy does not change and no work is done by external forces, so during the skid K i transforms entirely into thermal energy Eth . This energy transfer and transformation is shown on the energy bar chart. Note that 1 1 K i = mv 2 = (1500 kg)(20 m/s) 2 = 3.0 × 105 J 2 2
(b) The change in the thermal energy of the car and the road surface is 3.0 × 105 J.
11.25. Visualize:
1 K i = K 0 = mv02 = 0 J U i = U g0 = mgy0 = (20 kg)(9.8 m/s 2 )(3.0 m) = 5.9 × 102 J 2 1 1 Wext = 0 J K f = K1 = mv12 = (20 kg)(2.0 m/s) 2 = 40 J U f = U g1 = mgy1 = 0 J 2 2 At the top of the slide, the child has gravitational potential energy of 5.9 × 102 J. This energy is transformed into thermal energy of the child’s pants and the slide and the kinetic energy of the child. This energy transfer and transformation is shown on the energy bar chart. (b)
Solve:
(a)
The change in the thermal energy of the slide and of the child’s pants is 5.9 × 10 2 J − 40 J = 5.5 × 10 2 J.
11.26. Visualize: The system loses 400 J of potential energy. In the process of losing this energy, it does 400 J of work on the environment, which means Wext = −400 J. Since the thermal energy increases 100 J, we have ΔEth = 100 J, which must have been 100 J of kinetic energy originally. This is shown in the energy bar chart.
11.27. Visualize:
Note that the conservation of energy equation K i + U i + Wext = K f + U f + ΔEth requires that Wext be equal to +400 J.
11.28. Solve: Please refer to Figure EX11.28. The energy conservation equation yields K i + U i + Wext = K f + U f + ΔEth ⇒ 4 J + 1 J + Wext = 1 J + 2 J + 1 J ⇒ Wext = −1 J Thus, the work done to the environment is −1 J. In other words, 1 J of energy is transferred from the system into the environment. This is shown in the energy bar chart.
11.29. Visualize: The tension of 20.0 N in the cable is an external force that does work on the block Wext = (20.0 N)(2.00 m) = 40.0 J, increasing the gravitational potential energy of the block. We placed the origin of our
coordinate system on the initial resting position of the block, so we have U i = 0 J and U f = mgyf =
(1.02 kg)(9.8 m/s 2 )(2.00 m) = 20.0 J. Also, K i = 0 J, and ΔEth = 0 J. The energy bar chart shows the energy transfers and transformations.
Solve:
The conservation of energy equation is 1 K i + U i + Wext = K f + U f + ΔEth ⇒ 0 J + 0 J + 40.0 J = mvf2 + 20.0 J + 0 J 2 ⇒ vf = (20.0 J)(2) /1.02 kg = 6.26 m/s
11.30. Model: Model the elevator as a particle, and apply the conservation of energy equation. Solve: The tension in the cable does work on the elevator to lift it. Because the cable is pulled by the motor, we say that the motor does the work of lifting the elevator. (a) The energy conservation equation is K i + U i + Wext = K f + U f + ΔEth . Using K i = 0 J, K f = 0 J, and
ΔEth = 0 J gives
Wext = (U f − U i ) = mg ( yf − yi ) = (1000 kg)(9.8 m/s 2 )(100 m) = 9.80 × 105 J (b) The power required to give the elevator this much energy in a time of 50 s is
P=
Wext 9.8 × 105 J = = 1.96 × 104 W Δt 50 s
Assess: Since 1 horsepower (hp) is 746 W, the power of the motor is 26 hp. This is a reasonable amount of power to lift a mass of 1000 kg to a height of 100 m in 50 s.
11.31. Model: Model the steel block as a particle subject to the force of kinetic friction and use the energy conservation equation. Visualize:
G G G Solve: (a) The work done on the block is Wnet = Fnet ⋅ Δr where Δr is the displacement. We will find the displacement using kinematic equations and the force using Newton’s second law of motion. The displacement in the x-direction is 1 Δx = x1 = x0 + v0 x (t1 − t0 ) + ax (t1 − t0 ) 2 = 0 m + (1.0 m/s)(3.0 s − 0 s) + 0 m = 3.0 m 2
G Thus Δr = 3.0iˆ m. The equations for Newton’s second law along the x and y components are
( Fnet ) y = n − FG = 0 N ⇒ n = FG = mg = (10 kg)(9.8 m/s2 ) = 98.0 N G
G
( Fnet ) x = F − f k = 0 N ⇒ F = ⇒ Wnet
f k = μ k n = (0.6)(98.0 N) = 58.8 N G G = Fnet ⋅ Δr = F Δ x cos0° = (58.8 N)(3.0 m)(1) = 176 J
(b) The power required to do this much work in 3.0 s is P=
W 176 J = = 59 W 3.0 s t
11.32. Solve: The power of the solar collector is the solar energy collected divided by time. The intensity of the solar energy striking the earth is the power divided by area. We have ΔE 150 × 106 J W = = 41,667 W and intensity = 1000 2 3600 s m Δt 41,667 W 2 ⇒ Area of solar collector = = 41.7 m 1000 W/m 2
P=
11.33. Solve: The night light consumes more energy than the hair dryer. The calculations are 1.2 kW × 10 min = 1.2 × 103 × 10 × 60 J = 7.2 × 105 J 10 W × 24 hours = 10 × 24 × 60 × 60 J = 8.64 × 105 J
11.34. Solve: (a) A kilowatt hour is a kilowatt multiplied by 3600 seconds. It has the dimensions of energy. (b)
One kilowatt hour is energy
1 kwh = (1000 J/s)(3600 s) = 3.6 × 106 J Thus
⎛ 3.6 × 106 J ⎞ 9 500 kwh = (500 kwh) ⎜ ⎟ = 1.8 × 10 J ⎝ 1 kwh ⎠
11.35. Model: Model the sprinter as a particle, and use the constant-acceleration kinematic equations and the definition of power in terms of velocity. Visualize:
Solve: (a) We can find the acceleration from the kinematic equations and the horizontal force from Newton’s second law. We have
1 1 x = x0 + v0 x (t1 − t0 ) + ax (t1 − t0 ) 2 ⇒ 50 m = 0 m + 0 m + ax (7.0 s − 0 s) 2 ⇒ ax = 2.04 m/s 2 2 2 2 ⇒ Fx = max = (50 kg)(2.04 m/s ) = 102 N G G G (b) We obtain the sprinter’s power output by using P = F ⋅ v , where v is the sprinter’s velocity. At t = 2.0 s the power is
P = ( Fx )[v0 x + ax (t − t0 )] = (102 N)[0 m/s + (2.04 m/s 2 )(2.0 s − 0 s)] = 0.42 kW The power at t = 4.0 s is 0.83 kW, and at t = 6.0 s the power is 1.25 kW.
G
G
G
G
11.36. Model: Use the definition of work for a constant force F , W = F ⋅ Δs , where Δs is the displacement. G Visualize: Please refer to Figure P11.36. The force F = (6iˆ + 8 ˆj ) N on the particle is a constant. G G G G Solve: (a) WABD = WAB + WBD = F ⋅ (Δs ) AB + F ⋅ (Δs ) BD = (6iˆ + 8ˆ j ) N ⋅ (3iˆ) m + (6iˆ + 8 ˆj ) N ⋅ (4 ˆj ) m = 18 J + 32 J = 50 J G G G G (b) WACD = WAC + WCD = F ⋅ (Δs ) AC + F ⋅ (Δs )CD = (6iˆ + 8ˆ j ) N ⋅ (4 ˆj ) m + (6iˆ + 8 ˆj ) N ⋅ (3 ˆj ) m = 32 J + 18 J = 50 J G G (c) WAD = F ⋅ (Δs ) AD = (6iˆ + 8ˆ j ) N ⋅ (3iˆ + 4 ˆj ) m = 18 J + 32 J = 50 J The force is conservative because the work done is independent of the path.
11.37. Model:
The force is conservative, so it has a potential energy. Visualize: Please refer to Figure P11.37 for the graph of the force. Solve: The definition of potential energy is ΔU = −W (i → f ). In addition, work is the area under the force-versus-
displacement graph. Thus ΔU = U f − U i = − (area under the force curve). Since U i = 0 at x = 0 m, the potential energy at position x is U ( x) = − (area under the force curve from 0 to x). From 0 m to 3 m, the area increases linearly from 0 N m to −60 N m, so U increases from 0 J to 60 J. At x = 4 m, the area is −70 J. Thus U = 70 J at x = 4 m, and U doesn’t change after that since the force is then zero. Between 3 m and 4 m, where F changes linearly, U must have a quadratic dependence on x (i.e., the potential energy curve is a parabola). This information is shown on the potential energy graph below.
(b) Mechanical energy is E = K + U . From the graph, U = 20 J at x = 1.0 m. The kinetic energy is K = 12 mv 2 = 12 (0.100 kg) (25 m/s) 2 = 31.25 J. Thus E = 51.25 J. (c) The total energy line at 51.25 J is shown on the graph above. (d) The turning point occurs where the total energy line crosses the potential energy curve. We can see from the graph that this is at approximately 2.5 m. For a more accurate value, the potential energy function is U = 20x J. The TE line crosses at the point where 20 x = 51.25, which is x = 2.56 m.
11.38. Model: Use the relationship between force and potential energy and the work-kinetic energy theorem. Visualize:
Please refer to Figure P11.38. We will find the slope in the following x regions: 0 cm < x < 1 cm, 1 < x < 3 cm, 3 < x < 5 cm, 5 < x < 7 cm, and 7 < x < 8 cm. Solve: (a) Fx is the negative slope of the U-versus-x graph, for example, for 0 m < x < 2 m dU −4 J = = −400 N ⇒ Fx = +400 N dx 0.01 m Calculating the values of Fx in this way, we can draw the force-versus-position graph as shown. (b) Since W = ∫ xxif Fx dx = area of the Fx-versus-x graph between xi and xf, the work done by the force as the
particle moves from xi = 2 cm to xf = 6 cm is −2 J. (c) The conservation of energy equation is K f + U f = Ki + U i . We can see from the graph that U i = 0 J and
U f = 2 J in moving from x = 2 cm to x = 6 cm. The final speed is vf = 10 m/s, so
2 J + 12 (0.010 kg)(10.0 m/s) 2 = 0 J + 12 (0.010 kg)vi 2 ⇒ vi = 22.4 m/s
11.39. Model: Use the relationship between a conservative force and potential energy. Visualize: Please refer to Figure P11.39. We will obtain U as a function of x and Fx as a function of x by using the calculus techniques of integration and differentiation. Solve: (a) For the interval 0 m < x < 0.5 m, Fx = (4 x) N, where x is in meters. This means
dU = − Fx = −4 x ⇒ U = −2 x 2 + C1 = −2 x 2 dx where we have used U = 0 J at x = 0 m to obtain C1 = 0. For the interval 0.5 m < x < 1 m, Fx = ( −4 x + 4) N. Likewise, dU = 4 x − 4 ⇒ U = 2 x 2 − 4 x + C2 dx Since U should be continuous at the junction, we have the continuity condition
(−2 x 2 ) x = 0.5 m = (2 x 2 − 4 x + C2 ) x = 0.5 m ⇒ −0.5 = 0.5 − 2 + C2 ⇒ C2 = 1 U remains constant for x ≥ 1 m. (b) For the interval 0 m < x < 0.5 m, U = +4 x, and for the interval 0.5 m < x < 1.0 m, U = −4 x + 4, where x is in meters. The derivatives give Fx = −4 N and Fx = +4 N, respectively. The slope is zero for x ≥ 1 m.
dvx , x = ∫ vx dt , K = 12 mvx2 , and F = max . dt Visualize: Please refer to Figure P11.40. We know ax = slope of the vx-versus-t graph and x = area under the vx-versus-x graph between 0 and x. Solve: Using the above definitions and methodology, we can generate the following table: t(s) ax (m/s2) x(m) K(J) F(N)
11.40. Model: Use ax =
0 0.5 1.0 1.5 2.0 2.5
10 10 10 10 +10 or −10 −10 −10 or 0 0 0
3.0 3.5 4.0
0 1.25 5 11.25 20 28.75
0 6.25 25 56.25 100 56.25
35 40 45
25 25 25
5 5 5 5 −5 or +5 −5 −5 or 0 0 0
(e) Let J1 be the impulse from t = 0 s to t = 2 s and J 2 be the impulse from t = 2 s to t = 4 s. We have 2s
J1 =
∫F
x
4s
dt = (5 N)(2 s) = 10 N ⋅ s and J 2 =
0s
∫F
x
dt = (−5 N)(1 s) = −5 N ⋅ s
0s
(f) J = Δp = mvf − mvi ⇒ vf = vi + J / m
At t = 2 s, vx = 0 m/s + (10 N ⋅ s) / 0.5 kg = 0 m/s + 20 m/s = 20 m/s At t = 4 s, vx = 20 m/s + ( −5 N ⋅ s) / 0.5 kg = 20 m/s − 10 m/s = 10 m/s The vx-versus-t graph also gives vx = 20 m/s at t = 2 s and vx = 10 m/s at t = 4 s. (g)
(h) From t = 0 s to t = 2 s, W = ∫ Fx dx = (5 N)(20 m) = 100 J
From t = 2 s to t = 4 s, W = ∫ Fx dx = (−5 N)(15 m) = −75 J (i) At t = 0 s, vx = 0 m/s so the work-kinetic energy theorem for calculating vx at t = 2 s is 1 1 1 1 W = ΔK = mvf2 − mvi2 ⇒ 100 J = (0.5 kg)vx2 − (0.5 kg)(0 m/s) 2 ⇒ vx = 20 m/s 2 2 2 2
To calculate vx at t = 4 s, we use vx at t = 2 s as the initial velocity: 1 1 −75 J = (0.5 kg)vx2 − (0.5 kg)(20 m/s) 2 ⇒ vx = 10 m/s 2 2 Both of these values agree with the values on the velocity graph.
11.41. Model: Model the elevator as a particle. Visualize:
Solve:
(a) The work done by gravity on the elevator is
Wg = −ΔU g = mgy0 − mgy1 = − mg ( y1 − y0 ) = −(1000 kg)(9.8 m/s 2 )(10 m) = −9.8 × 104 J (b) The work done by the tension in the cable on the elevator is
WT = T (Δy )cos0° = T ( y1 − y0 ) = T (10 m) To find T we write Newton’s second law for the elevator:
∑F
y
= T − FG = ma y ⇒ T = FG + ma y = m( g + a y ) = (1000 kg)(9.8 m/s 2 + 1.0 m/s 2 ) = 1.08 × 104 N ⇒ WT = (1.08 × 104 N)(10 m) = 1.08 × 105 J
(c) The work-kinetic energy theorem is
1 1 Wnet = Wg + WT = ΔK = K f − K i = K f − mv02 ⇒ K f = Wg + WT + mv02 2 2 1 ⇒ K f = ( −9.8 × 104 J) + (1.08 × 105 J) + (1000 kg)(0 m/s) 2 = 1.0 × 104 J 2 (d)
1 1 K f = mvf2 ⇒ 1.0 × 104 J = (1000 kg)v 2f ⇒ vf = 4.5 m/s 2 2
11.42. Model: Model the rock as a particle, and apply the work-kinetic energy theorem. Visualize:
Solve:
(a) The work done by Bob on the rock is
1 1 1 1 WBob = ΔK = mv12 − mv02 = mv12 = (0.500 kg)(30 m / s) 2 = 225 J = 2.3 × 102 J 2 2 2 2 (b) For a constant force, WBob = FBob Δx ⇒ FBob = WBob / Δx = 2.3 × 102 N. (c) Bob’s power output is PBob = FBob vrock and will be a maximum when the rock has maximum speed. This is just
as he releases the rock with vrock = v1 = 30 m/s. Thus, Pmax = FBob v1 = ( 225 J )( 30 m/s ) = 6750 W = 6.8 kW.
11.43. Model: Model the crate as a particle, and use the work-kinetic energy theorem. Visualize:
Solve:
(a) The work-kinetic energy theorem is ΔK = 12 mv12 − 12 mv02 = 12 mv12 = Wtotal . Three forces act on the box,
so Wtotal = Wgrav + Wn + Wpush . The normal force is perpendicular to the motion, so Wn = 0 J. The other two forces do the following amount of work: G G Wpush = Fpush ⋅ Δr = Fpush Δx cos 20° = 137.4 J
G G Wgrav = FG ⋅ Δr = ( FG ) x Δ x = (− mg sin 20°)Δ x = −98.0 J
Thus, Wtotal = 39.4 J, leading to a speed at the top of the ramp equal to
v1 =
2Wtotal 2(39.4 J) = = 4.0 m/s 5.0 kg m
(b) The x-component of Newton’s second law is ax = a =
( Fnet ) x Fpush cos 20° − FG sin 20° Fpush cos 20° − mg sin 20° = = = 1.35 m/s 2 m m m
Constant-acceleration kinematics with x1 = h / sin 20° = 5.85 m gives the final speed v12 = v02 + 2a ( x1 − x0 ) = 2ax1 ⇒ v1 = 2ax1 = 2(1.35 m / s 2 )(5.85 m) = 4.0 m/s
11.44. Model: Model Sam strapped with skis as a particle, and apply the law of conservation of energy. Visualize:
Solve:
(a) The conservation of energy equation is
K1 + U g1 + Δ Eth = K 0 + U g0 + Wext The snow is frictionless, so ΔEth = 0 J. However, the wind is an external force doing work on Sam as he moves down the hill. Thus,
Wext = Wwind = ( K1 + U g1 ) − ( K 0 + U g0 ) 1 ⎛1 ⎞ ⎛1 ⎞ ⎛1 ⎞ = ⎜ mv12 + mgy1 ⎟ − ⎜ mv02 + mgy0 ⎟ = ⎜ mv12 + 0 J ⎟ − (0 J + mgy0 ) = mv12 − mgy0 2 ⎝2 ⎠ ⎝2 ⎠ ⎝2 ⎠ ⇒ v1 = 2 gy0 +
2Wwind m
We compute the work done by the wind as follows: G G Wwind = Fwind ⋅ Δr = Fwind Δr cos160° = (200 N)(146 m)cos160° = −27,400 J
where we have used Δr = h / sin 20° = 146 m. Now we can compute
v1 = 2(9.8 m / s 2 )(50 m) +
2( −27,400 J) = 15.7 m / s 75 kg
(b) We will use a tilted coordinate system, with the x-axis parallel to the slope. Newton’s second law for Sam is
( Fnet ) x FG sin 20° − Fwind cos 20° mg sin 20° − Fwind cos 20° = = m m m (75 kg)(9.8 m/s 2 )sin 20° − (200 N)cos 20° = = 0.846 m/s 2 75 kg
ax = a =
Now we can use constant-acceleration kinematics as follows:
v12 = v02 + 2a( x1 − x0 ) = 2ax1 ⇒ v1 = 2ax1 = 2(0.846 m/s 2 )(146 m) = 15.7 m/s Assess:
We used a vertical y-axis for energy analysis, rather than a tilted coordinate system, because U g is
determined by its vertical position.
11.45. Model: Model Paul and the mat as a particle, assume the mat to be massless, use the model of kinetic friction, and apply the work-kinetic energy theorem. Visualize:
We define the x-axis along the floor and the y-axis perpendicular to the floor. fk . Newton’s second Solve: We need to first determine
law
in
the
y-direction
is
n + T sin 30° = FG = mg ⇒ n = mg − T sin 30° = (10 kg)(9.8 m/s ) − (30 N)(sin 30°) = 83.0 N. Using n and the model of 2
kinetic friction,
f k = μ k n = (0.2)(83.0 N) = 16.60 N. The net force on Paul and the mat is therefore
Fnet = T cos30° − f k = ( 30 N ) cos30° − 16.6 N = 9.4 N . Thus,
Wnet = Fnet Δr = (9.4 N)(3.0 m) = 28 J
G G G The other forces n and FG make an angle of 90° with Δr and do zero work. We can now use the work-kinetic energy theorem to find the final velocity as follows: 1 Wnet = K f − K i = K f − 0 J = mvf2 ⇒ vf = 2Wnet / m = 2(28 J)/10 kg = 2.4 m/s 2
Assess:
A speed of 2.4 m/s or 5.4 mph is reasonable for the present problem.
11.46. Model: Assume an ideal spring that obeys Hooke’s law. Model the box as a particle and use the model of kinetic friction. Visualize:
Solve:
When the horizontal surface is frictionless, conservation of energy means
1 1 1 k ( x0 − xe ) 2 = mv12x = K1 ⇒ K1 = (100 N/m)(0.20 m − 0 m) 2 = 2.0 J 2 2 2 That is, the box is launched with 2.0 J of kinetic energy. It will lose 2.0 J of kinetic energy on the rough surface. G G The net force on the box is Fnet = − f k = − μ k mgiˆ. The work-kinetic energy theorem is G G Wnet = Fnet ⋅ Δr = K 2 − K1 = 0 J − 2.0 J = −2.0 J
⇒ (− μ k mg )( x2 − x1 ) = −2.0 J ⇒ ( x2 − x1 ) =
2.0 J 2.0 J = = 54 cm ( μ k )( mg ) (0.15)(2.5 kg)(9.8 m/s 2 )
Assess: Because the force of friction transforms kinetic energy into thermal energy, energy is transferred out of the box into the environment. In response, the box slows down and comes to rest.
11.47. Model: Model the suitcase as a particle, use the model of kinetic friction, and use the work-kinetic energy theorem. Visualize:
G G The net force on the suitcase is Fnet = f k . Solve: The work-kinetic energy theorem is
G 1 1 1 1 G G G Wnet = ΔK = mv12 − mv02 ⇒ Fnet ⋅ Δr = f k ⋅ Δr = 0 J − mv02 ⇒ ( f k )( x1 − x0 )cos180° = − mv02 2 2 2 2 1 2 v02 (1.2 m/s) 2 ⇒ − μ k mg ( x1 − x0 ) = − mv0 ⇒ μ k = = = 0.037 2 2 g ( x1 − x0 ) 2(9.8 m/s 2 )(2.0 m − 0 m) Assess: Friction transforms kinetic energy of the suitcase into thermal energy. In response, the suitcase slows down and comes to rest.
11.48. Model: Identify the truck and the loose gravel as the system. We need the gravel inside the system because friction increases the temperature of the truck and the gravel. We will also use the model of kinetic friction and the conservation of energy equation. Visualize:
We place the origin of our coordinate system at the base of the ramp in such a way that the x-axis is along the ramp and the y-axis is vertical so that we can calculate potential energy. The free-body diagram of forces on the truck is shown. Solve: The conservation of energy equation is K1 + U g1 + Δ Eth = K 0 + U g0 + Wext . In the present case,
Wext = 0 J, v1x = 0 m/s, U g 0 = 0 J, v0 x = 35 m/s. The thermal energy created by friction is
ΔEth = ( f k )( x1 − x0 ) = ( μ k n)( x1 − x0 ) = μ k ( mg cos6.0°)( x1 − x0 ) = (0.40)(15,000 kg)(9.8 m/s 2 )(cos6.0°)( x1 − x0 ) = (58,478 J/m)( x1 − x0 ) Thus, the energy conservation equation simplifies to 1 0 J + mgy1 + (58,478 J/m)( x1 − x0 ) = mv02x + 0 J + 0 J 2 1 (15,000 kg)(9.8 m/s 2 )[( x1 − x0 )sin 6.0°] + (58,478 J/m)( x1 − x0 ) = (15,000 kg)(35 m/s) 2 2 ⇒ ( x1 − x0 ) = 124 m Assess:
A length of 124 m at a slope of 6º seems reasonable.
11.49. Model: We will use the spring, the package, and the ramp as the system. We will model the package as a particle. Visualize:
We place the origin of our coordinate system on the end of the spring when it is compressed and is in contact with the package to be shot. Model: (a) The energy conservation equation is
K1 + U g1 + U s1 + Δ Eth = K 0 + U g 0 + U s0 + Wext 1 2 1 1 1 mv1 + mgy1 + k ( xe − xe ) 2 + Δ Eth = mv02 + mgy0 + k (Δx) 2 + Wext 2 2 2 2 Using y1 = 1 m, ΔEth = 0 J (note the frictionless ramp), v0 = 0 m/s, y0 = 0 m, Δx = 30 cm, and Wext = 0 J, we get
1 2 1 mv1 + mg (1 m) + 0 J + 0 J = 0 J + 0 J + k (0.30 m)2 + 0 J 2 2 1 1 2 2 (2.0 kg)v1 + (2.0 kg)(9.8 m/s )(1 m) = (500 N/m)(0.30 m) 2 2 2 ⇒ v1 = 1.70 m/s (b) How high can the package go after crossing the sticky spot? If the package can reach y1 ≥ 1.0 m before
stopping (v1 = 0), then it makes it. But if y1 < 1.0 m when v1 = 0, it does not. The friction of the sticky spot generates thermal energy
ΔEth = ( μ k mg )Δx = (0.30)(2.0 kg)(9.8 m/s 2 )(0.50 m) = 2.94 J The energy conservation equation is now 1 2
mv12 + mgy1 + ΔEth = 12 k ( Δx) 2
If we set v1 = 0 m/s to find the highest point the package can reach, we get
y1 = ( 12 k (Δx) 2 − ΔEth ) / mg = ( 12 (500 N/m)(0.30 m) 2 − 2.94 J ) /(2.0 kg)(9.8 m/s 2 ) = 0.998 m The package does not make it. It just barely misses.
11.50. Model: Model the two blocks as particles. The two blocks make our system. Visualize:
We place the origin of our coordinate system at the location of the 3.0 kg block. Solve: (a) The conservation of energy equation is K f + U gf + ΔEth = K i + U gi + Wext . Using ΔEth = 0 J and
Wext = 0 J we get 1 1 1 1 m2 (vf ) 22 + m3 (vf )32 + m2 g ( yf ) = m2 (vi ) 22 + m3 (vi )32 + m2 g ( yi ) 2 2 2 2 Noting that (vf ) 2 = (vf )3 = vf and (vi ) 2 = (vi )3 = 0 m / s, this becomes
1 ( m2 + m3 )vf2 = −m2 g ( yf − yi ) 2 vf =
2m2 g ( yi − yf ) 2(2.0 kg)(9.8 m/s 2 )(1.50 m) = = 3.4 m/s m2 + m3 (2.0 kg + 3.0 kg)
(b) We will use the same energy conservation equation. However, this time
ΔEth = μ k (m3 g )(Δ x) = (0.15)(3.0 kg)(9.8 m/s 2 )(1.50 m) = 6.615 J The energy conservation equation is now 1 1 1 1 m2vf2 + m3vf2 + m2 gyf + 6.615 J = m2 (vi ) 22 + m3 (v1 )32 + m2 gyi + 0 J 2 2 2 2 ⎛ 1 2 ⎞ (m2 + m3 )vf2 + 6.615 J = m2 g ( yi − yf ) ⇒ vf = ⎜ ⎟[ m2 g ( yi − yf ) − 6.615 J] m 2 + ⎝ 2 m3 ⎠ ⎛ 2 ⎞ 2 = ⎜ ⎟[(2.0 kg)(9.8 m/s )(1.50 m) − 6.615 J] = 3.0 m/s ⎝ 5.0 kg ⎠ Assess:
A reduced speed when friction is present compared to when there is no friction is reasonable.
G
G
11.51. Model: Use the particle model, the definition of work W = F ⋅ Δs , and the model of kinetic friction. Visualize: We place the coordinate frame on the incline so that its x-axis is along the incline.
Solve:
(a)
G G WT = T ⋅ Δr = (T )(Δx)cos18° = (120 N)(5.0 m)cos18° = 0.57 kJ
G G Wg = ( FG ⋅ Δr ) = mg (Δx)cos(120°) = (8 kg)(9.8 m/s 2 )(5.0 m)cos120° = −196 J G G Wn = (n ⋅ Δr ) = (n)(Δ x)cos90° = 0 J (b) The amount of energy transformed into thermal energy is ΔEth = ( f k )(Δx) = ( μ k n)(Δx). To find n, we write Newton’s second law as follows:
∑F
y
= n − FG cos30° + T sin18° = 0 N ⇒ n = FG cos30° − T sin18°
⇒ n = mg cos30° − T sin18° = (8.0 kg)(9.8 m/s 2 )cos30° − (120 N)sin18° = 30.814 N Thus, ΔEth = (0.25)(30.814 N)(5.0 m) = 38.5 J. Assess: Any force that acts perpendicular to the displacement does no work.
11.52. Model: Assume the spring is ideal so that Hooke’s law is obeyed, and model the weather rocket as a particle. Visualize:
The origin of the coordinate system is placed on the free end of the spring. Note that the bottom of the spring is anchored to the ground. Solve: (a) The rocket is initially at rest. The free-body diagram on the rocket helps us write Newton’s second law as
(∑ F ) = 0 N ⇒ F y
sp
⇒ Δy =
= FG = mg ⇒ k Δy = mg
mg (10.2 kg)(9.8 m/s 2 ) = = 20.0 cm k (500 N/m)
(b) The thrust does work. Using the energy conservation equation when y2 − y0 = 40 cm = 0.40 m:
K 2 + U g2 + U sp2 = K1 + U g1 + U sp1 + Wext 1 1 1 1 Wext = F ( y2 − y1 ) ⋅ mv22 + mgy2 + k ( y2 − ye ) 2 = mv12 + mgy1 + k ( y1 − ye ) 2 + (200 N)(0.60 m) 2 2 2 2 2 (5.10 kg)v2 + 40.0 J + 40.0 J = 0 − 20.0 J + 10.0 J + 120 J ⇒ v2 = 2.43 m/s If the rocket were not attached to the spring, the energy conservation equation would not involve the spring energy term U sp2 . That is,
K 2 + U g 2 = K1 + U g1 + U sp1 + Wext 1 (10.2 kg)v22 + (10.2 kg)(9.8 m/s 2 )(0.40 m) = 0 J − (10.2 kg)(9.8 m/s 2 )(0.20 m) 2 1 + (500 N/m)(0.20 m) 2 + (200 N)(0.60 m) 2
⇒ (5.10 kg)v22 = 70.0 J ⇒ v2 = 3.70 m/s
11.53. Model: Use the particle model for the ice skater, the model of kinetic/static friction, and the workkinetic energy theorem. Visualize:
Solve:
(a) The work-kinetic energy theorem is 1 1 ΔK = mv12 − mv02 = Wnet = Wwind 2 2
There is no kinetic friction along her direction of motion. Static friction acts to prevent her skates from slipping sideways on the ice, but this force is perpendicular to the motion and does not contribute to a change in thermal G G energy. The angle between Fwind and Δr is θ = 135°, so G G Wwind = Fwind ⋅ Δr = Fwind Δy cos135° = (4 N)(100 m)cos135° = −282.8 J
Thus, her final speed is v1 = v02 +
2Wwind = 2.16 m/s m
(b) If the skates don’t slip, she has no acceleration in the x-direction and so ( Fnet ) x = 0 N. That is:
f s − Fwind cos 45° = 0 N ⇒ fs = Fwind cos 45° = 2.83 N Now there is an upper limit to the static friction: f s ≤ ( fs ) max = μ s mg . To not slip requires
μs ≥
fs 2.83 N = = 0.0058 mg (50 kg)(9.8 m/s 2 )
Thus, the minimum value of μs is 0.0058. Assess: The work done by the wind on the ice skater is negative, because the wind slows the skater down.
11.54. Model: Model the ice cube as a particle, the spring as an ideal that obeys Hooke’s law, and the law of conservation of energy. Visualize:
Solve: (a) The normal force does no work and the slope is frictionless, so mechanical energy is conserved. We’ve drawn two separate axes: a vertical y-axis to measure potential energy and a tilted s-axis to measure distance along the slope. Both have the same origin which is at the point where the spring is not compressed. Thus, the two axes are related by y = s sin θ . Also, this choice of origin makes the elastic potential energy simply
U s = 12 k ( s − s0 ) 2 = 12 ks 2 . Because energy is conserved, we can relate the initial point—with the spring compressed—to the final point where the ice cube is at maximum height. We do not need to find the speed with which it leaves the spring. We have K 2 + U g 2 + U s2 = K1 + U g1 + U s1 1 2 1 1 1 mv2 + mgy2 + ks02 = mv12 + mgy1 + ks12 2 2 2 2 It is important to note that at the final point, when the ice cube is at y2, the end of the spring is only at s0. The spring does not stretch to s2 , so U s2 is not 12 ks22 . Three of the terms are zero, leaving 1 ks 2 mgy2 = + mgy1 + ks12 ⇒ y2 − y1 = Δy = height gained = 1 = 0.255 m = 25.5 cm 2 2mg
The distance traveled is Δs = Δy / sin 30° = 51.0 cm. (b) Using the energy equation and the expression for thermal energy:
K 2 + U g 2 + U s2 + ΔEth = K1 + U g1 + U s1 + Wext
ΔEth = f k Δs = μ k nΔs
From the free-body diagram,
G ( Fnet ) y = 0 N = n − mg cos30° ⇒ n = mg cos30° Now, having found ΔEth = μ k (mg cos30°) Δs, the energy equation can be written 1 0 J + mgy2 + 0 J + μ k (mg cos30°)Δs = 0 J + mgy1 + ks12 + 0 J 2 1 2 ⇒ mg ( y2 − y1 ) − ks1 + μ k mg cos30°Δs = 0 2 Using Δy = ( Δs )sin 30°, the above equation simplifies to
ks12 1 mg Δs sin 30° + μ k mg cos30°Δs = ks12 ⇒ Δs = = 0.379 m = 37.9 cm 2 2mg (sin 30° + μ k cos30°)
11.55. Model: Assume an ideal spring, so Hooke’s law is obeyed. Treat the box as a particle and apply the energy conservation law. Box, spring, and the ground make our system, and we also use the model of kinetic friction. Visualize: We place the origin of the coordinate system on the ground directly below the box’s starting position.
Solve:
(a) The energy conservation equation is
K1 + U g1 + U s1 + ΔEth = K 0 + U g0 + U s0 + Wext 1 2 1 1 mv1 + mgy1 + 0 J + 0 J = mv02 + mgy0 + 0 J + 0 J ⇒ mv12 + 0 J = 0 J + mgy0 2 2 2 ⇒ v1 = 2 gy0 = 2(9.8m/s 2 )(5.0 m) = 9.9 m/s (b) The friction creates thermal energy. The energy conservation equation for this part of the problem is
1 2 1 mv2 + 0 J + 0 J + μ k mg ( x2 − x1 ) = mv12 + 0 J + 0 J + 0 J 2 2 1 2 1 2 1 2 1 2 mv2 + μ k n( x2 − x1 ) = mv1 ⇒ mv2 + μ k mg ( x2 − x1 ) = mv1 2 2 2 2
K 2 + U g 2 + U s 2 + ΔEth = K1 + U g1 + U s1 + Wext
⇒ v2 = v12 − 2μ k g ( x2 − x1 ) = (9.9 m/s) 2 − 2(0.25)(9.8 m/s 2 )(2.0 m) = 9.4 m/s (c) To find how much the spring is compressed, we apply the energy conservation once again:
K 3 + U g3 + U s3 + ΔEth = K 2 + U g 2 + U s2 + Wext
1 1 0 J + 0 J + k ( x3 − x2 ) 2 + 0 J = mv22 + 0 J + 0 J + 0 J 2 2
Using v2 = 9.4 m/s, k = 500 N/m and m = 5.0 kg, the above equation yields ( x3 − x2 ) = Δx = 94 cm. (d) The initial energy = mgy0 = (5.0 kg)(9.8 m / s 2 )(5.0 m) = 254 J. The energy transformed to thermal energy during each passage is
μ k mg ( x2 − x1 ) = (0.25)(5.0 kg)(9.8 m / s 2 )(2.0 m) = 24.5 J The number of passages is equal to 245 J / 24.5 J or 10.
11.56. Model: Assume an ideal spring, so Hooke’s law is obeyed. Treat the physics student as a particle and apply the law of conservation of energy. Our system is comprised of the spring, the student, and the ground. We also use the model of kinetic friction. Visualize: We place the origin of the coordinate system on the ground directly below the end of the compressed spring that is in contact with the student.
Solve:
(a) The energy conservation equation is
K1 + U g1 + U s1 + ΔEth = K 0 + U g0 + U s0 + Wext 1 2 1 1 1 mv1 + mgy1 + k ( x1 − xe ) 2 + 0 J = mv02 + mgy0 + k ( x1 − x0 ) 2 + 0 J 2 2 2 2 Since y1 = y0 = 10 m, x1 = xe , v0 = 0 m/s, k = 80,000 N/m, m = 100 kg, and ( x1 − x0 ) = 0.5 m, k 1 2 1 mv1 = k ( x1 − x0 ) 2 ⇒ v1 = ( x1 − x0 ) = 14.14 m/s m 2 2 (b) Friction creates thermal energy. Applying the conservation of energy equation once again:
K 2 + U g 2 + U s2 + ΔEth = K 0 + U g 0 + U s0 + Wext 1 2 1 mv2 + mgy2 + 0 J + f k Δs = 0 J + mgy0 + k ( x1 − x0 ) 2 + 0 J 2 2 With v2 = 0 m/s and y2 = (Δs )sin 30°, the above equation is simplified to 1 mg (Δs )sin 30° + μ k nΔs = mgy0 + k ( x1 − x0 ) 2 2
From the free-body diagram for the physics student, we see that n = FG cos30° = mg cos30°. Thus, the conservation of energy equation gives 1 Δs (mg sin 30° + μ k mg cos30°) = mgy0 + k ( x1 − x0 ) 2 2
Using m = 100 kg, k = 80,000 N/m, ( x1 − x0 ) = 0.50 m, y0 = 10 m, and μ k = 0.15, we get
1 mgy0 + k ( x1 − x0 ) 2 2 Δs = = 32.1 m mg (sin 30° + μ k cos30°) Assess: y2 = (Δs )sin 30° = 16.05 m, which is greater than y0 = 10 m. The higher value is due to the transformation of the spring energy into gravitational potential energy.
11.57. Model: Treat the block as a particle, use the model of kinetic friction, and apply the energy conservation law. The block and the incline comprise our system. Visualize: We place the origin of the coordinate system directly below the block’s starting position at the same level as the horizontal surface. On the horizontal surface the model of kinetic friction applies.
Solve:
(a) For the first incline, the conservation of energy equation gives 1 2 K1 + U g1 + ΔEth = K 0 + U g0 + Wext mv1 + 0 J + 0 J = 0 J + mgy0 + 0 J ⇒ v1 = 2 gy0 = 2 gh 2 (b) The friction creates thermal energy. Applying once again the conservation of energy equation, we have K 3 + U g3 + ΔEth = K1 + U g1 + Wext
1 2 1 mv3 + mgy3 + μ k mg ( x2 − x1 ) = mv12 + mgy1 + Wext 2 2
Using v3 = 0 m/s, y1 = 0 m, Wext = 0 J, v1 = 2 gh , and ( x2 − x1 ) = L, we get 1 mgy3 + μ k mgL = m(2 gh) ⇒ y3 = h − μ k L 2
Assess:
For μ k = 0, y3 = h which is predicted by the law of the conservation of energy.
11.58. Model: Assume an ideal spring, so Hooke’s law is obeyed. Visualize:
Solve: For a conservative force the work done on a particle as it moves from an initial to a final position is independent of the path. We will show that WA →C → B = WA → B for the spring force. Work done by a spring force F = − kx is given by xf
W = ∫ Fdx = − ∫ kx dx xi
This means xB
WA → B = − ∫ kx dx = − xA
x
x
C B k 2 k k xB − xA2 ) , WA → C = − ∫ kx dx = − ( xC2 − xA2 ) , and WC → B = − ∫ kx dx = − ( xB2 − xC2 ) ( 2 2 2 xA xC
Adding the last two:
WA → C → B = WA → C + WC → B = −
k 2 ( xC − xA2 + xB2 − xC2 ) = WA → B 2
11.59. Model: A “sprong” that obeys the force law Fx = − q( x − xe )3 , where q is the sprong constant and xe is the equilibrium position. Visualize: We place the origin of the coordinate system on the free end of the sprong, that is, xe = xf = 0 m.
Solve: (a) The units of q are N/m3 . (b) A cubic curve rises more steeply than a parabola. The force increases by a factor of 8 every time x increases by a factor of 2.
x dU qx 4 , we have U ( x) = − ∫ Fx dx = − ∫ (− qx3 )dx = . 0 dx 4 (d) Applying the energy conservation equation to the ball and sprong system: K f + U f = Ki + U i
(c) Since Fx = −
1 2 qx 4 mvf + 0 J = 0 J + i 2 4 ⇒ vf =
q x4 (40,000 N/m3 ) ( −0.10 m) 4 = ⋅ = 10 m/s m 2 (0.020 kg) 2
11.60. Solve: (a) Because sin (cx) is dimensionless, F0 must have units of force in newtons. (b) The product cx is an angle because we are taking the sine of it. An angle has no real physical units. If x has units of m and the product cx is unitless, then c has to have units of m -1. (c) The force is a maximum when sin(cx) = 1. This occurs when cx = π /2, or for xmax = π /2c. (d) The graph is the first quarter of a sine curve.
(e) We can find the velocity vf at xf = xmax from the work-kinetic energy theorem:
1 1 1 1 2W ΔK = mvf2 − mvi2 = mvf2 − mv02 = W ⇒ vf = v02 + m 2 2 2 2 This is a variable force. As the particle moves from xi = 0 m to xf = xmax = π /2c, the work done on it is xf
π / 2c
xi
0
W = ∫ F ( x) dx = F0 ∫
sin(cx) dx = −
F0 F ⎛ π ⎞ F cos(cx)|π0 / 2 c = − 0 ⎜ cos − cos0 ⎟ = 0 2 c c ⎝ ⎠ c
Thus, the particle’s speed at xf = xmax = π /2c is vf = v02 + 2 F0 / mc .
11.61. Visualize: We place the origin of the coordinate system at the base of the stairs on the first floor.
Solve: (a) We might estimate y2 − y1 ≈ 4.0 m ≈ 12 ft ≈ y3 − y2 , thus, y3 − y1 ≈ 8.0 m. (b) We might estimate the time to run up these two flights of stairs to be 20 s. (c) Estimate your mass as m ≈ 70 kg ≈ 150 lb. Your power output while running up the stairs is
work done by you change in potential energy mg ( y3 − y1 ) = = time time time (70 kg)(9.8 m/s 2 )(8.0 m) ⎛ 1 hp ⎞ = ≈ 270 W = (270 W) ⎜ ⎟ ≈ 0.35 hp 20 s ⎝ 746 W ⎠ Assess:
Your estimate may vary, depending on your mass and how fast you run.
11.62. Solve: Power output during the push-off period is equal to the work done by the cat divided by the time the cat applied the force. Since the force on the floor by the cat is equal in magnitude to the force on the cat by the floor, work done by the cat can be found using the work-kinetic energy theorem during the push-off period, Wnet = Wfloor = ΔK . We do not need to explicitly calculate Wcat , since we know that the cat’s kinetic energy is transformed into its potential energy during the leap. That is, ΔU g = mg ( y2 − y1 ) = (5.0 kg)(9.8 m / s 2 )(0.95 m) = 46.55 J Thus, the average power output during the push-off period is P=
Wnet 46.55 J = = 0.23 kW 0.20 s t
11.63. Solve: Using the conversion 746 W = 1 hp, we have a power of 1492 J/s. This means W = Pt = (1492 J/s)(1 h) = 5.3712 × 106 J is the total work done by the electric motor in one hour. Furthermore,
Wmotor = −Wg = U gf − U gi = mg ( yf − yi ) = mg (10 m) m=
Wmotor 5.3712 × 106 J 1 liter = = 5.481 × 104 kg = 5.481 × 104 kg × = 5.5 × 104 liters g (10 m) (9.8 m/s 2 )(10 m) 1 kg
11.64. Solve: (a) The change in the potential energy of 1.0 kg of water in falling 25 m is ΔU g = − mgh = −(1.0 kg)(9.8 m/s 2 )(25 m) = −245 J ≈ −0.25 kJ (b) The power required of the dam is P=
W W = = 50 × 106 Watts ⇒ W = 50 × 106 J t 1s
That is, 50 × 106 J of energy is required per second for the dam. Out of the 245 J of lost potential energy, (245 J)(0.80) = 196 J is converted to electrical energy. Thus, the amount of water needed per second is (50 × 106 J)(1 kg /196 J) = 255,000 kg ≈ 2.6 × 105 kg.
11.65. Solve: The force required to tow a water skier at a speed v is Ftow = Av. Since power P = Fv, the power required to tow the water skier is Ptow = Ftow v = Av 2 . We can find the constant A by noting that a speed of v = 2.5 mph requires a power of 2 hp. Thus,
(2 hp) = A(2.5 mph) 2 ⇒ A = 0.32
hp (mph) 2
Now, the power required to tow a water skier at 7.5 mph is Ptow = Av 2 = 0.32 Assess:
hp ⋅ (7.5 mph) 2 = 18 hp (mph) 2
Since P ∝ v 2 , a three-fold increase in velocity leads to a nine-fold increase in power.
11.66. Solve: By definition, the maximum power output of a horse is P ≈ 1 hp = 746 W. At maximum G G speed, when the horse is running at constant speed, Fnet = 0. The propulsion force Fhorse , provided by the horse
pushing against the ground, is balanced by the drag force of air resistance: Fhorse = D. We learned in Chapter 6 that a reasonable model for drag is D ≈ 14 Av 2 , where A is the cross section area. Since the power needed for force F to push an object at velocity v is P = Fv, we have
P ≈ 1 hp = 746 W = Fhorsev = ( 14 Av 2 ) v = 14 Av3 1/ 3
⎛ 4P ⎞ v≈⎜ ⎟ ⎝ A⎠ Assess:
1/ 3
⎛ 4(746 W) ⎞ =⎜ ⎟ ⎝ (0.5 m)(1.8 m) ⎠
= 15 m/s
15 m/s ≈ 30 mph is a reasonable top speed for a well-trained horse.
11.67. Solve: The net force on a car moving at a steady speed is zero. The motion is opposed both by rolling friction and by air resistance. Thus the propulsion force provided by the drive wheels must be Fcar = μ r mg + 14 Av 2 , where μ r is the rolling friction, m is the mass, A is the cross-section area, and v is the car’s velocity. The power required to move the car at speed v is 1 3 Av 4 Since the maximum power output is 200 hp and 75% of the power reaches the drive wheels, P = (200 hp)(0.75) = 150 hp. Thus, P = Fcar v = μ r mgv +
⎛ 746 W ⎞ 1 2 3 (150 hp) ⎜ ⎟ = (0.02)(1500 kg)(9.8 m/s )v + (1.6 m)(1.4 m)v 1 hp 4 ⎝ ⎠ ⇒ 0.56 v3 + 294 v − 111,900 = 0 ⇒ v = 55.5 m/s The easiest way to solve this equation is through iterations by trial and error. Assess: A speed of 55.5 m/s ≈ 110 mph is very reasonable.
11.68. Model: Use the model of static friction, kinematic equations, and the definition of power.
Solve: (a) The rated power of the Porsche is 217 hp = 161,882 W and the gravitational force on the car is
(1480 kg)(9.8 m/s 2 ) = 14,504 N. The amount of that force on the drive wheels is (14,504)(2/3) = 9670 N. Because the static friction of the tires on road pushes the car forward, Fmax = fs max = μs n = μ s mg = (1.00)(9670 N) = mamax ⇒ amax =
9670 N = 6.53 m/s 2 1480 kg
(b) Only 70% of the power generated by the motor is applied at the wheels. P = Fvmax ⇒ vmax =
P (0.70)(161,882 W) = = 11.7 m/s 9670 N F
(c) Using the kinematic equation, vmax = v0 + amax (tmin − t0 ) with v0 = 0 m/s and t0 = 0 s, we obtain
tmin = Assess:
vmax 11.7 m/s = = 1.79 s amax 6.53 m/s 2
An acceleration time of 1.79 s for the Porsche to reach a speed of ≈ 26 mph from rest is reasonable.
11.69. (a) A student uses a string to pull her 2.0 kg physics book, starting from rest, across a 2.0-m-long lab bench. The coefficient of kinetic friction between the book and the lab bench is 0.15. If the book’s final speed is 4.0 m/s, what is the tension in the string? (b)
(c) The tension does external work Wext . This work increases the book’s kinetic energy and also causes an
increase ΔEth in the thermal energy of the book and the lab bench. Solving the equation gives T = 10.9 N.
11.70. (a) A 20 kg chicken crate slides down a 2.5-m-high, 40° ramp from the back of a truck to the ground. The coefficient of kinetic friction between the crate and the ramp bench is 0.15. How fast are the chickens going at the bottom of the ramp? (b)
(c) v1 = 6.34 m/s.
11.71. (a) If you expend 75 W of power to push a 30 kg sled on a surface where the coefficient of kinetic friction between the sled and the surface is μ k = 0.20, what speed will you be able to maintain? (b)
(c) Fpush = (0.20)(30 kg)(9.8 m/s 2 ) = 58.8 N ⇒ 75 W = (58.8 N)v ⇒ v =
75 W = 1.28 m/s 58.8 N
11.72. (a) A 1500 kg object is being accelerated upward at 1.0 m/s2 by a rope. How much power must the motor supply at the instant when the velocity is 2.0 m/s? (b)
(c) T = (1500 kg)(9.8 m/s 2 ) + 1500 kg(1.0 m/s 2 ) = 16,200 N=16.2 kN
P = T (2m/s) = (16,200 N)(2.0 m/s) = 32,400 W = 32 kW
11.73. Model: Model the water skier as a particle, apply the law of conservation of mechanical energy, and use the constant-acceleration kinematic equations. Visualize:
We placed the origin of the coordinate system at the base of the frictionless ramp. Solve: We’ll start by finding the smallest speed v1 at the top of the ramp that allows her to clear the shark tank. From the vertical motion for jumping the shark tank, 1 y2 = y1 + v1 y (t2 − t1 ) + a y (t2 − t1 ) 2 2 1 ⇒ 0 m = (2.0 m) + 0 m + (−9.8 m/s 2 )(t2 − t1 ) 2 ⇒ (t2 − t1 ) = 0.639 s 2 From the horizontal motion, 1 x2 = x1 + v1x (t2 − t1 ) + ax (t2 − t1 ) 2 2 5.0 m ⇒ ( x1 + 5.0 m) = x1 + v1 (0.639 s) + 0 m ⇒ v1 = = 7.825 m/s 0.639 s Having found the v1 that will take the skier to the other side of the tank, we now use the energy equation to find the minimum speed v0. We have 1 1 K1 + U g1 = K 0 + U g0 ⇒ mv12 + mgy1 = mv02 + mgy0 2 2
v0 = v12 + 2 g ( y1 − y0 ) = (7.825 m/s) 2 + 2(9.8 m/s 2 )(2.0 m) = 10.0 m/s
11.74. Model: Assume the spring to be ideal that obeys Hooke’s law, and model the block as a particle. Visualize: We placed the origin of the coordinate system on the free end of the compressed spring which is in contact with the block. Because the horizontal surface at the bottom of the ramp is frictionless, the spring energy appears as kinetic energy of the block until the block begins to climb up the incline.
Solve: Although we could find the speed v1 of the block as it leaves the spring, we don’t need to. We can use energy conservation to relate the initial potential energy of the spring to the energy of the block as it begins projectile motion at point 2. However, friction requires us to calculate the increase in thermal energy. The energy equation is K 2 + U g2 + ΔEth = K 0 + U g0 + Wext ⇒ 12 mv22 + mgy2 + f k Δs = 12 k ( x0 − xe ) 2 The distance along the slope is Δs = y2 / sin 45°. The friction force is f k = μ k n, and we can see from the freebody diagram that n = mg cos 45°. Thus 1/ 2
⎡k ⎤ v2 = ⎢ ( x0 − xe ) 2 − 2 gy2 − 2 μ k gy2 cos 45°/ sin45° ⎥ ⎣m ⎦
1/ 2
⎡1000 N/m ⎤ =⎢ (0.15 m) 2 − 2(9.8 m/s 2 )(2.0 m) − 2(0.20)(9.8 m/s 2 )(2.0 m) cos 45° / sin45° ⎥ = 8.091 m/s 0.20 kg ⎣ ⎦ Having found the velocity v2 , we can now find ( x3 − x2 ) = d using the kinematic equations of projectile motion:
1 y3 = y2 + v2 y (t3 − t2 ) + a2 y (t3 − t2 ) 2 2 1 2.0 m = 2.0 m + (v2 sin 45°)(t3 − t2 ) + (−9.8 m / s 2 )(t3 − t2 ) 2 2 ⇒ (t3 − t2 ) = 0 s and 1.168 s Finally, 1 x3 = x2 + v2 x (t3 − t2 ) + a2 x (t3 − t2 ) 2 2 d = ( x3 − x2 ) = (v2 cos 45°)(1.168 s) + 0 m = 6.68 m
11.75. Model: Assume an ideal spring that obeys Hooke’s law, the particle model for the ball, and the model of kinetic friction. The ball, the spring, and the barrel comprise our system. Visualize:
We placed the origin of the coordinate system on the free end of the spring which is in contact with the ball. G Solve: Force F does external work Wext . This work compresses the spring, gives the ball kinetic energy, and is partially dissipated to thermal energy by friction. (a) The energy equation for our system is K1 + U s1 + ΔEth = K 0 + U s0 + Wext
1 2 1 1 1 mv1 + k ( x1 − x0 ) 2 + μ k mg ( x1 − x0 ) = mv02 + k ( xe − xe ) 2 + F ( x1 − x0 ) 2 2 2 2 1 1 (1.0 kg)(2.0 m/s) 2 + (3000 N/m)(0.3 m) 2 + μ k mg (0.30 m) = 0 J + 0 J + F (0.30 m) 2 2 With μ k = 0.30 and m = 1.0 kg, we can solve this equation to obtain F = 460 N. (b) Using the energy equation for our system once again, we have 1 2 1 1 1 mv2 + k ( xe − xe ) 2 + μ k mg ( x2 − x1 ) = mv12 + k ( x1 − x0 ) 2 + 0 J 2 2 2 2 1 2 1 mv2 + μ k mg ( x2 − x1 ) = 0 J+ k ( x1 − x0 ) 2 2 2 1 1 2 2 (1.0 kg)v2 + (0.30)(1.0 kg)(9.8 m/s )(1.5 m) = (3000 N/m)(0.30 m) 2 2 2 (0.5 kg)v22 + 4.41 J=135 J ⇒ v2 = 16.2 m/s
K 2 + U s2 + ΔEth = K1 + U s1 + Wext
11.76. Solve: (a)
The graph is a hyperbola. (b) The separation for zero potential energy is r = ∞, since U =−
Gm1m2 → 0 J as r → ∞ r
This makes sense because two masses don’t interact at all if they are infinitely far apart. (c) Due to the absence of nonconservative forces in our system of two particles, the mechanical energy is conserved.
The equations of energy and momentum conservation are
K f + U gf = K i + U gi
⎛ Gm1m2 ⎞ 1 ⎛ Gm1m2 ⎞ 1 1 1 2 2 m1v1f2 + m2v2f2 + ⎜ − ⎟ = m1v1i + m2v2i + ⎜ − ⎟ 2 2 rf ⎠ 2 2 ri ⎠ ⎝ ⎝ ⎛ 1 1⎞ 1 1 m1v1f2 + m2v2f2 = Gm1m2 ⎜ − ⎟ 2 2 ⎝ rf ri ⎠
pf = pi ⇒ m1v1f + m2v2f = 0 kg m / s ⇒ v2f = −
m1 v1f m2
Substituting this expression for v2f into the energy equation, we get 2
⎞ ⎛ 1 1⎞ 1 1 ⎛m 2Gm2 ⎛ 1 1 ⎞ m1v1f2 + m2 ⎜ 1 v1f ⎟ = Gm1m2 ⎜ − ⎟ ⇒ v1f2 = ⎜ − ⎟ 2 2 ⎝ m2 ⎠ r r (1 m1 / m2 ) ⎝ rf ri ⎠ + i ⎠ ⎝ f With G = 6.67 × 10−11 N ⋅ m(kg) −2 , rf = R1 + R2 = 18 × 108 m, ri = 1.0 × 1014 m, m1 = 8.0 × 1030 kg, and m2 = 2.0 × 1030 kg, the above equation can be simplified to yield
v1f = 1.72 × 105 m/s, and v2f = −
⎛ 8.0 × 1030 kg ⎞ m1 5 5 v1f = ⎜ ⎟ × (1.72 × 10 m/s ) = 6.89 × 10 m/s 30 m2 ⎝ 2.0 × 10 kg ⎠
The speed of the heavier star is 1.7 × 105 m/s. That of the lighter star is 6.9 × 105 m/s.
11.77. Model: Model the lawnmower as a particle and use the model of kinetic friction. Visualize:
We placed the origin of our coordinate system on the lawnmower and drew the free-body diagram of forces. G G G Solve: The normal force n, which is related to the frictional force, is not equal to FG . This is due to the presence of F . G G The rolling friction is f r = μ r n, or n = f r μ r . The lawnmower moves at constant velocity, so Fnet = 0. The two components of Newton’s second law are
( ∑ F ) = n − F − F sin 37° = ma = 0 N ⇒ f / μ − mg − F sin 37° = 0 N ⇒ f ( ∑ F ) = F cos37° − f = 0 N ⇒ F cos37° − μ mg − μ F sin 37° = 0 N G
y
y
r
x
⇒F=
r
r
r
r
= μ r mg + μ r F sin 37°
r
μ r mg (0.15)(12 kg)(9.8 m/s 2 ) = = 29.4 N cos37° − μ r sin 37° (0.7986) − (0.15)(0.6018)
G G Thus, the power supplied by the gardener in pushing the lawnmower at a constant speed of 1.2 m/s is P = F ⋅ v = Fv cosθ = (24.9 N)(1.2 m/s)cos37° = 24 W.
11-1
12.1.
Model: A spinning skater, whose arms are outstretched, is a rigid rotating body.
Visualize:
Solve: The speed v = rω , where r = 140 cm/2 = 0.70 m. Also, 180 rpm = (180)2π /60 rad/s = 6π rad/s. Thus, v = (0.70 m)(6π rad/s) = 13.2 m/s. Assess: A speed of 13.2 m/s ≈ 26 mph for the hands is a little high, but reasonable.
12.2.
Model:
Assume constant angular acceleration.
⎛ 2π rad ⎞⎛ min ⎞ (a) The final angular velocity is ω f = ( 2000 rpm ) ⎜ ⎟⎜ ⎟ = 209.4 rad/s. The definition of ⎝ rev ⎠⎝ 60 s ⎠ angular acceleration gives us Solve:
α=
Δω ω f − ωi 209.4 rad/s − 0 rad/s = = = 419 rad/s Δt Δt 0.50 s
The angular acceleration of the drill is 4.2 × 102 rad/s. (b) 1 2
θ f = θ i + ωi Δt + α ( Δt ) = 0 rad + 0 rad + 2
⎛ rev ⎞ The drill makes (52.4 rad) ⎜ ⎟ = 8.3 revolutions. ⎝ 2π rad ⎠
1 2 ( 419 rad/s )( 0.50 s ) = 52.4 rad 2
12.3.
Model: Visualize:
Solve:
Assume constant angular acceleration.
(a) Since at = rα , find α first. With 90 rpm = 9.43 rad/s and 60 rpm = 6.28 rad/s,
α=
Δω 9.43 rad − 6.28 rad/s = = 0.314 rad/s 2 Δt 10 s
The angular acceleration of the sprocket and pedal are the same. So
at = rα = ( 0.18 m ) ( 0.314 rad/s 2 ) = 0.057 m/s2 (b) The length of chain that passes over the sprocket during this time is L = r Δθ . Find Δθ :
1 2
θ f = θ i + ωi Δt + α ( Δt )
2
θ f − θ i = Δθ = ( 6.28 rad/s )(10 s ) +
1 ( 0.314 rad/s2 ) (10 s )2 = 78.5 rad 2
The length of chain which has passed over the top of the sprocket is L = (0.10 m)(78.5 rad) = 7.9 m
12.4.
Model: Visualize:
Assume constant angular acceleration.
⎛ 2π rad ⎞⎛ min ⎞ The initial angular velocity is ωi = ( 60 rpm ) ⎜ ⎟⎜ ⎟ = 2π rad/s. ⎝ rev ⎠⎝ 60 s ⎠ The angular acceleration is Solve:
α=
ωf − ωi Δt
=
0 rad/s − 2π rad/s = −0.251 rad/s 2 25 s
The angular velocity of the fan blade after 10 s is
ω f = ωi + α ( t − t0 ) = 2π rad/s+ ( −0.251 rad/s 2 ) (10 s − 0 s ) = 3.77 rad/s The tangential speed of the tip of the fan blade is
vt = rω = ( 0.40 m )( 3.77 rad/s ) = 1.51 m/s (b) 1 2 2 The fan turns 78.6 radians = 12.5 revolutions while coming to a stop.
θ f = θ i + ω i Δt + α ( Δt ) = 0 rad + ( 2π rad/s )( 25 s ) +
1 ( −0.251 rad/s 2 ) ( 25 s )2 = 78.6 rad 2
12.5.
Model: Visualize:
The earth and moon are particles.
Choosing xE = 0 m sets the coordinate origin at the center of the earth so that the center of mass location is the distance from the center of the earth. Solve: 24 22 8 m x + mM xM ( 5.98 × 10 kg ) ( 0 m ) + ( 7.36 × 10 kg )( 3.84 × 10 m ) xcm = E E = mE + mM 5.98 × 1024 kg + 7.36 × 10 22 kg
= 4.67 × 106 m Assess: The center of mass of the earth-moon system is called the barycenter, and is located beneath the surface of the earth. Even though xE = 0 m the earth influences the center of mass location because mE is in the denominator of the expression for xcm .
12.6. Visualize: Please refer to Figure EX12.6. The coordinates of the three masses mA , mB , and mC are (0 cm, 0 cm), (0 cm, 10 cm), and (10 cm, 0 cm), respectively. Solve: The coordinates of the center of mass are
xcm =
mA xA + mB xB + mC xC (100 g)(0 cm) + (200 g)(0 cm) + (300 g)(10 cm) = = 5.0 cm mA + mB + mC (100 g + 200 g + 300 g)
ycm =
mA yA + mB yB + mC yC (100 g)(0 cm) + (200 g)(10 cm) + (300 g)(0 cm) = = 3.3 cm mA + mB + mC (100 g + 200 g + 300 g)
12.7. Visualize: Please refer to Figure EX12.7. The coordinates of the three masses mA , mB , and mC are (0 cm, 10 cm), (10 cm, 10 cm), and (10 cm, 0 cm), respectively. Solve: The coordinates of the center of mass are
xcm =
mA xA + mB xB + mC xC (200 g)(0 cm) + (300 g)(10 cm) + (100 g)(10 cm) = = 6.7 cm mA + mB + mC (200 g + 300 g + 100 g)
ycm =
mA yA + mB yB + mC yC (200 g)(10 cm) + (300 g)(10 cm) + (100 g)(0 cm) = = 8.3 cm mA + mB + mC (200 g + 300 g + 100 g)
12.8.
Model: Visualize:
Solve:
The balls are particles located at the ball’s respective centers.
The center of mass of the two balls measured from the left hand ball is xcm =
(100 g )( 0 cm ) + ( 200 g )( 30 cm ) = 20 cm 100 g + 200 g
The linear speed of the 100 g ball is ⎛ 2π rad ⎞⎛ min ⎞ v1 = rω = xcmω = ( 0.20 m )(120 rev/min ) ⎜ ⎟⎜ ⎟ = 2.5 m/s ⎝ rev ⎠⎝ 60 s ⎠
12.9.
Model: The earth is a rigid, spherical rotating body.
Solve:
The rotational kinetic energy of the earth is K rot = 12 Iω 2 . The moment of inertia of a sphere about its
diameter (see Table 12.2) is I = 52 M earth R 2 and the angular velocity of the earth is
ω=
2π rad = 7.27 × 10 −5 rad/s 24 × 3600 s
Thus, the rotational kinetic energy is 1⎛ 2 ⎞ K rot = ⎜ M earth R 2 ⎟ω 2 2⎝ 5 ⎠ 1 = (5.98 × 1024 kg)(6.37 × 106 m) 2 (7.27 × 10−5 rad/s) 2 = 2.57 × 1029 J 5
12.10. Model: The triangle is a rigid body rotating about an axis through the center. Visualize: Please refer to Figure EX12.10. Each 200 g mass is a distance r away from the axis of rotation, where r is given by 0.20 m 0.20 m = cos30° ⇒ r = = 0.2309 m r cos30° Solve: The moment of inertia of the triangle is I = 3 × mr 2 = 3(0.200 kg)(0.2309 m) 2 = 0.0320 kg m 2 . The frequency of rotation is given as 5.0 revolutions per s or 10π rad/s. The rotational kinetic energy is
K rot. =
1 2 1 Iω = (0.0320 kg m 2 )(10.0π rad/s) 2 = 15.8 J 2 2
12.11.
Model: The disk is a rigid body rotating about an axis through its center.
Visualize:
Solve:
The speed of the point on the rim is given by vrim = Rω . The angular velocity ω of the disk can be
determined from its rotational kinetic energy which is K = 12 Iω 2 = 0.15 J. The moment of inertia I of the disk about its center and perpendicular to the plane of the disk is given by 1 1 I = MR 2 = (0.10 kg)(0.040 m) 2 = 8.0 × 10−5 kg m 2 2 2 2(0.15 J) 0.30 J 2 ⇒ω = = ⇒ ω = 61.237 rad/s I 8.0 × 10−5 kg m 2 Now, we can go back to the first equation to find vrim . We get vrim = Rω = (0.040 m)(61.237 rad/s) = 2.4 m/s.
12.12.
Model:
The baton is a thin rod rotating about a perpendicular axis through its center of mass. 1 Solve: The moment of inertia of a thin rod rotating about its center is I = ML2 . For the baton, 12 1 2 I = ( 0.400 kg )( 0.96 m ) = 0.031 kg m 2 12 The rotational kinetic energy of the baton is 2
K rot =
1 2 1 ⎛ ⎛ 2π rad ⎞⎛ min ⎞ ⎞ Iω = ( 0.031 kg m 2 ) ⎜ (100 rev/min ) ⎜ ⎟⎜ ⎟ ⎟ = 1.68 J 2 2 ⎝ rev ⎠⎝ 60 s ⎠ ⎠ ⎝
12.13.
Model: The structure is a rigid body rotating about its center of mass.
Visualize:
We placed the origin of the coordinate system on the 300 g ball. Solve: First, we calculate the center of mass:
xcm =
(300 g)(0 cm) + (600 g)(40 cm) = 26.67 cm 300 g + 600 g
Next, we will calculate the moment of inertia about the structure’s center of mass: I = (300 g)( xcm ) 2 + (600 g)(40 cm − xcm ) 2 = (0.300 kg)(0.2667 m) 2 + (0.600 kg)(0.1333 m) 2 = 0.032 kg m 2 Finally, we calculate the rotational kinetic energy: 2 1 1 ⎛ 100 × 2π ⎞ K rot = Iω 2 = (0.032 kg m 2 ) ⎜ rad/s ⎟ = 1.75 J 2 2 ⎝ 60 ⎠
12.14. Model: The moment of inertia of any object depends on the axis of rotation. In the present case, the rotation axis passes through mass A and is perpendicular to the page. Visualize: Please refer to Figure EX12.14. ∑ mi xi = mA xA + mB xB + mC xC + mD xD Solve: (a) xcm = mA + mB + mC + mD ∑ mi (100 g)(0 m) + (200 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 m) = 0.057 m 100 g + 200 g + 200 g + 200 g m y + mB yB + mC yC + mD yD = A A mA + mB + mC + mD
= ycm
(100 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 cm) + (200 g )(0 m) = 0.057 m 700 g (b) The distance from the axis to mass C is 14.14 cm. The moment of inertia through A and perpendicular to the page is =
I A = ∑ mi ri 2 = mA rA2 + mB rB2 + mC rC2 + mD rD2 i
= (0.100 kg)(0 m) 2 + (0.200 kg)(0.10 m) 2 + (0.200 kg)(0.1414 m) 2 + (0.200 kg)(0.10 m) 2 = 0.0080 kg m 2
12.15.
Model: The moment of inertia of any object depends on the axis of rotation.
Visualize:
Solve:
(a) xcm = ∑
mi xi
∑m
i
ycm
=
mA xA + mB xB + mC xC + mD xD mA + mB + mC + mD
(100 g)(0 m) + (200 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 m) = = 0.057 m 100 g + 200 g + 200 g + 200 g m y + mB yB + mC yC + mD yD = A A mA + mB + mC + mD
(100 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 cm) + (200 g )(0 m) = 0.057 m 700 g (b) The moment of inertia about a diagonal that passes through B and D is =
I BD = mA rA2 + mC rC2 where rA = rC = (0.10 m)cos 45° = 7.07 cm and are the distances from the diagonal. Thus, I BD = (0.100 kg) rA2 + (0.200 kg)rC2 = 0.0015 kg m 2 Assess:
Note that the masses B and D, being on the axis of rotation, do not contribute to the moment of inertia.
12.16.
Model: The three masses connected by massless rigid rods is a rigid body.
Visualize: Solve:
(a)
Please refer to Figure EX12.16.
xcm =
∑m x ∑m
i i i
ycm =
∑m y ∑m
i i
=
=
(0.100 kg)(0 m) + (0.200 kg)(0.06 m) + (0.100 kg)(0.12 m) = 0.060 m 0.100 kg + 0.200 kg + 0.100 kg
(0.100 kg)(0 m) + (0.200 kg)
(
)
(0.10 m) 2 − (0.06 m) 2 + (0.100 kg)(0 m)
0.100 kg + 0.200 kg + 0.100 kg
i
= 0.040 m
(b) The moment of inertia about an axis through A and perpendicular to the page is
I A = ∑ mi ri 2 = mB (0.10 m) + mC (0.10 m) 2 = (0.100 kg)[(0.10 m) 2 + (0.10 m) 2 ] = 0.0020 kg m 2 2
(c) The moment of inertia about an axis that passes through B and C is
I BC = mA Assess:
(
(0.10 m) 2 − (0.06 m) 2
)
2
= 0.00128 kg m 2
Note that mass mA does not contribute to I A , and the masses mB and mC do not contribute to I BC .
12.17.
Model: The door is a slab of uniform density. (a) The hinges are at the edge of the door, so from Table 12.2, 1 2 I = ( 25 kg )( 0.91 m ) = 6.9 kg m 2 3 (b) The distance from the axis through the center of mass along the height of the door is ⎛ 0.91 m ⎞ d =⎜ − 0.15 m ⎟ = 0.305 m. Using the parallel–axis theorem, 2 ⎝ ⎠ 1 2 2 I = I cm + Md 2 = ( 25 kg )( 0.91 m ) + ( 25 kg )( 0.305 cm ) = 4.1 kg m 2 12 Assess: The moment of inertia is less for a parallel axis through a point closer to the center of mass. Solve:
12.18.
Model: The CD is a disk of uniform density. (a) The center of the CD is its center of mass. Using Table 12.2, 1 1 2 I cm = MR 2 = ( 0.021 kg )( 0.060 m ) = 3.8 × 10−5 kg m 2 2 2 (b) Using the parallel–axis theorem with d = 0.060 m, Solve:
I = I cm + Md 2 = 3.8 × 10−5 kg m 2 + ( 0.021 kg )( 0.060 m ) = 1.14 × 10−4 kg m 2 2
12.19. Visualize:
G Solve: Torque by a force is defined as τ = Fr sin φ where φ is measured counterclockwise from the r vector to G the F vector. The net torque on the pulley about the axle is the torque due to the 30 N force plus the torque due to the 20 N force:
(30 N)r1 sinφ1 + (20 N)r2 sinφ2 = (30 N)(0.02 m) sin ( − 90°) + (20 N)(0.02 m) sin (90°) = ( − 0.60 N m) + (0.40 N m) = −0.20 N m Assess:
A negative torque causes a clockwise acceleration of the pulley.
12.20. Visualize:
The two equal but opposite 50 N forces, one acting at point P and the other at point Q, make a couple that causes a net torque. Solve: The distance between the lines of action is l = d cos30°. The net torque is given by
τ = lF = (d cos30°)F = (0.10 m)(0.866)(50 N) = 4.3 N m
12.21. Visualize:
Solve:
The net torque on the spark plug is τ = Fr sin φ = −38 N m = F (0.25 m)sin(−120°) ⇒ F = 176 N
That is, you must pull with a force of 176 N to tighten the spark plug. Assess: The force applied on the wrench leads to its clockwise motion. That is why we have used a negative sign for the net torque.
12.22.
Model: The disk is a rotating rigid body.
Visualize:
The radius of the disk is 10 cm and the disk rotates on an axle through its center. Solve: The net torque on the axle is
τ = FA rA sin φA + FBrB sin φB + FC rC sin φC + FD rD sin φD = (30 N)(0.10 m)sin(−90°) + (20 N)(0.050 m)sin 90° + (30 N)(0.050 m)sin135° + (20 N)(0.10 m)sin 0° = −3 N m + 1 N m + 1.0607 N m = −0.94 N m Assess: A negative torque means a clockwise rotation of the disk.
12.23.
Model: The beam is a solid rigid body.
Visualize:
G The steel beam experiences a torque due to the gravitational force on the construction worker FG and the C G gravitational force on the beam FG . The normal force exerts no torque since the net torque is calculated about the
( )
( )
B
point where the beam is bolted into place. G Solve: The net torque on the steel beam about point O is the sum of the torque due to FG G FG . The gravitational force on the beam acts at the center of mass.
( )
( )
C
and the torque due to
B
τ = (( FG )C )(4.0 m)sin( −90°) + (( FG ) B )(2.0 m)sin(−90°) = −(70 kg)(9.80 m/s 2 )(4.0 m) − (500 kg)(9.80 m/s 2 )(2.0 m) = −12.5 kN m The negative torque means these forces would cause the beam to rotate clockwise. The magnitude of the torque is 12.5 kN m.
12.24.
Model: Model the arm as a uniform rigid rod. Its mass acts at the center of mass.
Visualize:
Solve:
(a) The torque is due both to the gravitational force on the ball and the gravitational force on the arm:
τ = τ ball + τ arm = ( mb g )rb sin 90° + (ma g )ra sin 90° = (3.0 kg)(9.8 m/s 2)(0.70 m)+(4.0 kg)(9.8 m/s 2)(0.35 m) = 34 N m (b) The torque is reduced because the moment arms are reduced. Both forces act at φ = 45° from the radial line, so
τ = τ ball + τ arm = (mb g )rb sin 45° + ( ma g ) ra sin 45° = (3.0 kg)(9.8 m/s 2 )(0.70 m)(0.707) + (4.0 kg)(9.8 m/s 2)(0.35 m)(0.707) = 24 N m
12.25. Solve: τ = Iα is the rotational analog 2 τ = (2.0 kg m )(4.0 rad/s 2 ) = 8.0 kg m 2 /s 2 = 8.0 N m.
of
Newton’s
second
law
F = ma.
We
have
12.26.
Visualize:
Since
α = τ /I , a graph of the angular acceleration looks just like the torque graph with
the numerical values divided by I = 4.0 kg m 2 .
Solve: From the discussion about Figure 4.47
ω f = ωi + area under the angular acceleration α curve between ti and tf The area under the curve between t = 0 s and t = 3 s is 0.75 rad/s. With ω1 = 0 rad/s, we have
ω f = 0 rad/s + 0.75 rad/s = 0.75 rad/s
12.27. Model: Two balls connected by a rigid, massless rod are a rigid body rotating about an axis through the center of mass. Assume that the size of the balls is small compared to 1 m. Visualize:
We placed the origin of the coordinate system on the 1.0 kg ball. Solve: The center of mass and the moment of inertia are
xcm =
(1.0 kg)(0 m) + (2.0 kg)(1.0 m) = 0.667 m and ycm = 0 m (1.0 kg + 2.0 kg)
I about cm = ∑ mi ri 2 = (1.0 kg)(0.667 m) 2 + (2.0 kg)(0.333 m) 2 = 0.667 kg m 2 We have ω f = 0 rad/s, tf − ti = 5.0 s, and ωi = −20 rpm = −20(2π rad/60 s) = − 32 π rad/s, so ω f = ωi + α (tf − ti ) becomes 2π ⎛ 2π ⎞ 0 rad/s = ⎜ − rad/s ⎟ + α (5.0 s) ⇒ α = rad/s 2 15 ⎝ 3 ⎠ Having found I and α , we can now find the torque τ that will bring the balls to a halt in 5.0 s: ⎛2 ⎝3
⎞⎛ 2π ⎞ 4π rad/s 2 ⎟ = N m = 0.28 N m 15 ⎠⎝ ⎠ 45
τ = I about cmα = ⎜ kg m 2 ⎟⎜
The magnitude of the torque is 0.28 N m, applied in the counterclockwise direction.
12.28. Model: A circular plastic disk rotating on an axle through its center is a rigid body. Assume axis is perpendicular to the disk. Solve: To determine the torque (τ) needed to take the plastic disk from ωi = 0 rad/s
to
ω f = 1800 rpm = (1800)(2π ) / 60 rad/s = 60π rad/s in tf − ti = 4.0 s, we need to determine the angular acceleration (α ) and the disk’s moment of inertia (I ) about the axle in its center. The radius of the disk is R = 10.0 cm. We have 1 1 I = MR 2 = (0.200 kg)(0.10 m) 2 = 1.0 × 10−3 kg m 2 2 2 ω − ωi 60π rad/s − 0 rad/s ω f = ωi + α (tf − ti ) ⇒ α = f = = 15π rad/s 2 t f − ti 4.0 s Thus, τ = Iα = (1.0 × 10−3 kg m 2 )(15π rad/s 2 ) = 0.047 N m.
12.29.
Model: The compact disk is a rigid body rotating about its center.
Visualize:
Solve:
(a) The rotational kinematic equation ω1 = ω 0 + α (t1 − t0 ) gives
200π ⎛ 2π ⎞ (2000 rpm) ⎜ rad/s 2 ⎟ rad/s = 0 rad + α (3.0 s − 0 s) ⇒ α = 9 ⎝ 60 ⎠ The torque needed to obtain this operating angular velocity is ⎛ 200π ⎞ rad/s 2 ⎟ = 1.75 × 10−3 N m ⎝ 9 ⎠
τ = Iα = (2.5 × 10−5 kg m 2 ) ⎜ (b) From the rotational kinematic equation,
1
1 ⎛ 200π
θ1 = θ 0 + ω 0 (t1 − t0 ) + α (t1 − t0 ) 2 = 0 rad + 0 rad + ⎜ 2 2⎝ 9 100π revolutions = 50 rev = 100π rad = 2π Assess:
Fifty revolutions in 3 seconds is a reasonable value.
2 ⎞ rad/s 2 ⎟ ( 3.0 s − 0 s ) ⎠
12.30. Model: The rocket attached to the end of a rigid rod is a rotating rigid body. Assume the rocket is small compared to 60 cm. Visualize: Please refer to Figure EX12.30. Solve: We can determine the rocket’s angular acceleration from the relationship τ = Iα . The torque τ can be found from the thrust (F) using τ = Fr sin φ . The moment of inertia (I) can be calculated from equations given in Table 12.2. Specifically, I = I rod about one end + I rocket becomes 1 1 M rod L2 + ML2 = (0.100 kg)(0.60 m) 2 + (0.200 kg)(0.60 m) 2 3 3 = 0.012 kg m 2 + 0.072 kg m 2 = 0.0840 kg m 2
⇒α = Assess:
τ I
=
Fr sin φ (4.0 N)(0.60 m)sin(45°) = = 20 rads/s 2 I 0.0840 kg m 2
The rocket will accelerate counterclockwise since α is positive.
12.31.
Model: The rod is in rotational equilibrium, which means that τ net = 0.
Visualize:
As the gravitational force on the rod and the hanging mass pull down (the rotation of the rod is exaggerated in the figure), the rod touches the pin at two points. The piece of the pin at the very end pushes down on the rod; the right end of the pin pushes up on the rod. To understand this, hold a pen or pencil between your thumb and forefinger, with your thumb on top (pushing down) and your forefinger underneath (pushing up). Solve: Calculate the torque about the left end of the rod. The downward force exerted by the pin acts through this G point, so it exerts no torque. To prevent rotation, the pin’s normal force npin exerts a positive torque (ccw about the left end) to balance the negative torques (cw) of the gravitational force on the mass and rod. The gravitational force on the rod acts at the center of mass, so
τ net = 0 N m = τ pin − (0.40 m)(2.0 kg)(9.8 m/s 2) − (0.80 m)(0.50 kg)(9.8 m/s 2) ⇒ τ pin = 11.8 N m
12.32.
Model: The massless rod is a rigid body.
Visualize:
Solve:
G To be in equilibrium, the object must be in both translational equilibrium ( Fnet = 0 N) and rotational
equilibrium (τ net = 0 Nm). We have ( Fnet ) y = (40 N) − (100 N) + (60 N) = 0 N, so the object is in translational equilibrium. Measuring τ net about the left end,
τ net = (60 N)(3.0 m)sin( +90°) + (100 N)(2.0 m)sin(−90°) = −20 N m The object is not in equilibrium.
12.33.
Model: The object balanced on the pivot is a rigid body.
Visualize:
Since the object is balanced on the pivot, it is in both translational equilibrium and rotational equilibrium. G Solve: There are three forces acting on the object: the gravitational force FG acting through the center of 1 G mass of the long rod, the gravitational force FG acting through the center of mass of the short rod, and the 2 G normal force P on the object applied by the pivot. The translational equilibrium equation ( Fnet ) y = 0 N is
( )
( )
− ( FG )1 − ( FG )2 + P = 0 N ⇒ P = ( FG )1 + ( FG )2 = (1.0 kg)(9.8 m/s 2 ) + (4.0 kg)(9.8 m/s 2 ) = 49 N
Measuring torques about the left end, the equation for rotational equilibrium τ net = 0 Nm is
Pd − w1 (1.0 m) − w2 (1.5 m) = 0 Nm ⇒ (49 N)d − (1.0 kg)(9.8 m/s 2 )(1.0 m) − (4.0 kg)(9.8 m/s2 )(1.5 m) = 0 N ⇒ d = 1.40 m Thus, the pivot is 1.40 m from the left end.
12.34.
Model: Visualize:
Solve:
The see-saw is a rigid body. The cats and bowl are particles.
The see-saw is in rotational equilibrium. Calculate the net torque about the pivot point.
τ net = 0 = ( FG )1 ( 2.0 m ) − ( FG )2 ( d ) − ( FG ) B ( 2.0 m )
m2 gd = m1 g ( 2.0 m ) − mB g ( 2.0 m ) d=
( m1 − mB )( 2.0 m ) = ( 5.0 kg − 2.0 kg )( 2.0 m ) = 1.5 m m2
4.0 kg
Assess: The smaller cat is close but not all the way to the end by the bowl, which makes sense since the combined mass of the smaller cat and bowl of tuna is greater than the mass of the larger cat.
12.35.
Solve:
(a) According to Equation 12.35, the speed of the center of mass of the tire is
vcm = Rω = 20 m/s ⇒ ω =
vcm 20 m/s ⎛ 60 ⎞ 2 = = 66.67 rad/s = ( 66.7 ) ⎜ ⎟ rpm = 6.4 × 10 rpm R 0.30 m 2 π ⎝ ⎠
(b) The speed at the top edge of the tire relative to the ground is vtop = 2vcm = 2(20 m/s) = 40 m/s. (c) The speed at the bottom edge of the tire relative to ground is vbottom = 0 m/s.
12.36. Model: The can is a rigid body rolling across the floor. Assume that the can has uniform mass distribution. Solve: The rolling motion of the can is a translation of its center of mass plus a rotation about the center of mass. The moment of inertia of the can about the center of mass is 12 MR 2, where R is the radius of the can. Also vcm = Rω , where ω is the angular velocity of the can. The total kinetic energy of the can is K = K cm + K rot =
1 1 1 1⎛ 1 ⎞⎛ v ⎞ 2 2 Mvcm + I cmω 2 = Mvcm + ⎜ MR 2 ⎟⎜ cm ⎟ 2 2 2 2⎝ 2 ⎠⎝ R ⎠
3 3 2 = Mvcm = (0.50 kg)(1.0 m/s) 2 = 0.38 J 4 4
2
12.37.
Model: The sphere is a rigid body rolling down the incline without slipping.
Visualize:
The initial gravitational potential energy of the sphere is transformed into kinetic energy as it rolls down. Solve: (a) If we choose the bottom of the incline as the zero of potential energy, the energy conservation equation will be K f = U i . The kinetic energy consists of both translational and rotational energy. This means
Kf =
1 1 1⎛ 2 1 ⎞ 2 I cmω 2 + Mvcm = Mgh ⇒ ⎜ MR 2 ⎟ω 2 + M ( Rω ) 2 = Mgh 2 2 2⎝ 5 2 ⎠ 7 ⇒ MR 2ω 2 = Mg (2.1 m)sin 25° 10
⇒ω =
10 7
g (2.1 m)(sin 25°) = R2
10 7
g (2.1 m)(sin 25°) = 88 rad/s (0.04 m) 2
(b) From part (a)
K total =
1 1 7 1 1⎛ 2 1 ⎞ 2 = MR 2ω 2 and K rot = I cmω 2 = ⎜ MR 2 ⎟ω 2 = MR 2ω 2 I cmω 2 + Mvcm 2 2 10 2 2⎝ 5 5 ⎠ 2 2 1 MR ω K 1 10 2 ⇒ rot = 75 = × = K total 10 MR 2ω 2 5 7 7
12.38. Visualize: Please refer to Figure EX12.38. To determine angle α , put the tails of the vectors together. G G G G Solve: (a) The magnitude of A × B is AB sin α = (6)(4)sin 45° = 17. The direction of A × B, using the right G G hand rule, is out of the page. Thus, A × B = (17, out of the page). G G G G G (b) The magnitude of C × D is CD sin α = (6)(4)sin180° = 0. Thus C × D = 0.
12.39.
Visualize: Please refer to Figure EX12.39. G G G G (a) The magnitude of A × B is AB sin α = (6)(4)sin 60° = 20.78. The direction of A × B is given by the G G G G right hand rule. To curl our fingers from A to B, we have to point our thumb out of the page. Thus, A × B = (21, out of the page). G G (b) C × D = ((6)(4)sin 90°, into the page) = (24, into the page).
Solve:
(a) (iˆ × ˆj ) × iˆ = kˆ × iˆ = ˆj (b) iˆ × ( ˆj × iˆ) = iˆ × (− kˆ) = −iˆ × kˆ = −(− ˆj ) = ˆj
12.40.
Solve:
(a) iˆ × (iˆ × ˆj ) = iˆ × kˆ = − ˆj G (b) (iˆ × ˆj ) × kˆ = kˆ × kˆ = 0
12.41.
Solve:
12.42.
(b)
Solve:
G G (a) A × B = (3iˆ + ˆj ) × (3iˆ − 2 ˆj + 2kˆ) = 9iˆ × iˆ − 6iˆ × ˆj + 6iˆ × kˆ + 3 ˆj × iˆ − 2 ˆj × ˆj + 2 ˆj × kˆ = 0 − 6kˆ + 6(− ˆj ) + 3(−kˆ) − 0 + 2iˆ = 2iˆ − 6 ˆj − 9kˆ
G
G
G
G
12.43. Solve: (a) C × D = G0 implies that D must also be in the same or opposite direction as the C vector or zero, because iˆ × iˆ = 0. Thus D = niˆ, where n could be any real number. G G G G (b) C × E = 6kˆ implies that E must be along the ˆj vector, because iˆ × ˆj = kˆ. Thus E = 2 ˆj. G G G G (c) C × F = −3 ˆj implies that F must be along the kˆ vector, because iˆ × kˆ = − ˆj. Thus F = 1kˆ.
12.44.
Solve:
G
G
G
τ = r × F = (5iˆ + 5 ˆj ) × (−10 ˆj ) N m
G = [−50(iˆ × ˆj ) − 50( ˆj × ˆj )] N m = [−50(+ kˆ) − 0] N m = −50kˆ N m
12.45.
Solve:
G
G
G
τ = r × F = (5 ˆj ) × (−10iˆ + 10 ˆj ) N m = (−5 ˆj × iˆ + 50 ˆj × ˆj ) N m G = [−50( −kˆ) + 0] N m = 50kˆ N m
12.46. Solve:
Visualize: Please refer to Figure EX12.46. G G G L = r × mv = (1.0iˆ + 2.0 ˆj ) m × (0.200 kg)(3.0 m/s) cos45°iˆ − sin 45° ˆj
(
)
= (0.42iˆ × iˆ − 0.42iˆ × ˆj + 0.85 ˆj × iˆ − 0.85 ˆj × ˆj ) kg m 2 /s = −(1.27 kˆ) kg m 2 /s or (1.27 kg m 2 /s, into page)
12.47.
Solve:
Visualize: Please refer to Figure EX12.47. G G G L = r × mv = (3.0iˆ + 2.0 ˆj ) m × (0.1 kg)(4.0ˆj ) m/s = 1.20(iˆ × ˆj ) kg m 2 /s + 0.8( ˆj × ˆj ) kg m 2 /s = 1.20kˆ kg m 2 /s + 0 kg m 2 /s = 1.20kˆ kg m 2 /s or (1.20 kg m 2 /s, out of page)
12.48. Model: The bar is a rotating rigid body. Assume that the bar is thin. Visualize: Please refer to Figure EX12.48. Solve: The angular velocity ω = 120 rpm = (120)(2π ) / 60 rad/s = 4π rad/s. From Table 12.2, the moment of inertial of a rod about its center is I = 121 ML2 . The angular momentum is ⎛1⎞ L = Iω = ⎜ ⎟ (0.50 kg)(2.0 m) 2 (4π rad/s) = 2.1 kg m 2 /s ⎝ 12 ⎠
If we wrap our fingers in the direction of the rod’s rotation, our thumb will point in the z direction or out of the page. Consequently, G L = (2.1 kg m 2 /s, out of the page)
12.49. Model: The disk is a rotating rigid body. Visualize: Please refer to Figure EX12.49. Solve: From Table 12.2, the moment of inertial of the disk about its center is
The
angular
1 1 I = MR 2 = (2.0 kg)(0.020 m) 2 = 4.0 × 10−4 kg m 2 2 2 ω is 600 rpm = 600 × 2π /60 rad/s = 20π rad/s. velocity
Thus,
L = Iω = (4.0 × 10 kg m )(20π rad/s) = 0.025 kg m /s. If we wrap our right fingers in the direction of the disk’s rotation, our thumb will point in the − x direction. Consequently, G L = −0.025 iˆ kg m 2 /s = (0.025 kg m 2 /s, into page) −4
2
2
12.50.
Model: The beach ball is a spherical shell. Solve: From Table 12.2, the moment of inertia about a diameter of a spherical shell is 2 2 1 2 I = MR 2 = ( 0.100 kg )( 0.50 m ) = kg m 2 3 3 60 Require ⎛ 1 ⎞ kg m 2 ⎟ω L = 0.10 kg m 2 /s = Iω = ⎜ 60 ⎝ ⎠ ⇒ ω = ( 60 rad/s )
⎛ rev ⎞ ⎛ 60 s ⎞ In rpm, this is ( 60 rad/s ) ⎜ ⎟⎜ ⎟ = 57 rpm. ⎝ 2π rad ⎠ ⎝ min ⎠
12.51.
Model: The wheel is a rigid rolling body.
Visualize:
Solve:
The front of the disk is moving forward at velocity vcm . Also, because of rotation the point is moving
downward at velocity vrel = Rω = vcm . So, this point has a speed 2 2 v = vcm + vcm = 2vcm = 2(20 m/s) = 28 m/s
Assess:
The speed v is independent of the radius of the wheel.
12.52
Model: The triangle is a rigid body rotating about its center of mass perpendicular to the plane of the triangle. The center of mass of any symmetrical object of uniform density is at the physical center of the object. Visualize:
The distance to one tip of the triangle from the center of mass is given by r cos30° = (5.0/2) cm, which yields r = (2.5 cm)/ cos30° = 2.9 cm. Solve: The speed of the tip is v = rω = (2.9 cm)(120 rpm) = (2.9 cm)(4π rad/s) = 36 cm/s
12.53. Visualize:
Solve: We will consider a vertical strip of width dx and of mass dm at a position x from the origin. The formula for the x component of the center of mass is 1 xcm = x dm M∫
The area of the steel plate is A = 12 (0.2 m)(0.3 m) = 0.030 m 2 . Mass dm in the strip is the same fraction of M as dA is of A. Thus dm dA M ⎛ 0.800 kg ⎞ = ⇒ dm = dA = ⎜ dA = (26.67 kg/m 2 )l dx 2 ⎟ M A A ⎝ 0.030 m ⎠
The relationship between l and x is l x 2 = ⇒l = x 0.20 m 0.30 m 3 Therefore, xcm
1 (17.78 kg/m 2 ) x3 2 ⎛ 2⎞ 2 (26.67 kg/m ) x dx = = ⎜ ⎟ M∫ M 3 ⎝ 3⎠
Due to symmetry ycm = 0 cm.
0.3 m
= 0m
(17.78 kg/m 2 ) (0.3 m)2 = 20 cm 0.8 kg 3
12.54.
Visualize:
Solve: Build the plate from the three shapes 1, 2, and 3. The center of mass of the complete shape is the center of mass of the center of masses of the three smaller shapes. m1 ( xcm )1 + m2 ( xcm )2 + m3 ( xcm )3 xcm = m1 + m2 + m3
ycm =
m1 ( ycm )1 + m2 ( ycm )2 + m3 ( ycm )3
m1 + m2 + m3 The masses of each of the smaller shapes are a fraction of the larger shape’s mass by the ratio of areas. The following table will be useful in the solution. Shape A mass xcm ycm 2 All M xcm ycm 32 cm 1
2 3
24 32
M
−1.0 cm
0 cm
2.0 cm
2
2.0 32
M
2.0 cm
−2.5 cm
6.0 cm
2
6.0 32
M
2.0 cm
1.5 cm
24 cm 2
Thus, ⎛ 24 ⎞ ⎛ 2.0 ⎞ ⎛ 6.0 ⎞ M ⎟ ( −1.0 cm ) + ⎜ M ⎟ ( 2.0 cm ) + ⎜ M ⎟ ( 2.0 cm ) ⎜ 32 ⎠ 32 32 ⎝ ⎝ ⎠ ⎝ ⎠ = −0.25 cm xcm = M Similarly, ycm = 0.125 cm. Assess: The plate’s center of mass of (−0.25 cm, 0.125 cm) is reasonable. The cutout means more of the mass is left of center and above the center.
12.55.
Model:
The disk is a rigid rotating body. The axis is perpendicular to the plane of the disk.
Visualize:
Solve:
(a) From Table 12.2, the moment of inertia of a disk about its center is
1 1 I = MR 2 = (2.0 kg)(0.10 m) 2 = 0.010 kg m 2 2 2 (b) To find the moment of inertia of the disk through the edge, we can make use of the parallel axis theorem:
I = I center + Mh 2 = (0.010 kg m 2 ) + (2.0 kg)(0.10 m) 2 = 0.030 kg m 2 Assess: The larger moment of inertia about the edge means there is more inertia to rotational motion about the edge than about the center.
12.56.
Model: The object is a rigid rotating body. Assume the masses m1 and m2 are small and the rod is thin. Visualize: Please refer to P12.56. Solve: The moment of inertia of the object is the sum of the moment of inertia of the rod, mass m1 , and mass m2 . Using Table 12.2 for the moment of inertia of the rod, we get 2
I rod = I rod about center + I m1 + I m2 = = Assess:
1 ⎛ L⎞ ⎛ L⎞ ML2 + m1 ⎜ ⎟ + m2 ⎜ ⎟ 12 2 ⎝ ⎠ ⎝ 4⎠
1 1 1 L2 ⎛ M m ⎞ ML2 + m1L2 + m2 L2 = ⎜ + m1 + 2 ⎟ 12 4 16 4⎝ 3 4 ⎠
With m1 = m2 = 0 kg, I rod 121 ML2 , as expected.
2
12.57. Visualize:
We chose the origin of the coordinate system to be on the axis of rotation, that is, at a distance d from one end of the rod. Solve: The moment of inertia can be calculated as follows: x2
I = ∫ x 2 dm x1
⇒I =
M L
L−d
∫
−d
3 ⎛M ⎞x x 2 dx = ⎜ ⎟ ⎝ L⎠3
L−d
−d
and
dm dx M = ⇒ dm = dx M L L
1⎛ M = ⎜ 3⎝ L
M ⎞ 3 3 3 3 ⎟ [( L − d ) − (− d ) ] = [( L − d ) + d ] 3L ⎠
For d = 0 m, I = 13 ML , and for d = 12 L, 2
I=
3 3 M ⎡⎛ L ⎞ ⎛ L ⎞ ⎤ 1 2 ⎢⎜ ⎟ + ⎜ ⎟ ⎥ = ML 3L ⎣⎢⎝ 2 ⎠ ⎝ 2 ⎠ ⎦⎥ 12
Assess: The special cases d = 0 m and d = L/2 of the general formula give the same results that are found in Table 12.2.
12.58. Visualize:
Solve: We solve this problem by dividing the disk between radii r1 and r2 into narrow rings of mass dm. Let dA = 2π rdr be the area of a ring of radius r. The mass dm in this ring is the same fraction of the total mass M as dA is of the total area A. (a) The moment of inertia can be calculated as follows:
I disk = ∫ r 2 dm and dm =
M M dA = ( 2π r ) dr A π ( r22 − r12 )
r
⇒ I disk =
r
2 M 2M 2 2M r 4 r 2 (2π r ) dr = 2 2 ∫ r 3dr = 2 2 2 2 ∫ π ( r2 − r1 ) r1 ( r2 − r1 ) r1 ( r2 − r1 ) 4
r2
r1
M 2M r 4 − r14 ) = ( r22 + r12 ) = 2 2 ( 2 2 4 ( r2 − r1 ) Replacing r1 with r and r2 with R, the moment of inertia of the disk through its center is I disk = 12 M ( R 2 + r 2 ). (b) For r = 0 m, I disk = 12 MR 2 . This is the moment of inertia for a solid disk or cylinder about the center.
Additionally, for r ≅ R, we have I = MR 2 . This is the expression for the moment of inertia of a cylindrical hoop or ring about the center. (c) The initial gravitational potential energy of the disk is transformed into kinetic energy as it rolls down. If we choose the bottom of the incline as the zero of potential energy, and use vcm = ω R, the energy conservation equation K f = U i is 1 2 1 1⎛ M ⎞ v2 1 2 2 Iω + Mvcm = Mgh ⇒ ⎜ ⎟ ( R 2 + r 2 ) cm2 + Mvcm = Mgyi = Mg (0.50 m)sin 20° 2 2 2⎝ 2 ⎠ R 2 2 2 1 2 1 r2 ⎞ 2 ⎛ R +r ⎞ 2 ⎛1 v v ⇒ vcm + = + + 2 ⎟ = 1.6759 m 2 /s 2 ⎜ ⎟ ⎜ cm cm 2 ⎝ 4R ⎠ 2 ⎝ 2 4 4R ⎠ (0.015 m) 2 ⎞ 2 ⎛3 vcm = 1.6759 m 2 /s 2 ⇒ vcm = 1.37 m/s ⎜ + 2 ⎟ 4 4(0.020 m) ⎝ ⎠ For a sliding particle on a frictionless surface K f = U i , so
1 2 v mvf = mgyi ⇒ vf = 2 gyi = 2 g (0.50 m)sin 20° = 1.83 m/s ⇒ cm = 0.75 2 vf That is, vcm is 75% of the speed of a particle sliding down a frictionless ramp.
12.59.
Model: Visualize:
Solve:
The plate has uniform density.
The moment of inertia is
I = ∫ r 2 dm. Let the mass of the plate be M. Its area is L2 . A region of area dA located at (x,y) has mass M M dm = dA = 2 dx dy. The distance from the axis of rotation to the point (x, y) is r = x 2 + y 2 . With A L L L L L − ≤ x ≤ and − ≤ y ≤ , 2 2 2 2 I=
L 2
L 2
∫ ∫ (x
L L − − 2 2
M = 2 L
2
⎛M + y2 )⎜ 2 ⎝L
M ⎞ ⎟ dx dy = 2 L ⎠
L 2
L 2
⎛ L3 y 2 L ⎛ − L3 y 2 L ⎞ ⎞ M ∫L ⎜⎜ 24 + 2 − ⎜⎝ 24 − 2 ⎟⎠ ⎟⎟dy = L2 ⎠ − ⎝
⎛ L3 L3 ⎞ ⎞ 1 2 M ⎛ L4 ⎜ + L ⎜ + ⎟ ⎟⎟ = ML 2 ⎜ L ⎝ 12 ⎝ 24 24 ⎠ ⎠ 6
⎞
∫ ⎜⎝ 3 + y x ⎟⎠ 2
L − 2
2
=
⎛ x3
L 2
dy −
L 2
L 2
⎛ ⎛ L3 M ⎜ L4 y3 2⎞ Ly dy L + = + ⎜ ⎟ ∫L ⎝ 12 L2 ⎜⎜ 12 3 ⎠ − ⎝ 2
⎞ ⎟ L ⎟ − ⎟ 2 ⎠ L 2
12.60.
Solve:
From Equation 12.16,
I = ∫ ( x 2 + y 2 )dm
A small region of area dA has mass dm, and M M dA = dx dy A A The area of the plate is 12 (0.20 m)(0.30 m) = 0.030 m 2 . So dm =
M ⎛ 0.800 kg ⎞ 2 =⎜ ⎟ = 26.67 kg m A ⎝ 0.030 m 2 ⎠ 1 1 1 The limits for x are 0 ≤ x ≤ 30 cm. For a particular value of x, − x ≤ y ≤ x. Note that ± is the slope of the 3 3 3 top and bottom edges of the triangle. Therefore, 30 cm
I=
1 x 3
∫ ∫ (x 0
2
+ y 2 )( 26.67 kg/m 2 ) dx dy
1 − x 3
1
= ( 26.67 kg/m
30 cm
2
)∫ 0
= ( 26.67 kg/m 2 )
30 cm
∫ 0
x
⎛ 2 y3 ⎞ 3 ⎜ x y + ⎟ dx 3 ⎠ −1 x ⎝ 3
⎛ 2 3 2 x3 ⎞ ⎜ x + ⎟dx 81 ⎠ ⎝3 30 cm
⎛ 56 ⎞⎛ 1 ⎞ = ( 26.67 kg/m 2 ) ⎜ ⎟⎜ x 4 ⎟ ⎝ 81 ⎠⎝ 4 ⎠ 0
= 0.037 kg m 2
12.61. G Model: The ladder is a rigid rod of length L. To not slip, it must be in both translational equilibrium G ( Fnet = 0 N) and rotational equilibrium (τ net = 0 N m). We also apply the model of static friction. Visualize:
G Since the wall is frictionless, the only force from the wall on the ladder is the normal force n2 . On the other G G G hand, the floor exerts both the normal force n1 and the static frictional force f s . The gravitational force FG on the ladder acts through the center of mass of the ladder. G G Solve: The x- and y-components of Fnet = 0 N are
∑F
x
= n2 − fs = 0 N ⇒ fs = n2
∑F
y
= n1 − FG = 0 N ⇒ n1 = FG
The minimum angle occurs when the static friction is at its maximum value f s max = μ s n1. Thus we have n2 = f s = μs n1 = μ s mg . We choose the bottom corner of the ladder as a pivot point to obtain τ net , because two forces pass through this point and have no torque about it. The net torque about the bottom corner is
τ net = d1mg − d 2 n2 = (0.5L cosθ min )mg − ( L sinθ min ) μ s mg = 0 N m ⇒ 0.5cosθ min = μs sin θ min ⇒ tanθ min =
0.5
μs
=
0.5 = 1.25 ⇒ θ min = 51° 0.4
12.62.
Model: The beam is a rigid body of length 3.0 m and the student is a particle.
Visualize:
Solve:
G G To stay in place, the beam must be in both translational equilibrium ( Fnet = 0 N) and rotational
equilibrium (τ net = 0 Nm). The first condition is
∑F
y
= − ( FG )beam − ( FG )student + F1 + F2 = 0 N
⇒ F1 + F2 = ( FG ) beam + ( FG )student = (100 kg + 80 kg)(9.80 m/s 2 ) = 1764 N Taking the torques about the left end of the beam, the second condition is −( FG ) beam (1.5 m) − ( FG )student (2.0 m) + F2 (3.0 m) = 0 N m −(100 kg)(9.8 m/s 2 )(1.5 m) − (80 kg)(9.8 m/s 2 )(2.0 m) + F2 (3.0 m) = 0 N m ⇒ F2 = 1013 N From F1 + F2 = 1764 N, we get F1 = 1764 N − 1013 N = 0.75 kN. Assess: To establish rotational equilibrium, the choice for the pivot is arbitrary. We can take torques about any point on the body of interest.
12.63.
Model: The structure is a rigid body.
Visualize:
Solve:
We pick the left end of the beam as our pivot point. We don’t need to know the forces Fh and Fv
because the pivot point passes through the line of application of Fh and Fv and therefore these forces do not exert a torque. For the beam to stay in equilibrium, the net torque about this point is zero. We can write
τ about left end = −( FG ) B (3.0 m) − ( FG ) W (4.0 m) + (T sin150°)(6.0 m) = 0 N m Using ( FG ) B = (1450 kg)(9.8 m/s 2 ) and ( FG ) W = (80 kg)(9.8 m/s 2 ), the torque equation can be solved to yield T = 15,300 N. The tension in the cable is slightly more than the cable rating. The worker should be worried.
12.64. Model: G Model the beam as a rigid body. For the beam not to fall over, it must be both in translational G equilibrium ( Fnet = 0 N) and rotational equilibrium (τ net = 0 N m). Visualize:
The boy walks along the beam a distance x, measured from the left end of the beam. There are four forces acting G G on the beam. F1 and F2 are from the two supports, FG is the gravitational force on the beam, and FG is
( )
b
( )
B
the gravitational force on the boy. Solve: We pick our pivot point on the left end through the first support. The equation for rotational equilibrium is −( FG ) b (2.5 m) + F2 (3.0 m) − ( FG ) B x = 0 N m −(40 kg)(9.80 m/s 2 )(2.5 m) + F2 (3.0 m) − (20 kg)(9.80 m/s 2 ) x = 0 N m The equation for translation equilibrium is
∑F
y
= 0 N = F1 + F2 − ( FG )b − ( FG )B
⇒ F1 + F2 = ( FG )b + ( FG )B = (40 kg + 20 kg)(9.8 m/s 2 ) = 588 N Just when the boy is at the point where the beam tips, F1 = 0 N. Thus F2 = 588 N. With this value of F2 , we can simplify the torque equation to: −(40 kg)(9.80 m/s 2 )(2.5 m) + (588 N)(3.0 m) − (20 kg)(9.80 m/s 2 ) x = 0 N m ⇒ x = 4.0 m Thus, the distance from the right end is 5.0 m − 4.0 m = 1.0 m.
12.65.
Visualize:
Please refer to Figure P12.65.
Solve: The bricks are stable when the net gravitational torque on each individual brick or combination of bricks is zero. This is true as long as the center of gravity of each individual brick and any combination is over a base of support. To determine the relative positions of the bricks, work from the top down. The top brick can extend past the second brick by L 2. For maximum extension, their combined center of gravity will be at the edge of the third brick, and the combined center of gravity of the three upper bricks will be at the edge of the fourth brick. The combined center of gravity of all four bricks will be over the edge of the table. Measuring from the left edge of the brick 2, the center of gravity of the top two bricks is ⎛L⎞ m ⎜ ⎟ + mL m1 x1 + m2 x2 3 2 ( x12 )com = = ⎝ ⎠ = L. 2m 4 m1 + m2
Thus the top two bricks can extend L 4 past the edge of the third brick. The top three bricks have a center of mass ⎛L⎞ ⎛ 3L ⎞ ⎛ 5L ⎞ m⎜ ⎟ + m⎜ ⎟ + m⎜ ⎟ m1 x1 + m2 x2 + m3 x3 2⎠ 4 ⎠ ⎝ ⎝ ⎝ 4 ⎠ = 5 L. = ( x123 )com = 3m 6 m1 + m2 + m3 Thus the top three bricks can extend past the edge of the fourth brick by L 6. Finally, the four bricks have a combined center of mass at ⎛L⎞ ⎛ 4L ⎞ ⎛ 11L ⎞ ⎛ 17 L ⎞ m⎜ ⎟ + m⎜ ⎟ + m⎜ ⎟ + m⎜ ⎟ 2⎠ 6 ⎠ 12 ⎠ ⎝ ⎝ ⎝ ⎝ 12 ⎠ = 7 L. ( x1234 )com = 4m 8 The center of gravity of all four bricks combined is 7 L 8 from the left edge of the bottom brick, so brick 4 can extend L 8 past the table edge. Thus the maximum distance to the right edge of the top brick from the table edge is L L L L 25 d max = + + + = L. 8 6 4 2 24 Thus, yes, it is possible that no part of the top brick is directly over the table because d max > L. Assess: As crazy as this seems, the center of gravity of all four bricks is stably supported, so the net gravitational torque is zero, and the bricks do not fall over.
12.66.
Model: Visualize:
The pole is a uniform rod. The sign is also uniform.
Solve:
The geometry of the rod and cable give the angle that the cable makes with the rod. ⎛ 250 ⎞ θ = tan −1 ⎜ ⎟ = 51.3° ⎝ 200 ⎠ The rod is in rotational equilibrium about its left-hand end. ⎛1⎞ ⎛1⎞ τ net = 0 = − (100 cm )( FG )P − ( 80 cm ) ⎜ ⎟ ( FG )S − ( 200 cm ) ⎜ ⎟ ( FG )S + ( 200 cm ) T sin 51.3° ⎝ 2⎠ ⎝ 2⎠
= − (100 cm )( 5.0 kg ) ( 9.8 m/s 2 ) − mS ( 9.8 m/s 2 ) (140 cm ) + (156 cm ) T
With T = 300 N , mS = 30.6 kg. Assess: A mass of 30.6 kg is reasonable for a sign.
12.67.
Model: The hollow cylinder is a rigid rotating body.
Visualize:
We placed the origin of the coordinate system on the ground. Solve: (a) Newton’s second law for the block is −( FG ) B + T = mB a y , where T is the tension in the string, ( FG ) B = mB g is the gravitational force on the block, and a y is the acceleration of block. The string tension exerts a negative (cw) torque on the cylinder, so the rotational form of Newton’s second law for the hollow cylinder is
τ = −TR = Iα = I
ay R
⇒T = −
Ia y R2
where we used the acceleration constraint a y = α R. With this expression for T, Newton’s second law for the block becomes
− mB g −
−mB g Ia = mB a y ⇒ a y = 2 R (mB + I / R 2 )
The moment of inertia of a hollow cylinder is I = mC R 2 , so the equation for a y simplifies to ay =
−mB g −(3.0 kg)(9.8 m/s 2 ) = = −5.88 m/s 2 mB + mC 3.0 kg + 2.0 kg
The speed of the block just before it hits the ground can now be found using kinematics: v12 = v02 + 2a y ( y1 − y0 ) = 0 m 2 /s 2 + 2(−5.88 m/s 2 )(0 m − 1.0 m) ⇒ v1 = 3.4 m/s
(b) The conservation of energy equation K1 + U g1 = K 0 + U g0 for the system (block + cylinder + earth) is
1 1 1 1 mBv12 + Iω12 + mB gy1 = mBv02 + Iω 02 + mB gy0 2 2 2 2 2 v m ⎞ 1 1 ⎛m mBv12 + ( mC R 2 ) 12 + 0 J = 0 J + 0 J + mB gy0 ⇒ v12 ⎜ B + C ⎟ = mB gy0 R 2 2 2 ⎠ ⎝ 2 ⇒ v12 =
2mB gy0 2(3.0 kg)(9.8 m/s 2 )(1.0 m) = = 11.76 m 2 /s 2 ⇒ v1 = 3.4 m/s mB + mC (3.0 kg) + (2.0 kg)
Assess: Newton’s second law and the conservation of energy method give the same result for the block’s final velocity.
12.68.
Model: The bar is a solid body rotating through its center.
Visualize:
Solve:
(a) The two forces form a couple. The net torque on the bar about its center is
Iα L where F is the force produced by one of the air jets. We can find I and α as follows:
τ net = LF = Iα ⇒ F =
1 1 ML2 = (0.50 kg)(0.60 m) 2 = 0.015 kg m 2 12 12 ω1 = ω 0 + α (t1 − t0 ) ⇒ 150 rpm = 5.0π rad/s = 0 rad + α (10 s − 0 s) ⇒ α = 0.50π rad/s 2 I=
⇒F=
(0.015 kg m 2 )(0.5π rad/s 2 ) = 0.0393 N (0.60 m)
The force F = 39 mN. (b) The torque of a couple is the same about any point. It is still τ net = LF . However, the moment of inertia has changed.
LF 1 1 where I = ML2 = (0.500 kg)(0.6 m) 2 = 0.060 kg m 2 I 3 3 (0.0393 N) × (0.60 m) ⇒α = = 0.393 rad/s 2 0.060 kg m 2
τ net = LF = Iα ⇒ α =
Finally,
ω1 = ω 0 + α (t1 − t0 ) = 0 rad/s + (0.393 rad/s 2 )(10 s − 0 s) = 3.93 rad/s =
(3.93)(60) rpm = 37.5 rpm 2π
The angular speed is 38 rpm. Assess: Note that ω ∝ α and α ∝ 1/ I . Thus, ω ∝ 1/ I . I about the center of the rod is 4 times smaller than I about one end of the rod. Consequently, ω is 4 times larger.
12.69.
Model: The flywheel is a rigid body rotating about its central axis.
Visualize:
Solve: (a) The radius of the flywheel is R = 0.75 m and its mass is M = 250 kg. The moment of inertia about the axis of rotation is that of a disk: 1 1 I = MR 2 = (250 kg)(0.75 m) 2 = 70.31 kg m 2 2 2 The angular acceleration is calculated as follows:
τ net = Iα ⇒ α = τ net / I = (50 N m)/(70.31 kg m 2 ) = 0.711 rad/s 2 Using the kinematic equation for angular velocity gives
ω1 = ω 0 + α (t1 − t0 ) = 1200 rpm = 40 π rad/s = 0 rad/s + 0.711 rad/s 2 (t1 − 0 s) ⇒ t1 = 177 s (b) The energy stored in the flywheel is rotational kinetic energy: 1 1 K rot = Iω12 = (70.31 kg m 2 )(40π rad/s) 2 = 5.55 × 105 J 2 2 5 The energy stored is 5.6 × 10 J. energy delivered (5.55 × 105 J)/2 = = 1.39 × 105 W = 139 kW (c) Average power delivered = time interval 2.0 s (d) Because τ = Iα , ⇒ τ = I
− ω half energy ⎞ ⎛ω Δω = I ⎜ full energy ⎟ . ω full energy =ω1 (from part (a)) = 40π rad/s. ω half energy Δt Δt ⎝ ⎠
can be obtained as: 1 2 1 Iω half energy = K rot ⇒ ω half energy = 2 2
5.55 × 105 J K rot = = 88.85 rad/s 70.31 kg m 2 I
Thus ⎛ 40 π rad/s − 88.85 rad/s ⎞ ⎟ = 1.30 kN m 2.0 s ⎝ ⎠
τ = (70.31 kg m 2 ) ⎜
12.70. Model: The pulley is a rigid rotating body. We also assume that the pulley has the mass distribution of a disk and that the string does not slip. Visualize:
Because the pulley is not massless and frictionless, tension in the rope on both sides of the pulley is not the same. Solve: Applying Newton’s second law to m1 , m2 , and the pulley yields the three equations:
T1 − ( FG )1 = m1a1
− ( FG )2 + T2 = m2 a2
T2 R − T1R − 0.50 N m = Iα
Noting that − a2 = a1 = a, I = 12 mp R 2 , and α = a/R, the above equations simplify to T1 − m1 g = m1a
m2 g − T2 = m2 a
0.50 N m ⎛1 ⎞⎛ a ⎞ 1 0.50 N m 1 T2 − T1 = ⎜ mp R 2 ⎟⎜ ⎟ + = mp a + R 2 0.060 m ⎝2 ⎠⎝ R ⎠ R
Adding these three equations, 1 ⎞ ⎛ (m2 − m1 ) g = a ⎜ m1 + m2 + mp ⎟ + 8.333 N 2 ⎠ ⎝ (m − m1 ) g − 8.333 N (4.0 kg − 2.0 kg)(9.8 m/s 2 ) − 8.333 N ⇒a= 2 = = 1.610 m/s 2 2.0 kg + 4.0 kg + (2.0 kg/2) m1 + m2 + 12 mp We can now use kinematics to find the time taken by the 4.0 kg block to reach the floor: 1 1 y1 = y0 + v0 (t1 − t0 ) + a2 (t1 − t0 ) 2 ⇒ 0 = 1.0 m + 0 + (−1.610 m/s 2 )(t1 − 0 s) 2 2 2 2(1.0 m) ⇒ t1 = = 1.11 s (1.610 m/s 2 )
12.71.
Model: Assume the string does not slip on the pulley.
Visualize:
The free-body diagrams for the two blocks and the pulley are shown. The tension in the string exerts an upward force on the block m2 , but a downward force on the outer edge of the pulley. Similarly the string exerts a force on block m1 to the right, but a leftward force on the outer edge of the pulley. Solve: (a) Newton’s second law for m1 and m2 is T = m1a1 and T − m2 g = m2 a2 . Using the constraint − a2 = + a1 = a, we have T = m1a and −T + m2 g = m2 a. Adding these equations, we get m2 g = (m1 + m2 )a, or
a=
m2 g mm g ⇒ T = m1a = 1 2 m1 + m2 m1 + m2
(b) When the pulley has mass m, the tensions (T1 and T2 ) in the upper and lower portions of the string are different. Newton’s second law for m1 and the pulley are:
T1 = m1a
and
T1R − T2 R = − Iα
We are using the minus sign with α because the pulley accelerates clockwise. Also, a = Rα . Thus, T1 = m1a and T2 − T1 =
I a aI = R R R2
Adding these two equations gives I ⎞ ⎛ T2 = a ⎜ m1 + 2 ⎟ R ⎝ ⎠
Newton’s second law for m2 is T2 − m2 g = m2 a2 = − m2 a. Using the above expression for T2 , I ⎞ m2 g ⎛ a ⎜ m1 + 2 ⎟ + m2 a = m2 g ⇒ a = R ⎠ m1 + m2 + I / R 2 ⎝ Since I = 12 mp R 2 for a disk about its center, a=
m2 g m1 + m2 + 12 mp
With this value for a we can now find T1 and T2 :
T1 = m1a = Assess:
m1m2 g m1 + m2 + 12 mp
T2 = a (m1 + I / R 2 ) =
1 m2 g 1 ⎞ m2 ( m1 + 2 mp ) g ⎛ + = m m 1 p ⎜ ⎟ ( m1 + m2 + 12 mp ) ⎝ 2 ⎠ m1 + m2 + 12 mp
For m = 0 kg, the equations for a, T1 , and T2 of part (b) simplify to a=
m2 g m1 + m2
These agree with the results of part (a).
and T1 =
m1m2 g m1 + m2
and
T2 =
m1m2 g m1 + m2
12.72. Model: The disk is a rigid spinning body. Visualize: Please refer to Figure P12.72. The initial angular velocity is 300 rpm or (300)(2π )/60 = 10π rad/s. After 3.0 s the disk stops. Solve: Using the kinematic equation for angular velocity, ω1 = ω 0 + α (t1 − t0 ) ⇒ α =
ω1 − ω 0 t1 − t0
=
(0 rad/s − 10π rad/s) −10π = rad/s 2 (3.0 s − 0 s) 3
Thus, the torque due to the force of friction that brings the disk to rest is
τ = Iα = − fR ⇒ f = −
( 12 mR 2 )α = − 1 (mR)α = − 1 (2.0 kg)(0.15 m) ⎛ −10 π rad/s2 ⎞ = 1.57 N Iα =− ⎜ ⎟ R R 2 2 3 ⎝ ⎠
The minus sign with τ = − fR indicates that the torque due to friction acts clockwise.
12.73.
Model: The entire structure is a rigid rotating body. The two thrust forces are a couple that exerts a torque on the structure about its center of mass. We will assume the thrust forces are perpendicular to the connecting tunnel. Visualize: Please refer to Figure 12.30. We chose a coordinate system in which m1 and m2 are on the x-axis
and m1 = 100,000 kg is at the origin. m3 is the mass of the tunnel, whose center is at x3 = 45 m. Solve: (a) Assuming the center of mass of the tunnel is at the center of the tunnel, we get
xcm =
m1 x1 + m2 x2 + m3 x3 m1 + m2 + m3
xcm =
(1.0 × 105 kg)(0 m) + (2.0 × 105 kg)(90 m) + (5.0 × 104 kg)(45 m) = 57.9 m 1.0 × 105 kg + 2.0 × 105 kg + 5.0 × 104 kg
The center of mass of the entire structure is 58 m from the 100,000 kg rocket. (b) Initially, the angular velocity is zero. The structure’s angular velocity after 30 s is
ω1 = ω 0 + α (t1 − t0 ) = 0 rad/s + α (30 s − 0 s) = α (30 s) The angular acceleration α can be found from τ = Iα , where τ is the net torque on the structure and I is its moment of inertia. The two thrusts form a couple with torque
τ = lF = (90 m)(50,000 N) = 4.50 × 106 N m 2 2⎞ ⎛1 I = I m1 + I m2 + I tunnel = m1 x12 + m2 x22 + ⎜ m3 ( 90 m ) + m3 ( 58 m − 45 m ) ⎟ 12 ⎝ ⎠ 1 2 5 2 5 2 4 = (1.0 × 10 kg)(57.9 m) + (2.0 × 10 kg)(32.1 m) + (5 × 10 kg)(90 m) 2 + ( 5 × 104 kg ) (13 m ) 12 = 5.83 × 108 kg m 2
⇒α =
τ I
=
4.5 × 106 N m = 7.71 × 10−3 rad/s 2 5.83 × 108 kg m 2
⇒ ω1 = (30 s)(7.71 × 10−3 rad/s 2 ) = 0.23 rad/s
Assess: Note that the parallel axis theorem was used in finding the moment of inertia of the tunnel.
12.74. Model: Assume that the hollow sphere is a rigid rolling body and that the sphere rolls up the incline without slipping. We also assume that the coefficient of rolling friction is zero. Visualize:
The initial kinetic energy, which is a combination of rotational and translational energy, is transformed in gravitational potential energy. We chose the bottom of the incline as the zero of the gravitational potential energy. Solve: The conservation of energy equation K f + U gf = K i + U gi is
1 1 1 1 2 2 + I cm (ω1 ) 2 + Mgy1 = M (v0 )cm + I cm (ω 0 ) 2 + Mgy0 M (v1 )cm 2 2 2 2 1 1 2 1 1 (v ) 2 ⎛ ⎞ 2 2 2 0 J + 0 J + Mgy1 = M (v0 )cm + ⎜ MR 2 ⎟ (ω 0 )cm + 0 J ⇒ Mgy1 = M (v0 )cm + MR 2 0 2cm R 2 2⎝ 3 2 3 ⎠ 5 (v ) 2 5 5 (5.0 m/s) 2 2 ⇒ gy1 = (v0 )cm ⇒ y1 = 6 0 cm = = 2.126 m g 6 6 9.8 m/s 2
The distance traveled along the incline is 2.126 m y1 = = 4.3 m sin 30° 0.5 Assess: This is a reasonable stopping distance for an object rolling up an incline when its speed at the bottom of the incline is approximately 10 mph. s=
12.75.
Model: Visualize:
Solve:
The masses are particles.
(a) The moment of inertia of the barbell is 2 I = I M + I m = Mx 2 + m ( L − x )
The rotational kinetic energy is therefore K rot =
(
)
1 2 1 2 Iω = Mx 2 + m ( L − x ) ω 2 2 2
To find x such that K rot is a minimum, set
1 dK rot = 0 = ( 2 Mx − 2m ( L − x ) )ω 2 dx 2 (b) This is the center of mass location measured from the mass M.
⇒x=
m L m+M
12.76. Model: The disk is a rigid body rotating on an axle passing through one edge. The gravitational potential energy is transformed into rotational kinetic energy as the disk is released. Visualize:
We placed the origin of the coordinate system at a distance R just below the axle. In the initial position, the center of mass of the disk is at the same level as the axle. The center of mass of the disk in the final position is coincident with the origin of the coordinate system. Solve: (a) The torque is due to the gravitational force on the disk acting at the center of mass. Thus
τ = (mg ) R = (5.0 kg)(9.8 m/s 2 )(0.30 m) = 14.7 N m The moment of inertia about the disk’s edge is obtained using the parallel-axis theorem: 1 3 ⎛ 3⎞ I = I cm + mR 2 = mR 2 + mR 2 = mR 2 = ⎜ ⎟ (5.0 kg)(0.30 m)2 = 0.675 kg m 2 2 2 ⎝ 2⎠ τ 14.7 N m ⇒α = = = 22 rad/s 2 I 0.675 kg m 2 (b) The energy conservation equation K f + U gf = K i = U gi is
1 2 1 1 Iω1 + mgy1 = Iω 02 + mgy0 ⇒ Iω12 + 0 J = 0 J + mgR 2 2 2
ω1 =
2mgR 2(5.0 kg)(9.8 m/s 2 )(0.30 m) = = 6.6 rad/s I 0.675 kg m 2
Assess: An angular velocity of 6.6 rad/s (or 1.05 revolutions/s) as the center of mass of the disk reaches below the axle is reasonable.
12.77. Model: The hoop is a rigid body rotating about an axle at the edge of the hoop. The gravitational torque on the hoop causes it to rotate, transforming the gravitational potential energy of the hoop’s center of mass into rotational kinetic energy. Visualize:
We placed the origin of the coordinate system at the hoop’s edge on the axle. In the initial position, the center of mass is a distance R above the origin, but it is a distance R below the origin in the final position. Solve: (a) Applying the parallel-axis theorem, I edge = I cm + mR 2 = mR 2 + mR 2 = 2mR 2 . Using this expression in
the energy conservation equation K f + U gf = K i + U gi yields:
1 1 I edgeω12 + mgy1 = I edgeω 02 + mgy0 2 2
1 2g (2mR 2 )ω12 − mgR = 0 J + mgR ⇒ ω1 = 2 R
(b) The speed of the lowest point on the hoop is
2g (2 R ) = 8 gR R Note that the speed of the lowest point on the loop involves a distance of 2R instead of R. v = (ω1 )(2 R) =
Assess:
12.78. Model: The long, thin rod is a rigid body rotating about a frictionless pivot on the end of the rod. The gravitational torque on the rod causes it to rotate, transforming the gravitational potential energy of the rod’s center of mass into rotational kinetic energy. Visualize:
We placed the origin of the coordinate system at the pivot point. In the initial position, the center of mass is a distance 1 L above the origin. In the final position, the center of mass is at y = 0 m and thus has zero gravitational potential 2 energy. Solve: (a) The energy conservation equation for the rod K f + U gf = K i + U gi is
1 2 1 Iω1 + mgy1 = Iω 02 + mgy0 2 2
1⎛1 2⎞ 2 ⎜ mL ⎟ω1 + 0 J = 0 J + mg ( L /2) ⇒ ω1 = 3 g / L 2⎝3 ⎠
(b) The speed at the tip of the rod is vtip = (ω1 ) L = 3 gL .
12.79. Model: The sphere attached to a thin rod is a rigid body rotating about the rod. Assume the rod is vertical and the sphere solid. Visualize: Please refer to Figure P12.79. The sphere rotates because the string wrapped around the rod exerts a torque τ. Solve: The torque exerted by the string on the rod is τ = Tr. From the parallel-axis theorem, the moment of inertia of the sphere about the rod’s axis is 2
MR 2 13 ⎛R⎞ 2 I off center = I cm + M ⎜ ⎟ = MR 2 + = MR 2 4 20 ⎝2⎠ 5 From Newton’s second law,
α=
τ I
=
Tr 20Tr = (13MR 2 /20) 13MR 2
12.80.
Model: The pulley is a rigid rotating body.
Visualize:
We placed the origin of the coordinate system on the floor. The pulley rotates about its center. Solve: Using kinematics for the physics book (mass = m1 ), 1 1 y1 = y0 + v0 (t1 − t0 ) + a (t1 − t0 ) 2 0 m = 1.0 m + 0 m + a(0.71 s − 0 s) 2 ⇒ a = −3.967 m/s 2 2 2 Since the torque acting on the pulley is τ = −TR = Iα , we have
I =−
TR
α
=−
TR TR 2 =− a /R a
We can compare the measured value of I for the pulley with the theoretical value. We first must find the tension T. From the free-body diagram, Newton’s second law of motion is T − m1 g = m1a. This means T = m1 ( g + a ) = (1.0 kg)(9.80 m/s 2 − 3.967 m/s 2 ) = 5.833 N With these values of T and a, we can now find I as: I =−
TR 2 (5.833 N)(0.06 m) 2 =− = 5.3 × 10−3 kg m 2 a (−3.967 m/s 2 )
Let us now calculate the theoretical values of I:
Since I disk
hoop about center: I = MR 2 = (2.0 kg)(0.06 m) 2 = 7.2 × 10 −3 kg m 2 1 1 disk about center: I disk = MR 2 = (2.0 kg)(0.06 m) 2 = 3.6 × 10 −3 kg m 2 2 2 < I < I hoop , the mass of the disk is not uniformly distributed. The mass is concentrated near the rim.
12.81.
Model: The angular momentum of the satellite in the elliptical orbit is a constant.
Visualize:
Solve: (a) Because the gravitational force is always along the same direction as the direction of the moment G G G arm vector, the torque τ = r × Fg is zero at all points on the orbit. (b) The angular momentum of the satellite at any point on the elliptical trajectory is conserved. The velocity is G perpendicular to r at points a and b, so β = 90° and L = mvr. Thus
⎛r ⎞ Lb = La ⇒ mvb rb = mva ra ⇒ vb = ⎜ a ⎟ va ⎝ rb ⎠ 30,000 km 30,000 km − 9000 km = 6000 km and rb = + 9000 km = 24,000 km ra = 2 2 ⎛ 6000 km ⎞ ⇒ vb = ⎜ ⎟ (8000 m/s) = 2000 m/s ⎝ 24,000 km ⎠ (c) Using the conservation of angular momentum Lc = La , we get
⎛r ⎞ mvc rc sin β c = mva ra ⇒ vc = ⎜ a ⎟ va /sin β c ⎝ rc ⎠
rc = (9000 km)2 + (12,000 km) 2 = 1.5 × 107 m
From the figure, we see that sin β c = 12,000 15,000 = 0.80. Thus ⎛ 6000 km ⎞ (8000 m/s) vc = ⎜ = 4000 m/s ⎟ 0.80 ⎝ 15,000 km ⎠
12.82.
Model: The clay balls are particles and undergo a totally inelastic collision. Linear momentum is conserved during the collision. Visualize:
Solve:
(a)
The
angle
is
measured counterclockwise ⎛1⎞ β1 = 180° + tan −1 (1) = 225°, and β 2 = tan −1 ⎜ ⎟ = 26.6°. So ⎝ 2⎠ L = L1 + L2 + m1r1v1 sin β1 + m2 r2v2 sin β 2
= ( 0.015 kg )
(
)
2 m ( 2 m/s ) sin 225° + ( 0.025 kg )
G r
from
(
to
G v.
)
5 m ( 2.0 m/s ) sin 26.6°
= 0.020 kg m /s. 2
Note that the signs of L1 and L2 agree with those determined by the right-hand rule. (b) At the instant before the clay balls collide they are located at (1.5 m, 1.0 m). Here, ⎛ 2⎞ β 2 = tan −1 ⎜ ⎟ = 33.7° β1 = 180° + β 2 = 146.3° ⎝ 3⎠ So
( (1.0 m) + (1.5 m) )( 2.0 m/s) sin 213.7 + ( 0.025 kg ) ( (1.0 m ) + (1.5 m ) ) ( 2.0 m/s ) sin 33.7°
L = ( 0.015 kg )
2
2
2
2
= 0.020 kg m 2 /s (c) The clay balls have a final speed v after the collision. Linear momentum is conserved. p1i + p2i = p(1+ 2) f
( 0.015 kg )( 2.0 m/s ) + ( 0.025 kg )( −2.0 m/s ) = ( 0.015 kg + 0.025 kg ) v ⇒ v = −0.50 m/s. The balls are moving to the left. The angle β = β 2 from part (b). The angular momentum after the collision is L = ( 0.040 kg ) Assess:
(
(1.0 m )
2
+ (1.5 m )
2
)(0.50 m/s )sin 33.7° = 0.020 kg m /s
G Angular momentum is also conserved since τ net = 0.
From
2
geometry,
Model: For the (bullet + block + rod) system, angular momentum is conserved. After the bullet is stuck in the block, the mechanical energy of the system is conserved. Assume the block is small. Visualize:
12.83.
The origin of the coordinate system was placed at the center-of-mass of the block as it freely hangs from the bottom of the rod. Solve: The initial angular momentum of the system about the pivot is due only to the bullet:
Li = mb vb L = (0.010 kg)vb (1.0 m) = (0.010 kg m)vb The angular momentum immediately after the bullet hits and sticks in the block is equal to Iω . The moment of inertia is 1 ⎛1 ⎞ I = I rod + I block + I bullet = mR L2 + mB L2 + mb L2 = ⎜ mR + mB + mb ⎟ L2 3 ⎝3 ⎠ ⎡1 ⎤ = ⎢ (1.0 kg) + 2.0 kg + 0.010 kg ⎥ (1.0 m) 2 = 2.343 kg m 2 ⎣3 ⎦ Angular momentum is conserved in the collision:
Lf = (2.343 kg m 2 )ω = Li = (0.010 kg m)vb We need to determine ω before we can find vb . To find ω we use the conservation of mechanical energy equation K f + U gf = K i + U gi as the pendulum swings out, which is 1 2 Iω + 0 J 2 1 L (0.010 kg + 2.0 kg)(9.8 m/s 2 ) L(1 − cos30°) + (1.0 kg)(9.8 m/s 2 ) (1 − cos30°) = (2.343 kg m 2 )ω 2 2 2 0 J + (mb + mB ) gyB + mR gyR =
The energy equation can be further simplified to
2.6390 kg m 2 /s 2 + 0.6565 kg m 2 /s 2 = (1.1715 kg m 2 )ω 2 ⇒ ω = 1.677 rad/s Finally, we can use the conservation of angular momentum equation to obtain the speed of the bullet: vb =
(2.343 kg m 2 )(1.677 rad/s) = 3.9 × 102 m/s 0.010 kg m
Assess: A speed of 390 m/s for a bullet is reasonable.
12.84.
Model: For the (bullet + door) system, the angular momentum is conserved in the collision.
Visualize:
G G Solve: As the bullet hits the door, its velocity v is perpendicular to r . Thus the initial angular momentum about the rotation axis, with r = L, is
Li = mBvB L = (0.010 kg)(400 m/s)(1.0 m) = 4.0 kg m 2 /s After the collision, with the bullet in the door, the moment of inertia about the hinges is 1 1 I = I door + I bullet = mD L2 + mB L2 = (10.0 kg)(1.0 m) 2 + (0.010 kg)(1.0 m) 2 = 3.343 kg m 2 3 3
Therefore,
Lf = Iω = (3.343 kg m 2 )ω .
Using
the
angular
Lf = Li (3.343 kg m )ω = 4.0 kg m /s and thus ω = 1.20 rad/s. 2
2
momentum
conservation
equation
12.85.
Model: The mechanical energy of both the hoop (h) and the sphere (s) is conserved. The initial gravitational potential energy is transformed into kinetic energy as the objects roll down the slope. The kinetic energy is a combination of translational and rotational kinetic energy. We also assume no slipping of the hoop or of the sphere. Visualize:
The zero of gravitational potential energy is chosen at the bottom of the slope. Solve: (a) The energy conservation equation for the sphere or hoop K f + U gf = K i + U gi is 1 1 1 1 I (ω1 ) 2 + m(v1 ) 2 + mgy1 = I (ω 0 ) 2 + m(v0 ) 2 + mgy0 2 2 2 2
For the sphere, this becomes 2 1⎛ 2 1 2 ⎞ (v1 )s 2 ⎜ mR ⎟ 2 + m(v1 )s + 0 J = 0 J + 0 J + mghs 2⎝ 5 2 ⎠ R
⇒
7 (v1 )s2 = gh ⇒ (v1 )s = 10 gh /7 = 10(9.8 m/s 2 )(0.30 m)/7 = 2.05 m/s 10
For the hoop, this becomes 1 (v ) 2 1 (mR 2 ) 1 2 h + m(v1 ) 2h + 0 J = 0 J + 0 J + mghhoop 2 R 2 (v ) 2 ⇒ hhoop = 1 h g For the hoop to have the same velocity as that of the sphere, hhoop =
(v1 )s2 (2.05 m/s) 2 = = 42.9 cm g 9.8 m/s 2
The hoop should be released from a height of 43 cm. (b) As we see in part (a), the speed of a hoop at the bottom depends only on the starting height and not on the mass or radius. So the answer is No.
12.86.
Model: Model the turntable as a rigid disk rotating on frictionless bearings. As the blocks fall from above and stick on the turntable, the turntable slows down due to increased rotational inertia of the (turntable + blocks) system. Any torques between the turntable and the blocks are internal to the system, so angular momentum of the system is conserved. Visualize: The initial moment of inertia is I1 and the final moment of inertia is I 2 .
Solve: The initial moment of inertia is I1 = I disk = 12 mR 2 = 12 (2.0 kg)(0.10 m) 2 = 0.010 kg m 2 and the final moment of inertia is
I 2 = I1 + 2mR 2 = 0.010 kg m 2 + 2(0.500 kg) × (0.10 m) 2 = 0.010 kg m2 + 0.010 kg m 2 = 0.020 kg m2 Let ω1 and ω 2 be the initial and final angular velocities. Then Lf = Li ⇒ ω 2 I 2 = ω1I1 ⇒ ω 2 =
I1ω1 (0.010 kg m 2 )(100 rpm) = = 50 rpm I2 0.020 kg m 2
12.87.
Model: Model the turntable as a rigid disk rotating on frictionless bearings. For the (turntable + block) system, no external torques act as the block moves outward towards the outer edge. Angular momentum is thus conserved. Visualize: The initial moment of inertia of the turntable is I1 and the final moment of inertia is I 2 . Solve: The initial moment of inertia is I1 = I disk = 12 mR 2 = 12 (0.2 kg)(0.2 m) 2 = 0.0040 kg m 2 . As the block reaches the outer edge, the final moment of inertia is
I 2 = I1 + mB R 2 = 0.0040 kg m 2 + (0.020 kg)(0.20 m) 2 = 0.0040 kg m 2 + 0.0008 kg m 2 = 0.0048 kg m 2 Let ω1 and ω 2 be the initial and final angular velocities, then the conservation of angular momentum equation is Lf = Li ⇒ ω 2 I 2 = ω1I1 ⇒ ω 2 =
I1ω1 (0.0040 kg m 2 )(60 rpm) = = 50 rpm I2 (0.0048 kg m 2 )
Assess: A change of angular velocity from 60 rpm to 50 rpm with an increase in the value of the moment of inertia is reasonable.
12.88. Model: Model the merry-go-round as a rigid disk rotating on frictionless bearings about an axle in the center and John as a particle. For the (merry-go-round + John) system, no external torques act as John jumps on the merry-go-round. Angular momentum is thus conserved. Visualize: The initial angular momentum is the sum of the angular momentum of the merry-go-round and the angular momentum of John. The final angular momentum as John jumps on the merry-go-round is equal to I finalω final . Solve: John’s initial angular momentum is that of a particle: LJ = mJ vJ R sin β = mJ vJ R. The angle β = 90° since John runs tangent to the disk. The conservation of angular momentum equation Lf = Li is ⎛1 ⎞ I finalω final = Ldisk + LJ = ⎜ MR 2 ⎟ωi + mJ vJ R ⎝2 ⎠ 2π ⎛ rad ⎞ ⎛1⎞ 2 = ⎜ ⎟ (250 kg)(1.5 m) 2 (20 rpm) ⎜ ⎟ + (30 kg)(5.0 m/s)(1.5 m) = 814 kg m /s 60 ⎝ rpm ⎠ ⎝ 2⎠ ⇒ ω final =
814 kg m 2 /s I final
1 1 I final = I disk + I J = MR 2 + mJ R 2 = (250 kg)(1.5 m) 2 + (30 kg)(1.5 m)2 = 349 kg m 2 2 2 814 kg m 2 /s ω final = = 2.33 rad/s = 22 rpm 349 kg m 2
12.89.
Model: The toy car is a particle located at the rim of the track. The track is a cylindrical hoop rotating about its center, which is an axis of symmetry. No net torques are present on the track, so the angular momentum of the car and track is conserved. Visualize:
Solve:
The toy car’s steady speed of 0.75 m/s relative to the track means that vc − vt = 0.75 m/s ⇒ v c = vt + 0.75 m/s,
where vt is the velocity of a point on the track at the same radius as the car. Conservation of angular momentum implies that Li = Lf 0 = I cω c + I tω t = ( mr 2 )ω c + ( Mr 2 )ω t = mω c + M ω t
The initial and final states refer to before and after the toy car was turned on. Table 12.2 was used for the track. v v Since ω c = c , ω t = t , we have r r 0 = mvc + Mvt ⇒ m ( vt + 0.75 m/s ) + Mvt = 0 ⇒ vt = −
M ( 0.200 kg ) ( 0.75 m/s ) = − ( 0.75 m/s ) = −0.125 m/s m+M ( 0.200 kg + 1.0 kg )
The minus sign indicates that the track is moving in the opposite direction of the car. The angular velocity of the track is v ( 0.125 m/s ) ωt = t = = 0.417 rad/s clockwise. r 0.30 m In rpm, ⎛ rev ⎞⎛ 60 s ⎞ ω t = ( 0.417 rad/s ) ⎜ ⎟⎜ ⎟ ⎝ 2π rad ⎠⎝ min ⎠ = 4.0 rpm Assess: The speed of the track is less than that of the car because it is more massive.
12.90.
Model: Model the skater as a cylindrical torso with two rod-like arms that are perpendicular to the axis of the torso in the initial position and collapse into the torso in the final position. Visualize:
Solve: For the initial position, the moment of inertia is I1 = I Torso + 2 I Arm . The moment of inertia of each arm is that of a 66 cm long rod rotating about a point 10 cm from its end, and can be found using the parallel-axis theorem. In the final position, the moment of inertia is I 2 = 12 MR 2 . The equation for the conservation of angular
momentum Lf = Li can be written I 2ω 2 = I1ω1 ⇒ ω 2 = ( I1 / I 2 )ω1. Calculating I1 and I 2 , 1 ⎡1 ⎤ I1 = M T R 2 + 2 ⎢ M A L2A + M A d 2 ⎥ 2 12 ⎣ ⎦ 1 2⎤ ⎡1 2 = (40 kg)(0.10 m) + 2 ⎢ (2.5 kg)(0.66 m) 2 + (2.5 kg) ( 0.33 m + 0.10 m ) ⎥ = 1.306 kg m 2 2 ⎣12 ⎦ 1 1 (1.306 kg m 2 ) I 2 = MR 2 = (45 kg)(0.10 m) 2 = 0.225 kg m 2 ⇒ ω 2 = (1.0 rev/s) = 5.8 rev /s 2 2 (0.225 kg m 2 )
12.91. Model: Assume that the marble does not slip as it rolls down the track and around a loop-the-loop. The mechanical energy of the marble is conserved. Visualize:
Solve: The ball’s center of mass moves in a circle of radius R − r. The free-body diagram on the marble at its highest position shows that Newton’s second law for the marble is
mv12 R−r The minimum height (h) that the track must have for the marble to make it around the loop-the-loop occurs when the normal force of the track on the marble tends to zero. Then the weight will provide the centripetal acceleration needed for the circular motion. For n → 0 N, mg + n =
mg =
mv 2 ⇒ v12 = g ( R − r ) (R − r)
Since rolling motion requires v12 = r 2ω12 , we have
ω12 r 2 = g ( R − r ) ⇒ ω12 =
g (R − r) r2
The conservation of energy equation is 1 1 ( K f + U gf ) top of loop = ( K i + U gi )initial ⇒ mv12 + Iω12 + mgy1 = mgy0 = mgh 2 2 Using the above expressions and I = 52 mr 2 the energy equation simplifies to 1 1⎛ 2⎞ ⎛ g(R − r) ⎞ mg ( R − r ) + ⎜ ⎟ mr 2 ⎜ ⎟ + mg 2( R − r ) = mgh ⇒ h = 2.7 ( R − r ) 2 2 2⎝ 5⎠ ⎝ r ⎠
12.92.
Model: The Swiss cheese wedge is of uniform density—or at least uniform enough that its center of mass is at the same location as that of a solid piece. To find the angle at which the cheese starts sliding, the cheese will be treated as a particle, and the model of static friction will be used. Visualize:
Solve: The angle at which the cheese starts sliding, θS , will be compared to the critical angle θ c for stability. Use Newton’s second law with the free body diagram. ( Fnet ) x = 0 = fs − FG sinθ s
( Fnet ) y = 0 = n − FG cosθ s
With FG = mg , the y-direction equation gives n = mg cosθ . The cheese starts sliding when μ s is at its maximum value. Combining that with the x-direction equation and fs = μ s n, 0 = μs ( mg cosθ s ) − mg sin θ s ⇒ θ s = tan −1 ( μ s ) = tan −1 ( 0.90 ) = 42° The cheese will start sliding at an angle of 42°. The center of mass of the cheese wedge can be found using the result of problem 12.53. There, the center of mass of a triangle with the same proportions as the cheese wedge was found. So xcm is at the center of the cheese wedge (by symmetry). The ycm can be found by proportional reasoning.
ycm ( 30 cm − 20 cm ) ⇒ y = 4.0 cm = cm 12 cm 30 cm Note that here we have measured ycm from the base of the wedge. Stability considerations require that the center of mass be no further than the left corner of the wedge. At the critical angle geometry shown in the figure above, the right triangle formed by the wedge’s center of mass, lower left corner, and center point of the base is a 45°-45°-90° triangle. So θ c = 45°. Assess: The cheese will slide first as the incline reaches 42°. It would not topple until the angle reaches 45°. So Emily is correct.
12.93.
Model: Define the system as the rod and cube. Energy and angular momentum are conserved in a perfectly elastic collision in the absence of a net external torque. The rod is uniform. Visualize: Please refer to Figure CP12.93. Solve: Let the final speed of the cube be vf , and the final angular velocity of the rod be ω . Energy is conserved, and angular momentum around the rod’s pivot point is conserved. 1 1 1 Ei = Ef ⇒ mv02 = mvf2 + I rodω 2 2 2 2 ⎛d⎞ ⎛d⎞ Li = Lf ⇒ mv0 ⎜ ⎟ = mvf ⎜ ⎟ + I rodω ⎝ 2⎠ ⎝ 2⎠ This is two equations in the two unknowns vf and ω . From Table 12.2,
I rod =
1 1 1 Md 2 = ( 2m ) d 2 = md 2 12 12 6
From the angular momentum equation, d v0 = vf + ω 3
⇒ω =
3 ( v0 − vf ) d
Substituting into the energy equation, 1 1 1⎛1 2 ⎞⎛ 9 ⎞ mv0 2 = mvf 2 + ⎜ md 2 ⎟⎜ 2 ⎟ ( v0 − vf ) 2 2 2⎝ 6 ⎠⎝ d ⎠ 3 2 v0 2 = vf 2 + ( v0 − vf ) 2 6 1 0 = vf 2 − v0vf + v0 2 5 5 This is a quadratic equation in vf . The roots are 2
⎛v 2⎞ 6 ⎛6 ⎞ v0 ± ⎜ v0 ⎟ − 4 ⎜ 0 ⎟ 5 ⎝5 ⎠ 2 ⎝ 5 ⎠ 3 vf = = v0 ± v0 2 5 5 ⎧1 ⎪ v = ⎨5 0 ⎪⎩ v0 1 The answer vf = v0 means the ice cube missed the rod. So vf = v0 to the right. 5
12.94.
Model: The clay ball is a particle. The rod is a uniform thin rod rotating about its center. Angular momentum is conserved in the collision. Visualize:
Solve: This is a two-part problem. Angular momentum is conserved in the collision, and energy is conserved as the ball rises like a pendulum. The angular momentum conservation equation about the rod’s pivot point is Li = Lf ⇒ mv0 r = ( I ball+rod )ω
L 1 = 15 cm. The rod and ball are a composite object. From Table 12.2, I rod = ML2 , so 2 12 1 L2 1 L2 ⎛ M⎞ I ball+rod = I ball + I rod = mr 2 + ML2 = m + ML2 = ⎜ m + ⎟ 12 4 12 4⎝ 3 ⎠ v 2v If vf is the final velocity of the clay ball, ω = f = f since the ball sticks to the rod. Thus r L 2 mv0 L L ⎛ M ⎞⎛ 2v ⎞ = ⎜ m + ⎟⎜ f ⎟ 2 4⎝ 3 ⎠⎝ L ⎠ Note r =
mv0 ( 0.010 kg )( 2.5 m/s ) = 0.714 m/s = M ( 0.075 kg ) m+ ( 0.010 kg ) + 3 3 Energy is conserved as the clay ball rises. Compare the energy of the ball-rod system just after the collision to when the ball reaches the maximum height. Note that the center of mass of the rod does not change position. 1 Ei = Ef ⇒ ( I rod + ball )ω 2 = mgh 2 Thus ⇒ vf =
2
1 L2 ⎛ M ⎞⎛ 2vf ⎞ M⎞ 2⎛ ⎜ m + ⎟⎜ ⎟ = mgh ⇒ vf ⎜ m + ⎟ = mgL (1 − cosθ ) 2 4⎝ 3 ⎠⎝ L ⎠ 3 ⎠ ⎝ 2 v ⎛ M⎞ ⇒ 1 − f ⎜ m + ⎟ = cosθ mgL ⎝ 3 ⎠ Using the various values, cosθ = 0.393 ⇒ θ = 67°. L Assess: The clay ball rises h = (1 − cosθ ) = 9.1 cm. This is about 2/3 of the height of the pivot point, and is 2 reasonable.
12.95. Model: Because no external torque acts on the star during gravitational collapse, its angular momentum is conserved. Model the star as a solid rotating sphere. Solve: (a) The equation for the conservation of angular momentum is ⎛2 ⎞ ⎛2 ⎞ Li = Lf ⇒ I iωi = I f ω f ⇒ ⎜ mRi2 ⎟ωi = ⎜ mRf2 ⎟ω f 5 5 ⎝ ⎠ ⎝ ⎠ ⇒ Rf = Ri
ωi ωf
The angular velocity is inversely proportional to the period T. We can write
Rf = Ri
0.10 s Tf = ( 7.0 × 108 m ) = 1.3749 × 105 m = 137 km 2.592 × 106 s Ti
(b) A point on the equator rotates with r = Rf . Its speed is v=
2π Rf 2π (137,490 m) = = 8.6 × 106 m/s 0.10 s T
Model: For the (turntable + bicycle wheel + professor) system the angular momentum is conserved because the turntable is frictionless and no external torques act on the system. Solve: (a) Nothing happens. The bicycle wheel already has an angular momentum and nothing changes for the wheel when it is handed to the professor. So, nothing happens to the professor. (b) The initial angular momentum is 12.96.
⎛ 180 × 2π rad ⎞ 2 ( Lwheel )i = Iω = (mR 2 )ω = (4.0 kg)(0.32 m) 2 ⎜ ⎟ = 7.72 kg m /s s ⎠ ⎝ 60 When the wheel is turned upside down, the angular momentum of the wheel becomes
( Lwheel )f = −7.72 kg m 2 /s Since the initial and final total angular momentum should be equal, the professor must acquire some angular momentum. From the angular momentum conservation equation, Lprof + ( Lwheel ) f = ( Lwheel )i ⇒ Lprof = ( Lwheel )i − ( Lwheel )f = 7.72 kg m 2 /s − (−7.72 kg m 2 /s) = 15.44 kg m 2 /s
Let us now calculate the angular velocity of the turntable by using Lprof = I totalω , where I total = I turntable + I body + I arms +
I wheel in hand . We assume that the axis of the professor and turntable also become the axis of the arms and wheel (i.e., no leaning). 1 1 I turntable = mT RT2 = (5.0 kg)(0.25 m) 2 = 0.156 kg m 2 2 2 1 2 I body = mbody R = (70 kg)(0.125 m) 2 = 0.547 kg m 2 2 1 ⎛ ⎞ 2 I arms = 2 ⎜ marms R 2 ⎟ = (2.5 kg)(0.45 m) 2 = 0.338 kg m 2 ⎝3 ⎠ 3 I wheel in hand = mwheel r 2 = (4.0 kg)(0.45 m) 2 = 0.810 kg m 2 Thus, I total = (0.156 + 0.547 + 0.338 + 0.810) kg m 2 = 1.851 kg m 2 . Using this value for I total , we can now find ω to be
ω=
Lprof I total
=
15.44 kg m 2 /s = 8.34 rad/s = 79.7 rpm 1.851 kg m 2
12-1
13.1.
Model: Model the sun (s) and the earth (e) as spherical masses. Due to the large difference between your size and mass and that of either the sun or the earth, a human body can be treated as a particle. GM s M y GM e M y and Fe on you = Solve: Fs on you = 2 rs − e re2 Dividing these two equations gives
Fs on y Fe on y
2
2
⎛ M ⎞⎛ r ⎞ ⎛ 1.99 × 1030 kg ⎞⎛ 6.37 × 106 m ⎞ −4 = ⎜ s ⎟⎜ e ⎟ = ⎜ ⎟⎜ ⎟ = 6.00 × 10 24 11 M r × × 5.98 10 kg 1.50 10 m ⎠⎝ ⎠ ⎝ e ⎠⎝ s − e ⎠ ⎝
13.2.
Model:
Assume the two lead balls are spherical masses. Gm1m2 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(10 kg)(0.100 kg) = = 6.7 × 10−9 N Solve: (a) F1 on 2 = F2 on 1 = r2 (0.10 m) 2 (b) The ratio of the above gravitational force to the gravitational force on the 100 g ball is
6.67 × 10−9 N = 6.81 × 10−9 (0.100 kg)(9.8 m/s 2 ) Assess: The answer in part (b) shows the smallness of the gravitational force between two lead balls separated by 10 cm compared to the gravitational force on the 100 g ball.
13.3.
Model:
Model the sun (s), the moon (m), and the earth (e) as spherical masses. GM s M m GM e M m and Fe on m = Solve: Fs on m = rs2− m re2− m Dividing the two equations and using the astronomical data from Table 13.2, 2
2
Fs on m ⎛ M s ⎞⎛ re − m ⎞ ⎛ 1.99 × 1030 kg ⎞⎛ 3.84 × 108 m ⎞ =⎜ ⎟⎜ ⎟ =⎜ ⎟⎜ ⎟ = 2.18 Fe on m ⎝ M e ⎠⎝ rs − m ⎠ ⎝ 5.98 × 1024 kg ⎠⎝ 1.50 × 1011 m ⎠ Note that the sun-moon distance is not noticeably different from the tabulated sun-earth distance.
13.4.
Solve:
Fsphere on particle =
GM s M p rs2− p
and
Fearth on particle =
GM e M p re2
Dividing the two equations,
Fsphere on particle Fearth on particle
2
2
⎛ M ⎞ ⎛ r ⎞ ⎛ 5900 kg ⎞ ⎛ 6.37 × 106 m ⎞ −7 = ⎜ s ⎟⎜ e ⎟ = ⎜ ⎟ = 1.60 × 10 ⎟⎜ 24 ⎜ ⎟ M r 5.98 10 kg 0.50 m × ⎠⎝ ⎠ ⎝ e ⎠ ⎝ s−p ⎠ ⎝
13.5.
Model:
Solve:
Fw on m
Model the woman (w) and the man (m) as spherical masses or particles. GM w M m (6.67 × 10−11 N ⋅ m 2 /kg 2 )(50 kg)(70 kg) = Fm on w = = = 2.3 × 10−7 N rm2 − w (1.0 m) 2
13.6.
Model: Visualize:
Model the earth (e) as a sphere.
The space shuttle or a 1.0 kg sphere (s) in the space shuttle is Re + rs = 6.37 × 106 m + 0.30 × 106 m = 6.67 × 106 m away from the center of the earth. GM e M s (6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)(1.0 kg) = = 9.0 N Solve: (a) Fe on s = ( Re + rs ) 2 (6.67 × 106 m) 2 (b) Because the sphere and the shuttle are in free fall with the same acceleration around the earth, there cannot be any relative motion between them. That is why the sphere floats around inside the space shuttle.
13.7. Solve:
Model:
Model the sun (s) as a spherical mass. GM (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg) = 274 m/s 2 (a) gsun surface = 2 s = Rs (6.96 × 108 m) 2
(b) gsun at earth =
GM s (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg) = = 5.90 × 10−3 m/s 2 rs2− e (1.50 × 1011 m) 2
13.8. Solve:
Model:
Model the moon (m) and Jupiter (J) as spherical masses. GM m (6.67 × 10−11 N ⋅ m 2 /kg 2 )(7.36 × 1022 kg) = = 1.62 m/s 2 (a) g moon surface = Rm2 (1.74 × 106 m) 2
(b) g Jupiter surface =
GM J (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.90 × 1027 kg) = = 25.9 m/s 2 RJ2 (6.99 × 107 m) 2
13.9.
Model: Model the earth (e) as a spherical mass. Visualize: The acceleration due to gravity at sea level is 9.83 m/s 2 (see Table 13.1) and Re = 6.37 × 106 m (see Table 13.2).
Solve:
g observatory =
GM e = ( Re + h) 2
GM e ⎛ h⎞ Re2 ⎜1 + ⎟ ⎝ Re ⎠
2
=
g earth ⎛ h⎞ ⎜1 + ⎟ ⎝ Re ⎠
2
= (9.83 − 0.0075) m/s 2
Here g earth = GM e Re2 is the acceleration due to gravity on a non-rotating earth, which is why we’ve used the value 9.83 m/s 2 . Solving for h, ⎛ 9.83 ⎞ h = ⎜⎜ − 1⎟⎟ Re = 2.43 km ⎝ 9.8225 ⎠
13.10. Model: Model the earth (e) as a spherical mass. Solve:
Let the acceleration due to gravity be 3gsurface when the earth is shrunk to a radius of x. Then,
gsurface = ⇒3
GM e Re2
and
3 gsurface =
GM e x2
GM e GM e R = 2 ⇒ x = e = 0.577 Re Re2 x 3
The earth’s radius would need to be 0.577 times its present value.
13.11.
Model:
Model Planet Z as a spherical mass. GM Z (6.67 × 10−11 N ⋅ m 2 /kg 2 ) M Z ⇒ 8.0 m/s 2 = ⇒ M Z = 3.0 × 1024 kg Solve: (a) g Z surface = 2 RZ (5.0 × 106 m) 2 (b) Let h be the height above the north pole. Thus,
g above N pole =
GM Z = ( RZ + h) 2
GM Z ⎛ h ⎞ RZ2 ⎜1 + ⎟ ⎝ RZ ⎠
2
=
g Z surface ⎛ h ⎞ ⎜1 + ⎟ ⎝ RZ ⎠
2
=
8.0 m/s 2 ⎛ 10.0 × 106 m ⎞ ⎜1 + ⎟ 5.0 × 106 m ⎠ ⎝
2
= 0.89 m/s 2
13.12.
Model: Model Mars (M) as a spherical mass. Ignore air resistance. Also consider Mars and the ball as an isolated system, so mechanical energy is conserved. Visualize:
Solve: A height of 15 m is very small in comparison with the radius of earth or Mars. We can use flat-earth gravitational potential energy to find the speed with which the astronaut can throw the ball. On earth, with yi = 0 m and vf = 0 m/s, energy conservation is 1 2
mvf2 + mgyf = 12 mvi2 + mgyi ⇒ vi = 2 g e yf = 2(9.80 m/s 2)(15 m) = 17.1 m/s
Energy is also conserved on Mars, but the acceleration due to gravity is different. GM M (6.67 × 10−11 N ⋅ m 2 /kg 2 )(6.42 × 1023 kg) g Mar’s surface = = = 3.77 m/s 2 RM2 (3.37 × 106 m) 2 On Mars, with yi = 0 m and vf = 0 m/s, energy conservation is 1 2
mvf2 + mgyf = 12 mvi2 + mgyi ⇒ yf =
vi2 (17.1 m/s) 2 = = 39 m 2 g m 2(3.77 m/s 2)
13.13.
Model: Model Jupiter as a spherical mass and the object as a point particle. The object and Jupiter form an isolated system, so mechanical energy is conserved. The minimum launch speed for escape, which is called the escape speed, causes an object to stop only as the distance approaches infinity. Visualize:
Solve:
The energy conservation equation K 2 + U 2 = K1 + U1 is
1 GM J mo 1 GM J mo mo v22 − = mo v12 − 2 r2 2 RJ where RJ and M J are the radius and mass of Jupiter. Using the asymptotic condition v2 = 0 m/s as r2 → ∞, 1 2GM J 2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.90 × 1027 kg) GM J mo 0 J = mo v12 − ⇒ v1 = = = 6.02 × 104 m/s 2 6.99 × 107 m RJ RJ
Thus, the escape velocity from Jupiter is 60.2 km/s.
13.14.
Model: Model the earth (e) as a spherical mass. Compared to the earth’s size and mass, the rocket (r) is modeled as a particle. This is an isolated system, so mechanical energy is conserved. Visualize:
Solve:
The energy conservation equation K 2 + U 2 = K1 + U1 is
1 GM e mr 1 GM e mr mr v22 − = mr v12 − 2 r2 2 Re In the present case, r2 → ∞, so 1 1 GM e mr 2GM e mr v22 = mr v12 − ⇒ v22 = v12 − Re Re 2 2 ⇒ v2 = (1.50 × 104 m/s) 2 − = 9.99 × 103 m/s
2(6.67 × 10−11 N m 2 /kg 2 )(5.98 × 1024 kg) 6.37 × 106 m
13.15. Model: The probe and the sun form an isolated system, so mechanical energy is conserved. The minimum launch speed for escape, which is called the escape speed, causes an object to stop only as the distance approaches infinity. Visualize:
We denote by mp the mass of the probe. M S is the sun’s mass, and RS − p is the separation between the centers of the sun and the probe. Solve: The conservation of energy equation K 2 + U 2 = K1 + U1 is
GM Smp 1 GM Smp 1 mp v22 − = mp v12 − 2 ( r2 ) 2 RS− p Using the condition v2 = 0 m/s asymptotically as r2 → ∞, GM Smp 1 2GM S 2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg) 2 mp vescape = ⇒ vescape = = = 4.21× 104 m/s 2 (1.50 × 1011 m) RS − p RS − p
13.16.
Model: Model the distant planet (p) and the earth (e) as spherical masses. Because both are isolated, the mechanical energy of the object on both the planet and the earth is conserved. Visualize:
Let us denote the mass of the planet by M p and that of the earth by M e . Your mass is m0 . Also, acceleration due to gravity on the surface of the planet is g p and on the surface of the earth is g e . Rp and Re are the radii of the planet and the earth, respectively. Solve: (a) We are given that M P = 2M e and g p = 14 g e .
Since g p =
GM p Rp2
and g e = GM p 2 p
R
=
GM e , we have Re2
1 GM e 1 G ( M p /2) = ⇒ Rp = 8Re = 8(6.37 × 106 m) = 1.80 × 107 m Re2 4 Re2 4
(b) The conservation of energy equation K 2 + U 2 = K1 + U1 is
GM p mo 1 GM p mo 1 mov22 − = mo v12 − 2 r2 2 Rp Using v2 = 0 m/s as r2 → ∞, we have GM p mo 1 2 = mo vescape 2 Rp ⇒ vescape =
2GM p Rp
=
2G (2M e ) 4(6.67 × 10 −11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg) = = 9.41 km/s 1.80 × 107 m Rp
13.17.
Model: Model the sun (s) as a spherical mass and the asteroid (a) as a point particle. Visualize: The asteroid, having mass ma and velocity va , orbits the sun in a circle of radius ra . The asteroid’s
time period is Ta = 5.0 earth years = 1.5779 × 108 s. Solve: The gravitational force between the sun (mass = M S ) and the asteroid provides the centripetal acceleration required for circular motion. 2
1/ 3
⎛ GM sTa2 ⎞ GM s ma ma va2 GM s ⎛ 2π ra ⎞ = ⇒ =⎜ ⎟ ⇒ ra = ⎜ ⎟ 2 2 ra ra ra ⎝ 4π ⎠ ⎝ Ta ⎠
Substituting G = 6.67 × 10−11 N ⋅ m 2 /kg 2 , M s = 1.99 × 1030 kg, and the time period of the asteroid, we obtain
ra = 4.37 × 1011 m. The velocity of the asteroid in its orbit will therefore be va =
2π ra (2π )(4.37 × 1011 m) = = 1.74 × 104 m/s Ta 1.5779 × 108 s
13.18.
Model: Model the sun (s) and the earth (e) as spherical masses. Visualize: The earth orbits the sun with velocity ve in a circular path with a radius denoted by rs-e . The sun’s
and the earth’s masses are denoted by M s and me . Solve: The gravitational force provides the centripetal acceleration required for circular motion. GM s me meve2 me (2π rs − e ) 2 = = rs2− e rs − e rs − eTe2 ⇒ Ms =
4π 2 rs3− e 4π 2 (1.50 × 1011 m)3 = = 2.01 × 1030 kg 2 −11 GTe (6.67 × 10 N ⋅ m 2 /kg 2 )(365 × 24 × 3600 s) 2
Assess: The tabulated value is 1.99 × 1030 kg. The slight difference can be ascribed to the fact that the earth’s orbit isn’t exactly circular.
13.19. Solve:
Model: Model the star (s) and the planet (p) as spherical masses. A planet’s acceleration due to gravity is
gp = ⇒ Mp =
g p Rp2 G
=
GM p Rp2
(12.2 m/s 2 )(9.0 × 106 m) 2 = 1.48 × 1025 kg 6.67 × 10−11 N ⋅ m 2 /kg 2
(b) A planet’s orbital period is
⇒ Ms =
4π 2 r 3 = GT 2 (6.67 × 10−11
⎛ 4π 2 ⎞ 3 T2 =⎜ ⎟r ⎝ GM s ⎠ 4π 2 (2.2 × 1011 m)3 = 5.2 × 1030 kg N ⋅ m 2 /kg 2 )(402 × 24 × 3600 s) 2
13.20.
Model: Model the planet and satellites as spherical masses. Visualize: Please refer to Figure EX13.20. Solve: (a) The period of a satellite in a circular orbit is T = [(4π 2 /GM ) r 3 ]1/ 2 . This is independent of the satellite’s mass, so we can find the ratio of the periods of two satellites a and b:
⎛r ⎞ Ta = ⎜ a⎟ Tb ⎝ rb ⎠
3
Satellite 2 has r2 = r1 , so T2 = T1 = 250 min. Satellite 3 has r3 = (3/2) r1 , so T3 = (3/2)3/ 2 T1 = 459 min. (b) The force on a satellite is F = GMm/r 2 . Thus the ratio of the forces on two satellites a and b is 2
Fa ⎛ rb ⎞ ⎛ ma ⎞ =⎜ ⎟ ⎜ ⎟ Fb ⎝ ra ⎠ ⎝ mb ⎠ Satellite 2 has r2 = r1 and m2 = 2m1 , so F2 = (1) 2(2) F1 = 20,000 N. Similarly, satellite 3 has r3 = (3/2) r1 and m3 = m1 , so F3 = (2/3) 2 (1) F1 = 4440 N. (c) The speed of a satellite in a circular orbit is v = (GM/r ) 2 , so its kinetic energy is K = 12 mv 2 = GMm/2r. Thus the ratio of the kinetic energy of two satellites a and b is
K a ⎛ rb ⎞⎛ ma ⎞ = ⎜ ⎟⎜ ⎟ K b ⎝ ra ⎠⎝ mb ⎠ Satellite 3 has r3 = (3/2) r1 and m3 = m1 , so K1/K 3 = (3/2)(1/1) = 3/2 = 1.50.
13.21.
Model: Model the sun (S) as a spherical mass and the satellite (s) as a point particle. Visualize: The satellite, having mass ms and velocity vs , orbits the sun with a mass M S in a circle of radius
rs . Solve: The gravitational force between the sun and the satellite provides the necessary centripetal acceleration for circular motion. Newton’s second law is GM Sms ms vs2 = rs2 rs
Because vs = 2π rs / Ts where Ts is the period of the satellite, this equation simplifies to GM S (2π rs ) 2 GM STs2 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)(24 × 3600 s) 2 3 r = ⇒ = = ⇒ rs = 2.9 × 109 m s rs2 Ts2 rs 4π 2 4π 2
13.22.
Model: Model the earth (e) as a spherical mass and the shuttle (s) as a point particle. Visualize: The shuttle, having mass ms and velocity vs , orbits the earth in a circle of radius rs . We will
denote the earth’s mass by M e . Solve: The gravitational force between the earth and the shuttle provides the necessary centripetal acceleration for circular motion. Newton’s second law is
GM e ms msvs2 GM e GM e = ⇒ vs2 = ⇒ vs = rs2 rs rs rs Because rs = Re + 350 km = 6.72 × 106 m, vs = Ts =
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg) = 7704 m/s (6.72 × 106 m) 2π rs 2π (6.72 × 106 m) = = 5580 s = 91.3 minutes vs 7.704 × 103 m/s
13.23.
Model: Model the earth (e) as a spherical mass and the satellite (s) as a point particle. Visualize: The satellite has a mass is ms and orbits the earth with a velocity vs . The radius of the circular orbit
is denoted by rs and the mass of the earth by M e . Solve: The satellite experiences a gravitational force that provides the centripetal acceleration required for circular motion: (6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg) GM e ms ms vs2 GM = ⇒ rs = 2 e = = 1.32 × 107 m 2 (5500 m/s) 2 rs rs vs
⇒ Ts =
2π Rs (2π )(1.32 × 107 m) = = 1.51 × 104 s = 4.2 hr (5500 m/s) vs
13.24.
Model: Model Mars (m) as a spherical mass and the satellite (s) as a point particle. Visualize: The geosynchronous satellite whose mass is ms and velocity is vs orbits in a circle of radius rs
around Mars. Let us denote mass of Mars by M m . Solve: The gravitational force between the satellite and Mars provides the centripetal acceleration needed for circular motion: 1/ 3
⎛ GM mTs2 ⎞ GM m ms ms vs2 ms (2π rs ) 2 = = ⇒ = r ⎜ ⎟ s 2 rs2 rs rs (Ts ) 2 ⎝ 4π ⎠ Using G = 6.67 × 10−11 N ⋅ m 2 /kg 2 ,
M m = 6.42 × 1023 kg, and Ts = (24.8 hrs) = (24.8)(3600) s = 89,280 s, we
obtain rs = 2.05 × 107 m. Thus, altitude = rs − Rm = 1.72 × 107 m.
13.25.
Solve: We are given M 1 + M 2 = 150 kg which means M 1 = 150 kg − M 2 . We also have
GM 1M 2
( 0.20 m ) Thus,
2
= 8.00 × 10−6 N ⇒ M 1M 2 =
(8.00 × 10−6 N)(0.20 m) 2 = 4798 kg 2 6.67 × 10−11 N ⋅ m 2 /kg 2
(150 kg − M 2 ) M 2 = 4798 kg 2 or M 22 − (150 kg) M 2 + (4798 kg 2 ) = 0.
M 2 = 103.75 kg and 46.25 kg. The two masses are 104 kg and 46 kg.
Solving
this
equation
gives
13.26. Visualize: Because of the gravitational force of attraction between the lead spheres, the cables will make an angle of θ with the vertical. The distance between the sphere centers is therefore going to be less than 1 m. The free-body diagram shows the forces acting on the lead sphere.
Solve: We can see from the diagram that the distance between the centers is d = 1.000 m − 2 L sin θ . Each sphere is in static equilibrium, so Newton’s second law is
∑F = F x
grav
− T sin θ = 0 ⇒ T sin θ = Fgrav
∑F
y
= T cosθ − mg = 0 ⇒ T cosθ = mg
Dividing these two equations to eliminate the tension T yields
F Gmm / d 2 Gm sin θ = tan θ = grav = = 2 cosθ mg mg d g We know that d is going to be only very, very slightly less than 1.00 m. The very slight difference is not going to be enough to affect the value of Fgrav , the gravitational attraction between the two masses, so we’ll evaluate the right side of this equation by using 1.00 m for d. This gives tan θ =
(6.67 × 10−11 N m 2 /kg 2 ) (100 kg) = 6.81 × 10−10 ⇒ θ = (3.90 × 10−8 )° (1.00 m) 2 (9.80 m/s 2 )
This small angle causes the two spheres to move closer by 2 L sin θ = 1.4 × 10−7 m = 0.00000014 m. Consequently, the distance between their centers is d = 0.99999986 m.
13.27. Visualize:
We placed the origin of the coordinate system on the 20 kg sphere (m1 ). The sphere (m2 ) with a mass of 10 kg is 20 cm away on the x-axis. The point at which the net gravitational force is zero must lie between the masses m1
and m2 . This is because on such a point, the gravitational forces due to m1 and m2 are in opposite directions. As the gravitational force is directly proportional to the two masses and inversely proportional to the square of distance between them, the mass m must be closer to the 10-kg mass. The small mass m, if placed either to the left of m1 or to the right of m2 , will experience gravitational forces from m1 and m2 pointing in the same direction, thus always leading to a nonzero force. mm m2 m 20 10 Solve: Fm1 on m = Fm2 on m ⇒ G 12 = G ⇒ 2 = 2 x x (0.20 − x ) (0.20 − x) 2 ⇒ 10 x 2 − 8 x + 0.8 = 0 ⇒ x = 0.683 m and 0.117 m The value x = 68.3 cm is unphysical in the current situation, since this point is not between m1 and m2 . Thus, the point ( x, y ) = (11.7 cm, 0 cm) is where a small mass is to be placed for a zero gravitational force.
13.28.
We placed the origin of the coordinate system on the 20.0 kg mass (m1 ) so that the 5.0 kg
Visualize:
mass (m3 ) is on the x-axis and the 10.0 kg mass (m2 ) is on the y-axis.
Solve:
(a) The forces acting on the 20 kg mass (m1 ) are
G Gm1m2 ˆ (6.67 × 10−11 N ⋅ m 2 /kg 2 )(20.0 kg)(10.0 kg) ˆ Fm2 on m1 = j= j = 3.335 × 10−7 ˆj N r122 (0.20 m) 2 G Gm1m3 ˆ (6.67 × 10−7 N ⋅ m 2 /kg 2 )(20.0 kg)(5.0 kg) ˆ Fm3 on m1 = i= i = 6.67 × 10−7 iˆ N r132 (0.10 m) 2 G Fon m = 6.67 × 10−7 iˆ N + 3.335 × 10−7 ˆj N ⇒ Fon m = 7.46 × 10−7 N 1
1
⎛F θ = tan −1 ⎜ m3 on m1 ⎜ ⎝ Fm2 on m1
⎞ ⎛ 6.67 × 10−7 N ⎞ ⎟ = tan −1 ⎜ ⎟ = 63.4° −7 ⎟ ⎝ 3.335 × 10 N ⎠ ⎠
G Thus the force is Fon m1 = ( 7.5 × 10−7 N, 63° cw from the y -axis ). (b) The forces acting on the 5 kg mass (m3 ) are G G Fm1 on m3 = − Fm3 on m1 = −6.67 × 10−7 iˆ N
Gm2 m3 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(10.0 kg)(5.0 kg) = = 6.67 × 10−8 N r232 [(0.20) 2 + (0.10) 2 ] m = −(6.67 × 10−8 N)cos φ iˆ + (6.67 × 10−8 N)sin φ ˆj
Fm2 on m3 = G Fm2 on m3
G Fon m3
⎛ 10 cm ⎞ ˆ ⎛ 20 cm ⎞ ˆ −8 = −(6.67 × 10−8 N) ⎜ ⎟ i + (6.67 × 10 N) ⎜ ⎟j ⎝ 22.36 cm ⎠ ⎝ 22.36 cm ⎠ = −(2.983 × 10−8 N)iˆ + (5.966 × 10−8 N) ˆj = −6.968 × 10−7 iˆ N + 5.966 × 10−8 ˆj N
Fon m3 = (−6.968 × 10−7 N) 2 + (5.966 × 10−8 N)2 = 6.99 × 10−7 N ⎛ 6.968 × 10−7 N ⎞ ⎟ = 85.1° −8 ⎝ 5.966 × 10 N ⎠
θ ′ = tan −1 ⎜
G Thus Fon m3 = ( 7.0 × 10−7 N, 85° ccw from the y -axis ).
13.29. Visualize:
⎛ 5 ⎞ The angle θ = tan −1 ⎜ ⎟ = 14.04°. The distance r1 = r2 = ⎝ 20 ⎠ forces on the 20.0 kg mass are G Mm F1 = G 2 1 − sinθ iˆ + cosθ ˆj r1 G Mm F2 = G 2 2 sin θ iˆ + cosθ ˆj r2
( 0.050 m )
Solve:
(
(
2
+ ( 0.200 m ) = 0.206 m. The 2
)
)
Note m1 = m2 and r1 = r2 . Thus the net force on the 20.0 kg mass is
G G G Mm Fnet = F1 + F2 = 2G 2 1 cosθ ˆj r1 =
2 ( 6.67 × 10−11 N ⋅ m 2 /kg 2 ) ( 20.0 kg )( 5.0 kg ) cos14.04°
= 3.0 × 10−7 ˆj N
( 0.206 m )
2
ˆj
13.30. Visualize:
Solve: The total gravitational potential energy is the sum of the potential energies due to the interactions of the pairs of masses. U = U12 + U13 + U 23
= −G
m1m2 mm mm −G 1 3 −G 2 3 r12 r13 r23
With m1 = 20.0 kg, m2 = 10.0 kg, m3 = 5.0 kg, r12 = 0.20 m, r13 = 0.10 m, and r23 = 0.2236 m, U = −1.48 × 10 −7 J
Assess: The gravitational potential energy is negative because the masses attract each other. It is a scalar, so there are no vector calculations required.
13.31. Visualize:
Solve: The total gravitational potential energy is the sum of the potential energies due to the interactions of the pairs of masses.
U = U12 + U13 + U 23 = −G With m1 = 20.0 kg, m2 = m3 = 10.0 kg, r12 = r13 =
m1m2 mm mm −G 1 3 −G 2 3 r12 r13 r23
( 0.050 m )
2
+ ( 0.200 m ) = 0.206 m, and r23 = 0.100 m, 2
U = −1.96 × 10−7 J Assess: The gravitational potential energy is negative because the masses attract each other. It is a scalar, so there are no vector calculations required.
13.32.
Model: Model the earth (e) as a spherical mass and the satellite (s) as a point particle. Visualize: Let h be the height from the surface of the earth where the acceleration due to gravity ( g altitude ) is
10% of the surface value ( gsurface ). Solve:
(a) Since g altitude = (0.10) gsurface , we have
GM e GM = (0.10) 2 e ⇒ ( Re + h) 2 = 10Re2 2 ( Re + h) Re ⇒ h = 2.162 Re ⇒ h = (2.162)(6.37 × 106 m) = 1.377 × 107 m= 1.38 × 107 m (b) For a satellite orbiting the earth at a height h above the surface of the earth, the gravitational force between the earth and the satellite provides the centripetal acceleration necessary for circular motion. For a satellite orbiting with velocity vs ,
GM e ms ms vs2 = ⇒ vs = 2 ( Re + h) ( Re + h)
GM e (6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg) = = 4.45 km/s Re + h (6.37 × 106 m + 1.377 × 107 m)
13.33.
Model: Model the earth as a spherical mass and the object (o) as a point particle. Ignore air resistance. This is an isolated system, so mechanical energy is conserved. Visualize:
Solve:
(a) The conservation of energy equation K 2 + U g2 = K1 + U g1 is
1 GM e mo 1 GM e mo mov22 − = mo v12 − Re 2 2 ( Re + y1 ) ⎛ 1 1 ⎞ ⇒ v2 = 2GM e ⎜ − ⎟ ⎝ Re Re + y1 ⎠ 1 1 ⎛ ⎞ = 2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg) ⎜ − ⎟ = 3.02 km/s 6 6 ⎝ 6.37 × 10 m 6.87 × 10 m ⎠ (b) In the flat-earth approximation, U g = mgy. The energy conservation equation thus becomes
1 1 mov22 + mo gy2 = mov12 + mo gy1 2 2 ⇒ v2 = v12 + 2 g ( y1 − y2 ) = 2(9.80 m/s 2 )(5.00 × 105 m − 0 m) = 3.13 km/s (c) The percent error in the flat-earth calculation is
3130 m/s − 3020 m/s ≈ 3.6% 3020 m/s
13.34. Model: Model the earth and the projectile as spherical masses. Ignore air resistance. This is an isolated system, so mechanical energy is conserved. Visualize:
A pictorial representation of the before-and-after events is shown. Solve: After using v2 = 0 m/s, the energy conservation equation K 2 + U 2 = K1 + U1 is
0 J−
GM e mp
GM e mp 1 = mpv12 − Re + h 2 Re
The projectile mass cancels. Solving for h, we find −1
⎡1 v2 ⎤ h = ⎢ − 1 ⎥ − Re ⎣ Re 2GM e ⎦ = 4.18 × 105 m = 418 km
13.35.
Model: Model the planet (p) as a spherical mass and the projectile as a point mass. This is an isolated system, so mechanical energy is conserved. Visualize: The projectile of mass m was launched on the surface of the planet with an initial velocity v0 .
Solve:
(a) The energy conservation equation K1 + U1 = K 0 + U 0 is
1 2 GM p m 1 2 GM p m = mv0 − mv1 − 2 Rp + y1 2 Rp −1
⎡1 v02 ⎤ 6 ⇒ y1 = ⎢ − ⎥ − Rp = 2.8 × 10 m 2 R GM ⎥ p⎦ ⎣⎢ p (b) Using the energy conservation equation K1 + U1 = K 0 + U 0 with y1 = 1000 km = 1.000 × 106 m:
1 2 GM p m 1 2 GM p m mv1 − = mv0 − 2 Rp + y1 2 Rp 12
⎡ ⎛ 1 1 ⎞⎤ ⇒ v1 = ⎢v02 + 2GM p ⎜ − ⎟⎥ ⎜ ⎟ ⎢⎣ ⎝ Rp + y1 Rp ⎠ ⎥⎦
⎡ ⎛ ⎞⎤ 1 1 = ⎢(5000 m/s) 2 + 2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(2.6 × 1024 kg) ⎜ − ⎟⎥ 6 6 6 ⎝ (5.0 × 10 m + 1.000 × 10 m) 5.0 × 10 m ⎠ ⎦ ⎣ = 3.7 km/s
12
13.36. Model: The object is a particle. The planet is a spherical mass. Solve:
Conservation of mechanical energy of the object gives
−G
Mm 1 2 Mm = mv − G R+h 2 R
The object’s mass drops out. Solving for the speed as it hits the ground,
⎛ R+h−R⎞ 1 ⎞ ⎛1 v = 2GM ⎜ − ⎟⎟ = ⎟ = 2GM ⎜⎜ ⎝ R R+h⎠ ⎝ R ( R + h) ⎠ Assess:
Compare this to v = 2 gh =
2GMh R ( R + h)
2GMh , which is the result if the potential U = mgh is used. R2
13.37.
Model: Visualize:
Solve:
Model the earth as a spherical mass and the meteoroids as point masses.
(a) The energy conservation equation K 2 + U 2 = K1 + U1 is 1/ 2
1 2
⎡ ⎛ 1 1 ⎞⎤ GM e m 1 2 GM e m = 2 mv1 − ⇒ v2 = ⎢ v12 + 2GM e ⎜ − ⎟ ⎥ mv − Re rm ⎢⎣ ⎝ Re rm ⎠ ⎦⎥ = 1.13 × 104 m/s =11.3 km/s 2 2
The speed does not depend on the meteoroid’s mass. (b) This part differs in that r2 = Re + 5000 km = 1.137 × 107 m. The shape of the meteoroid’s trajectory is not important for using energy conservation. Thus 1 2
⎡ ⎛ GM e m GM e m 1 1 ⎞⎤ mv − = 1 mv12 − ⇒ v2 = ⎢v12 + 2GM e ⎜ − ⎟⎥ + 5000 km Re + 5000 km 2 rm R r ⎢⎣ ⎥ m ⎠⎦ ⎝ e 3 = 8.94 × 10 m/s = 8.94 km/s 2 2
1/ 2
13.38.
Model: Model the two stars as spherical masses, and the comet as a point mass. This is an isolated system, so mechanical energy is conserved. Visualize:
In the initial state, the comet is far away from the two stars and thus it has neither kinetic energy nor potential energy. In the final state, as the comet passes through the midpoint connecting the two stars, it possesses both kinetic energy and potential energy. Solve: The conservation of energy equation K f + U f = K i + U i is
1 2 GMm GMm mvf − − =0 J +0 J 2 rf1 rf2 ⇒ vf = Assess:
4GM 4(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg) = = 32,600 m/s rf (0.50 × 1012 m)
Note that the final velocity of 33 km/s does not depend on the mass of the comet.
13.39.
Model: Model the asteroid as a spherical mass and yourself as a point mass. This is an isolated system, so mechanical energy is conserved. Visualize: The radius of the asteroid is M a and its mass is Ra .
Solve:
The conservation of energy equation K f + U f = K i + U i for the asteroid is
1 2 GM a m 1 2 GM a m mvf − = mvi − 2 Ra + r 2 Ra The minimum speed for escape is the one that will cause you to stop only when the separation between you and the asteroid becomes very large. Noting that vf → 0 m/s as r → ∞, we have vi2 =
2GM a 2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.0 × 1014 kg) = ⇒ vi = 2.58 m/s Ra (2.0 × 103 m)
That is, you need a speed of 2.58 m/s to escape from the asteroid. We can now calculate your jumping speed on the earth. The conservation of energy equation is 1 0 J − mvi2 = − mg (0.50 m) ⇒ vi = 2(9.8 m/s 2 )(0.50 m) = 3.13 m/s 2
This means you can escape from the asteroid.
13.40. Model: The projectile is a particle. The earth and moon are spherical masses. Solve: The projectile is attracted to both the moon and earth. Its final velocity and potential energy are zero. Since the projectile is fired from the far side of the moon, its initial distance from the center of the earth is the earth-moon distance Re-m plus the radius of the moon Rm . Let the projectile have mass m. The conservation of mechanical energy equation is
1 2 M em M m mvescape − G −G m = 0 J +0 J R R Rm 2 + ( e-m m ) ⎛ Me M ⎞ 2 ⇒ vescape = 2G ⎜ + m⎟ R R Rm ⎠ + m ⎝ e-m From
Table
13.2,
Re-m = 3.84 × 108 m, Rm = 1.74 × 106 m, M e = 5.98 × 1024 kg,
and
M m = 7.36 × 1022 kg.
vescape = 2.78 km/s. Assess:
The escape velocity does not depend on the mass of the object which is trying to escape.
So
13.41. Model: The earth and sun are spherical masses. The earth is in a circular orbit around the sun. The projectile is a particle. Visualize:
Solve: At the earth’s distance from the sun, to escape the sun’s gravitational pull a projectile must have speed vescape . The energy conservation equation for the projectile is
M m 1 2 mvescape − G 2 = 0 J Re − s 2 ⇒ vescape = 2G
2 ( 6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg ) Ms = = 4.207 × 104 m/s Re − s (1.50 ×1011 m )
The earth’s speed in its orbit is found by considering its period T = 1 year × 365.25 × 24 × 3600 = 3.156 × 107 s.
ve =
11 2π Re − s 2π (1.50 × 10 m ) = = 2.987 × 104 m/s T 3.156 × 107 s
The projectile’s total speed is the sum of its launch speed and the earth’s speed. vescape = ve + vlaunch ⇒ vlaunch = vescape − ve = 1.22 × 104 m/s Assess: The projectile need only attain a launch speed of 12.2 km/s if launched in the direction of earth’s motion. This is about a factor of 3.5 times less than what is required from rest.
13.42.
Model: Visualize:
Gravity is a conservative force, so we can use conservation of energy.
The planets will be pulled together by gravity and each will have speed v2 as they crash and the separation between their centers will be 2R. Solve: The planets begin with only gravitational potential energy. When they crash, they have both potential and kinetic energy. Thus,
⎡ 1 1 GMM ⎤ ⎡ GMM ⎤ 2 2 ⎢ K 2 + U 2 = Mv2 + Mv2 − ⎥ = ⎢ K1 + U1 = 0 J − ⎥ 2 2 r2 ⎦ ⎣ r1 ⎦ ⎣ ⎛ 1 1⎞ ⇒ v2 = GM ⎜ − ⎟ ⎝ r2 r1 ⎠ Because the planet is “Jupiter-size,” we’ll use M = M Jupiter = 1.9 × 1027 kg and r2 = 2 RJupiter = 1.4 × 108 m. The
crash speed of each planet is v2 = 3.0 × 104 m/s. Assess: Note that the force is not constant, because it varies with distance, so the motion is not constant acceleration motion. The formulas from constant-acceleration kinematics do not work for problems such as this.
13.43.
Model: The two asteroids make an isolated system, so mechanical energy is conserved. We will also use the law of conservation of momentum for our system. Visualize:
Solve:
The conservation of momentum equation pfx = pix is
M (vfx )1 + 2 M (vfx ) 2 = 0 kg m/s ⇒ (vfx )1 = −2(vfx ) 2 The equation for mechanical energy conservation K f + U f = K i + U i is
1 1 G ( M )(2M ) G ( M )(2M ) 1 0.8GM M (vfx )12 + (2M )(vfx ) 22 − =− ⇒ [2(vfx ) 2 ]2 + (vfx ) 22 = 2 2 2R 10 R 2 R GM GM ⇒ (vfx ) 2 = −0.516 ⇒ (vfx )1 = −2(vfx ) 2 = 1.032 R R The heavier asteroid has a speed of 0.516(GM/R)1/ 2 and the lighter one a speed of 1.032(GM/R)1/ 2 .
13.44. Solve:
Model: Model the distant planet (P) as a spherical mass. The acceleration at the surface of the planet and at the altitude h are
gsurface =
GM P R2
and g altitude =
1 MP 1 GM P = g surface ⇒ G 2 2 ( R + h) 2 R2
⇒ ( R + h) 2 = 2 R 2 ⇒ R + h = 2 R ⇒ h = ( 2 − 1) R = 0.414 R That is, the starship is orbiting at an altitude of 0.414R.
13.45. Model: The stars are spherical masses. Visualize:
Solve: The starts are identical, so their final speeds vf are the same. They collide when their centers are 2R apart. From energy conservation,
⎛ ⎞ ⎛ M2 ⎞ M2 ⎛1 2⎞ 3 ⎜ −G ⎟ = 3 ⎜ Mvf ⎟ − 3⎜ G ⎟ 9 (5.0 × 10 m) ⎠ ⎝2 ⎠ ⎝ 2R ⎠ ⎝ 1 ⎛ 1 ⎞ vf2 = 2GM ⎜ − ⎟ 9 ⎝ 2 R 5.0 × 10 m ⎠ ⇒ vf = 3.71 × 105 m/s
13.46.
Model: Model the moon (m) as a spherical mass and the lander (l) as a particle. This is an isolated system, so mechanical energy is conserved. Visualize: The initial position of the lunar lander (mass = m1 ) is at a distance r1 = Rm + 50 km from the center of the moon. The final position of the lunar lander is the orbit whose distance from the center of the moon is r2 = Rm + 300 km.
Solve:
The external work done by the thrusters is
Wext = ΔEmech = 12 ΔU g where we used Emech = 12 U g for a circular orbit. The change in potential energy is from the initial orbit at r1 = Rm + 50 km to the final orbit rf = Rm + 300 km. Thus
1 ⎛ −GM m m −GM m m ⎞ GM m m ⎛ 1 1 ⎞ Wext = ⎜ − ⎟= ⎜ − ⎟ 2⎝ rf ri 2 ⎝ ri rf ⎠ ⎠ (6.67 × 10−11 N m 2 / kg 2 )(7.36 × 1022 kg)(4000 kg) ⎛ 1 1 ⎞ = − ⎜ ⎟ 6 6 2 1.79 10 m 2.04 10 m × × ⎝ ⎠ = 6.72 × 108 J
13.47.
Model: Model the earth (e) as a spherical mass and the space shuttle (s) as a point particle. This is an isolated system, so the mechanical energy is conserved. Visualize: The space shuttle (mass = ms ) is at a distance of r1 = Re + 250 km.
Solve:
The external work done by the thrusters is
Wext = ΔEmech = 12 ΔU g where we used Emech = 12 U g for a circular orbit. The change in potential energy is from the initial orbit at
r1 = Re + 250 km to the final orbit rf = Re + 610 km. Thus 1 ⎛ −GM e m −GM e m ⎞ GM e m ⎛ 1 1 ⎞ Wext = ⎜ − ⎟= ⎜ − ⎟ 2 ⎝ rf ri 2 ⎝ ri rf ⎠ ⎠ (6.67 × 10−11 N m 2 / kg 2 )(5.98 × 1024 kg)(75,000 kg) ⎛ 1 1 ⎞ = − ⎜ ⎟ 6 6 2 ⎝ 6.62 × 10 m 6.98 × 10 m ⎠ = 1.17 × 1011 J This much energy must be supplied by burning the on-board fuel.
13.48.
Solve:
(a) Using Equation 13.31 for a satellite in a circular orbit,
⎛ ⎞ 1 1 1 1 ΔEmech = ΔU g = (−GM e m) ⎜ − ⎟ + + 2 2 R 300 km R 500 km e ⎝ e ⎠ =
−1 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)(5.00 × 104 kg) 2
1 1 ⎛ ⎞ 10 ×⎜ − ⎟ = −4.35 × 10 J 6 5 6 5 ⎝ 6.37 × 10 m + 3 × 10 m 6.37 × 10 m + 5 × 10 m ⎠ The negative sign indicates that 4.35 × 1010 J of energy is lost. (b)
The shuttle would fire retro-rockets to lose enough energy to go into an elliptical orbit. When halfway around, at rmin , the shuttle would fire retro-rockets again to lose the rest of the energy to stay at rmin .
13.49. Model: Planet Physics is a spherical mass. The cruise ship is in a circular orbit. Solve:
(a) At the surface, the free-fall acceleration is g = G
M . From kinematics, R2
y f = yi + v0 Δt − 2 g ( Δt ) ⇒ 0 m = 0 m + (11 m/s )( 2.5 s ) − 2 g ( 2.5 s ) 2
2
⇒ g = 2.20 m/s 2 The period of the cruise ship’s orbit is 230 × 60 = 13,800 s. For the circular orbit of the cruise ship, ⎛ 4π 2 ⎞ ⎛ R2 ⎞ ⎛1⎞ T2 3 T2 =⎜ R = ⎟ ( 2R ) ⇒ ⎜ ⎟ = R⎜ ⎟ 2 32π ⎝g⎠ ⎝ GM ⎠ ⎝ GM ⎠
( 2.20 m/s ) (13,800 s ) 2
⇒R=
R2 g = 5.8 × 1022 kg. G (b) From part (a), R = 1.33 × 106 m. The mass is thus M =
32π 2
2
= 1.327 × 106 m.
13.50.
Model: Assume a spherical asteroid and a point mass model for the satellite. This is an isolated system, so mechanical energy is conserved. Visualize:
The orbital radius of the satellite is
r = Ra + h = 8,800 m + 5,000 m = 13,800 m Solve:
(a) The speed of a satellite in a circular orbit is 12
GM ⎡ (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.0 × 1016 kg) ⎤ =⎢ v= ⎥ = 7.0 m/s r (13,800 m) ⎣ ⎦ (b) The minimum launch speed for escape (vi ) will cause the satellite to stop asymptotically (vf = 0 m/s) as rf → ∞. Using the energy conservation equation K 2 + U 2 = K1 + U1 , we get GM a ms 1 GM a ms GM a 1 1 2 ms vf2 − = msvi2 − ⇒ 0 J − 0 J = vescape − rf Ra Ra 2 2 2 ⇒ vescape =
2GM a 2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.0 × 1016 kg) = = 12.3 m/s Ra 8800 m
13.51.
Model: Model the moon as a spherical mass and the satellite as a point mass. Visualize: The rotational period of the satellite is the same as the rotational period of the moon around its own axis. This time happens to be 27.3 days. Solve: The gravitational force between the moon and the satellite provides the centripetal acceleration necessary for circular motion around the moon. Therefore, 2
GM m m ⎛ 2π ⎞ = mrω 2 = mr ⎜ ⎟ r2 ⎝ T ⎠ GM mT 2 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(7.36 × 1022 kg)(27.3 × 24 × 3600 s) 2 ⇒ r3 = = 4π 2 4π 2 ⇒ r = 8.84 × 107 m Since r = Rm + h, then h = r − Rm = 8.84 × 107 m − 1.74 × 106 m = 8.67 × 107 m.
13.52.
Model: Model the earth as a spherical mass and the satellite as a point mass. Visualize: The satellite is directly over a point on the equator once every two days. Thus, T = 2Te = 2 × 24 ×
3600 s = 1.728 × 105 s. Solve: A satellite’s period is
⎛ 4π 2 ⎞ 3 T2 = ⎜ ⎟r ⎝ GM e ⎠ GM eT 2 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)(1.728 × 105 ) 2 ⇒ r3 = = 4π 2 4π 2 7 ⇒ r = 6.71 × 10 m Assess:
The radius of the orbit is larger than the geosynchronous orbit.
13.53.
Solve:
(a) Taking the logarithm of both sides of v p = Cu q gives
[log(v p ) = p log v] = [log(Cu q ) = log C + q log u ] ⇒ log v =
q log C log u + p p
But x = log u and y = log v, so x and y are related by ⎛q⎞ log C y = ⎜ ⎟x + p p ⎝ ⎠ (b) The previous result shows there is a linear relationship between x and y, hence there is a linear relationship between log u and log v. The graph of a linear relationship is a straight line, so the graph of log v-versus-log u will be a straight line. (c) The slope of the straight line represented by the equation y = (q / p ) x + log C / p is q/p. Thus, the slope of the log v-versus-log u graph will be q/p. (d) From Newton’s theory, the period T and radius r of an orbit around the sun are related by
⎛ 4π 2 ⎞ 3 T2 =⎜ ⎟r ⎝ GM ⎠ This equation is of the form T p = Cr q , with p = 2, q = 3, and C = 4π 2 /GM . If the theory is correct, we expect a graph of log T-versus-log r to be a straight line with slope q/p = 3/2 = 1.500. The experimental measurements of actual planets yield a straight line graph whose slope is 1.500 to four significant figures. Note that the graph has nothing to do with theory—it is simply a graph of measured values. But the fact that the shape and slope of the graph agree precisely with the prediction of Newton’s theory is strong evidence for its correctness. (e) The predicted y-intercept of the graph is log C/p, and the experimentally determined value is 9.264. Equating these, we can solve for M. Because the planets all orbit the sun, the mass we are finding is M = M sun .
⎛ 4π 2 ⎞ 1 1 4π 2 1 log C = log ⎜ = 10−18.528 = 18.528 ⎟ = −9.264 ⇒ 2 2 GM sun 10 ⎝ GM sun ⎠ ⇒ M sun =
4π 2 ⋅ 1018.528 = 1.996 × 1030 kg G
The tabulated value, to three significant figures, is M sum = 1.99 × 1030 kg. We have used the orbits of the planets to “weigh the sun!”
13.54. Solve:
Visualize: Please refer to Figure P13.54. The gravitational force on one of the masses is due to the star and the other planet. Thus 2
G
Mm Gmm mv 2 m ⎛ 2π r ⎞ GM Gm 4π 2 r 2 + = = ⇒ + = ⎜ ⎟ r2 (2r ) 2 r r⎝ T ⎠ r 4r T2 12
⎡ 4π 2 r 3 ⎤ G⎛ m ⎞ 4π 2 r 2 1 T ⇒ = ⎢ ⎥ ⎜M + ⎟ = 2 r⎝ 4⎠ T ⎣ G ( M + m /4) ⎦
13.55.
Solve:
(a) Dividing the circumference of the orbit by the period,
2π Rs 2π (1.0 × 104 m) = = 6.3 × 104 m/s T 1.0 s (b) Using the formula for the acceleration at the surface, v=
gsurface =
GM s (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg) = = 1.33 × 1012 m/s 2 Rs2 (1 × 104 m) 2
(c) The mass of an object on the earth will be the same as its mass on the star. The gravitational force is
( FG )star = mg surface = 1.33 × 1012 N (d) The radius of the orbit of the satellite is r = 1 × 104 m + 1.0 × 103 m = 1.1 × 104 m. The period is
T2 =
4π 2 r 3 4π 2 (1.1 × 104 m)3 = ⇒ T = 6.29 × 10−4 s −11 GM s (6.67 × 10 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
This means there are 1589 revolutions per second or 9.5 × 104 orbits per minute. (e) Applying Equation 13.25 for a geosynchronous orbit, r3 =
GM s 2 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)(1.0 s) T = ⇒ r = 1.50 × 106 m 4π 2 4π 2
13.56. Solve:
Model: Assume the solar system is a point particle. (a) The radius of the orbit of the solar system in the galaxy is 25,000 light years. This means
r = 25,000 light years = 2500(3 × 108 )(365)(24)(3600) m = 2.36 × 1020 m T=
2π r (2π )(2.36 × 1020 m) = = 6.64 × 1015 s = 2.05 × 108 years v (2.30 × 105 m/s)
5.0 × 109 years = 24 orbits. 2.05 × 108 years (c) Applying Newton’s second law yields GM g center mss mss v 2 v 2 r (2.30 × 105 m/s) 2 (2.36 × 1020 m) = ⇒ M = = = 1.87 × 1041 kg g center r2 r G 6.67 × 10−11 N ⋅ m 2 /kg 2 (b) The number of orbits =
(d) The number of stars in the center of the galaxy is
1.87 × 1041 kg = 9.4 × 1010 1.99 × 1030 kg
13.57.
Model: Assume the two stars are spherical masses. Visualize: The gravitational force between the two stars provides the centripetal acceleration required for circular motion about the center of mass.
Solve:
Newton’s second law is 2
Fgravitation
1
⎛ GT 2 M ⎞ 3 GMM ⎛ 2π ⎞ = = MRω 2 = MR ⎜ ⎟ ⇒ R=⎜ 2 2 ⎟ (2 R ) ⎝ T ⎠ ⎝ 16π ⎠
Using T = 90 days = 90 × 24 × 3600 s and M = 2M sun = 3.98 × 1030 kg, we get R = 4.667 × 1010 m. Thus the star separation is 2 R = 9.33 × 1010 m.
13.58.
Model: Visualize:
Assume the three stars are spherical masses.
The stars rotate about the center of mass, which is the center of the triangle and equal distance r from all three stars. The gravitational force between any two stars is the same. On a given star the two forces from the other stars make an angle of 60°. Solve: The value of r can be found as follows:
L /2 L 1.0 × 1012 m = cos30° ⇒ r = = = 0.577 × 1012 m r 2cos30° 2cos30° The gravitational force between any two stars is Fg =
GM 2 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg) 2 = = 2.64 × 1026 N L2 (1.0 × 1012 m) 2
The component of this force toward the center is Fc = Fg cos30° = (2.64 × 10 26 N)cos30° = 2.29 × 1026 N
The net force on a star toward the center is twice this force, and that force equals MRω 2 . This means ⎛ 2π ⎞ 2 × 2.29 × 1026 N = MRω 2 = MR ⎜ ⎟ ⎝ T ⎠ ⇒T =
2
4π 2 MR 4π 2 (1.99 × 1030 kg)(0.577 × 1012 m) = = 3.15 × 108 s = 10 years 26 4.58 × 10 N 4.58 × 1026 N
13.59.
Model: Visualize:
Angular momentum is conserved for a particle following a trajectory of any shape.
For a particle in an elliptical orbit, the particle’s angular momentum is L = mrvt = mrv sin β , where v is the G G velocity tangent to the trajectory and β is the angle between r and v . Solve: At the distance of closest approach (rmin ) and also at the most distant point, β = 90°. Since there is no tangential force (the only force being the radial force), angular momentum must be conserved:
mPluto v1rmin = mPlutov2 rmax ⇒ v2 = v1 (rmin / rmax ) = (6.12 × 103 m/s)(4.43 × 1012 m/7.30 × 1012 m) = 3.71 km/s
13.60.
Model: Visualize:
Angular momentum is conserved for a particle following a trajectory of any shape.
For a particle in an elliptical orbit, the particle’s angular momentum is L = mrvt = mrv sin β , where v is the velocity G G tangent to the trajectory, and β is the angle between r and v . Solve: At the distance of closest approach (rmin ) and also at the most distant point, β = 90°. Since there is no tangential force (the only force being the radial force), angular momentum must be conserved:
mMercuryv1rmin = mMercury v2 rmax
⇒ rmin = rmax (v2 / v1 ) =
(6.99 × 1010 m)(38.8 km/s) = 4.60 × 1010 m 59.0 km/s
13.61. Model: Visualize:
Solve:
For the sun + comet system, the mechanical energy is conserved.
The conservation of energy equation K f + U f = K i + U i is
1 GM s M c 1 GM s M c M cv22 − = M cv12 − 2 r2 2 r1 Using G = 6.67 × 10−11 Nm 2 /kg 2 , M s = 1.99 × 1030 kg, r1 = 8.79 × 1010 m, r2 = 4.50 × 1012 m, and v1 = 54.6 km/s, we get v2 = 4.49 km/s.
13.62.
Model: Visualize:
Solve:
Model the planet (p) as a spherical mass and the spaceship (s) as a point mass.
(a) For the circular motion of the spaceship around the planet, GM p ms 2 0
r
=
GM p mv02 ⇒ v0 = r0 r0
Immediately after the rockets were fired v1 = v0 /2 and r1 = r0 . Therefore,
v1 =
1 GM p 2 r0
(b) The spaceship’s maximum distance is rmax = r0 . Its minimum distance occurs at the other end of the ellipse. The energy at the firing point is equal to the energy at the other end of the elliptical trajectory. That is,
GM p ms 1 GM p ms 1 ms v12 − = ms v22 − 2 r1 2 r2 Since the angular momentum at these two ends is conserved, we have mv1r1 = mv2 r2 ⇒ v2 = v1 (r1 / r2 ) With this expression for v2 , the energy equation simplifies to GM p 1 2 GM p 1 2 v1 − = v1 (r1 / r2 ) 2 − 2 r1 2 r2 Using r1 = r0 and v1 = v0 /2 =
1 GM p , 2 r0
1 ⎛ 1 GM p ⎞ GM p 1 ⎛ 1 GM p ⎞ r02 GM p 1 1 r 1 = ⎜ ⇒ − = 0 − ⎜ ⎟− ⎟ − 2 ⎝ 4 r0 ⎠ r0 2 ⎝ 4 r0 ⎠ r22 r2 8r0 r0 8r22 r2 ⎛ 7 ⎞ 2 7 r 1 r0 ⇒ + 02 − = 0 ⇒ ⎜ ⎟ r2 − r2 + = 0 8r0 8r2 r2 8 ⎝ 8r0 ⎠ The solutions are r2 = r0 (the initial distance) and r2 = r0 / 7. Thus the minimum distance is rmin = r0 / 7.
1 2
13.63. Solve: (a) The satellite is in a circular orbit if K = − Ug. Calculate K and U g : 2 1 1 K = mv 2 = m ( 5.5 × 103 m/s ) = 1.51 × 107 m. 2 2 ( 5.98 × 1024 kg ) m Mm U g = −G e = − ( 6.67 × 10−11 N ⋅ m 2 /kg 2 ) 1.10 × 107 m r 7 = −3.63 × 10 m
1 The quantity − U g = 1.81× 107 m ≠ K , so no, the satellite is not in a circular orbit. 2 (b) The orbit is bound if the total mechanical energy is negative. Emech = K + U g = 1.51 × 107 m − 3.63 × 107 m < 0.
Yes, the orbit is bound. Assess: Even without knowing the mass of the satellite we could answer questions about its orbit.
13.64. Visualize:
Solve:
We choose two equal time intervals tb − ta and tc − tb . A constant velocity and equal time intervals
means that xb − xa = xc − xb . The area swept from t = 0 s to t = ta is Rxa /2 and the area swept from t = 0 s to t = tb is Rxb /2. Thus, the area swept between t = ta and t = tb is R( xb − xa ) /2. In the same way, the area
swept between t = tb and t = t is R ( xc − xb )/2. Since xc − xb = xb − xa , the area swept during time (tb − ta ) is the same as the area swept during an equal time (tc − tb ). Kepler’s second law is obeyed.
13.65. Solve: (a) At what distance from the center of Saturn is the acceleration due to gravity the same as on the surface of the earth? (b)
(c) The distance is 6.21 × 107 m. This is 1.06RSaturn
13.66.
Solve:
(a) A 1000 kg satellite orbits the earth with a speed of 1997 m/s. What is the radius of the
orbit? (b)
(c) The radius of the orbit is
r=
GM E 6.67 × 10−11 N ⋅ m 2 /kg 2 (5.98 × 1024 kg) = = 1.00 × 108 m (vpayload ) 2 (1997 m/s) 2
13.67. Solve: (a) A 100 kg object is released from rest at an altitude above the moon equal to the moon’s radius. At what speed does it impact the moon’s surface? (b)
(c) The speed is v2 = 1680 m/s.
13.68.
Solve:
(a) A 2.0 × 1030 kg star and a 4.0 × 1030 kg star, each 1.0 × 109 m in diameter, are at rest
1.0 × 1012 m apart. What are their speeds as they crash together? (b)
(c) The first equation is the conservation of momentum. It can be used in the conservation of energy equation to give vf1 = 596 km/s and vf2 = −298 km/s. That is, the speeds of the two stars are 6.0 × 102 km/s and
3.0 × 102 km/s.
13.69. Solve: (a) Kepler’s third law for circular orbits is ⎛ 4π 2 ⎞ 3 4π 2 32 T2 =⎜ r ⎟r ⇒ T = GM ⎝ GM ⎠
Letting a =
4π 2 3 , the first satellite obeys T = ar 2 . For the second satellite, which orbits the same mass, GM 3
T + ΔT = a ( r + Δr ) Since
3 2
⎛ Δr ⎞ 2 = ar ⎜1 + ⎟ r ⎠ ⎝ 3 2
Δr n << 1, we can use the approximation (1 ± x ) ≈ 1 ± nx. r
Thus 3 ⎛ 3 Δr ⎞ T + ΔT ≈ ar 2 ⎜ 1 + ⎟ ⎝ 2 r ⎠
3
Subtracting the equation T = ar 2 for the first satellite from this, 3 3 ⎛ Δr ⎞ ΔT = ar 2 ⎜ ⎟ 2 ⎝ r ⎠
Dividing this by the equation for the first satellite, ΔT 3 Δr = T 2 r (b) The satellites orbit the earth. The fractional difference in their periods is
ΔT 3 1 km = = 2.24 × 10−4 T 2 6700 km After
1 = 4467 periods they will meet again. For the inner satellite, 2.24 × 10−4
T=
3 4π 2 6.700 × 106 m ) 2 ( 24 G ( 5.98 × 10 kg )
= 5456 s = 1.52 hrs So the satellites will meet again in 4467 × 1.52 hrs = 6770 hrs = 282 days. Assess: A communications satellite has an orbital period of around 1.5 h. The surprising length of time between the two satellites meeting is due to the small differences in their periods.
13.70.
Model: Model the earth and sun as spherical masses, the satellite as a point mass. Assume the satellite’s distance from the earth is very small compared to the earth’s (and satellite’s) distance to the sun. Visualize:
Solve: The net force on the satellite is the sum of the gravitational force toward the sun and the gravitational force toward the earth. This net force is responsible for circular motion around the sun. We want to chose the distance d to make the period T match the period Te with which the earth orbits the sun. The earth’s orbital period is given by T e2 = (4π 2 / GM s ) Re3 . Thus 2
Fnet =
GM s m GM e m mv 2 m ⎛ 2π r ⎞ 4π 2 GM ma mr − = = = = = mr 3 s ⎜ ⎟ centripetal 2 2 2 r d r r ⎝ Te ⎠ Te Re
Using r = Re − d and canceling the Gm term gives
Ms M M − 2e = 3s ( Re − d ) 2 ( Re − d ) d Re This equation can’t be solved exactly, but we can make use of the fact that d Re to use the binomial approximation. Factor the Re out of the expressions ( Re − d ) to get Ms M M − e = 2s (1 − d / Re ) Re2 (1 − d / Re ) 2 d 2 Re If we think of d/Re = x 1, we can simplify the first term by using (1 ± x )n ≈ 1 ± nx. Here n = −2, so we get
Ms M M M M (1 − (−2)d / Re ) − 2e = 2s (1 − d / Re ) ⇒ 2e = 3 2s Re2 d Re d Re Thus d = ( M e /3M s )1/3 Re = 1.50 × 109 m. Assess:
d/Re = 0.010, so our assumption that d/Re 1 is justified.
13.71.
Model: Model the earth as a spherical mass and the shuttle and payload as point masses. We’ll assume mpayload mshuttle .
Visualize:
Solve: (a) The payload, in steady state, is undergoing uniform circular motion. This means that the net force is directed toward the center of the earth. There is no tangential force component, since such a force would cause the payload to speed up or slow down and the motion wouldn’t be uniform. Since the net force is due to gravity and tension, and the gravitational force is radial, the tension force cannot have any tangential component. Thus the rope is radially outward at angle 0°. (b) To move with the shuttle, the payload’s period Tp in an orbit of radius rp must exactly match the period Ts
of the shuttle in an orbit of radius rs . The shuttle’s period around the earth is given by T s2 = (4π 2 /GM e ) rs3 , where we’ve used the assumption mp ms to infer that the rope’s tension will be too small to have any influence on the shuttle’s motion. Newton’s second law for the payload’s motion is ( Fnet ) r =
GM e mp rp 2
− T = mp ar =
mp v 2 rp
2
m ⎛ 2π rp ⎞ 4π 2 = p⎜ = mp rp 2 ⎟ rp ⎜⎝ Tp ⎟⎠ Tp
Using Tp = Ts and the above expression for Ts , this becomes
GM e mp r 2p ⇒T =
− T = mp rp
GM e mp ⎡ ⎛ rp ⎞ ⎢1 − ⎜ ⎟ r 2p ⎢ ⎝ rs ⎠ ⎣
3
3 GM e GM e mp r p = rs3 r p2 rs3
⎤ GM e mp ⎡ ⎛ R + 290 km ⎞3 ⎤ ⎥= ⎢1 − ⎜ e ⎟ ⎥ = 4.04 N r p2 ⎢ ⎝ Re + 300 km ⎠ ⎥ ⎥⎦ ⎣ ⎦
Assess: The fact that the tension is so small justifies our assumption that it will have no effect on the motion of the shuttle.
13.72.
Model: Model the 400 kg satellite and the 100 kg satellite as point masses and model the earth as a spherical mass. Momentum is conserved during the inelastic collision of the two satellites. Visualize:
Solve:
For the given orbit, r0 = Re + 1× 106 m = 7.37 × 106 m. The speed of a satellite in this orbit is
GM e = 7357 m/s r0
v0 =
The two satellites collide, stick together, and move with velocity v1. The equation for momentum conservation for the perfectly inelastic collision is (400 kg + 100 kg)v1 = (400 kg)(7357 m/s) − (100 kg)(7357 m/s) ⇒ v1 = 4414 m/s The new satellite’s radius immediately after the collision is still r1 = r0 = 7.37 × 106 m. Now it is moving in an elliptical orbit. We need to determine if the minimum distance r2 is larger or smaller than the earth’s radius Re = 6.37 × 106 m. The combined satellites will continue moving in an elliptical orbit. The momentum of the combined satellite is L = mrv sin β (see Equation 13.26) and is conserved in a trajectory of any shape. The angle β is 90° when v1 = 4414 m/s and when the satellite is at its closest approach to the earth. From the conservation of angular momentum, we have r1v1 = r2v2 = (7.37 × 106 )(4414 m/s) = 3.253 × 1010 m 2 /s
⇒ r2 =
3.253 × 1010 m 2 /s v2
Using the conservation of energy equation at positions 1 and 2,
GM e (500 kg) 1 GM e (500 kg) 1 = (500 kg)v22 − (500 kg)(4414 m/s) 2 − r2 2 7.37 × 106 m 2 Using the above expression for r2 , we can simplify the energy equation to v22 − (2.452 × 104 m/s)v2 + (8.876 × 107 m 2 /s 2 ) = 0 m 2 /s 2 ⇒ v2 = 20,107 m/s and 4414 m/s A velocity of 20,107 m/s for v2 yields r2 =
3.253 × 1010 m 2 /s = 1.62 × 106 m 20,107 m/s
Since r2 < Re = 6.37 × 106 m, the combined mass of the two satellites will crash into the earth.
13.73.
Model: The stars are spherical masses. They each rotate about the system’s center of mass.
Visualize:
Solve: (a) The stars rotate about the system’s center of mass with the same period: T1 = T2 = T . We can locate the center of mass by letting the origin be at the smaller-mass star. Then r1 = rcm =
(2.0 × 1030 kg)(0 m) + (6.0 × 1030 kg)(2.0 × 1012 m) = 1.5 × 1012 m 2.0 × 1030 kg + 6.0 × 1030 kg
Mass m2 undergoes uniform circular motion with radius r2 = 0.5 × 1012 m due to the gravitational force of mass m1 at distance R = 2.0 × 1012 m. The gravitational force is responsible for the centripetal acceleration, so 2
Fgrav = 1/ 2
⎡ 4π 2 ⎤ ⇒ T =⎢ r2 R 2 ⎥ ⎣ Gm1 ⎦
Gm1m2 m2v22 m2 ⎛ 2π r2 ⎞ 4π 2 m2 r2 = = = m a 2 centripetal ⎜ ⎟ = R2 r2 r2 ⎝ T ⎠ T2 1/ 2
⎡ 4π 2 (0.5 × 1012 m)(0.5 × 1012 m) 2 ⎤ =⎢ ⎥ −11 2 2 30 ⎣ (6.67 × 10 N m /kg ) (2.0 × 10 kg) ⎦
(b) The speed of each star is v = (2π r )/T . Thus
2π r1 2π (1.5 × 1012 m) = = 12.3 km/s T 7.693 × 108 s 2π r2 2π (0.5 × 1012 m) v2 = = = 4.1 km/s T 7.693 × 108 s
v1 =
= 7.693 × 108 s = 24 years
13.74. Model: The moon is a spherical mass. The moon lander is originally in a circular orbit. Visualize:
Solve: Energy and momentum are conserved between points 1 and 2 in the elliptical orbit. Also, at both points β = 90°, so L = mrv sin β = mrv. Let h = 1000 km. The original speed of the lander is
GM m v r R +h . The = 1338 m/s. Conservation of angular momentum requires mr1v1 = mr2v2 ⇒ 2 = 1 = m Rm + h v1 r2 Rm energy conservation equation is v0 =
1 M m 1 M m mv12 − G m = mv22 − G m 2 Rm + h 2 Rm ⎛ R +h⎞ Substituting v2 = ⎜ m ⎟ v1 , ⎝ Rm ⎠ 2
1 2 Mm 1 ⎛ R +h⎞ Mm v1 − G = v12 ⎜ m ⎟ −G 2 Rm + h 2 ⎝ Rm ⎠ Rm ⎛ ⎛ R +h⎞ v12 ⎜1 − ⎜ m ⎟ ⎜ Rm ⎠ ⎝ ⎝
2
⎞ ⎛ 1 1 ⎞ ⎟ = 2GM m ⎜ − ⎟ ⎟ ⎝ Rm + h Rm ⎠ ⎠
2GM m h 1 2GM m ⎛ Rm ⎞ 6 2 2 = ⎜ ⎟ = 1.392 × 10 m /s Rm ( Rm + h ) ⎛ R + h ⎞ 2 Rm + h ) ⎝ 2 Rm + h ⎠ ( m ⎜ ⎟ −1 ⎝ Rm ⎠ v1 = 1180 m/s v12 =
The fractional change in speed required to just graze the moon at point 2 is 1338 m/s − 1180 m/s = 11.8% 1338 m/s Assess:
A reduction in speed by almost 12% is reasonable.
13.75.
Model: Model the earth as a spherical mass and the satellite as a point mass. This is an isolated system, so mechanical energy is conserved. Also, the angular momentum of the satellite is conserved. Visualize: Please refer to Figure CP13.75. Solve: (a) Angular momentum is L = mrv sin β . The angle β = 90° at points 1 and 2, so conservation of angular momentum requires
⎛r ⎞ mr1v1′ = mr2v′2 ⇒ v1′ = ⎜ 2 ⎟ v′2 ⎝ r1 ⎠ The energy conservation equation is
⎛ 1 1⎞ 1 GMm 1 GMm m(v2′ ) 2 − = m(v1′ ) 2 − ⇒ (v′2 ) 2 − (v1′ ) 2 = 2GM ⎜ − ⎟ 2 r2 2 r1 ⎝ r2 r1 ⎠ Using the angular momentum result for v1′ gives 2
⎛r ⎞ ⎛r −r ⎞ ⎡ r2 − r2 ⎤ ⎛r −r ⎞ (v2′ ) 2 − (v2′ ) 2 ⎜ 2 ⎟ = 2GM ⎜ 1 2 ⎟ ⇒ (v′2 ) 2 ⎢ 1 2 2 ⎥ = 2GM ⎜ 1 2 ⎟ r r r r ⎝ 1⎠ ⎝ 12 ⎠ ⎣ 1 ⎦ ⎝ r1r2 ⎠ v′2 =
2GM ( r1 / r2 ) r1 + r2
⎛r ⎞ 2GM (r2 / r1 ) and v1′ = ⎜ 2 ⎟ v′2 = r r1 + r2 ⎝ 1⎠
(b) For the circular orbit,
v1 =
GM (6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg) = = 7730 m/s r1 (6.37 × 106 m + 3 × 105 m)
For the elliptical orbit, r1 = Re + 300 km = 6.37 × 106 m + 3 × 105 m = 6.67 × 106 m r2 = Re + 35,900 km = 6.37 × 106 m + 3.59 × 107 m = 4.23 × 107 m v1′ =
2GM ( r2 / r1 ) ⇒ v1′ = 10,160 m/s ( r1 + r2 )
1 1 (c) From the work-kinetic energy theorem, W = ΔK = mv1′2 − mv12 = 2.17 × 1010 J 2 2 (d) v2′ =
2GM (r1 / r2 ) r1 + r2
Using the same values of r1 and r2 as in (b), v′2 = 1600 m/s. For the circular orbit, v2 =
GM = 3070 m/s r2
1 1 W = mv22 − mv2′2 = 3.43 × 109 J 2 2 (f) The total work done is 2.513 × 1010 J. This is the same as in Example 13.6, but here we’ve learned how the work has to be divided between the two burns. (e)
13.76.
Model: The rod is thin and uniform.
Visualize:
Solve: (a) The rod is not spherical so must be divided into thin sections each dr wide and having mass dm. Since the rod is uniform,
dm dr M = ⇒ dm = dr M L L The width of the rod is small enough so that all of dm is distance r away from m. The gravitational potential energy of dm and m is
m dm dU = −G = −G r
⎛M ⎞ m ⎜ ⎟ dr ⎝ L⎠ r
The total potential energy of the rod and mass m is found by adding the contributions dU from every point along the rod in an integral:
GMm U = ∫ dU = − L
L⎞ ⎛ x+ ⎟ dr GMm ⎜ ∫L r = − L ln ⎜⎜ L2 ⎟⎟ x− x− 2 2⎠ ⎝
x+
L 2
Note r is increasing with the limits chosen as they are. (b) The force on m when at x is ⎛ ⎞ 1 ⎟ dU GMm d ⎛ ⎛ L⎞ L ⎞ ⎞ GMm ⎜ 1 ⎛ = − F =− ⎜ ⎟ ⎜ ln ⎜ x + ⎟ − ln ⎜ x − ⎟ ⎟ = dx L dx ⎝ ⎝ L ⎜ x+ L x− L⎟ 2⎠ 2 ⎠⎠ ⎝ ⎝ 2 2⎠ L 4 ⎛ ⎞ = −GMm ⎜ 2 , x≥ 2 ⎟ 2 ⎝ 4x − L ⎠ Assess: The direction of the force is towards the –x direction, as expected. The force magnitude approaches ∞ 1 as the mass m approaches the end of the rod, but goes to zero like 2 as x gets large. This is expected since from x far away the rod looks like a point mass.
13.77.
Model: The ring is uniform and is so thin that every point on it is the same distance r from m.
Visualize:
Solve: (a) We must determine the gravitational potential (dU) between m and an arbitrary part of the ring dm, then add using an integral all the contributions to U. Since the ring is uniform,
dm dl M dl = ⇒ dm = M 2π r 2π r The distance from m to dm is r = x 2 + R 2 . The gravitational potential between m and dm is
dU = −G
m dm mM = −G r 2π r
dl x + R2 2
The total gravitational potential is
U = ∫ dU = −
GmM
∫ dl
2π r x 2 + R 2 ring
Note that x and R do not change for any location of dm. The integral is just the length of the ring.
∫ dl = 2π R
ring
Thus U =−
GmM x2 + R2
(b) The force on m when at x is
(
)
−1 −3 dU d ⎛ 1⎞ x 2 + R 2 ) 2 = GmM ⎜ − ⎟ ( x 2 + R 2 ) 2 ( 2 x ) = GmM ( dx dx ⎝ 2⎠ x = −GmM 3 2 ( x + R2 ) 2
Fx = −
Thus the magnitude of the force is
x . ( x 2 + R 2 )3 2 Assess: The force is zero at the center of the ring. Elsewhere its direction is toward the origin. As x gets large, the force 1 decreases like 2 . This is expected since from far away the ring looks like a point mass. x F = GmM
13-1
14.1. Solve: The frequency generated by a guitar string is 440 Hz. The period is the inverse of the frequency, hence
T=
1 1 = = 2.27 × 10−3 s = 2.27 ms f 440 Hz
14.2. Model: The air-track glider oscillating on a spring is in simple harmonic motion. Solve: The glider completes 10 oscillations in 33 s, and it oscillates between the 10 cm mark and the 60 cm mark. 33 s (a) T= = 3.3 s oscillation = 3.3 s 10 oscillations 1 1 (b) f = = = 0.303 Hz ≈ 0.30 Hz T 3.3 s ω = 2π f = 2π ( 0.303 Hz ) = 1.90 rad s (c) (d) The oscillation from one side to the other is equal to 60 cm − 10 cm = 50 cm = 0.50 m. Thus, the amplitude is A = 12 ( 0.50 m ) = 0.25 m. (e) The maximum speed is
⎛ 2π ⎞ vmax = ω A = ⎜ ⎟ A = (1.90 rad s )( 0.25 m ) = 0.48 m s ⎝ T ⎠
14.3. Model: The air-track glider attached to a spring is in simple harmonic motion. Visualize: The position of the glider can be represented as x (t ) = A cos ω t. Solve: The glider is pulled to the right and released from rest at t = 0 s. It then oscillates with a period T = 2.0 s and a maximum speed vmax = 40 cm s = 0.40 m s.
2π 2π v 0.40 m s = = π rad s ⇒ A = max = = 0.127 m = 12.7 cm T 2.0 s ω π rad s (b) The glider’s position at t = 0.25 s is (a) vmax = ω A and ω =
x0.25 s = ( 0.127 m ) cos ⎡⎣(π rad s )( 0.25 s ) ⎤⎦ = 0.090 m = 9.0 cm
14.4. Model: The oscillation is the result of simple harmonic motion. Visualize: Please refer to Figure EX14.4. Solve: (a) The amplitude A = 10 cm. (b) The time to complete one cycle is the period, hence T = 2.0 s and
f = (c)
The
position
of
an
object
1 1 = = 0.50 Hz T 2.0 s
undergoing
simple
harmonic
motion
is
At t = 0 s, x0 = −5 cm, thus
−5 cm = (10 cm ) cos ⎣⎡ω ( 0 s ) + φ0 ⎦⎤ ⇒ cos φ0 =
−5 cm 1 2π ⎛ 1⎞ = − ⇒ φ0 = cos −1 ⎜ − ⎟ = ± rad or ± 120° 10 cm 2 3 ⎝ 2⎠
Since the oscillation is originally moving to the left, φ0 = +120°.
x ( t ) = A cos (ω t + φ0 ) .
14.5. Model: The oscillation is the result of simple harmonic motion. Visualize: Please refer to Figure EX14.5. Solve: (a) The amplitude A = 20 cm. (b) The period T = 4.0 s, thus f =
1 1 = = 0.25 Hz T 4.0 s
(c) The position of an object undergoing simple harmonic motion is
x ( t ) = A cos (ω t + φ0 ) . At
t = 0 s, x0 = 10 cm. Thus, 10 cm =
10 cm ⎞ π −1 ⎛ 1 ⎞ ⎟ = cos ⎜ ⎟ = ± rad = ±60° 3 ⎝ 20 cm ⎠ ⎝ 2⎠
( 20 cm ) cos φ0 ⇒ φ0 = cos −1 ⎛⎜
Because the object is moving to the right at t = 0 s, it is in the lower half of the circular motion diagram and thus must have a phase constant between π and 2π radians. Therefore, φ0 = −
π 3
rad = −60°.
14.6. Visualize: The phase constant 32 π has a plus sign, which implies that the object undergoing simple harmonic motion is in the second quadrant of the circular motion diagram. That is, the object is moving to the left. Solve: The position of the object is
x ( t ) = A cos (ω t + φ0 ) = A cos ( 2π ft + φ0 ) = ( 4.0 cm ) cos ⎡⎣( 4π rad s ) t + 32 π rad ⎤⎦ The amplitude is A = 4 cm and the period is T = 1 f = 0.50 s. A phase constant φ0 = 2π 3 rad =120° (second
quadrant) means that x starts at − 12 A and is moving to the left (getting more negative).
Assess:
We can see from the graph that the object starts out moving to the left.
14.7. Visualize: A phase constant of −
π
implies that the object that undergoes simple harmonic motion is 2 in the lower half of the circular motion diagram. That is, the object is moving to the right. Solve: The position of the object is given by the equation
⎡⎛ π ⎤ ⎞ π x ( t ) = A cos (ω t + φ0 ) = A cos ( 2π ft + φ0 ) = ( 8.0 cm ) cos ⎢⎜ rad s ⎟ t − rad ⎥ 2 2 ⎠ ⎣⎝ ⎦ The amplitude is A = 8.0 cm and the period is T = 1 f = 4.0 s. With φ0 = −π 2 rad, x starts at 0 cm and is moving to the right (getting more positive).
Assess:
As we see from the graph, the object starts out moving to the right.
14.8. Solve: The position of the object is given by the equation x ( t ) = A cos (ω t + φ0 ) = A cos ( 2π ft + φ0 ) We can find the phase constant φ0 from the initial condition: 0 cm = ( 4.0 cm ) cos φ0 ⇒ cos φ0 = 0 ⇒ φ0 = cos −1 ( 0 ) = ± 12 π rad Since the object is moving to the right, the object is in the lower half of the circular motion diagram. Hence, φ0 = − 12 π rad. The final result, with f = 4.0 Hz, is x ( t ) = ( 4.0 cm ) cos ⎡⎣( 8.0π rad s ) t − 12 π rad ⎤⎦
14.9. Solve: The position of the object is given by the equation x ( t ) = A cos (ω t + φ0 ) The amplitude A = 8.0 cm. The angular frequency ω = 2π f = 2π ( 0.50 Hz ) = π rad/s. Since at t = 0 it has its most negative velocity, it must be at the equilibrium point x = 0 cm and moving to the left, so φ0 = x ( t ) = ( 8.0 cm ) cos[(π rad/s)t +
π 2
rad]
π 2
. Thus
14.10. Model: The air-track glider is in simple harmonic motion. Solve:
(a) We can find the phase constant from the initial conditions for position and velocity:
x0 = A cos φ0
v0 x = −ω A sin φ0
Dividing the second by the first, we see that
sin φ0 v = tan φ0 = − 0 x cos φ0 ω x0 The glider starts to the left ( x0 = −5.00 cm) and is moving to the right (v0 x = +36.3 cm/s). With a period of 1.5 s =
3 2
s, the angular frequency is ω = 2π /T = 34 π rad/s. Thus ⎛
⎞ 1 36.3 cm/s 2 ⎟ = 3 π rad (60°) or – 3 π rad (–120°) π (4 /3 rad/s)( 5.00 cm) − ⎝ ⎠
φ0 = tan −1 ⎜ −
The tangent function repeats every 180°, so there are always two possible values when evaluating the arctan function. We can distinguish between them because an object with a negative position but moving to the right is in the third quadrant of the corresponding circular motion. Thus φ0 = − 23 π rad, or −120°. (b) At time t, the phase is φ = ω t + φ0 = ( 34 π rad/s) t − 32 π rad. This gives φ = − 23 π rad, 0 rad, 4 3
π rad at, respectively, t = 0 s, 0.5 s, 1.0 s, and 1.5 s. This is one period of the motion.
2 3
π rad, and
14.11. Model: The block attached to the spring is in simple harmonic motion. Solve:
The period of an object attached to a spring is
T = 2π
m = T0 = 2.0 s k
where m is the mass and k is the spring constant. (a) For mass = 2m, T = 2π
(b) For mass
1 2
2m = k
( 2 )T
0
= 2.8 s
m, T = 2π
1 2
m = T0 k
2 = 1.41 s
(c) The period is independent of amplitude. Thus T = T0 = 2.0 s (d) For a spring constant = 2k ,
T = 2π
m = T0 2k
2 = 1.41 s
14.12. Model: The air-track glider attached to a spring is in simple harmonic motion. Solve:
Experimentally, the period is T = (12.0 s ) (10 oscillations ) = 1.20 s. Using the formula for the period, 2
T = 2π
2
m ⎛ 2π ⎞ ⎛ 2π ⎞ ⇒k =⎜ ⎟ m=⎜ ⎟ ( 0.200 kg ) = 5.48 N m k ⎝ T ⎠ ⎝ 1.20 s ⎠
14.13. Model: The mass attached to the spring oscillates in simple harmonic motion. Solve: (a) The period T = 1 f = 1 2.0 Hz = 0.50 s. (b) The angular frequency ω = 2π f = 2π (2.0 Hz) = 4π rad/s. (c) Using energy conservation 1 2
kA2 = 12 kx02 + 12 mv02x
Using x0 = 5.0 cm, v0 x = −30 cm/s and k = mω 2 = (0.200 kg)(4π rad/s) 2 , we get A = 5.54 cm. (d) To calculate the phase constant φ0 ,
A cos φ0 = x0 = 5.0 cm ⎛ 5.0 cm ⎞ ⇒ φ0 = cos −1 ⎜ ⎟ = 0.45 rad ⎝ 5.54 cm ⎠ (e) The maximum speed is vmax = ω A = ( 4π rad s )( 5.54 cm ) = 70 cm s. (f) The maximum acceleration is
amax = ω 2 A = ω (ω A ) = ( 4π rad s )( 70 cm s ) = 8.8 m s 2 2 (g) The total energy is E = 12 mvmax =
1 2
( 0.200 kg )( 0.70 m s )
2
= 0.049 J.
(h) The position at t = 0.40 s is
x0.4 s = ( 5.54 cm ) cos ⎡⎣( 4π rad s )( 0.40 s ) + 0.45 rad ⎤⎦ = +3.8 cm
14.14. Model: The oscillating mass is in simple harmonic motion. Solve: (a) The amplitude A = 2.0 cm. (b) The period is calculated as follows:
ω=
2π 2π = 10 rad s ⇒ T = = 0.63 s T 10 rad s
(c) The spring constant is calculated as follows:
ω=
k 2 ⇒ k = mω 2 = ( 0.050 kg )(10 rad s ) = 5.0 N m m
(d) The phase constant φ0 = − 14 π rad. (e) The initial conditions are obtained from the equations
x ( t ) = ( 2.0 cm ) cos (10t − 14 π ) and vx ( t ) = − ( 20.0 cm s ) sin (10t − 14 π ) At t = 0 s, these equations become
x0 = ( 2.0 cm ) cos ( − 14 π ) = 1.41 cm and v0 x = − ( 20 cm s ) sin ( − 14 π ) = 14.1 cm s In other words, the mass is at +1.41 cm and moving to the right with a velocity of 14.1 cm/s. (f) The maximum speed is vmax = Aω = ( 2.0 cm )(10 rad s ) = 20 cm s. (g) The total energy E = 12 kA2 = 12 ( 5.0 N m )( 0.020 m ) = 1.00 × 10−3 J. 2
(h) At t = 0.41 s, the velocity is
v0 x = − ( 20 cm s ) sin ⎡⎣(10 rad s )( 0.40 s ) − 14 π ⎤⎦ = 1.46 cm s
14.15. Model: The block attached to the spring is in simple harmonic motion. Visualize:
Solve:
(a) The conservation of mechanical energy equation K f + U sf = K i + U si is 1 2
mv12 + 12 k ( Δx ) = 12 mv02 + 0 J ⇒ 0 J + 12 kA2 = 12 mv02 + 0 J
⇒ A=
2
m 1.0 kg v0 = ( 0.40 m s ) = 0.10 m = 10.0 cm k 16 N m
(b) We have to find the velocity at a point where x = A/2. The conservation of mechanical energy equation K 2 + U s2 = K i + U si is 2
1 2 1 ⎛ A⎞ 1 2 1 1 1⎛1 1⎛ 1 ⎞ 1 ⎞ mv2 + k ⎜ ⎟ = mv0 + 0 J ⇒ mv22 = mv02 − ⎜ kA2 ⎟ = mv02 − ⎜ mv02 ⎟ = 2 2 ⎝2⎠ 2 2 2 4⎝ 2 2 4 2 ⎠ ⎝ ⎠ ⇒ v2 = The velocity is 35 cm/s.
3 3 v0 = ( 0.40 m s ) = 0.346 m s 4 4
3⎛ 1 2⎞ ⎜ mv0 ⎟ 4⎝ 2 ⎠
14.16. Model: The vertical oscillations constitute simple harmonic motion. Visualize:
Solve:
(a) At equilibrium, Newton’s first law applied to the physics book is
(F )
sp y
− mg = 0 N ⇒ −k Δy − mg = 0 N
⇒ k = − mg Δy = − ( 0.500 kg ) ( 9.8 m s 2 ) ( −0.20 m ) = 24.5 N m (b) To calculate the period:
ω=
k 24.5 N m 2π 2π rad = = 7.0 rad s and T = = = 0.90 s m 0.500 kg ω 7.0 rad s
(c) The maximum speed is
vmax = Aω = ( 0.10 m )( 7.0 rad s ) = 0.70 m s Maximum speed occurs as the book passes through the equilibrium position.
14.17. Model: The vertical oscillations constitute simple harmonic motion. Solve: To find the oscillation frequency using ω = 2π f = k m , we first need to find the spring constant k. In equilibrium, the weight mg of the block and the spring force k ΔL are equal and opposite. That is, mg = k ΔL ⇒ k = mg ΔL . The frequency of oscillation f is thus given as f =
1 2π
k 1 = m 2π
mg ΔL 1 = m 2π
g 1 = ΔL 2π
9.8 m s 2 = 3.5 Hz 0.020 m
14.18. Model: The vertical oscillations constitute simple harmonic motion. Visualize:
Solve:
The period and angular frequency are
T=
20 s 2π 2π = 0.6667 s and ω = = = 9.425 rad s 30 oscillations T 0.6667 s
(a) The mass can be found as follows:
ω=
k k 15 N/m ⇒m= 2 = = 0.169 kg 2 ω m ( 9.425 rad s )
(b) The maximum speed vmax = ω A = ( 9.425 rad s )( 0.060 m ) = 0.57 m/s.
14.19. Model: Assume a small angle of oscillation so there is simple harmonic motion. Solve:
The period of the pendulum is
T0 = 2π
L0 = 4.0 s g
(a) The period is independent of the mass and depends only on the length. Thus T = T0 = 4.0 s. (b) For a new length L = 2 L0 ,
T = 2π
2 L0 = 2T0 = 5.7 s g
(c) For a new length L = L0 /2,
T = 2π
L0 2 1 = T0 = 2.8 s g 2
(d) The period is independent of the amplitude as long as there is simple harmonic motion. Thus T = 4.0 s.
14.20. Model: The pendulum undergoes simple harmonic motion. Solve: (a) The amplitude is 0.10 rad. (b) The frequency of oscillations is f =
ω 5 = Hz = 0.796 Hz 2π 2π
(c) The phase constant φ = π rad. (d) The length can be obtained from the period: 2
2
ω = 2π f =
⎛ ⎞ ⎛ 1 ⎞ g 1 2 ⇒ L=⎜ ⎟⎟ ( 9.8 m s ) = 0.392 m ⎟ g = ⎜⎜ π L 2 0.796 Hz ( ) ⎝ 2π f ⎠ ⎝ ⎠
(e) At t = 0 s, θ 0 = ( 0.10 rad ) cos (π ) = −0.10 rad. To find the initial condition for the angular velocity we take
the derivative of the angular position:
θ ( t ) = ( 0.10 rad ) cos ( 5t + π ) ⇒ At t = 0 s,
( dθ
dt )0 = ( −0.50 rad ) sin (π ) = 0 rad s.
dθ ( t ) = − ( 0.10 rad )( 5 ) sin ( 5t + π ) dt
(f) At t = 2.0 s, θ 2.0 = ( 0.10 rad ) cos ( 5 ( 2.0 s ) + π ) = 0.084 rad.
14.21. Model: Assume the small-angle approximation so there is simple harmonic motion. Solve:
The period is T = 12 s 10 oscillations = 1.20 s and is given by the formula 2
T = 2π
2
L ⎛ T ⎞ ⎛ 1.20 s ⎞ 2 ⇒L=⎜ ⎟ g =⎜ ⎟ ( 9.8 m s ) = 36 cm g ⎝ 2π ⎠ ⎝ 2π ⎠
14.22. Model: Assume a small angle of oscillation so there is simple harmonic motion. Solve:
(a) On the earth the period is
Tearth = 2π
1.0 m L = 2π = 2.0 s 9.80 m s 2 g
(b) On Venus the acceleration due to gravity is
g Venus =
GM Venus ( 6.67 × 10 = 2 RVenus ⇒ TVenus = 2π
−11
N ⋅ m 2 kg 2 )( 4.88 × 1024 kg )
( 6.06 ×106 m )
2
L 1.0 m = 2π = 2.1 s g Venus 8.86 m s 2
= 8.86 m s 2
14.23. Model: Assume the pendulum to have small-angle oscillations. In this case, the pendulum undergoes simple harmonic motion. Solve: Using the formula g = GM R 2 , the periods of the pendulums on the moon and on the earth are
Tearth = 2π
L L R2 L R2 = 2π earth earth and Tmoon = 2π moon moon g GM earth GM moon
Because Tearth = Tmoon , 2
2π
2 ⎛M ⎞⎛ R ⎞ Learth Rearth L R2 = 2π moon moon ⇒ Lmoon = ⎜ moon ⎟⎜ earth ⎟ Learth GM earth GM moon ⎝ M earth ⎠⎝ Rmoon ⎠ 2
⎛ 7.36 × 1022 kg ⎞⎛ 6.37 × 106 m ⎞ =⎜ ⎟⎜ ⎟ ( 2.0 m ) = 33 cm 24 6 ⎝ 5.98 × 10 kg ⎠⎝ 1.74 × 10 m ⎠
14.24. Model: Assume a small angle of oscillation so that the pendulum has simple harmonic motion. Solve:
The time periods of the pendulums on the earth and on Mars are
Tearth = 2π
L g earth
and TMars = 2π
L g Mars
Dividing these two equations, Tearth = TMars
2
2
⎛T ⎞ g Mars ⎛ 1.50 s ⎞ 2 ⇒ g Mars = g earth ⎜ earth ⎟ = ( 9.8 m s 2 ) ⎜ ⎟ = 3.67 m s g earth ⎝ 2.45 s ⎠ ⎝ TMars ⎠
14.25. Visualize: Please refer to Figure Ex14.25. Solve: The mass of the wrench can be obtained from the length that it stretches the spring. From Equation 14.41,
ΔL =
mg k ΔL 360 N/m ( 0.030 m ) ⇒m= = = 1.10 kg k g 9.8 m/s 2
When swinging on a hook the wrench is a physical pendulum. From Equation 14.52, 2π f =
mgl 2π mgl ⇒ = I T I
From the figure, l = 0.14 m. Thus 2
2
⎛ T ⎞ ⎛ 0.90 s ⎞ −2 2 2 I =⎜ ⎟ mgl = ⎜ ⎟ (1.10 kg ) ( 9.8 m/s ) ( 0.14 m ) = 3.1 × 10 kg m ⎝ 2π ⎠ ⎝ 2π ⎠
14.26. Model: The spider is in simple harmonic motion. Solve:
Your tapping is a driving frequency. Largest amplitude at f ext = 1.0 Hz means that this is the resonance
frequency, so f 0 = f ext = 1.0 Hz. That is, the spider’s natural frequency of oscillation f 0 is 1.0 Hz and
ω 0 = 2π f 0 = 2π rad s. We have ω0 =
k 2 ⇒ k = mω 02 = ( 0.0020 kg )( 2π rad s ) = 0.079 N m m
14.27. Model: The motion is a damped oscillation. Solve:
The amplitude of the oscillation at time t is given by Equation 14.58: A ( t ) = A0e− t 2τ , where τ = m b is
the time constant. Using x = 0.368 A and t = 10.0 s, we get
0.368 A = Ae−10.0 s 2τ ⇒ ln ( 0.368 ) =
−10 s 10.0 s ⇒τ = − = 5.00 s 2τ 2ln ( 0.368 )
14.28. Model: The motion is a damped oscillation. The position of a damped oscillator is x ( t ) = Ae− ( t 2τ ) cos (ω t + φ0 ) . The frequency is 1.0 Hz and the
Solve:
damping time constant τ is 4.0 s. Let us assume φ0 = 0 rad and A = 1 with arbitrary units. Thus, x ( t ) = e − t (8.0 s ) cos ⎣⎡ 2π (1.0 Hz ) t ⎦⎤ ⇒ x ( t ) = e −0.125 t cos ( 2π t ) where t is in s. Values of x(t) at selected values of t are displayed in the following table: t(s)
x(t)
t(s)
x(t)
t(s)
x(t)
0
1
2.00
0.779
6.00
0.472
0.25
0
2.50
−0.732
6.50
− 0.444
0.50
−0.939
3.00
0.687
7.00
0.417
0.75
0
3.50
−0.646
7.50
− 0.392
1.00
0.882
4.00
0.607
8.00
0.368
1.25
0
4.50
−0.570
8.50
− 0.346
1.50
−0.829
5.00
0.535
9.00
0.325
1.75
0
5.50
−0.503
9.50
− 0.305
10.00
0.286
14.29. Model: The pendulum is a damped oscillator. Solve:
The period of the pendulum and the number of oscillations in 4 hours are calculated as follows:
T = 2π
4 ( 3600 s ) 15.0 m L = 2π = 7.773 s ⇒ N osc = = 1853 2 9.8 m s 7.773 s g
The amplitude of the pendulum as a function of time is A ( t ) = Ae − bt / 2 m . The exponent of this expression can be calculated to be −
bt ( 0.010 kg s )( 4 × 3600 s ) = −0.6545 =− 2m 2 (110 kg )
We have A ( t ) = (1.50 m ) e −0.6545 = 0.780 m.
14.30. Model: The vertical oscillations are damped and follow simple harmonic motion. Solve: The position of the ball is given by x ( t ) = Ae − ( t 2τ ) cos (ω t + φ0 ) . The amplitude A ( t ) = Ae − ( t 2τ ) is a function of time. The angular frequency is
ω=
k = m
(15.0
N m) 2π = 5.477 rad s ⇒ T = = 1.147 s 0.500 kg ω
Because the ball’s amplitude decreases to 3.0 cm from 6.0 cm after 30 oscillations, that is, after 30 × 1.147 s = 34.41 s, we have 3.0 cm = ( 6.0 cm ) e − (34.414 s 2τ ) ⇒ 0.50 = e − ( 34.41 s 2τ ) ⇒ ln ( 0.50 ) =
−34.41 s ⇒ τ = 25 s 2τ
14.31. Visualize: Please refer to Figure P14.31. Solve:
The position and the velocity of a particle in simple harmonic motion are
x ( t ) = A cos (ω t + φ0 ) and vx ( t ) = − Aω sin (ω t + φ0 ) = −vmax sin (ω t + φ0 ) (a) At t = 0 s, the equation for x yields
( −5.0 cm ) = (10.0 cm ) cos (φ0 ) ⇒ φ0 = cos−1 ( −0.5) = ± 23 π
rad
Because the particle is moving to the left at t = 0 s, it is in the upper half of the circular motion diagram, and the phase constant is between 0 and π radians. Thus, φ0 = 23 π rad. (b) The period is 4.0 s. At t = 0 s,
⎛ 2π ⎞ ⎛ 2π ⎞ v0 x = − Aω sin φ0 = − (10.0 cm ) ⎜ ⎟ sin ⎜ ⎟ = −13.6 cm/s ⎝ T ⎠ ⎝ 3 ⎠ (c) The maximum speed is
⎛ 2π ⎞ vmax = ω A = ⎜ ⎟ (10.0 cm ) = 15.7 cm s ⎝ 4.0 s ⎠ Assess: The negative velocity at t = 0 s is consistent with the position-versus-time graph and the positive sign of the phase constant.
14.32. Visualize: Please refer to Figure P14.32. Solve:
The position and the velocity of a particle in simple harmonic motion are
x ( t ) = A cos (ω t + φ0 ) and vx ( t ) = − Aω sin (ω t + φ0 ) = −vmax sin (ω t + φ0 ) From the graph, T = 12 s and the angular frequency is
ω=
2π 2π π = = rad s T 12 s 6
(a) Because vmax = Aω = 60 cm s, we have
A=
60 cm s
ω
=
60 cm s = 115 cm π 6 rad s
(b) At t = 0 s, v0x = − Aω sin φ0 = −30 cm/s ⇒ − ( 60 cm s ) sin φ0 = −30 cm/s ⇒ φ0 = sin −1 ( 0.5 rad ) = 16 π rad ( 30° ) or 56 π rad (150° )
Because the velocity at t = 0 s is negative and the particle is slowing down, the particle is in the second quadrant of the circular motion diagram. Thus φ0 = 56 π rad.
(c) At t = 0 s, x0 = (115 cm ) cos ( 56 π rad ) = −100 cm.
14.33. Model: The vertical mass/spring systems are in simple harmonic motion. Visualize: Please refer to Figure P14.33. Solve: (a) For system A, the maximum speed while traveling in the upward direction corresponds to the maximum positive slope, which is at t = 3.0 s. The frequency of oscillation is 0.25 Hz. (b) For system B, all the energy is potential energy when the position is at maximum amplitude, which for the first time is at t = 1.5 s. The time period of system B is thus 6.0 s. (c) Spring/mass A undergoes three oscillations in 12 s, giving it a period TA = 4.0 s. Spring/mass B undergoes 2
oscillations in 12 s, giving it a period TB = 6.0 s. We have
TA = 2π
⎛ m ⎞⎛ k ⎞ 4.0 s 2 mA mB T and TB = 2π ⇒ A = ⎜ A ⎟⎜ B ⎟ = = kA kB TB ⎝ mB ⎠⎝ kA ⎠ 6.0 s 3
If mA = mB , then
kB 4 k 9 = ⇒ A = = 2.25 kA 9 kB 4
14.34. Solve: The object’s position as a function of time is x ( t ) = A cos (ω t + φ0 ) . Letting x = 0 m at t = 0 s, gives 0 = A cos φ0 ⇒ φ0 = ± 12 π Since the object is traveling to the right, it is in the lower half of the circular motion diagram, giving a phase constant between −π and 0 radians. Thus, φ0 = − 12 π and x ( t ) = A cos (ω t − 12 π ) ⇒ x ( t ) = A sin ω t = ( 0.10 m ) sin ( 12 π t )
where we have used A = 0.10 m and
ω=
2π 2π rad π = = rad s T 4.0 s 2
Let us now find t where x = 0.60 m:
2 ⎛π ⎞ ⎛ 0.060 m ⎞ 0.060 m = ( 0.10 m ) sin ⎜ t ⎟ ⇒ t = sin −1 ⎜ ⎟ = 0.41 s π ⎝2 ⎠ ⎝ 0.10 m ⎠ Assess:
The answer is reasonable because it is approximately
1 8
of the period.
14.35. Model: The block attached to the spring is in simple harmonic motion. Visualize:
The position and the velocity of the block are given by the equations
x ( t ) = A cos (ω t + φ0 ) and vx ( t ) = − Aω sin (ω t + φ0 ) Solve:
To graph x(t) we need to determine ω , φ0 , and A. These quantities will be found by using the initial
(t = 0 s) conditions on x(t) and vx (t ). The period is
T = 2π
1.0 kg 2π 2π rad m = 2π = 1.405 s ⇒ ω = = = 4.472 rad s 20 N m k T 1.405 s
At t = 0 s, x0 = A cos φ0 and v0 x = − Aω sin φ0 . Dividing these equations, tan φ0 = −
v0 x ( −1.0 m s ) =− = 1.1181 ⇒ φ0 = 0.841 rad ω x0 ( 4.472 rad s )( 0.20 m )
From the initial conditions, 2
⎛v ⎞ A = x + ⎜ 0x ⎟ = ⎝ω ⎠ 2 0
2
⎛ −1.0 m s ⎞ ( 0.20 m ) + ⎜ ⎟ = 0.300 m ⎝ 4.472 rad s ⎠ 2
The position-versus-time graph can now be plotted using the equation x ( t ) = ( 0.300 m ) cos ⎡⎣( 4.472 rad s ) t + 0.841 rad ⎤⎦
14.36. Model: The astronaut attached to the spring is in simple harmonic motion. Visualize: Please refer to Figure P14.36. Solve: (a) From the graph, T = 3.0 s, so we have 2
T = 2π
2
m ⎛ T ⎞ ⎛ 3.0 s ⎞ ⇒m=⎜ ⎟ k =⎜ ⎟ ( 240 N m ) = 55 kg k ⎝ 2π ⎠ ⎝ 2π ⎠
(b) Oscillations occur about an equilibrium position of 1.0 m. From the graph, A =
1 2
( 0.80 m ) = 0.40 m, φ0 =
0 rad, and
ω=
2π 2π = = 2.1 rad s T 3.0 s
The equation for the position of the astronaut is
x ( t ) = A cos ω t + 1.0 m = ( 0.4 m ) cos ⎡⎣( 2.1 rad s ) t ⎤⎦ + 1.0 m ⇒ 1.2 m= ( 0.4 m ) cos ⎡⎣( 2.1 rad s ) t ⎤⎦ + 1.0 m ⇒ cos ⎡⎣( 2.1 rad s ) t ⎤⎦ = 0.5 ⇒ t = 0.50 s The equation for the velocity of the astronaut is vx ( t ) = − Aω sin (ω t ) ⇒ v0.5 s = − ( 0.4 m )( 2.1 rad s ) sin ⎡⎣( 2.1 rad s )( 0.50 s ) ⎤⎦ = −0.73 m s Thus her speed is 0.73 m/s.
14.37. Model: The particle is in simple harmonic motion. Solve:
The equation for the velocity of the particle is
vx ( t ) = − ( 25 cm )(10 rad s ) sin (10 t ) Substituting into K = 2U gives 2 2 1 2 ⎛1 ⎞ 1 mvx ( t ) = 2 ⎜ kx 2 ( t ) ⎟ ⇒ m ⎣⎡ − ( 250 cm s ) sin (10 t ) ⎦⎤ = k ⎣⎡( 25 cm ) cos (10 t ) ⎦⎤ 2 2 2 ⎝ ⎠
sin 2 (10 t ) ⎛ k ⎞ ( 25 cm ) ⎛ 1 ⎞ 2 = 2⎜ ⎟ = 2ω 2 ⎜ ⎟s 2 cos 2 (10 t ) ⎝ m ⎠ ( 250 cm s ) ⎝ 100 ⎠ 2
⇒
1 2⎛ 1 ⎞ 2 −1 ⇒ tan 2 (10 t ) = 2 (10 rad s ) ⎜ 2.0 = 0.096 s ⎟ s = 2.0 ⇒ t = tan 10 ⎝ 100 ⎠
14.38. Model: The spring undergoes simple harmonic motion. Solve:
(a) Total energy is E = 12 kA2 . When the displacement is x = 12 A, the potential energy is
U = 12 kx 2 = 12 k ( 12 A) 2 =
1 4
(
1 2
kA2 ) = 14 E ⇒ K = E − U = 34 E
One quarter of the energy is potential and three-quarters is kinetic. (b) To have U = 12 E requires U = 12 kx 2 = 12 E = 12 ( 12 kA2 ) ⇒ x =
A 2
14.39. Solve: Average speed is vavg = Δx/Δt. During half a period (Δt = 12 T ), the particle moves from x = − A to x = + A( Δx = 2 A). Thus vavg =
Δx 2 A 4 A 4A 2 2 π = = = = (ω A) = vmax ⇒ vmax = vavg Δt T /2 T 2π /ω π π 2
14.40. Model: The ball attached to a spring is in simple harmonic motion. Solve:
(a) Let t = 0 s be the instant when x0 = −5.0 cm and v0 = 20 cm/s. The oscillation frequency is
k 2.5 N m = = 5.0 rad / s m 0.100 kg
ω=
Using Equation 14.27, the amplitude of the oscillation is 2
⎛v ⎞ A = x02 + ⎜ 0 ⎟ = ⎝ω ⎠
2
⎛ 20 cm / s ⎞ 2 ( −5.0 cm ) + ⎜ ⎟ = 6.4 cm ⎝ 5.0 rad / s ⎠
(b) The maximum acceleration is amax = ω 2 A = 160 cm s 2 .
(c) For an oscillator, the acceleration is most positive ( a = amax ) when the displacement is most negative
( x = − xmax = − A) .
So the acceleration is maximum when x = −6.4 cm.
(d) We can use the conservation of energy between x0 = −5.0 cm and x1 = 3.0 cm: 1 2
mv02 + 12 kx02 = 12 mv12 + 12 kx12 ⇒ v1 = v02 +
k 2 ( x0 − x12 ) = 0.283 m s m
The speed is 28 cm/s. Because k is known in SI units of N/m, the energy calculation must be done using SI units of m, m/s, and kg.
14.41. Model: The block on a spring is in simple harmonic motion. Solve:
(a) The position of the block is given by x ( t ) = A cos (ω t + φ0 ) . Because x ( t ) = A at t = 0 s, we have
φ0 = 0 rad, and the position equation becomes x ( t ) = A cos ω t. At t = 0.685 s, 3.00 cm = A cos ( 0.685ω ) and at t = 0.886 s, −3.00 cm = A cos ( 0.886ω ) . These two equations give
cos ( 0.685ω ) = − cos ( 0.886ω ) = cos (π − 0.886ω ) ⇒ 0.685ω = π − 0.886ω ⇒ ω = 2.00 rad s (b) Substituting into the position equation,
3.00 cm = A cos ( ( 2.00 rad s )( 0.685 s ) ) = A cos (1.370 ) = 0.20 A ⇒ A =
3.00 cm = 15.0 cm 0.20
14.42. Model: The oscillator is in simple harmonic motion. Energy is conserved. Solve:
The energy conservation equation E1 = E2 is 1 2
mv12 + 12 kx12 = 12 mv22 + 12 kx22
1 1 1 1 2 2 2 2 ( 0.30 kg )( 0.954 m s ) + k ( 0.030 m ) = ( 0.30 kg )( 0.714 m s ) + k ( 0.060 m ) 2 2 2 2 ⇒ k = 44.48 N m The total energy of the oscillator is 1 1 1 1 2 2 Etotal = mv12 + kx12 = ( 0.30 kg )( 0.954 m s ) + ( 44.48 N m )( 0.030 m ) = 0.1565 J 2 2 2 2 2 , Because Etotal = 12 mvmax
0.1565 J = Assess:
1 2 ⇒ v max = 1.02 m s ( 0.300 kg ) vmax 2
A maximum speed of 1.02 m/s is reasonable.
14.43. Model: The transducer undergoes simple harmonic motion. Solve:
Newton’s second law for the transducer is
Frestoring = ma max ⇒ 40,000 N = ( 0.10 × 10−3 kg ) amax ⇒ amax = 4.0 × 108 m s 2 Because amax = ω 2 A,
A=
amax
ω
2
=
4.0 × 108 m s 2 ⎡ 2π (1.0 × 10 Hz ) ⎤ ⎣ ⎦ 6
2
= 1.01 × 10−5 m = 10.1 μ m
(b) The maximum velocity is
vmax = ω A = 2π (1.0 × 106 Hz )(1.01 × 10−5 m ) = 64 m s
14.44. Model: The block attached to the spring is in simple harmonic motion. Solve:
(a) The frequency is
f =
1 2π
1 k = m 2π
2000 N m = 3.183 Hz 5.0 kg
The frequency is 3.2 Hz. (b) From energy conservation, 2
2
⎛v ⎞ ⎛ 1.0 m/s ⎞ A = x02 + ⎜ 0 ⎟ = (0.050 m) 2 + ⎜ ⎟ = 0.0707 m ω ⎝ ⎠ ⎝ 2π ⋅ 3.183 Hz ⎠ The amplitude is 7.1 cm. (c) The total mechanical energy is 2 E = 12 kA2 = 12 ( 2000 N m )( 0.0707 m ) = 5.0 J
14.45. Model: The tips of the tuning fork are in simple harmonic oscillation. Solve:
(a) The maximum speed is related to the amplitude.
vmax = ω A = 2π fA = 2π ( 440 Hz ) ( 5.0 × 10−4 m ) = 1.38 m/s (b) The acceleration of the flea can not be greater than that allowed by the maximum force with which it can hold on. From Newton’s second law, the maximum acceleration that the flea can withstand is
aflea =
F 1.0 × 10−3 N = = 100 m/s 2 m 10 × 10−6 kg
The maximum acceleration at the tip of the prong is amax = ω 2 A = ( 2π f ) A = ( 2π ( 440 Hz ) ) ( 5.0 × 10−4 m ) = 382 m/s 2 2
The flea will not be able to hold on to the tuning fork.
2
14.46. Model: The block undergoes simple harmonic motion. Visualize:
Solve:
(a) The frequency of oscillation is
f = The frequency is 1.13 Hz. (b) Using conservation of energy,
1 2
1 2π
k 1 = m 2π
10 N/m = 1.125 Hz 0.20 kg
mv12 + 12 kx12 = 12 mv02 + 12 kx02 , we find
m 2 2 0.20 kg (v0 − v1 ) = (−0.20 m) 2 + ((1.00 m/s) 2 − (0.50 m/s) 2 ) k 10 N/m = 0.2345 m or 23 cm
x1 = x02 +
(c) At time t, the displacement is x = A cos(ω t + φ0 ). The angular frequency is ω = 2π f = 7.071 rad/s. The amplitude is 2
2
⎛v ⎞ ⎛ 1.00 m/s ⎞ A = x02 + ⎜ 0 ⎟ = (−0.20 m) 2 + ⎜ ⎟ = 0.245 m ω ⎝ ⎠ ⎝ 7.071 rad/s ⎠ The phase constant is ⎛ x0 ⎞ −1 ⎛ −0.200 m ⎞ ⎟ = cos ⎜ ⎟ = ±2.526 rad or ± 145° ⎝ A⎠ ⎝ 0.245 m ⎠
φ0 = cos −1 ⎜
A negative displacement (below the equilibrium point) and positive velocity (upward motion) indicate that the corresponding circular motion is in the third quadrant, so φ0 = −2.526 rad. Thus at t = 1.0 s,
x = (0.245 m)cos ( (7.071 rad/s)(1.0 s) − 2.526 rad ) = −0.0409 m = −4.09 cm The block is 4.1 cm below the equilibrium point.
14.47. Model: The mass is in simple harmonic motion. Visualize:
The high point of the oscillation is at the point of release. This conclusion is based on energy conservation. Gravitational potential energy is converted to the spring’s elastic potential energy as the mass falls and stretches the spring, then the elastic potential energy is converted 100% back into gravitational potential energy as the mass rises, bringing the mass back to exactly its starting height. The total displacement of the oscillation—high point to low point—is 20 cm. Because the oscillations are symmetrical about the equilibrium point, we can deduce that the equilibrium point of the spring is 10 cm below the point where the mass is released. The mass oscillates about this equilibrium point with an amplitude of 10 cm, that is, the mass oscillates between 10 cm above and 10 cm below the equilibrium point. Solve: The equilibrium point is the point where the mass would hang at rest, with Fsp = FG = mg . At the
equilibrium point, the spring is stretched by Δy = 10 cm = 0.10 m. Hooke’s law is Fsp = k Δy, so the equilibrium condition is k g 9.8 m s 2 ⎡⎣ Fsp = k Δy ⎤⎦ = [ FG = mg ] ⇒ = = = 98 s −2 m Δy 0.10 m
The ratio k m is all we need to find the oscillation frequency:
f =
1 2π
k 1 98 s −2 = 1.58 Hz = m 2π
14.48. Model: The spring is ideal, so the apples undergo SHM. Solve: The spring constant of the scale can be found by considering how far the pan goes down when the apples are added. ΔL =
mg mg 20 N ⇒k= = = 222 N/m k ΔL 0.090 m
The frequency of oscillation is
f =
1 2π
1 k = m 2π
222 N/m = 1.66 Hz ( 20 N 9.8 m/s 2 )
Assess: An oscillation of fewer than twice per second is reasonable.
14.49. Model: The compact car is in simple harmonic motion. Solve:
(a) The mass on each spring is (1200 kg ) 4 = 300 kg. The spring constant can be calculated as follows:
ω2 =
2 k 2 ⇒ k = mω 2 = m ( 2π f ) = ( 300 kg ) ⎡⎣ 2π ( 2.0 Hz ) ⎤⎦ = 4.74 × 104 N m m
The spring constant is
4.7 × 104 N/m. (b) The car carrying four persons means that each spring has, on the average, an additional mass of 70 kg. That is, m = 300 kg + 70 kg = 370 kg. Thus,
f =
ω 1 = 2π 2π
1 k = m 2π
4.74 × 104 N m = 1.80 Hz 370 kg
Assess: A small frequency change from the additional mass is reasonable because frequency is inversely proportional to the square root of the mass.
14.50. Model: Hooke’s law for the spring. The spring’s compression and decompression constitutes simple harmonic motion. Visualize:
Solve: (a) The spring’s compression or decompression is one-half of the oscillation cycle. This means the contact time is Δt = 12 T , where T is the period. The period is calculated as follows:
ω=
k 50 N m 1 2π 2π = = 10 rad s ⇒ T = = = = 0.628 s m 0.500 kg f ω 10 rad s ⇒ Δt =
T = 0.31 s 2
(b) There is no change in contact time, because period of oscillation is independent of the amplitude or the maximum speed.
14.51. Model: The two blocks are in simple harmonic motion, without the upper block slipping. We will also apply the model of static friction between the two blocks. Visualize:
Solve: The net force acting on the upper block m1 is the force of friction due to the lower block m2 . The model of static friction gives the maximum force of static friction as
( fs )max = μs n = μs ( m1g ) = m1amax ⇒ amax = μs g Using μ s = 0.5, amax = μS g = ( 0.5) ( 9.8 m s 2 ) = 4.9 m s 2 . That is, the two blocks will ride together if the maximum acceleration of the system is equal to or less than amax . We can calculate the maximum value of A as follows: amax = ω 2 Amax =
2 a ( m + m2 ) ( 4.9 m s ) (1.0 kg + 5.0 kg ) k Amax ⇒ Amax = max 1 = = 0.59 m m1 + m2 k 50 N m
14.52. Model: Assume simple harmonic motion for the two-block system without the upper block slipping. We will also use the model of static friction between the two blocks. Visualize:
Solve: The net force on the upper block m1 is the force of static friction due to the lower block m2 . The two blocks ride together as long as the static friction doesn’t exceed its maximum possible value. The model of static friction gives the maximum force of static friction as
( fs )max = μs n = μs ( m1g ) = m1amax ⇒ amax = μs g 2
⇒ μs =
2
amax ω 2 Amax ⎛ 2π ⎞ ⎛ Amax ⎞ ⎛ 2π ⎞ ⎛ 0.40 m ⎞ = =⎜ = 0.72 ⎟=⎜ ⎟ ⎜ ⎟ ⎜ 2 ⎟ g g ⎝ T ⎠ ⎝ g ⎠ ⎝ 1.5 s ⎠ ⎝ 9.8 m s ⎠
Assess: Because the period is given, we did not need to use the block masses or the spring constant in our calculation.
14.53. Model: The DNA and cantilever undergo SHM. Visualize: Please refer to figure P14.53. Solve: The cantilever has the same spring constant with and without the DNA molecule. The frequency of oscillation without the DNA is k ω1 = 1 3M With the DNA, the frequency of oscillation is
ω2 =
1 3
k M +m
where m is the mass of the DNA. Divide the two equations, and express ω 2 = ω1 − Δω , where Δω = 2πΔf = 2π ( 50 Hz ) .
ω1 f1 f1 = = = ω 2 f 2 f1 − Δf Thus
k ( 13 M ) = k ( 13 M + m)
f12 ( 13 M ) = ( f1 − Δf )
m=
Since Δf f1 , (50 Hz 12MHz),
1 3
(f M
2 1
− ( f1 − Δf )
( f1 − Δf )
( f1 − Δf )
−2
2
2
)=
2
1 3
1 3
M +m 1 M 3
( 13 M + m ) ⎛ ⎞ f12 − 1⎟ M⎜ 2 ⎜ ( f − Δf ) ⎟ ⎝ 1 ⎠ −2
Δf ⎛ Δf ⎞ −2 ⎛ = f1−2 ⎜ 1 − ⎟ ≈ f1 ⎜ 1 + 2 f f 1 ⎠ 1 ⎝ ⎝
⎞ ⎟ . Thus ⎠
⎛ ⎞ Δf Δf m = 13 M ⎜1 + 2 − 1⎟ = 23 M f f1 1 ⎝ ⎠ The mass of the cantilever M = (2300 kg/m3 )(4000 × 10−9 m)(100 × 10−9 m) = 3.68 × 10−16 kg Thus the mass of the DNA molecule is 2 ⎛ 50 Hz ⎞ −21 m = ( 3.68 × 10−16 kg ) ⎜ ⎟ = 1.02 × 10 kg 6 3 ⎝ 12 × 10 Hz ⎠ Assess: The mass of the DNA molecule is about 6.2 × 105 atomic mass units, which is reasonable for such a large molecule.
14.54. Model: Assume that the swinging lamp makes a small angle with the vertical so that there is simple harmonic motion. Visualize:
Solve:
(a) Using the formula for the period of a pendulum, 2
T = 2π
2
L ⎛ T ⎞ 2 ⎛ 5.5 s ⎞ ⇒ L = g⎜ ⎟ = ( 9.8 m s ) ⎜ ⎟ = 7.5 m π g 2 ⎝ ⎠ ⎝ 2π ⎠
(b) The conservation of mechanical energy equation K 0 + U g0 = K1 + U g1 for the swinging lamp is 1 2
2 mv02 + mgy0 = 12 mv12 + mgy1 ⇒ 0 J + mgh = 12 mvmax +0 J
⇒ vmax = 2 gh = 2 g ( L − L cos3° ) = 2 ( 9.8 m s 2 ) ( 7.5 m )(1 − cos3° ) = 0.45 m s
14.55. Model: Assume that the angle with the vertical that the pendulum makes is small enough so that there is simple harmonic motion. Solve: The angle θ made by the string with the vertical as a function of time is
θ ( t ) = θ max cos (ω t + φ0 ) The pendulum starts from maximum displacement, thus φ0 = 0. Thus, θ ( t ) = θ max cos ω t. To find the time t when
the pendulum reaches 4.0° on the opposite side:
( −4.0° ) = (8.0° ) cosω t ⇒ ω t = cos −1 ( −0.5) = 2.094 rad Using the formula for the angular frequency,
ω=
g 9.8 m s 2 2.0944 rad 2.094 rad = = 3.130 rad s ⇒ t = = = 0.669 s L 1.0 m 3.130 rad s ω
The time t = 0.67 s. Assess: Because T = 2π ω = 2.0 s, a value of 0.67 s for the pendulum to cover a little less than half the oscillation is reasonable.
14.56. Model: Assume a small angle oscillation of the pendulum so that it has simple harmonic motion. Solve:
(a) At the equator, the period of the pendulum is
Tequator = 2π
1.000 m = 2.009 s 9.78 m s 2
The time for 100 oscillations is 200.9 s. (b) At the north pole, the period is Tpole = 2π
1.000 m = 2.004 s 9.83 m s 2
The time for 100 oscillations is 200.4 s. (c) The difference between the two answers is 0.5 s, and this difference is quite measurable with a hand-operated stopwatch. (d) The period on the top of the mountain is 2.010 s. The acceleration due to gravity can be calculated by rearranging the formula for the period: 2
2
⎛ 2π ⎞ ⎛ 2π ⎞ 2 g mountain = L ⎜ ⎟ = (1.000 m ) ⎜ ⎟ = 9.772 m s T 2.010 s ⎝ ⎠ mountain ⎝ ⎠ Assess:
This last result is reasonable because g decreases with altitude.
14.57. Model: The mass is a particle and the string is massless.
Solve:
Equation 14.52 is ω=
Mgl I
The moment of inertia of the mass on a string is I = Ml 2 , where l is the length of the string. Thus
ω=
Mgl = Ml 2
g l
This is Equation 14.49 with L = l. Assess: Equation 14.49 is really a specific case of the more general physical pendulum described by Equation 14.52.
14.58. Model: The rod is thin and uniform with moment of inertia described in Table 12.2. The clay ball is a particle located at the end of the rod. The ball and rod together form a physical pendulum. The oscillations are small. Visualize:
Solve:
The moment of inertia of the composite pendulum formed by the rod and clay ball is
1 I rod + ball = I rod + I ball = mrod L2 + mball L2 3 1 2 2 = ( 0.200 kg )( 0.15 m ) + ( 0.020 kg )( 0.15 m ) = 1.95 × 10−3 kg m 2 3 The center of mass of the rod and ball is located at a distance from the pivot point of
ycm =
( 0.200 kg ) ⎛⎜
0.15 ⎞ m ⎟ + ( 0.020 kg )( 0.15 m ) ⎝ 2 ⎠ = 8.18 × 10−2 m ( 0.200 kg + 0.020 kg )
The frequency of oscillation of a physical pendulum is 1 2π
Mgl 1 = I 2π
The period of oscillation T =
1 = 0.66 s. f
f =
( 0.220 kg ) ( 9.8 m/s 2 )(8.18 × 10−2 m ) 1.95 × 10−3 kg m 2
= 1.51 Hz
14.59. Model: The circular hoop can be modeled as a cylindrical hoop and its moment of inertia about the point of rotation found with the parallel-axis theorem. Visualize: Please refer to Figure P14.59. Solve: Using the parallel-axis theorem, the moment of inertia of the cylindrical hoop about the rotation point is
I = MR 2 + MR 2 = 2MR 2 The frequency of small oscillations is given by Equation 14.52. f =
1 2π
Mgl I
The center of mass of the hoop is its center, so l = R. Thus f =
1 2π
MgR 1 = 2 2 MR 2π
g 2R
14.60. Model: The motion is a damped oscillation. Solve:
The position of the air-track glider is x ( t ) = Ae − ( t 2τ ) cos (ω t + φ0 ) , where τ = m b and
k b2 − m 4m 2
ω= Using A = 0.20 m, φ0 = 0 rad, and b = 0.015 kg/s,
4.0 N m ( 0.015 kg s ) − = 16 − 9 × 10−4 rad s = 4.0 rad s 0.250 kg 4 ( 0.250 kg )2 2
ω= Thus the period is
T=
2π
ω
=
2π rad = 1.57 s 4.0 rad s
The amplitude at t = 0 s is x0 = A and the amplitude will be equal to e −1 A at a time given by 1 m A = Ae− ( t 2τ ) ⇒ t = 2τ = 2 = 33.3 s e b The number of oscillations in a time of 33.3 s is (33.3 s)/(1.57 s) = 21.
14.61. Model: A completely inelastic collision between the two gliders resulting in simple harmonic motion. Visualize:
Let us denote the 250 g and 500 g masses as m1 and m2 , which have initial velocities vi1 and vi2 . After m1 collides with and sticks to m2 , the two masses move together with velocity vf . Solve:
The momentum conservation equation pf = pi for the completely inelastic collision is ( m1 + m2 ) vf =
m1vi1 + m2vi2 . Substituting the given values,
( 0.750 kg ) vf = ( 0.250 kg )(1.20
m s ) + ( 0.500 kg )( 0 m s ) ⇒ vf = 0.400 m s
We now use the conservation of mechanical energy equation:
( K + U s )compressed = ( K + U s )equilibrium ⇒ 0 J + 12 kA2 = 12 ( m1 + m2 ) vf2 + 0 J ⇒ A=
m1 + m2 0.750 kg vf = ( 0.400 m s ) = 0.110 m k 10 N m
The period is
T = 2π
m1 + m2 0.750 kg = 2π = 1.72 s k 10 N m
14.62. Model: The block attached to the spring is oscillating in simple harmonic motion. Solve: (a) Because the frequency of an object in simple harmonic motion is independent of the amplitude and/or the maximum velocity, the new frequency is equal to the old frequency of 2.0 Hz. (b) The speed v0 of the block just before it is given a blow can be obtained by using the conservation of mechanical energy equation as follows: 1 2
⇒ v0 =
2 = 12 mv02 kA2 = 12 mvmax
k A = ω A = ( 2π f ) A = ( 2π )( 2.0 Hz )( 0.02 m ) = 0.25 m s m
The blow to the block provides an impulse that changes the velocity of the block:
( −20 N ) (1.0 × 10
J x = Fx Δt = Δp = mvf − mv0
−3
s ) = ( 0.200 kg ) vf − ( 0.200 kg )( 0.25 m s ) ⇒ vf = 0.150 m s
Since vf is the new maximum velocity of the block at the equilibrium position, it is equal to Aω . Thus, A=
Assess:
0.150 m s
ω
=
0.150 m s = 0.012 m = 1.19 cm 2π ( 2.0 Hz )
Because vf is positive, the block continues to move to the right even after the blow.
14.63. Model: The pendulum falls, then undergoes small-amplitude oscillations in simple harmonic motion. Visualize:
We placed the origin of the coordinate system at the bottom of the arc. Solve: We need to find the length of the pendulum. The conservation of mechanical energy equation for the pendulum’s fall is ( K + U g ) = ( K + U g ) : top
1 2
bottom
mv + mgy0 = mv12 + mgy1 ⇒ 0 J + mg ( 2 L ) = 12 m ( 5.0 m s ) + 0 J 2 0
2
1 2
1 ( 5.0 m s ) = 0.6377 m 4 g Using L = 0.6377 m, we can find the frequency f as 2
⇒L=
f =
1 2π
g 1 = L 2π
9.8 m s 2 = 0.62 Hz 0.6377 m
14.64. Model: Assume the small-angle approximation. Visualize:
Solve:
The tension in the two strings pulls downward at angle θ . Thus Newton’s second law is
∑ Fy = −2T sinθ = ma y From the geometry of the figure we can see that sin θ =
y L + y2 2
If the oscillation is small, then y L and we can approximate sin θ ≈ y / L. Since y/L is tanθ , this approximation is equivalent to the small-angle approximation sinθ ≈ tanθ if θ 1 rad. With this approximation, Newton’s second law becomes 2T d2y d2y 2T y = ma y = m 2 ⇒ y −2T sin θ ≈ − =− 2 L dt dt mL This is the equation of motion for simple harmonic motion (see Equations 14.33 and 14.47). The constants 2T/mL are equivalent to k/m in the spring equation or g/L in the pendulum equation. Thus the oscillation frequency is f =
1 2π
2T mL
14.65. Visualize: Please refer to Figure P14.65. Solve:
The potential energy curve of a simple harmonic oscillator is described by U = 12 k ( Δx ) , where 2
Δx = x − x0 is the displacement from equilibrium. From the graph, we see that the equilibrium bond length is x0 = 0.13 nm. We can find the bond’s spring constant by reading the value of the potential energy U at a displacement Δx and using the potential energy formula to calculate k. x (nm)
0.11
Δx (nm)
U (J)
k (N/m)
0.02
0.8 × 10−19 J
400
0.03
0.10
0.04
0.09
1.9 × 10
−19
J
422
3.4 × 10−19 J
425
The three values of k are all very similar, as they should be, with an average value of 416 N/m. Knowing the spring constant, we can now calculate the oscillation frequency of a hydrogen atom on this “spring” to be
f =
1 2π
1 416 N m k = = 7.9 × 1013 Hz m 2π 1.67 × 10−27 kg
14.66. Model: Visualize:
Solve: is
Assume that the size of the ice cube is much less than R and that θ is a small angle.
The ice cube is like an object on an inclined plane. The net force on the ice cube in the tangential direction
− ( FG ) sin θ = ma = mRα = mR
d 2θ d 2θ θ ⇒ − mg sin = mR dt 2 dt 2
where α is the angular acceleration. With the small-angle approximation sinθ ≈ θ , this becomes d 2θ g = − θ = −ω 2θ dt 2 R
This is the equation of motion of an object in simple harmonic motion with a period of
T=
2π
ω
= 2π
R g
14.67.
Visualize:
Solve:
(a) Newton’s second law applied to the penny along the y-axis is
Fnet = n − mg = ma y
G Fnet is upward at the bottom of the cycle (positive a y ), so n > mg . The speed is maximum when passing G through equilibrium, but a y = 0 so n = mg . The critical point is the highest point. Fnet points down and a y is negative. If a y becomes sufficiently negative, n drops to zero and the penny is no longer in contact with the surface. (b) When the penny loses contact (n = 0), the equation for Newton’s law becomes amax = g . For simple harmonic motion,
9.8 m s 2 g = = 15.65 rad 0.040 m A ω 15.65 rad s ⇒ f = = = 2.5 Hz 2π 2π
amax = Aω 2 ⇒ ω =
14.68. Model: The vertical oscillations constitute simple harmonic motion. Visualize:
Solve:
At the equilibrium position, the net force on mass m on Planet X is: Fnet = k ΔL − mg X = 0 N ⇒
k gX = m ΔL
For simple harmonic motion k m = ω 2 , thus 2
ω2 =
2
⎛ 2π ⎞ gX g X 2π ⎛ 2π ⎞ 2 ⇒ω = = ⇒ gX = ⎜ ⎟ ( 0.312 m ) = 5.86 m s ⎟ ΔL = ⎜ 14.5 s 10 ΔL ΔL T ⎝ T ⎠ ⎝ ⎠
14.69. Model: The doll’s head is in simple harmonic motion and is damped. Solve:
(a) The oscillation frequency is
f =
1 2π
k 2 2 2 ⇒ k = m ( 2π f ) = ( 0.015 kg )( 2π ) ( 4.0 Hz ) = 9.475 N m m
The spring constant is 9.5 N/m. (b) The maximum speed is
vmax = ω A =
k 9.475 N m A= ( 0.020 m ) = 0.50 m s m 0.015 kg
(c) Using A ( t ) = A0e− bt / 2 m , we get
( 0.5 cm ) = ( 2.0 cm ) e−b(4.0 s)/(2×0.015 kg) ⇒ 0.25 = e− (133.3 s/kg)b ⇒ − (133.33 s kg ) b = ln 0.25 ⇒ b = 0.0104 kg / s
14.70. Model: The oscillator is in simple harmonic motion. Solve:
(a) The maximum displacement at time t of a damped oscillator is
xmax ( t ) = Ae −t 2τ ⇒ −
⎛ x (t ) ⎞ t = ln ⎜ max ⎟ 2τ ⎝ A ⎠
Using xmax = 0.98 A at t = 0.50 s, we can find the time constant τ to be
τ =−
0.50 s = 12.375 s 2ln ( 0.98 )
25 oscillations will be completed at t = 25T = 12.5 s. At that time, the amplitude will be xmax, 12.5 s = (10 cm ) e−12.5 s ( 2)(12.375 s ) = 6.0 cm (b) The energy of a damped oscillator decays more rapidly than the amplitude: E (t ) = E0e−1/ τ . When the energy
is 60% of its initial value, E (t )/E0 = 0.60. We can find the time this occurs as follows: ⎛ E (t ) ⎞ ⎛ E (t ) ⎞ t − = ln ⎜ ⎟ ⇒ t = −τ ln ⎜ ⎟ = − (12.375 s ) ln ( 0.60 ) = 6.3 s τ ⎝ E0 ⎠ ⎝ E0 ⎠
14.71. Model: The oscillator is in simple harmonic motion. Solve:
The maximum displacement, or amplitude, of a damped oscillator decreases as xmax (t ) = Ae− t 2τ , where
τ is the time constant. We know xmax A = 0.60 at t = 50 s, so we can find τ as follows: −
⎛ x (t ) ⎞ t 50 s = ln ⎜ max ⎟ ⇒ τ = − = 48.9 s 2τ 2ln ( 0.60 ) ⎝ A ⎠
Now we can find the time t30 at which xmax A = 0.30 :
⎛ x (t ) ⎞ t30 = −2τ ln ⎜ max ⎟ = −2 ( 48.9 s ) ln ( 0.30 ) = 118 s ⎝ A ⎠ The undamped oscillator has a frequency f = 2 Hz = 2 oscillations per second. Damping changes the oscillation frequency slightly, but the text notes that the change is negligible for “light damping.” Damping by air, which allows the oscillations to continue for well over 100 s, is certainly light damping, so we will use f = 2.0 Hz. Then the number of oscillations before the spring decays to 30% of its initial amplitude is N = f ⋅ t30 = ( 2 oscillations s ) ⋅ (118 s ) = 236 oscillations
14.72. Solve: The solution of the equation d 2 x b dx k + + x=0 dt 2 m dt m is x ( t ) = Ae − bt 2 m cos (ω t + φ0 ) . The first and second derivatives of x(t) are
dx Ab −bt / 2 m e =− cos (ω t + φ0 ) − Aω e − bt / 2 m sin (ω t + φ0 ) dt 2m ⎤ ⎞ d 2 x ⎡⎛ Ab 2 ω Ab = ⎢⎜ − Aω 2 ⎟ cos (ω t + φ0 ) + sin (ω t + φ0 ) ⎥ e −bt / 2 m dt 2 ⎣⎝ 4m 2 m ⎠ ⎦ Substituting these expressions into the differential equation, the terms involving sin(ω t + φ0 ) cancel and we obtain the simplified result
⎛ −b 2 k⎞ 2 ⎜ 2 − ω + ⎟ cos (ω t + φ0 ) = 0 m⎠ ⎝ 4m Because cos (ω t + φ0 ) is not equal to zero in general, k k b2 −b 2 −ω2 + = 0 ⇒ ω = − 2 4m m m 4m 2
14.73. Model: The two springs obey Hooke’s law. Visualize:
Solve: There are two restoring forces on the block. If the block’s displacement x is positive, both restoring forces—one pushing, the other pulling—are directed to the left and have negative values:
( Fnet ) x = ( Fsp 1 ) x + ( Fsp 2 ) x = −k1x − k2 x = − ( k1 + k2 ) x = −keff x where keff = k1 + k2 is the effective spring constant. This means the oscillatory motion of the block under the influence of the two springs will be the same as if the block were attached to a single spring with spring constant keff . The frequency of the blocks, therefore, is
f =
1 2π
keff 1 = m 2π
k1 + k2 k1 k = + 22 = 2 m 4π m 4π m
f12 + f 22
14.74. Model: The two springs obey Hooke’s law. Assume massless springs. Visualize:
Solve:
Each spring is shown separately. Note that Δx = Δx1 + Δx2 .
Only spring 2 touches the mass, so the net force on the mass is Fm = F2 on m . Newton’s third law tells us
that F2 on m = Fm on 2 and that F2 on 1 = F1 on 2 . From Fnet = ma, the net force on a massless spring is zero. Thus Fw on 1 = F2 on 1 = k1Δx1 and Fm on 2 = F1 on 2 = k2 Δx2 . Combining these pieces of information, Fm = k1Δx1 = k2 Δx2 The net displacement of the mass is Δx = Δx1 + Δx2 , so
Δx = Δx1 + Δx2 =
Fm Fm ⎛ 1 1 ⎞ k +k + = ⎜ + ⎟ Fm = 1 2 Fm k1 k2 ⎝ k1 k2 ⎠ k1k2
Turning this around, the net force on the mass is
Fm =
k1k2 kk Δx = keff Δx where keff = 1 2 k1 + k2 k1 + k2
keff , the proportionality constant between the force on the mass and the mass’s displacement, is the effective spring constant. Thus the mass’s angular frequency of oscillation is
ω=
keff 1 k1k2 = m m k1 + k2
Using ω12 = k1 / m and ω 22 = k2 / m for the angular frequencies of either spring acting alone on m, we have
ω=
( k1 / m) ( k2 / m) ω12ω 22 = ( k1 / m) + ( k2 / m) ω12 + ω 22
Since the actual frequency f is simply a multiple of ω , this same relationship holds for f: f =
f 12 f 22 f 12 + f 22
14.75. Model: The blocks undergo simple harmonic motion. Visualize:
The length of the stretched spring due to a block of mass m is ΔL1. In the case of the two block system, the spring is further stretched by an amount ΔL2 . Solve: The equilibrium equations from Newton’s second law for the single block and double block systems are
( ΔL1 ) k = mg and ( ΔL1 + ΔL2 ) k = ( 2m ) g Using ΔL2 = 5.0 cm, and subtracting these two equations, gives us
( ΔL1 + ΔL2 ) k − ΔL1k = ( 2m ) g − mg ⇒ ( 0.05 m ) k = mg ⇒
k 9.8 m/s 2 = = 196 s –2 m 0.05 m
With both blocks attached, giving total mass 2m, the angular frequency of oscillation is
ω=
k 1k 1 196 s –2 = 9.90 rad/s = = 2m 2m 2
Thus the oscillation frequency is f = ω /2π = 1.58 Hz.
14.76. Model: A completely inelastic collision between the bullet and the block resulting in simple harmonic motion. Visualize:
Solve:
(a) The equation for conservation of energy after the collision is
1 2 1 2500 N m k kA = ( mb + mB ) vf2 ⇒ vf = A= ( 0.10 m ) = 5.0 m s 2 2 1.010 kg mb + mB The momentum conservation equation for the perfectly inelastic collision pafter = pbefore is
(1.010 kg )( 5.0
( mb + mB ) vf = mbvb + mBvB m s ) = ( 0.010 kg ) vb + (1.00 kg )( 0 m s ) ⇒ vb = 5.0 × 102
ms
(b) No. The oscillation frequency k ( mb + mB ) depends on the masses but not on the speeds.
14.77. Model: The block undergoes SHM after sticking to the spring. Energy is conserved throughout the motion. Visualize:
It’s essential to carefully visualize the motion. At the highest point of the oscillation the spring is stretched upward. Solve: We’ve placed the origin of the coordinate system at the equilibrium position, where the block would sit on the spring at rest. The spring is compressed by ΔL at this point. Balancing the forces requires k ΔL = mg . The angular frequency is w2 = k/m = g/ΔL, so we can find the oscillation frequency by finding ΔL. The block hits the spring (1) with kinetic energy. At the lowest point (3), kinetic energy and gravitational potential energy have been transformed into the spring’s elastic energy. Equate the energies at these points: K1 + U1g = U 3s + U 3g ⇒ 12 mv12 + mg ΔL = 12 k (ΔL + A) 2 + mg (− A)
We’ve used y1 = ΔL as the block hits and y3 = − A at the bottom. The spring has been compressed by Δy = ΔL + A. Speed v1 is the speed after falling distance h, which from free-fall kinematics is v12 = 2 gh. Substitute this expression for v12 and mg/ΔL for k, giving
mgh + mg ΔL =
mg (ΔL + A) 2 + mg (− A) 2(ΔL)
The mg term cancels, and the equation can be rearranged into the quadratic equation (ΔL) 2 + 2h(ΔL) − A2 = 0 The positive solution is ΔL = h 2 + A2 − h = (0.030 m)2 + (0.100 m) 2 − 0.030 m = 0.0744 m Now that ΔL is known, we can find
ω=
g 9.80 m/s 2 ω = = 11.48 rad/s ⇒ f = = 1.83 Hz 0.0744 m 2π ΔL
14.78. Model: Model the bungee cord as a spring. The motion is damped SHM. Visualize:
Solve:
(a) For light damping, the oscillation period is 2
2
m ⎛ 2π ⎞ ⎛ 2π ⎞ ⇒ k = m⎜ ⎟ = (75 kg) ⎜ ⎟ = 185 N/m k ⎝ T ⎠ ⎝ 4.0 s ⎠
T = 2π
(b) The maximum speed is vmax = ω A = (2π /4.0 s)(11.0 m) = 17.3 m/s. (c) Jose oscillates about the equilibrium position at which he would hang at rest. Balancing the forces, ΔL = mg / k = (75 kg)(9.80 m/s 2 )(185 N/m) = 3.97 m. Jose’s lowest point is 11.0 m below this point, so the
bungee cord is stretched by Δymax = ΔL + A = 14.97 m. Choose this lowest point as y = 0. Because Jose is instantaneously at rest at this point, his energy is entirely the elastic potential energy of the stretched bungee cord. Initially, his energy was entirely gravitational potential energy. Equating his initial energy to his energy at the lowest point,
U lowest point = U highest point ⇒ 12 k (Δymax ) 2 = mgh h=
k (Δymax ) 2 (185 N/m)(14.97 m)2 = = 28.2 m 2mg 2(75 kg)(9.80 m/s 2)
Jose jumped 28 m above the lowest point. (d) The amplitude decreases due to damping as A(t ) = Ae −bt/2m . At the time when the amplitude has decreased from 11.0 m to 2.0 m, 2.0 m 2m ⎛ 2 ⎞ 2(75 kg) = e − bt / 2 m ⇒ t = − ln ⎜ ⎟ = − (−1.705) = 42.6 s 11.0 m b ⎝ 11 ⎠ 6.0 kg/s
With a period of 4.0 s, the number of oscillations is N osc = (42.6 s)/(4.0 s) = 10.7 oscillations.
14.79. Model: The vertical movement of the car is simple harmonic motion. Visualize:
The fact that the car has a maximum oscillation amplitude at 5 m/s implies a resonance. The bumps in the road provide a periodic external force to the car’s suspension system, and a resonance will occur when the “bump frequency” f ext matches the car’s natural oscillation frequency f 0 . Solve: Now the 5.0 m/s is not a frequency, but we can convert it to a frequency because we know the bumps are spaced every 3.0 meters. The time to drive 3.0 m at 5.0 m/s is the period:
T=
Δx 3.0 m = = 0.60 s v 5.0 m s
The external frequency due to the bumps is thus f ext = 1 T = 1.667 Hz. This matches the car’s natural frequency f 0 , which is the frequency the car oscillates up and down if you push the car down and release it. This is enough information to deduce the spring constant of the car’s suspension: 1.667 Hz =
1 2π
N k 2 ⇒ k = m ( 2π f ext ) = 131,600 m m
where we used m = mtotal = mcar + 2mpassenger = 1200 kg. When at rest, the car is in static equilibrium with Fnet = 0 N. The downward weight mtotal g of the car and passengers is balanced by the upward spring force k Δy of the suspension. Thus the compression Δy of the suspension is Δy =
mtotal g k
Initially mtotal = mcar + 2mpassenger = 1200 kg, causing an initial compression Δyi = 0.0894 m = 8.94 cm. When three
additional passengers get in, the mass increases to mtotal = mcar + 5mpassenger = 1500 kg. The final compression is Δyf = 0.1117 m = 11.17 cm. Thus the three new passengers cause the suspension to “sag” by 11.17 cm − 8.94 cm = 2.23 cm.
14.80. Model: The rod is thin and uniform. Visualize: Please refer to Figure CP14.80.
G Solve: We must derive our own equation for this combination of a pendulum and spring. For small oscillations, Fs remains horizontal. The net torque around the pivot point is ⎛ L⎞ ⎝ ⎠
τ net = Iα = − Fs L cosθ − FG ⎜ ⎟ sinθ 2 With α =
d 2θ 1 , FG = mg , Fs = k Δx = kL sin θ , and I = mL2 , dt 2 3 d 2θ 3k 3g = − sinθ cosθ − sinθ dt 2 m 2L
1 We can use sin θ cosθ = sin 2θ . For small angles, sinθ ≈ θ and sin 2θ ≈ 2θ . So 2
d 2θ ⎛ 3k 3 g ⎞ = −⎜ + ⎟θ dt 2 ⎝ m 2L ⎠ This is the same as Equations 14.33 and 14.47 with
ω=
3k 3 g + m 2L
The frequency of oscillation is thus f =
1 2π
2 3 ( 3.0 N/m ) 3 ( 9.8 m/s ) + = 1.73 Hz ( 0.200 kg ) 2 ( 0.20 m )
1 = 0.58 s. f Assess: Fewer than two oscillations per second is reasonable. The rod’s angle from the vertical must be small enough that sin 2θ ≈ 2θ . This is more restrictive than other examples, which only require that sinθ ≈ θ .
The period T =
14-1
15.1. Solve: The density of the liquid is ρ= Assess:
0.240 kg m 0.240 kg = = = 960 kg m3 V 250 mL 250 × 10−3 × 10−3 m3
The liquid’s density is near that of water (1000 kg/m3 ) and is a reasonable number.
15.2. Solve: The volume of the helium gas in container A is equal to the volume of the liquid in container B. That is, VA = VB . Using the definition of mass density ρ = m V , the above relationship becomes mA
ρA
=
mB
ρB
⇒
mHe
ρ He
=
7000 mHe
ρB
⇒ ρ B = ( 7000 ) ρ He = ( 7000 ) ( 0.18 kg m3 ) = 1260 kg m3
Referring to Table 15.1, we find that the liquid is glycerine.
15.3. Model: The density of water is 1000 kg/m3 . Visualize:
Solve:
Volume of water in the swimming pool is V = 6 m × 12 m × 3 m − 12 ( 6 m × 12 m × 2 m ) = 144 m3
The mass of water in the swimming pool is
m = ρV = (1000 kg m3 )(144 m3 ) = 1.44 × 105 kg
15.4. Model: The densities of gasoline and water are given in Table 15.1. Solve:
(a) The total mass is
mtotal = mgasoline + mwater = 0.050 kg + 0.050 kg = 0.100 kg The total volume is
Vtotal = Vgasoline + Vwater =
mgasoline
ρ gasoline
⇒ ρ avg =
+
mwater
ρ water
=
0.050 kg 0.050 kg + = 1.24 × 10−4 m3 680 kg m3 1000 kg m3
mtotal 0.100 kg = = 8.1× 102 kg m3 Vtotal 1.24 × 10−4 m3
(b) The average density is calculated as follows:
mtotal = mgasoline + mwater = ρ waterVwater + ρ gasolineVgasoline ⇒ ρ avg = Assess:
ρ waterVwater + ρgasolineVgasoline Vwater + Vgasoline
( 50 cm )(1000 kg/m 3
=
3
100 cm
3
+ 680 kg/m3 )
= 8.4 × 102 kg/m3
The above average densities are between those of gasoline and water, and are reasonable.
15.5. Model: The density of sea water is 1030 kg/m3 . Solve:
The pressure below sea level can be found from Equation 15.6 as follows: p = p0 + ρ gd = 1.013 × 105 Pa + (1030 kg m3 )( 9.80 m s 2 )(1.1 × 104 m ) = 1.013 × 105 Pa + 1.1103 × 108 Pa = 1.1113 × 108 Pa = 1.10 × 103 atm
where we have used the conversion 1 atm = 1.013 × 105 Pa. Assess: The pressure deep in the ocean is very large.
15.6. Model: The density of water is 1000 kg/m3 and the density of ethyl alcohol is 790 kg/m3 . Solve:
(a) The volume of water that has the same mass as 8.0 m 3 of ethyl alcohol is
Vwater =
mwater
ρ water
=
malcohol
ρ water
=
ρ alcoholValcohol ⎛ 790 kg/m3 ⎞ 8.0 m3 ) = 6.3 m3 =⎜ 3 ⎟( ρ water ⎝ 1000 kg/m ⎠
(b) The pressure at the bottom of the cubic tank is p = p0 + ρ water gd :
p = 1.013 × 105 Pa + (1000 kg/m3 )( 9.80 m s 2 ) ( 6.3) = 1.19 × 105 Pa 13
where we have used the relation d = (Vwater ) . 13
15.7.
Visualize:
Solve:
The pressure at the bottom of the vat is p = p0 + ρ gd = 1.3 atm. Substituting into this equation gives
1.013 × 105 Pa + ρ ( 9.8 m s 2 ) ( 2.0 m ) = (1.3) (1.013 × 105 ) Pa ⇒ ρ = 1550.5 kg m3 The mass of the liquid in the vat is
m = ρV = ρπ ( 0.50 m ) d = (1550.5 kg m3 )π ( 0.50 m ) ( 2.0 m ) = 2.4 × 103 kg 2
2
15.8. Model: Visualize:
The density of oil ρoil = 900 kg m 3 and the density of water ρ water = 1000 kg m3 .
Solve: The pressure at the bottom of the oil layer is p1 = p0 + ρ oil gd1 , and the pressure at the bottom of the water layer is
p2 = p1 + ρ water gd 2 = p0 + ρoil gd1 + ρ water gd 2
⇒ p2 = (1.013 × 10 Pa ) + ( 900 kg m 3 )( 9.80 m s 2 ) ( 0.50 m ) + (1000 kg m3 )( 9.80 m s 2 ) (1.20 m ) = 1.18 × 105 Pa 5
Assess:
A pressure of 1.18 × 105 Pa = 1.16 atm is reasonable.
15.9. Model: The density of seawater ρseawater = 1030 kg m 2 . Visualize:
The pressure outside the submarine’s window is pout = p0 + ρseawater gd , where d is the maximum safe
Solve:
depth for the window to withstand a force F. This force is F A = pout − pin , where A is the area of the window. With pin = p0 , we simplify the pressure equation to pout − p0 = Assess:
F F = ρseawater gd ⇒ d = A Aρseawater g
d=
1.0 × 106 N
π ( 0.10 m ) (1030 kg m 2 )( 9.8 m s 2 ) 2
A force of 1.0 × 106 N corresponds to a pressure of
ρ= A depth of 3 km is therefore reasonable.
F 1.0 × 106 N = = 314 atm A π ( 0.10 m )2
= 3.2 km
15.10.
Visualize:
We assume that the seal is at a radius of 5 cm. Outside the seal, atmospheric pressure presses on both sides of the cover and the forces cancel. Thus, only the 10 cm diameter opening inside the seal is relevant, not the 20 cm diameter of the cover. Solve: Within the 10 cm diameter area where the pressures differ,
Fto left = patmos A
Fto right = pgas A
where A = π r 2 = 7.85 × 10−3 m 2 is the area of the opening. The difference between the forces is
Fto left − Fto right = ( patmos − pgas ) A = (101,300 Pa − 20,000 Pa ) ( 7.85 × 10−3 m 2 ) = 0.64 kN Normally, the rubber seal exerts a 0.64 kN force to the right to balance the air pressure force. To pull the cover off, an external force must pull to the right with a force ≥ 0.64 kN.
15.11. Model: The density of water is ρ = 1000 kg m3 . Visualize: Please refer to Figure 15.17. Solve: From the figure and the equation for hydrostatic pressure, we have
p0 + ρ gh = patmos
Using p0 = 0 atm, and patmos = 1.013 × 105 Pa, we get 0 Pa + (1000 kg m3 )( 9.8 m s 2 ) h = 1.013 × 105 Pa ⇒ h = 10.3 m Assess:
This large value of h is due to water having a much smaller density than mercury.
15.12. Model: Assume that the oil is incompressible. Its density is 900 kg/m3 . Visualize: Please refer to Figure 15.19. Because the liquid is incompressible, the volume displaced in the left cylinder of the hydraulic lift is equal to the volume displaced in the right cylinder. Solve: Equating the two volumes, 2
2
⎛r ⎞ ⎛ 0.04 m ⎞ A1d1 = A2 d 2 ⇒ (π r12 ) d1 = (π r22 ) d 2 ⇒ d1 = ⎜ 2 ⎟ d 2 = ⎜ ⎟ ( 0.20 m ) = 3.2 m r 0.01 m ⎠ ⎝ ⎝ 1⎠
15.13.
Visualize:
Solve: By sucking on the top and reducing the pressure, the straw is essentially a barometer. Atmospheric pressure pushes the liquid up the straw. The length of the longest straw can be obtained from the formula p = p0 + ρ gd . If one could reduce the mouth pressure to zero, the length of the straw would be
1.013 × 105 Pa = 0 Pa + (1000 kg m3 )( 9.8 m s 2 ) d ⇒ d = 10.3 m
15.14.
Model: Visualize:
The vacuum cleaner can create zero pressure.
Solve: The gravitational force on the dog is balanced by the force resulting from the pressure difference between the atmosphere and the vacuum ( phose = 0) in the hose. The force applied by the hose is
F = ( patmos − phose ) A = patmos A = mg
(10 kg ) ( 9.8 m/s2 )
= 9.7 × 10−4 m 2 1.013 × 105 Pa 2 A ⎛d⎞ = 0.035 m = 3.5 cm. Since A = π ⎜ ⎟ , the diameter of the hose is d = 2 π ⎝2⎠ Assess: The dog will have a terrible skin rash. It’s a good thing the miscreant holding the vacuum does not have a younger sibling. ⇒ A=
15.15. Model: The buoyant force on the sphere is given by Archimedes’ principle. Visualize:
Solve:
The sphere is in static equilibrium because it is neutrally buoyant. That is,
∑F
y
= FB − FG = 0 N ⇒ ρ lVl g − ms g = 0 N
The sphere displaces a volume of liquid equal to its own volume, Vl = Vs , so
ρl =
ms m 0.0893 kg = 4 s3 = = 7.9 × 102 kg m3 3 4 Vs 3 π rs 0.030 m π ( ) 3
A density of 790 kg/m3 in Table 15.1 identifies the liquid as ethyl alcohol. Assess: If the density of the fluid and an object are equal, we have neutral buoyancy.
15.16. Model: The buoyant force on the cylinder is given by Archimedes’ principle. Visualize:
Vcyl is the volume of the cylinder and Vw is the volume of the water displaced by the cylinder. Note that the volume displaced is only from the part of the cylinder that is immersed in water. Solve: The cylinder is in static equilibrium, so FB = FG . The buoyant force is the weight ρ wVw g of the displaced water. Thus
FB = ρ wVw g = FG = mg = ρcylVcyl g ⇒ ρ wVw = ρcylVcyl ⇒ ρ cyl = ρ w ⇒ ρ cyl = (1000 kg m3 ) Assess:
A ( 0.040 m ) = 6.7 × 102 kg m3 A ( 0.060 m )
ρcyl < ρ w for a cylinder floating in water is an expected result.
Vw Vcyl
15.17. Model: The buoyant force on the sphere is given by Archimedes’ principle. Visualize:
Solve:
The sphere is in static equilibrium. The free body diagram on the sphere shows that
1 4 = FB − T − FG = 0 N ⇒ FB = T + FG = FG + FG = FG 3 3 4 3 3 3 ⇒ ρ wVsphere g = ρsphereVsphere g ⇒ ρsphere = ρ w = (1000 kg m ) = 750 kg m3 3 4 4
∑F
y
15.18. Model: The buoyant force on the rock is given by Archimedes’ principle. Visualize:
Solve:
Because the rock is in static equilibrium, Newton’s first law is
Fnet = T + FB − ( FG )rock = 0 N ⎞ 1 1 ρ ⎛1 ⎞ ⎛ ⎞ ⎛ ⎞⎛ m g ⎞ ⎛ ⇒ T = ρ rockVrock g − ρ water ⎜ Vrock ⎟ g = ⎜ ρ rock − ρ water ⎟Vrock g = ⎜ ρ rock − ρ water ⎟ ⎜ rock ⎟ = ⎜ 1 − water ⎟ mrock g 2 2 ⎝2 ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ρ rock ⎠ ⎝ 2ρ rock ⎠ Using ρ rock = 4800 kg m3 and mrock = 5.0 kg, we get T = 44 N.
15.19. Model: The buoyant force on the aluminum block is given by Archimedes’ principle. The density of aluminum and ethyl alcohol are ρ Al = 2700 kg m3 and ρ ethyl alcohol = 790 kg m3 . Visualize:
The buoyant force FB and the tension due to the string act vertically up, and the gravitational force on the aluminum block acts vertically down. The block is submerged, so the volume of displaced fluid equals VAl , the volume of the block. Solve: The aluminum block is in static equilibrium, so
∑F
y
= FB + T − FG = 0 N ⇒ ρ fVAl g + T − ρ AlVAl g = 0 N ⇒ T = VAl g ( ρ Al − ρ f )
T = (100 × 10−6 m3 )( 9.80 m s 2 )( 2700 kg m3 − 790 kg m3 ) = 1.87 N
where we have used the conversion 100 cm3 = 100 × (10−2 m ) = 10−4 m3 . 3
Assess: The gravitational force on the aluminum block is ρ AlVAl g = 2.65 N. A similar order of magnitude for T is reasonable.
15.20. Model: The buoyant force on the steel cylinder is given by Archimedes’ principle. Visualize:
The length of the cylinder above the surface of mercury is d. Solve: The cylinder is in static equilibrium with FB = FG . Thus
FB = ρ HgVHg g = FG = mg = ρ cylVcyl g ⇒ ρ HgVHg = ρcylVcyl ⇒ ρ Hg A ( 0.20 m − d ) = ρ cyl A ( 0.20 m ) ⇒ d = 0.20 m −
ρ cyl ⎛ 7900 kg m3 ⎞ = 0.084 m = 8.4 cm ( 0.20 m ) = ( 0.20 m ) ⎜1 − 3 ⎟ ρ Hg ⎝ 13,600 kg m ⎠
That is, the length of the cylinder above the surface of the mercury is 8.4 cm.
15.21.
Model: The buoyant force is determined by Archimedes’ principle. Ignore any compression the air in the beach ball may undergo as a result of submersion. Solve: The mass of the beach ball is negligible, so the force needed to push it below the water is equal to the buoyant force. 3⎞ ⎛4 ⎞ ⎛4 FB = pw ⎜ π R 3 ⎟ g = (1000 kg/m3 ) ⎜ π ( 0.30 m ) ⎟ ( 9.8 m/s 2 ) = 1.11 kN ⎝3 ⎠ ⎝3 ⎠ Assess: It would take a 113 kg (250 lb) person to push the ball below the water. Two people together could do it. This seems about right.
15.22. Model: The buoyant force on the sphere is given by Archimedes’ principle. Visualize:
Solve: For the Styrofoam sphere and the mass not to sink, the sphere must be completely submerged and the buoyant force FB must be equal to the sum of the gravitational force on the Styrofoam sphere and the attached mass. The volume of displaced water equals the volume of the sphere, so
FB = ρ waterVwater g = (1000 kg m3 ) 43π ( 0.25 m ) ( 9.80 m s 2 ) = 641.4 N 3
( FG )Styrofoam = ρStyrofoamVStyrofoam g = (150
3 kg m3 ) ⎡ 34 π ( 0.25 m ) ⎤ ( 9.80 m s 2 ) = 96.2 N ⎣ ⎦
Because ( FG )Styrofoam + mg = FB ,
m= The mass is 56 kg.
FB − ( FG )Styrofoam g
=
641.4 N − 96.2 N = 55.6 kg 9.80 m s 2
15.23. Model: Treat the water as an ideal fluid. The pipe is a flow tube, so the equation of continuity applies. Solve:
The volume flow rate is
Q=
300 L 300 × 10−3 m3 = = 1.0 × 10−3 m3 s 5.0 min 5.0 × 60 s
Using the definition Q = vA, we get
v=
Q 1.0 × 10−3 m3 s = = 3.2 m s A π ( 0.010 m )2
15.24. Model: Treat the water as an ideal fluid. The pipe itself is a flow tube, so the equation of continuity applies.
Visualize:
Note that A1 , A2 , and A3 and v1 , v2 , and v3 are the cross-sectional areas and the speeds in the first, second, and third segments of the pipe. Solve: (a) The equation of continuity is
A1v1 = A2v2 = A3v3 ⇒ π r12v1 = π r22v2 = π r32v3 ⇒ r12v1 = r22v2 = r32v3 ⇒ ( 0.0050 m ) ( 4.0 m s ) = ( 0.010 m ) v2 = ( 0.0025 m ) v3 2
2
2
⎛ 0.0050 m ⎞ ⇒ v2 = ⎜ ⎟ ( 4.0 m s ) = 1.00 m s ⎝ 0.010 m ⎠
2
2
⎛ 0.0050 m ⎞ v3 = ⎜ ⎟ ( 4.0 m s ) = 16.0 m s ⎝ 0.0025 m ⎠
(b) The volume flow rate through the pipe is
Q = A1v1 = π ( 0.0050 m ) ( 4.0 m s ) = 3.1 × 10−4 m3 s 2
15.25. Model: Treat the water as an ideal fluid so that the flow in the tube follows the continuity equation. Visualize:
Solve:
The equation of continuity is v0 A0 = v1 A1 , where A0 = L2 and A1 = π ( 12 L ) . The above equation simplifies 2
to 2
⎛ L⎞ ⎛ 4⎞ v0 L2 = v1π ⎜ ⎟ ⇒ v1 = ⎜ ⎟ v0 = 1.27v0 2 ⎝ ⎠ ⎝π ⎠
15.26. Model: Treat the oil as an ideal fluid obeying Bernoulli’s equation. Consider the path connecting point 1 in the lower pipe with point 2 in the upper pipe a streamline. Visualize: Please refer to Figure EX15.26. Solve: Bernoulli’s equation is
p2 + 12 ρ v22 + ρ gy2 = p1 + 12 ρ v12 + ρ gy1 ⇒ p2 = p1 + 12 ρ ( v12 − v22 ) + ρ g ( y1 − y2 ) Using p1 = 200 kPa = 2.00 × 105 Pa, ρ = 900 kg m3 , y2 − y1 = 10.0 m, v1 = 2.0 m s, and v2 = 3.0 m s, we
get p2 = 1.096 × 105 Pa = 110 kPa.
15.27. Model: Turning the tuning screws on a guitar string creates tensile stress in the string. Solve: The tensile stress in the string is given by T/A, where T is the tension in the string and A is the crosssectional area of the string. From the definition of Young’s modulus,
Y=
T/A T ⎛ L⎞ ⇒ ΔL = ⎜ ⎟ ΔL / L A⎝Y ⎠
Using T = 2000 N, L = 0.80 m, A = π (0.00050 m) 2 , and Y = 20 × 1010 N/m 2 (from Table 15.3), we obtain ΔL = 0.010 m = 1.02 cm. Assess: 1.02 cm is a large stretch for a length of 80 cm, but 2000 N is a large tension.
15.28. Model: The dangling mountain climber creates tensile stress in the rope. Solve:
Young’s modulus for the rope is
Y=
F / A stress = ΔL / L strain
The tensile stress is
( 70 kg ) ( 9.8 m/s 2 ) = 8.734 × 106 Pa 2 π ( 0.0050 m ) and the strain is 0.080 m 50 m = 0.00160. Dividing the two quantities yields Y = 5.5 × 109 N m 2 .
15.29. Model: The hanging mass creates tensile stress in the wire. Solve: The force (F) pulling on the wire, which is simply the gravitational force (mg) on the hanging mass, produces tensile stress given by F/A, where A is the cross-sectional area of the wire. From the definition of Young’s modulus, we have
(π r 2 )Y ΔL = π ( 2.50 × 10−4 m ) ( 20 × 1010 N/m2 )(1.0 × 10−3 m ) = 2.0 kg mg / A ⇒m= ΔL / L gL ( 9.80 m/s 2 ) ( 2.0 m ) 2
Y=
15.30. Model: The load supported by a concrete column creates compressive stress in the concrete column. Solve: The gravitational force on the load produces tensile stress given by F/A, where A is the cross-sectional area of the concrete column and F equals the gravitational force on the load. From the definition of Young’s modulus,
Y= Assess:
2 F/A 3.0 m ⎛ F ⎞⎛ L ⎞ ⎛ 200,000 kg × 9.8 m/s ⎞ ⎛ ⎞ ⎟⎜ ⇒ ΔL = ⎜ ⎟⎜ ⎟ = ⎜ = 1.0 mm 2 10 2 ⎟ ⎜ ⎟ ΔL / L × 3 10 N/m ⎝ A ⎠⎝ Y ⎠ ⎝ ⎝ ⎠ π ( 0.25 m ) ⎠
A compression of 1.0 mm of the concrete column by a load of approximately 200 tons is reasonable.
15.31. Model: Water is almost incompressible and it applies a volume stress. Solve:
(a) The pressure at a depth of 5000 m in the ocean is
p = p0 + ρsea water g ( 5000 m ) = 1.013 × 105 Pa + (1030 kg m3 )( 9.8 m s 2 ) ( 5000 m ) = 5.057 × 107 Pa (b) Using the bulk modulus of water,
ΔV p 5.057 × 107 Pa =− =− = −0.025 V B 0.2 × 1010 Pa (c) The volume of a mass of water decreases from V to 0.975V. Thus the water’s density increases from ρ to ρ /0.975. The new density is
ρ5000m =
1030 kg/m3 = 1056 kg/m3 0.975
15.32. Solve: The pressure p at depth d in a fluid is p = p0 + ρ gd . Using 1.29 kg/m3 for the density of air, pbottom = ptop + ρ air gd ⇒ pbottom − ptop = 202 Pa = 1.99 × 10−3 atm Assuming pbottom = 1 atm, pbottom − ptop pbottom
=
1.99 × 10−3 atm = 0.2% 1 atm
15.33. Model: We assume that there is a perfect vacuum inside the cylinders with p = 0 Pa. We also assume that the atmospheric pressure in the room is 1 atm. Visualize: Please refer to Figure P15.33. 2 Solve: (a) The flat end of each cylinder has an area A = π r 2 = π ( 0.30 m ) = 0.283 m 2 . The force on each end is thus
Fatm = p0 A = (1.013 × 105 Pa )( 0.283 m 2 ) = 2.86 × 104 N The force on each end is 2.9 × 104 N. (b) The net vertical force on the lower cylinder when it is on the verge of being pulled apart is
∑F
y
= Fatm − ( FG )players = 0 N ⇒ ( FG )players = Fatm = 2.86 × 104 N ⇒ number of players =
2.86 × 104 N = 29.2 (100 kg ) ( 9.8 m s 2 )
That is, 30 players are needed to pull the two cylinders apart.
15.34. Model: Assume that the oil is incompressible and its density is 900 kg/m3 . Visualize: Please refer to Figure P15.34. Solve: (a) The pressure at depth d in a fluid is p = p0 + ρ gd . Here, pressure p0 at the top of the fluid is due
both to the atmosphere and to the gravitational force on the piston. That is, p0 = patm + ( FG ) p / A. At point A, pA = patm +
( FG )P A
+ ρ g (1.00 m − 0.30 m )
(10 kg ) ( 9.8 m s 2 ) + ( 900 kg m3 )( 9.8 2 π ( 0.02 m ) 2 ⇒ FA = pA A = (185,460 Pa )π ( 0.10 m ) = 5.8 kN = 1.013 × 105 Pa +
m s 2 ) ( 0.70 m ) = 185,460 Pa
(b) In the same way,
pB = patm + Assess:
( FG )P A
+ ρ g (1.30 m ) = 190,752 Pa ⇒ FB = 6.0 kN
FB is larger than FA , because pB is larger than pA .
15.35. Model: The tire flattens until the pressure force against the ground balances the upward normal force of the ground on the tire. Solve: The area of the tire in contact with the road is A = ( 0.15 m )( 0.13 m ) = 0.0195 m 2 . The normal force on each tire is
n=
2 FG (1500 kg ) ( 9.8 m s ) = = 3675 N 4 4
Thus, the pressure inside each tire is pinside =
3675 N 14.7 psi n = = 188,500 Pa = 1.86 atm × = 27 psi 1 atm A 0.0195 m 2
15.36.
Visualize:
Solve: (a) Because the patient’s blood pressure is 140/100, the minimum fluid pressure needs to be 100 mm of Hg above atmospheric pressure. Since 760 mm of Hg is equivalent to 1 atm and 1 atm is equivalent to 1.013 × 105 Pa, the minimum pressure is 100 mm = 1.333 × 104 Pa. The excess pressure in the fluid is due to force F pushing on the internal 6.0-mm-diameter piston that presses against the liquid. Thus, the minimum force the nurse needs to apply to the syringe is 2 F = fluid pressure × area of plunger = (1.333 × 104 Pa ) ⎡π ( 0.0030 m ) ⎤ = 0.38 N ⎣ ⎦
(b) The flow rate is Q = vA, where v is the flow speed of the medicine and A is the cross-sectional area of the needle. Thus,
v=
Q 2.0 × 10−6 m3 2.0 s = = 20 m s A π ( 0.125 × 10−3 m )2
Assess: Note that the pressure in the fluid is due to F that is not dependent on the size of the plunger pad. Also note that the syringe is not drawn to scale.
15.37. Solve: The fact that atmospheric pressure at sea level is 101.3 kPa = 101,300 N/m 2 means that the weight of the atmosphere over each square meter of surface is 101,300 N. Thus the mass of air over each square meter is m = (101,300 N)/g = (101,300 N)/(9.80 m/s 2 ) = 10,340 kg per m 2 . Multiplying by the earth’s surface
area will give the total mass. Using Re = 6.27 × 106 m for the earth’s radius, the total mass of the atmosphere is
M air = Aearth m = (4π Re2 )m = 4π (6.37 × 106 m) 2 (10,340 kg/m 2 ) = 5.27 × 1018 kg
15.38. Visualize: Let d be the atmosphere’s thickness, p the atmospheric pressure on the earth’s surface, and p0 (= 0 atm) the pressure beyond the earth’s atmosphere. Solve:
The pressure at a depth d in a fluid is p = p0 + ρ gd . This equation becomes
1 atm = 0 atm + ρair gd ⇒ d =
1 atm 1.013 × 105 Pa = = 7.95 km ρair g (1.3 kg/m3 )( 9.8 m/s 2 )
15.39. Solve: (a) We can measure the atmosphere’s pressure by measuring the height of the liquid column in a barometer, because patmos = ρ gh. In the case of the water barometer, the height of the column at a pressure of 1 atm is h=
patmos 1.013 × 105 Pa = = 10.337 m ρ water g (1000 kg/m3 )( 9.8 m/s 2 )
Because the pressure of the atmosphere can vary by 5 percent, the height of the barometer must be at least be 1.05 greater than this amount. That is, hmin = 10.85 m. (b) Using the conversion rate 1 atm = 29.92 inches of Hg, we have 29.55 inches of Hg =
29.55 × 1 atm = 0.9876 atm 29.92
The height of the water in your barometer will be
h=
patmos 0.9876 × 1.013 × 105 Pa = = 10.21 m ρ water g (1000 kg/m3 )( 9.8 m/s 2 )
15.40. Model: Oil is incompressible and has a density 900 kg/m3 . Visualize: Please refer to Figure P15.40. Solve: (a) The pressure at point A, which is 0.50 m below the open oil surface, is
pA = p0 + ρ oil g (1.00 m − 0.50 m ) = 101,300 Pa + ( 900 kg m3 )( 9.8 m s 2 ) ( 0.50 m ) = 106 kPa (b) The pressure difference between A and B is
pB − pA = ( p0 + ρ gd B ) − ( p0 + ρ gd A ) = ρ g (d B − d A ) = (900 kg/m3 )(9.8 m/s 2 )(0.50 m) = 4.4 kPa Pressure depends only on depth, and C is the same depth as B. Thus pC − pA = 4.4 kPa also, even though C isn’t directly under A.
15.41. Model: Assume that oil is incompressible and its density is 900 kg/m3 . Visualize: Please refer to Figure P15.41. Solve: (a) The hydraulic lift is in equilibrium and the pistons on the left and the right are at the same level. Equation 15.11, therefore, simplifies to Fleft piston Fright piston ( FG )student ( FG )elephant = ⇒ = 2 2 Aleft piston Aright piston π ( rstudent ) π ( relephant )
⎛ (F ) ⇒ rstudent = ⎜ G student ⎜ ( FG )elephant ⎝
⎞ ⎟ ( relephant ) = ⎟ ⎠
( 70 kg ) g 1.0 m = 0.2415 m ( ) (1200 kg ) g
The diameter of the piston the student is standing on is therefore 2 × 0.2415 m = 0.48 m. (b) From Equation 15.13, we see that an additional force ΔF is required to increase the elephant’s elevation through a distance d 2 . That is,
ΔF = ρ g ( Aleft piston + Aright piston ) d 2
2 2 ⇒ ( 70 kg ) ( 9.8 m s 2 ) = ( 900 kg m 3 )( 9.8 m s 2 )π ⎡( 0.2415 m ) + (1.0 m ) ⎤ d 2 ⎣ ⎦ ⇒ d 2 = 0.0234 m The elephant moves 2.3 cm.
15.42. Model: Assume that the oil is incompressible and its density is 900 kg/m3 . Visualize:
Solve:
The pressures p1 and p2 are equal. Thus,
p0 +
F1 F F F = p0 + 2 + ρ gh ⇒ 1 = 2 + ρ gh A1 A2 A1 A2
With F1 − m1 g , F2 = 4m2 g , A1 = π r12 , and A2 = π r2 2 , we have 12
⎞ m1 g 4m2 g ⎛ 4m g ⎞ ⎛ m g = + ρ gh ⇒ r2 = ⎜ 2 ⎟ ⎜ 1 2 − ρ gh ⎟ r π r12 π r22 π π ⎝ ⎠ ⎝ 1 ⎠ Using
m1 = 55 kg, m2 = 110 kg, r1 = 0.08 m, ρ = 900 kg/m3 ,
and
−1 2
h = 1.0 m,
the
r2 = 0.276 m. The diameter is 55 cm. Assess: Both pistons are too small to hold the people as shown, but the ideas are correct.
calculation
yields
15.43. Model: Assume that the oil is incompressible. Visualize:
Solve:
When the force F1 balances the gravitational force mg, the pressure p2 is related to the pressure p1 as
p1 = p2 + ρ gh When F1 is increased to F1′, the weight is raised higher through a distance d 2 and the left piston is lowered through a distance d1. The pressures p′2 and p1′ are now related through
p1′ = p′2 + ρ g ( h + d1 + d 2 ) Subtracting these two equations,
p1′ − p1 = ( p′2 − p2 ) + ρ g ( d1 + d 2 )
Because p1 = F1 A1 , p1′ = F1′ A1 , and p2 = p′2 = mg A2 , the above equation simplifies to
( F1′ − F1 )
A1 = ρ g ( d1 + d 2 ) ⇒ ΔF = ρ g ( A1d1 + A1d 2 )
Since the oil is incompressible, A1d1 = A2 d 2 . The equation for ΔF thus becomes ΔF = ρ g ( A2 d 2 + A1d 2 ) = ρ g ( A1 + A2 ) d 2
15.44. Model: Water and mercury are incompressible and immiscible liquids. Visualize:
The water in the left arm floats on top of the mercury and presses the mercury down from its initial level. Because points 1 and 2 are level with each other and the fluid is in static equilibrium, the pressure at these two points must be equal. If the pressures were not equal, the pressure difference would cause the fluid to flow, violating the assumption of static equilibrium. Solve: The pressure at point 1 is due to water of depth d w = 10 cm:
p1 = patmos + ρ w gd w Because mercury is incompressible, the mercury in the left arm goes down a distance h while the mercury in the right arm goes up a distance h. Thus, the pressure at point 2 is due to mercury of depth d Hg = 2h :
p2 = patmos + ρ Hg gd Hg = patmos + 2 ρ Hg gh Equating p1 and p2 gives patmos + ρ w gd w = patmos + 2 ρ Hg gh ⇒ h =
1 ρw 1 1000 kg m3 dw = (10 cm ) = 3.7 mm 2 ρ Hg 2 13,600 kg m3
The mercury in the right arm rises 3.7 mm above its initial level.
15.45. Model: Glycerin and ethyl alcohol are incompressible and do not mix. Visualize:
Solve: The alcohol in the left arm floats on top of the denser glycerin and presses the glycerin down distance h from its initial level. This causes the glycerin to rise distance h in the right arm. Points 1 and 2 are level with each other and the fluids are in static equilibrium, so the pressures at these two points must be equal:
p1 = p2 ⇒ p0 + ρ eth gd eth = p0 + ρ gly gd gly ⇒ ρ eth g (20 cm) = ρ gly g (2h) h=
1 ρ eth 1 790 kg/m3 (20 cm) = (20 cm) = 6.27 cm 2 ρgly 2 1260 kg/m3
You can see from the figure that the difference between the top surfaces of the fluids is Δy = 20 cm − 2h = 20 cm − 2(6.27 cm) = 7.46 cm ≈ 7.5 cm
15.46.
Model: The water is in hydrostatic equilibrium. Visualize: Please refer to figure P15.46. Solve: (a) Can 2 has moved down with respect to Can 1 since the water level in Can 2 has risen. Since the total volume of water stays constant, the water level in Can 1 has fallen by the same amount. The water level is equalized in the two cans at the middle of the height change, so the change in height of the water is half the relative change in height of the cans. Can 2 has moved relative to Can 1 (6.5 cm − 5.0 cm) × 2 = 3.0 cm down. (b) The water level in Can 1 has fallen by the same amount. The new level is 5.0 − 1.5 cm = 3.5 cm Assess: The two cans are an inexpensive method of measuring relative changes in height.
15.47.
Visualize:
The figure shows a dam with water height d. We chose a coordinate system with the origin at the bottom of the dam. The horizontal slice has height dy, width w, and area dA = wdy. The slice is at the depth of d − y. Solve: (a) The water exerts a small force dF = pdA = pwdy on this small piece of the dam, where p = ρ g (d − y ) is the pressure at depth d − y. Altogether, the force on this small horizontal slice at position y is dF = ρ gw(d − y ) dy. Note that a force from atmospheric pressure is not included. This is because atmospheric pressure exerts a force on both sides of the dam. The total force of the water on the dam is found by adding up all the small forces dF for the small slices dy between y = 0 m and y = d . This summation is expressed as the integral d
Ftotal
d
d
1 ⎞ 1 ⎛ = ∫ dF = ∫ dF = ρ gw∫ ( d − y ) dy = ρ gw ⎜ yd − y 2 ⎟ = ρ gwd 2 2 ⎝ ⎠0 2 all slices 0 0
(b) The total force is
Ftotal =
1 2
(1000 kg
m3 )( 9.8 m s 2 ) (100 m )( 60 m ) = 1.76 × 109 N 2
15.48.
Visualize:
The figure shows an aquarium tank with water at a height d. We chose a coordinate system with the origin at the bottom of the tank. The slice has a height dy, length l, and area dA = ldy. It is at depth d − y. Solve: (a) Pressure in excess of atmospheric pressure of the water on the bottom is pbottom = ρ water gd . Atmospheric pressure is ignored because it exerts an equal force on both sides of the bottom. Therefore, the force of the water on the bottom is
Fbottom = pbottom ( lw ) = ( ρ water gd )( lw ) = (1000 kg m3 )( 9.8 m s 2 ) ( 0.40 m )(1.0 m )( 0.35 m ) = 1.37 kN (b) The water exerts a small force dF = pdA = pldy on a small piece of the window, where p = ρ g (d − y ) is the excess pressure at a depth of d − y. The force on this small horizontal slice at position y is
dF = ρ gl ( d − y ) dy. The total force on the front window is d
Ftotal = ∫ dF = ∫ ρ water gL ( d − y ) dy = ρ water gl ( yd − 12 y 2 ) = 0
=
1 2
(1000 kg
d
0
1 ρ water gld 2 2
m3 )( 9.8 m s 2 ) (1.00 m )( 0.40 m ) = 0.78 kN 2
15.49.
Visualize:
The figure shows a small column of air of thickness dz, of cross-sectional area A = 1 m 2 , and of density ρ ( z ). The column is at a height z above the surface of the earth. Solve: (a) The atmospheric pressure at sea level is 1.013 × 105 Pa. That is, the weight of the air column with a
1 m 2 cross section is 1.013 × 105 N. Consider the weight of a 1 m 2 slice of thickness dz at a height z. This slice has volume dV = Adz = (1 m 2 ) dz, so its weight is dw = ( ρ dV ) g = ρ g (1 m 2 )dz = ρ 0e− z/z0 g (1 m 2 )dz. The total weight of the 1 m 2 column is found by adding all the dw. Integrating from z = 0 to z = ∞, ∞
w = ∫ ρ 0 g (1 m 2 )e− z z0 dz 0
= ( − ρ 0 g (1 m 2 ) z0 ) ⎡⎣e− z z0 ⎤⎦
∞ 0
= ρ 0 g (1 m 2 ) z0 Because w = 101,300 N = ρ 0 g (1 m 2 ) z0 ,
z0 =
101,300 N = 8.08 × 103 m (1.28 kg/m3 )( 9.8 m/s2 ) (1.0 m2 )
(b) Using the density at sea level from Table 15.1,
ρ = (1.28 kg m3 ) e − z /(8.08×10
3
This is 82% of ρ 0 .
m)
= (1.28 kg m3 ) e −1600 m/(8.08×10
3
m)
= 1.05 kg m3
15.50. Model: The buoyant force on the ceramic statue is given by Archimedes’ principle. Visualize:
Solve: The gravitational force on the statue is the 28.4 N registered on the scale in air. In water, the gravitational force on the statue is balanced by the sum of the buoyant force FB and the spring’s force on the statue. That is,
( FG )statue = FB + Fspring on statue ⇒ 28.4 N = ρ wVstatue g + 17.0 N ⇒ Vstatue = ⇒ ρstatue =
( mstatue g ) ρ w = ( 28.4 N ) (1000 kg (11.4 N ) (11.4 N )
m3 )
11.4 N mstatue = g ρw ρstatue
= 2.49 × 103 kg m3
15.51. Model: The buoyant force on the cylinder is given by Archimedes’ principle. Visualize:
Solve: (a) Initially, as it floats, the cylinder is in static equilibrium, with the buoyant force balancing the gravitational force on the cylinder. The volume of displaced liquid is Ah, so
FB = ρ liq ( Ah) g = FG Force F pushes the cylinder down distance x, so the submerged length is h + x and the volume of displaced liquid is A(h + x ). The cylinder is again in equilibrium, but now the buoyant force balances both the gravitational force and force F. Thus
FB = ρ liq ( A(h + x)) g = FG + F Since ρ liq ( Ah) g = FG , we’re left with
F = ρ liq Agx (b) The amount of work dW done by force F to push the cylinder from x to x + dx is dW = Fdx = ( ρ liq Agx)dx.
To push the cylinder from xi = 0 m to xf = 10 cm = 0.10 m requires work xf
xf
xi
xi
W = ∫ F dx = ρliq Ag ∫ x dx = 12 ρ liq Ag ( xi2 − xf2 ) = 12 (1000 kg/m3 )π (0.020 m) 2 (9.8 m/s 2 )(0.10 m) 2 = 0.62 J
15.52. Model: The buoyant force on the cylinder is given by Archimedes’ principle. Visualize:
Let d1 be the length of the cylinder in the less-dense liquid with density ρ1 , and d 2 be the length of the cylinder in the more-dense liquid with density p2 . Solve: The cylinder is in static equilibrium, so
∑F
y
= FB1 + FB2 − FG = 0 N ⇒ ρ1 ( Ad1 ) g + ρ 2 ( Ad 2 ) g = ρ A ( d1 + d 2 ) g ⇒
ρ1 ρ d1 + d 2 = ( d1 + d 2 ) ρ2 ρ2
Since l = d1 + d 2 ⇒ d1 = l − d 2 , we can simplify the above equation to obtain
⎛ ρ − ρ1 ⎞ d2 = ⎜ ⎟l ⎝ ρ 2 − ρ1 ⎠ That is, the fraction of the cylinder in the more dense liquid is f = ( ρ − ρ1 ) ( ρ 2 − ρ1 ) . Assess:
As expected f = 0 when ρ is equal to ρ1 , and f = 1 when ρ = ρ 2 .
15.53. Model: The buoyant force on the cylinder is given by Archimedes’ principle. Visualize:
Solve:
The tube is in static equilibrium, so
∑F
y
= FB − ( FG ) tube − ( FG )Pb = 0 N ⇒ ρ liquid A ( 0.25 m ) g = ( 0.030 kg ) g + ( 0.250 kg ) g ⇒ ρ liquid =
Assess:
( 0.280 kg ) ( 0.280 kg ) = = 8.9 × 102 A ( 0.25 m ) π ( 0.020 m )2 ( 0.25 m )
This is a reasonable value for a liquid.
kg m 3
15.54. Model: Archimedes’ Principle determines the buoyant force. Visualize:
Solve: The plastic hemisphere will hold the most weight when its rim is at the surface of the water. The buoyant force balances the gravitational force on the bowl and rock.
∑F
y
= FB − ( FG )rock − ( FG )bowl = 0 N
Thus 3⎞ ⎛ 1 ⎞⎛ 4 mg = pwVbowl g − m bowl g = (1000 kg/m3 ) ⎜ ⎟⎜ π ( 0.040 m ) ⎟ ( 9.8 m/s 2 ) − ( 0.021 kg ) ( 9.8 m/s 2 ) ⇒ m = 0.155 kg 2 3 ⎝ ⎠⎝ ⎠
Assess:
Putting a rock as big as 155 g in an 8 cm diameter bowl before it sinks is reasonable.
15.55. Model: The buoyant force is determined by Archimedes’ principle. The spring is ideal. Visualize:
Solve: The spring is stretched by the same amount that the cylinder is submerged. The buoyant force and spring force balance the gravitational force on the cylinder.
∑F
y
= FB + FS − mg = 0 N ⇒ pw Ayg + ky = mg
y=
(1.0 kg ) ( 9.8 m/s 2 ) mg = pw Ag + k (1000 kg/m3 )π ( 0.025 m )2 ( 9.8 m/s 2 ) + 35 N/m
= 0.181 m = 18.1 cm Assess: This is difficult to assess because we don’t know the height h of the cylinder and can’t calculate it without the density of the metal material.
15.56. Model: The balloon displaces air, so the air exerts an upward buoyant force on the balloon. The buoyant force on the balloon is given by Archimedes’ principle.
Visualize:
FB is the buoyant force on the balloon, ( FG ) b is the gravitational force on the balloon, ( FG ) He is the gravitational force on the helium gas, and FG is the gravitational force on the mass tied to the balloon. Solve:
The volume of the balloon Vballoon = 34 π ( 0.10 m ) = 4.1888 × 10−3 m3 . We have 3
FB = ρ air gVballoon = (1.28 kg m3 )( 9.8 m s 2 )( 4.188 × 10−3 m3 ) = 0.05254 N
( FG )He = ρ He gVballoon = ( 0.18 kg m3 )( 9.8 m s 2 )( 4.188 × 10−3 m3 ) = 0.007387 N ( FG )b = ( 0.0010 kg ) g = 0.0098 N For the balloon to be in static equilibrium,
FB = ( FG )He + ( FG )b + FG = 0.05254 N=0.01719 N + mg ⇒ m = 0.0036 kg = 3.6 g
15.57. Model: The buoyant force on the can is given by Archimedes’ principle. Visualize:
The length of the can above the water level is d, the length of the can is L, and the cross-sectional area of the can is A. Solve: The can is in static equilibrium, so
∑F
y
= FB − ( FG )can − ( FG ) water = 0 N ⇒ ρ water A ( L − d ) g = ( 0.020 kg ) g + mwater g
The mass of the water in the can is
355 × 10 ⎛V ⎞ mwater = ρ water ⎜ can ⎟ = (1000 kg m3 ) 2 2 ⎝ ⎠
−6
m3
= 0.1775 kg
⇒ ρ water A ( L − d ) = 0.020 kg + 0.1775 kg = 0.1975 kg ⇒ d − L = − Because
Vcan = π ( 0.031 m ) L = 355 × 10−6 m3 , 2
L = 0.1176 m.
Using
0.1975 kg = 0.0654 m ρ water A this
value
of
L,
we
get
d = 0.0522 m ≈ 5.2 cm. Assess: d L = 5.22 cm 11.76 cm = 0.444, thus 44.4% of the length of the can is above the water surface. This is reasonable.
15.58. Model: The buoyant force on the boat is given by Archimedes’ principle. Visualize:
The minimum height of the boat that will enable the boat to float in perfectly calm water is h. Solve: The boat barely floats if the water comes completely to the top of the sides. In this case, the volume of displaced water is the volume of the boat. Archimedes’ principle in equation form is FB = ρ wVboat g . For the boat
to float FB = ( FG )boat . Let us first calculate the gravitational force on the boat:
( FG )boat = ( FG )bottom + 2 ( FG )side 1 + 2 ( FG )side 2 , where
( FG )bottom = ρsteelVbottom g = ( 7900 kg m3 )( 5 × 10 × 0.02 m3 ) g = 7900g N ( FG )side 1 = ρsteelVside 1g = ( 7900 kg m3 )( 5 × h × 0.005 m3 ) g = 197.5gh N ( FG )side 2 = ρsteelVside 2 g = ( 7900 kg m3 )(10 × h × 0.005 m3 ) g = 395gh ⇒ ( FG )boat = ⎡⎣7900g + 2 (197.5gh ) + 2 ( 395gh ) ⎤⎦ N = ( 7900+1185h ) g N Going back to the Archimedes’ equation and remembering that h is in meters, we obtain
ρ wVboat g = ( 7900 + 1185h ) g N ⇒ (1000 ) (10 × 5 × ( h + 0.020 ) ) = 7900 + 1185h ⇒ h = 14.1 cm
15.59. Model: Treat the water as an ideal fluid obeying Bernoulli’s equation. There is a streamline connecting point 1 in the wider pipe with point 2 in the narrower pipe.
Visualize:
Solve:
Bernoulli’s equation, Equation 15.28, relates the pressure, water speed, and heights at points 1 and 2:
p1 + 12 ρ v12 + ρ gy1 = p2 + 12 ρ v22 + ρ gy2 For no height change y1 = y2 . The flow rate is given as 5.0 L/s, which means
Q = v1 A1 = v2 A2 = 5.0 × 10−3 m3 s ⇒ v1 =
5.0 × 10−3 m3 s
π ( 0.05 m )
2
= 0.6366 m s
2
⇒ v2 = v1
A1 ⎛ 0.050 m ⎞ = ( 0.6366 m s ) ⎜ ⎟ = 2.546 m s A2 ⎝ 0.025 m ⎠
Bernoulli’s equation now simplifies to
p1 + 12 ρ v12 = p2 + 12 ρ v22 2 2 ⇒ p1 = p2 + 12 ρ ( v22 − v12 ) = 50 × 103 Pa + 12 (1000 kg m3 ) ⎡( 2.546 m s ) − ( 0.6366 m s ) ⎤ = 53 kPa ⎣ ⎦
Assess:
Reducing the pipe size reduces the pressure because it makes v2 > v1.
15.60. Model: The two pipes are identical. Visualize:
Solve:
The water speed is the same in both pipes. The flow rate Q = 3.0 × 106 L/min = 2 ( vA )
( 3.0 ×10 )(10 ) ⎛⎜⎝ 601 ⎞⎟⎠ m /s 6
Q ⇒v= = 2A
−3
2π (1.5 m )
3
2
= 3.5 m/s
15.61. Model: Treat the water as an ideal fluid obeying Bernoulli’s equation. A streamline begins in the bigger size pipe and ends at the exit of the narrower pipe.
Visualize: Please see Figure P15.61. Let point 1 be beneath the standing column and point 2 be where the water exits the pipe. Solve: (a) The pressure of the water as it exits into the air is p2 = patmos . (b) Bernoulli’s equation, Equation 15.28, relates the pressure, water speed, and heights at points 1 and 2:
p1 + 12 ρ v12 + ρ gy1 = p2 + 12 ρ v22 + ρ gy2 ⇒ p1 − p2 = 12 ρ ( v22 − v12 ) + ρ g ( y2 − y1 ) From the continuity equation,
v1 A1 = v2 A2 = ( 4 m s ) ( 5 × 10−4 m 2 ) ⇒ v1 (10 × 10−4 m 2 ) = 20 × 10−4 m3 s ⇒ v1 = 2.0 m s Substituting into Bernoulli’s equation, 2 2 m3 ) ⎡( 4.0 m s ) − ( 2.0 m s ) ⎤ + (1000 kg m 3 ) ( 9.8 m s )( 4.0 m ) ⎣ ⎦ = 6000 Pa + 39,200 Pa = 45 kPa
p1 − p2 = p1 − patmos =
1 2
(1000 kg
But p1 − p2 = ρ gh, where h is the height of the standing water column. Thus h=
45 × 103 Pa = 4.6 m (1000 kg/m3 )( 9.8 m/s2 )
15.62. Model: Treat the water as an ideal fluid obeying Bernoulli’s equation. A streamline begins at the faucet and continues down the stream.
Visualize:
The pressure at point 1 is p1 and the pressure at point 2 is p2 . Both p1 and p2 are atmospheric pressure. The velocity and the area at point 1 are v1 and A1 and they are v2 and A2 at point 2. Let the distance of point 2 below point 1 be d. Solve: The flow rate 3
Q = v1 A =
2.0 × 10−4 m 3 2.0 × 1000 × 10−6 m3 2 = 1.0 m/s = 2.0 × 10−4 m ⇒ v1 = 2 s 10 s π ( 0.0080 m )
Bernoulli’s equation at points 1 and 2 is
p1 + 12 ρ v12 + ρ gy1 = p2 + 12 ρ v22 + ρ gy2 ⇒ ρ gd = 12 ρ ( v22 − v12 )
From the continuity equation,
v1 A1 = v2 A2 ⇒ (1.0 m s )π ( 8.0 × 10−3 m ) = v2π ( 5.0 × 10−3 m ) ⇒ v2 = 2.56 m s 2
2
Going back to Bernoulli’s equation, we have 2 2 gd = 12 ⎡( 2.56 m s ) − (1.0 m s ) ⎤ ⇒ d = 0.283 m ≈ 28 cm ⎣ ⎦
15.63. Model: Treat the air as an ideal fluid obeying Bernoulli’s equation. Solve: (a) The pressure above the roof is lower due to the higher velocity of the air. (b) Bernoulli’s equation, with yinside ≈ youtside , is 2
pinside = poutside + 12 ρair v 2 ⇒ Δp =
1 1 ⎛ 130 × 1000 m ⎞ ρ air v 2 = (1.28 kg m3 ) ⎜ ⎟ = 835 Pa 2 2 ⎝ 3600 s ⎠
The pressure difference is 0.83 kPa (c) The force on the roof is ( Δp ) A = ( 835 Pa )( 6.0 m × 15.0 m ) = 7.5 × 104 N. The roof will blow up, because pressure inside the house is greater than pressure on the top of the roof.
15.64. Model: The ideal fluid obeys Bernoulli’s equation. Visualize: Please refer to Figure P15.64. There is a streamline connecting point 1 in the wider pipe on the left with point 2 in the narrower pipe on the right. The air speeds at points 1 and 2 are v1 and v2 and the crosssectional area of the pipes at these points are A1 and A2 . Points 1 and 2 are at the same height, so y1 = y2 . Solve:
The volume flow rate is Q = A1v1 = A2v2 = 1200 × 10−6 m3 s. Thus v2 =
1200 × 10−6 m3 s = 95.49 m s π (0.0020 m) 2
v1 =
3 1200 × 10−6 m
π ( 0.010 m )
2
s = 3.82 m s
Now we can use Bernoulli’s equation to connect points 1 and 2:
p1 + 12 ρ v12 + ρ gy1 = p2 + 12 ρ v22 + ρ gy2 ⇒ p1 − p2 = 12 ρ ( v22 − v12 ) + ρ g ( y2 − y1 ) =
1 2
(1.28 kg
2 2 m3 ) ⎡( 95.49 m s ) − ( 3.82 m s ) ⎤ + 0 Pa = 5.83 kPa ⎣ ⎦
Because the pressure above the mercury surface in the right tube is p2 and in the left tube is p1 , the difference in the pressures p1 and p2 is ρ Hg gh. That is, p1 − p2 = 5.83 kPa = ρ Hg gh ⇒ h =
5.83 × 103 Pa = 4.4 cm (13,600 kg m3 )(9.8 m s 2 )
15.65. Model: The ideal fluid (that is, air) obeys Bernoulli’s equation. Visualize: Please refer to Figure P15.65. There is a streamline connecting points 1 and 2. The air speeds at points 1 and 2 are v1 and v2 , and the cross-sectional areas of the pipes at these points are A1 and A2 . Points 1 and 2 are at the same height, so y1 = y2 . Solve: (a) The height of the mercury is 10 cm. So, the pressure at point 2 is larger than at point 1 by
ρ Hg g ( 0.10 m ) = (13,600 kg m3 )( 9.8 m s 2 ) ( 0.10 m ) = 13,328 Pa ⇒ p2 = p1 + 13,328 Pa
Using Bernoulli’s equation,
p1 + 12 ρair v12 + ρ air gy1 = p2 + 12 ρ air v22 + ρ air gy2 ⇒ p2 − p1 = 12 ρair ( v12 − v22 ) ⇒ v12 − v22 =
2 ( p2 − p1 )
ρ air
=
2 (13,328 Pa )
(1.28 kg/m ) 3
= 20,825 m 2 s 2
From the continuity equation, we can obtain another equation connecting v1 and v2 :
π ( 0.005 m ) A A1v1 = A2v2 ⇒ v1 = 2 v2 = v = 25 v2 2 2 A1 π ( 0.001 m ) 2
Substituting v1 = 25v2 in the Bernoulli equation, we get
( 25 v2 )
2
− v22 = 20,825 m 2 s 2 ⇒ v2 = 5.78 m s
Thus v2 = 5.8 m/s and v1 = 25v2 = 144 m/s. (b) The volume flow rate A2v2 = π ( 0.0050 m ) ( 5.78 m s ) = 4.5 × 10−3 m3 s. 2
15.66. Model: Treat the water as an ideal fluid that obeys Bernoulli’s equation. There is a streamline from the top of the water to the hole. Visualize: Please refer to Figure P15.66. The top of the water (at y1 = h) and the hole (at y2 = y ) are at atmospheric pressure. The speed of the water at the top is zero because the tank is kept filled. Solve: (a) Bernoulli’s equation connecting the two points is
0 + ρ gh = 12 ρ v 2 + ρ gy ⇒ v = 2 g ( h − y ) (b) For a particle shot horizontally from a height y with speed v, the range can be found using kinematic equations. For the y-motion, using t0 = 0 s, we have y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) ⇒ 0 m = y + ( 0 m s ) t1 + 12 ( − g ) t12 ⇒ t1 = 2 y g 2
For the x-motion, x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) ⇒ x = 0 m + vt1 + 0 m ⇒ x = v 2 y g 2
(c) Combining the results of (a) and (b), we obtain x = 2g ( h − y ) 2 y g = 4 y ( h − y )
To find the maximum range relative to the vertical height, 1 dx h =0⇒ ⎡⎣ 4 ( h − y ) − 4 y ⎤⎦ = 0 ⇒ y = 2 dy 4y(h − y)
With y = 12 h, the maximum range is h⎞ ⎛ h ⎞⎛ xmax = 4 ⎜ ⎟⎜ h − ⎟ = h 2⎠ ⎝ 2 ⎠⎝
15.67. Model: Treat water as an ideal fluid that obeys Bernoulli’s equation. There is a streamline connecting the top of the tank with the hole. Visualize: Please refer to Figure P15.67. We placed the origin of the coordinate system at the bottom of the tank so that the top of the tank (point 1) is at a height of h + 1.0 m and the hole (point 2) is at a height h. Both points 1 and 2 are at atmospheric pressure. Solve: (a) Bernoulli’s equation connecting points 1 and 2 is
p1 + 12 ρ v12 + ρ gy1 = p2 + 12 ρ v22 + ρ gy2 ⇒ patmos + 12 ρ v12 + ρ g ( h + 1.0 m ) = patmos + 12 ρ v22 + ρ gh ⇒ v22 − v12 = 2g (1.0 m ) = 19.6 m 2 s 2 Using the continuity equation A1v1 = A2v2 ,
π ( 2.0 × 10 m ) ⎛A ⎞ v2 v1 = ⎜ 2 ⎟ v2 = v2 = 2 A 250,000 π (1.0 m ) ⎝ 1⎠ −3
2
Because v1 v2 , we can simply put v1 ≈ 0 m/s. Bernoulli’s equation thus simplifies to
v22 = 19.6 m 2 s 2 ⇒ v2 = 4.43 m s Therefore, the volume flow rate through the hole is Q = A2v2 = π ( 2.0 × 10−3 m ) ( 4.43 m s ) = 5.56 × 10−5 m3 s = 3.3 L min 2
(b) The rate at which the water level will drop is v2 4.43 m s = = 1.77 × 10−2 mm s = 1.06 mm min v1 = 250,000 250,000 Assess: Because the hole through which water flows out of the tank has a diameter of only 4.0 mm, a drop in the water level at the rate of 1.06 mm/min is reasonable.
15.68. Model: The aquarium creates tensile stress. Solve:
Weight of the aquarium is
FG = mg = ρ waterVg = (1000 kg m3 )(10 m3 )( 9.8 m s 2 ) = 9.8 × 104 N
where we have used the conversion 1 L = 10−3 m 3 . The weight supported by each wood post is 14 (9.8 × 104 N) = 2.45 × 104 N. The cross-sectional area of each post is A = ( 0.040 m ) = 1.6 × 10−3 m 2 . Young’s modulus for the 2
wood is F/A FL = ΔL / L AΔL 4 2.45 10 N 0.80 m × ( ) ( ) FL ⇒ ΔL = = = 1.23 × 10−3 m = 1.23 mm 2 10 −3 AY (1.6 × 10 m )(1× 10 N/m 2 ) Y = 1 × 1010 N m 2 =
Assess:
A compression of 1.23 mm due to a weight of 2.45 × 104 N is reasonable.
15.69. Model: The water pressure applies a volume stress to the sphere. Solve:
The volume change is ΔV = −1.0 × 10−3 V . The volume stress is
F ΔV −1.0 × 10−3V = p = −B = − ( 7 × 1010 N m 2 ) = 7.0 × 107 Pa A V V Using p = p0 + ρ gh, we get
7.0 × 107 Pa = 1.013 × 105 Pa + (1030 kg m3 )( 9.8 m s 2 ) d ⇒ d = 6.9 km Assess: A pressure of 7.0 × 107 Pa = 690 atm causes only a volume change of 0.1%. This is reasonable because liquids and solids are nearly incompressible.
15.70. Model: Pressure applies a volume stress to water in the cylinder. Solve:
The volume strain of water due to the pressure applied is
ΔV p 2 × 106 Pa =− =− = −1.0 × 10−3 V B 0.20 × 1010 Pa ⇒ ΔV = V ′ − V = − (1.0 × 10−3 )(1.30 m 3 ) = −1.30 × 10−3 m3 = −1.3 L As the safety plug on the top of the cylinder bursts, the water comes back to atmospheric pressure. The volume of water that comes out is 1.30 L.
15.71. Model: Air is an ideal gas and obeys Boyle’s law. Visualize: Please refer to Figure CP15.71. h is the length of the air column when the mercury fills the cylinder to the top. A is the cross-sectional area of the cylinder. Solve: For the column of air, Boyle’s law is p0V0 = p1V1 , where p0 and V0 are the pressure and volume before any mercury is poured, and p1 and V1 are the pressure and volume when mercury fills the cylinder above the air. Using p1 = p0 + ρ Hg g (1.0 m − h ) , Boyle’s law becomes
p0V0 = ⎡⎣ p0 + ρ Hg g (1.0m − h ) ⎤⎦ V1 ⇒ p0 A (1.0 m ) = ⎡⎣ p0 + ρ Hg g (1.0m − h ) ⎤⎦ Ah p 1.013 × 105 Pa p0 (1.0m − h ) = ρ Hg g (1.0m − h ) h ⇒ h = 0 = = 0.76 m = 76 cm ρ Hg g (13,600 kg/m3 )( 9.8 m/s 2 )
15.72. Model: The buoyant force on the cone is given by Archimedes’ principle. Visualize:
Solve: It may seem like we need the formula for the volume of a cone. You can use that formula if you know it, but it isn’t essential. The volume is clearly the area of the base multiplied by the height multiplied by some constant. That is, the cone shown above has V = aAd , where a is some constant. But the radius of the base is r = d tan α , where α is the angle of the apex of the cone, and A = π r 2 , making A proportional to d 2 . Thus the volume of a cone of height d is V = cd 3 , where c is a constant. Because the cone is floating in static equilibrium, we must have FB = FG . The cone’s density is ρ 0 , so the gravitational force on it is FG = ρ0Vg = ρ0cl 3 g. The buoyant force is the gravitational force on the displaced fluid. The volume of displaced fluid is the full volume of the cone minus the volume of the cone of height h above the water, or Vdisp = cl 3 = ch3 . Thus
FB = ρf Vdisp g = ρ f c(l 3 − h3 ) g , and the equilibrium condition is 1/ 3
⎛ h3 ⎞ h ⎛ ρ ⎞ FB = FG ⇒ ρ f c(l 3 − h3 ) g = ρ 0cl 3 g ⇒ ρ f ⎜1 − 3 ⎟ = ρ 0 ⇒ = ⎜1 − 0 ⎟ l ⎠ l ⎝ ρf ⎠ ⎝
15.73. Model: The grinding wheel is a uniform disk. We will use the model of kinetic friction and hydrostatics. Visualize: Please refer to figure CP15.73. Solve: This is a three-part problem. First find the desired angular acceleration, then use that to find the force applied by each brake pad, then finally the needed oil pressure. The angular acceleration required to stop the wheel is found using rotational kinematics. 1 Δω ωf − ωi 0 rad/s − 900 × 2π × 60 rad/s α= = = = −18.85 rad/s 2 5.0 s Δt Δt Each brake pad applies a frictional force f k = μ k n to the wheel. The normal force is equal to the force applied by the piston by Newton’s third law. Rotational dynamics can be used to find the magnitude of the force. f k is applied 12 cm from the rotation axis on both sides of the disk. τ net = Iα
⎛1 ⎞ −2 f k ( 0.12 m ) = ⎜ MR 2 ⎟α ⎝2 ⎠
(15 kg )( 0.13 m ) ( −18.85 rad/s2 ) MR 2α ⇒n= = = 16.6 N −4μ k ( 0.12 m ) −4 ( 0.60 )( 0.12 m ) 2
The oil pressure required to generate this much force at each brake pad is F 16.6 N = 53 kPa p= = A π ( 0.010 m )2 relative to atmospheric pressure. Assess: The required oil pressure is about half an atmosphere, which is quite reasonable.
15.74. Model: The buoyant force on the cylinder is given by Archimedes’ principle. Visualize:
A is cross-sectional area of the cylinder. Solve: (a) The gravitational force on the cylinder is FG = ρ0 ( Al ) g and the weight of the displaced fluid is
FB = ρf ( Ah ) g. In static equilibrium, FB = FG and we can write
ρf Ahg = ρ0 Alg ⇒ h = ( ρ0 ρf ) l (b) Now the volume of the displaced liquid is A(h − y ). Applying Newton’s second law in the y-direction,
∑F
y
= − FG + FB = − ρ0 Alg + ρf A ( h − y ) g
Using ρf Ahg = ρ0 Alg from part (a), we find
( Fnet ) y = − ρf Agy
(c) The result in part (b) is F = −ky, where k = ρ f Ag . This is Hooke’s law. (d) Since the cylinder’s equation of motion is determined by Hooke’s law, the angular frequency for the resulting simple harmonic motion is ω = k m , and the period is
T=
2π
ω
= T = 2π
m m ρ0 Al h = 2π = 2π = 2π k ρf Ag ρf Ag g
where we have used the expression for h from part (a). (e) The oscillation period for the 100-m-tall iceberg ( pice = 917 kg/m3 ) in sea water is T = 2π
ρ 0l h = 2π = 2π ρf g g
( 917 kg/m ) (100 m ) = 18.9 s (1030 kg/m )( 9.8 m/s ) 3
3
2
15.75. Model: A streamline connects every point on the surface of the liquid to a point in the drain. The drain diameter is much smaller than the tank diameter (r R ). Visualize:
Solve: The pressures at the surface and drain (points 1 and 2) are equal to one atmosphere. When the liquid is a depth y, Bernoulli’s principle comparing points 1 and 2 is 1 1 p1 + ρ v12 + ρ gy = p2 + ρ v22 + ρ g ( 0 ) 2 2 ⇒ v22 = v12 + 2 gy The flow rate through the drain is the same as through a horizontal layer in the tank: A Q = v1 A1 = v2 A2 ⇒ v1 = v2 2 A1 Thus the velocity of the liquid through the drain is 2
⎛ A ⎞ 2 gy v22 = ⎜ v2 2 ⎟ + 2 gy ⇒ v2 = π r2 ⎝ A1 ⎠ 1− π R2 Since r R, v2 ≈ 2 gy , and the flow rate through the drain is
Q ≈ π r 2 2 gy The flow rate gives the volume of liquid flowing out per unit time. The inverse gives the time needed for a unit volume of liquid to flow out. For the volume of liquid to decrease by dV requires a time 2 − R 2 dy dV − (π R dy ) = = dt = Q π r 2 2 gy 2g r 2 y Note that dV < 0 implies the volume of liquid is decreasing. Integrating both sides from the initial condition (t = 0, y = d ) to the final condition (t , y = 0) yields 0
R2 2 R 2 1 2 0 R 2 2d − 12 = − = 2 y dy y d r g 2 gr 2 ∫d 2 gr 2 Assess: The time for the tank to drain depends on the ratio of the cross-sectional areas of the tank to drain, which makes sense, as well as on the strength of the free-fall acceleration. Note that if g were larger (say we were on Jupiter) the time to drain would be shorter, while if the tank were taller (larger d) the time would be longer. t=−
15.76.
Visualize:
The column of air has a cross-sectional area A. We chose the origin of the coordinate system at z = 0 m and denote the vertical axis as the z-axis. Solve: (a) The pressure at height z is pz . The pressure at a height z + dz is
pz + dz = pz −
weight of air column of height dz ρ Adzg = pz − = pz − ρ gdz A A
(b) The change in pressure is dp = pz + dz − pz = − ρ gdz. The sign is minus because pressure decreases as z increases. (c) From the ideal gas law,
p
ρ
=
p0
ρ0
⇒ρ=
p ρ0 p0
The result for part (b) thus becomes dp = − ( p ρ 0 p0 ) gdz. (d) To find the pressure p at a height z, we rewrite the result of part (c) as
⎛ρ ⎞ dp = − ⎜ 0 g ⎟ dz p ⎝ p0 ⎠ Integrating both sides p
⎛ p⎞ dp ρ0 g z ρg = − dz ⇒ ln ⎜ ⎟ = − 0 z ⇒ p = p0e − ρ0 gz ∫p p ∫ p p p0 0 0 ⎝ 0⎠ 0 where z0 = p0 ρ 0 g . (e) The scale height is z0 =
p0
= p0e− z z0
(1.013 × 105 Pa ) = 8076 m. p0 = ρ0 g (1.28 kg m3 )( 9.8 m s 2 )
(f) A table showing values of p = p0e − z z0 = e − z (8076 m ) atm at selected values of z follows.
z (m) 0 1000 2000 4000 6000 8000 10,000 12,000 14,000 15,000
p (atm) 1 0.884 0.781 0.609 0.476 0.371 0.290 0.226 0.177 0.156
15-1
Fluids and Elasticity
15-2
16.1. Model: Recall the density of water is 1000 kg/m3. Solve: The mass of lead mPb = ρ PbVPb = (11,300 kg m3 )( 2.0 m3 ) = 22,600 kg . For water to have the same mass its volume must be
Vwater = Assess:
mwater
ρ water
=
22,600 kg = 22.6 m3 1000 kg m3
Since the lead is 11.3 times as dense we expect the water to take 11.3 times the volume.
16.2. Model: Assume the nucleus is spherical. Solve:
The volume of the uranium nucleus is
V = 34 π R 3 = 43 π ( 7.5 × 10−15 m ) = 1.767 × 10−42 m3 3
The density of the uranium nucleus is
ρ nucleus = Assess:
mnucleus 4.0 × 10−25 kg = = 2.3 × 1017 kg m3 Vnucleus 1.767 × 10−42 m3
This density is extremely large compared to the typical density of materials.
16.3. Solve: The volume of the aluminum cube is 10−3 m3 and its mass is mAl = ρ AlVAl = ( 2700 kg m3 )(1.0 × 10−3 m3 ) = 2.7 kg
The volume of the copper sphere with this mass is
VCu =
m 4π 2.7 kg 3 = 3.027 × 10−4 m3 ( rCu ) = Cu = 3 ρ Cu 8920 kg m3 ⎡ 3 ( 3.027 × 10−4 m3 ) ⎤ ⎥ ⇒ rCu = ⎢ 4π ⎢⎣ ⎥⎦
1
3
= 0.042 m
The diameter of the copper sphere is 0.0833 m = 8.33 cm. Assess: The diameter of the sphere is a little less than the length of the cube, and this is reasonable considering the density of copper is greater than the density of aluminum.
16.4. Model: The volume of a hollow sphere is V= Solve:
4p 3 ( rout − rin3 ) 3
We are given m = 0.690 kg, rout = 0.050 m, and we know that for aluminum r = 2700 kg/m3 .
Solve the above equation for rin . 3 rin = 3 rout −
3 = 3 rout −
V p
4 3
m/r 4 p 3
= 3 0.050 m3 −
0.690 kg/2700 kg/m3 4 p 3
= 0.040 m So the inner diameter is 8.0 cm. Assess: We are happy that the inner diameter is less than the outer diameter, and in a reasonable range.
16.5. Solve: The volume of the aluminum cube V = 8.0 × 10−6 m3 and its mass is M = ρV = (2700 kg/m3 )(8.0 × 10−6 m3 ) = 0.0216 kg = 21.6 g One mole of aluminum (27Al) has a mass of 27 g. The number of atoms is ⎛ 6.02 × 1023 atoms ⎞⎛ 1 mol ⎞ 23 N =⎜ ⎟⎜ ⎟ ( 21.6 g ) = 4.8 × 10 atoms 1 mol 27 g ⎠ ⎝ ⎠⎝ Assess:
This is almost one mole of atoms, which is a reasonable value.
16.6. Solve: The volume of the copper cube is 8.0 × 10−6 m3 and its mass is M = ρV = (8920 kg/m3 )(8.0 × 10−6 m3 ) = 0.07136 kg = 71.36 g Because the atomic mass number of Cu is 64, one mole of Cu has a mass of 64 g. The number of moles in the cube is ⎛ 1 mol ⎞ n=⎜ ⎟ ( 71.36 g ) = 1.1 mol ⎝ 64 g ⎠ Assess:
This answer is in the same ballpark as the previous exercise.
16.7. Solve: (a) The number density is defined as N V , where N is the number of particles occupying a volume V. Because Al has a mass density of 2700 kg/m3, a volume of 1 m3 has a mass of 2700 kg. We also know that the molar mass of Al is 27 g/mol or 0.027 kg/mol. So, the number of moles in a mass of 2700 kg is
⎛ 1 mol ⎞ 5 n = ( 2700 kg ) ⎜ ⎟ = 1.00 × 10 mol ⎝ 0.027 kg ⎠ The number of Al atoms in 1.00 × 105 mols is
N = nN A = (1.00 × 105 mol )( 6.02 × 1023 atoms mol ) = 6.02 × 1028 atoms Thus, the number density is N 6.02 × 1028 atoms = = 6.02 × 1028 atoms m3 V 1 m3 (b) Pb has a mass of 11,300 kg in a volume of 1 m3. Since the atomic mass number of Pb is 207, the number of moles in 11,300 kg is
⎛ 1 mole ⎞ n = (11,300 kg ) ⎜ ⎟ ⎝ 0.207 kg ⎠ The number of Pb atoms is thus N = nNA, and hence the number density is atoms N nN A ⎛ 11,300 kg ⎞ ⎛ 23 atoms ⎞ (1 mol ) = =⎜ = 3.28 ×1028 ⎟⎜ 6.02 ×10 ⎟ 3 V V mol ⎠ 1 m m3 ⎝ 0.207 kg ⎠ ⎝ Assess:
We expected to get very large numbers like this.
16.8. Solve: The mass density is ρ = M/V . The mass M of the sample is related to the number of atoms N and the mass m of each atom by M = Nm. Combining these, the atomic mass is
m=
M ρV ρ 1750 kg/m3 = = = = 3.986 × 10−26 kg/atom N N N / V 4.39 × 1028 atoms/m3
the atomic mass in m = A u, where A is the atomic mass number. Thus A=
m 3.986 × 10−26 kg = = 24 1 u 1.661 × 10−27 kg
The element’s atomic mass number is 24. Assess: This is a reasonable answer for an isotope of neon (although neon is a gas at normal temperatures), sodium, or magnesium.
4 16.9 Model: Assume the gold is shaped into a solid sphere of volume V = π r 3 . 3 Visualize: We want to know D = 2r. Because the atomic mass number of gold is 197, 1.0 mol of gold has a mass of 197 g or 0.197 kg. Table 16.1 gives ρ = 19,300 kg/m3 . Solve:
4 3 M 3 0.197 kg =3 = 0.0135 m = 1.35 cm V = π r3 = M / ρ ⇒ r = 3 3 4π ρ 4π 19,300 kg/m3 D = 2r = 2(1.35 cm) = 2.70 cm. Assess: This is about the right size for a chunk that contains one mole of material.
16.10. Solve: The mass of mercury is ⎛ 10−6 m3 ⎞ M = ρV = (13,600 kg m3 )(10 cm3 ) ⎜ = 0.136 kg = 136 g 3 ⎟ ⎝ 1 cm ⎠ and the number of moles is
n=
M 0.136 g = = 0.6766 mol M mol 201 g mol
The mass of aluminum with 0.6766 mol of Al is ⎛ 27 g ⎞ M = ( 0.6766 mol ) M mol = ( 0.6766 mol ) ⎜ ⎟ = 18.27 g = 0.01827 kg ⎝ mol ⎠ This mass M of aluminum corresponds to a volume of V=
M
ρ
=
0.01827 kg = 6.8 × 10−6 m3 = 6.8 cm3 2700 kg m3
Assess: We expected an answer in the same order of magnitude. The size of atoms doesn’t vary as much as the density of atoms from element to element.
16.11. Solve: The lowest temperature is TF = 95 TC + 32° ⇒ −127°F = 95 TC + 32° ⇒ TC = −88°C ⇒ Tk = ( −88.3 + 273) K = 185 K In the same way, the highest temperature is 136°F = 95 TC + 32° ⇒ TC = 58°C = 331 K Assess:
On the absolute scale the highest recorded temperature is not quite twice the lowest.
16.12. Solve: Let TF = TC = T : TF = 95 TC + 32° ⇒ T = 95 T + 32° ⇒ T = −40° That is, the Fahrenheit and the Celsius scales give the same numerical value at −40° . Assess: It is usually unnecessary to specify the scale when the temperature is reported as −40° .
16.13. Model: A temperature scale is a linear scale.
Solve: (a) We need a conversion formula for °C to °Z, analogous to the conversion of °C to °F. Since temperature scales are linear, TC = aTZ + b, where a and b are constants to be determined. We know the boiling
point of liquid nitrogen is 0°Z and –196°C. Similarly, the melting point of iron is 1000°Z and 1538°C. Thus −196 = 0a + b 1538 = 1000a + b
From the first, b = −196°. Then from the second, a = (1538 + 196)/1000 = 1734/1000. Thus the conversion is TC = (1734/1000)TZ − 196°. Since the boiling point of water is TC = 100°C, its temperature in °Z is ⎛ 1000 ⎞ TZ = ⎜ ⎟ (100° + 196°) = 171°Z ⎝ 1734 ⎠
(b) A temperature TZ = 500°Z is
⎛ 1734 ⎞ TC = ⎜ ⎟ 500° − 196° = 671°C = 944 K ⎝ 1000 ⎠
16.14. Solve: (a) The triple point of water is T = 0.01°C and p = 0.006 atm. Thus TF = 95 TC + 32° = 95 ( 0.01) + 32 = 32.02°F p = 0.006 atm ×
1.013 × 105 Pa = 608 Pa 1 atm
(b) The triple point of carbon dioxide is T = −56°C and p = 5 atm. Thus
TF = 95 TC + 32° = 95 ( −56° ) + 32° = −68.8°F
⎛ 1.013 × 105 Pa ⎞ 5 p = 5.0 atm = ( 5.0 atm ) ⎜ ⎟ = 5.06 × 10 Pa 1 atm ⎝ ⎠
16.15. Model: Treat the gas in the container as an ideal gas. Solve:
From the ideal-gas law pV = nRT , the pressure of the gas is
p= Assess:
nRT ( 3.0 mol )( 8.31 J mol K ) ⎡⎣( 273 − 120 ) K ⎤⎦ = = 1.9 × 106 Pa = 19 atm V ( 2.0 × 10−3 m3 )
19 atm is a high pressure, but not unreasonable.
16.16. Model: Treat the nitrogen gas in the closed cylinder as an ideal gas. Solve: (a) The density before and after the compression are ρ before = m1 V1 and ρ after = m2 V2 . Noting that m1 = m2 and V2 = 12 V1 ,
ρ after m V1 = = 2 ⇒ ρ after = 2 ρ before ρ before V2 m The mass density has changed by a factor of 2. (b) The number of atoms in the gas is unchanged, implying that the number of moles in the gas remains the same; hence the number density is unchanged.
16.17. Model: Treat the gas in the sealed container as an ideal gas. Solve:
(a) From the ideal gas law equation pV = nRT , the volume V of the container is
V=
nRT ( 2.0 mol )( 8.31 J mol K ) ⎡⎣( 273 + 30 ) K ⎤⎦ = = 0.050 m3 1.013 × 105 Pa p
Note that pressure must be in Pa in the ideal-gas law. (b) The before-and-after relationship of an ideal gas in a sealed container (constant volume) is p1V p2V T ( 273 + 130 ) K = 1.3 atm = ⇒ p2 = p1 2 = (1.0 atm ) T1 T2 T1 ( 273 + 30 ) K
Note that gas-law calculations must use T in kelvins.
16.18. Model: Treat the gas as an ideal gas in the sealed container. Solve:
(a) For p1 = p0 , the before-and-after relationship of an ideal gas in a sealed container is
V1 V0 T 473 K = ⇒ V1 = V0 1 = V0 = 1.27V0 T1 T0 T0 373 K where T0 = (273 + 100) K and T1 = (273 + 200) K. (b) When the Kelvin temperature is doubled, T1 = 2T0 = 2(373 K) = 746 K and the above equation becomes
V1 = V0
T1 746 K = V0 = 2V0 T0 373 K
Assess: When we use the Kelvin scale we expect the volume to double when the temperature doubles (if the pressure is kept constant).
16.19. Model: Treat the air in the compressed-air tank as an ideal gas. Solve:
(a) From the ideal-gas law pV = nRT , the number of moles n is
pV p (π r h ) = = RT RT 2
n=
⎛ 1.013 × 105 Pa ⎞ ⎡ 2 ⎟ ⎣π ( 0.075 m ) ( 0.50 m ) ⎤⎦ 1 atm ⎝ ⎠ = 55.1 mol ≈ 55 mol ( 8.31 J mol K ) ⎣⎡( 273 + 20 ) K ⎦⎤
(150 atm ) ⎜
(b) At STP, the ideal-gas law yields V=
nRT ( 55.1 mol )( 8.31 J mol K )( 273 K ) = = 1.234 m3 ≈ 1.2 m3 p 1.013 × 105 Pa
Assess: Because the volume of the compressed air tank is (π r 2 )h = 8.8357 × 10−3 m3 , the volume at STP is 140 times the volume of the tank.
16.20. Model: Treat the oxygen gas in the cylinder as an ideal gas. Solve:
(a) The number of moles of oxygen is
n=
M 50 g = = 1.563 mol ≈ 1.6 mol M mol 32 g mol
(b) The number of molecules is
N = nN A = (1.563 mol ) ( 6.02 × 1023 mol−1 ) = 9.41× 1023 ≈ 9.4 ×1023 (c) The volume of the cylinder V = π r 2 L = π ( 0.10 m ) ( 0.40 m ) = 1.257 × 10−2 m3 . Thus, 2
N 9.41 × 1023 = = 7.5 × 1025 m −3 V 1.257 × 10−2 m3 (d) From the ideal-gas law pV = nRT we can calculate the absolute pressure to be
p=
nRT (1.563 mol )( 8.31 J mol K )( 293 K ) = = 303 kPa V 1.257 × 10−2 m3
where we used T = 20°C = 293 K. But a pressure gauge reads gauge pressure:
pg = p − 1 atm = 303 kPa − 101 kPa = 202 kPa ≈ 200 kPa
16.21. Model: Treat the helium gas in the sealed cylinder as an ideal gas. 2 Solve: The volume of the cylinder is V = π r 2 h = π ( 0.05 m ) ( 0.30 m ) = 2.356 × 10−3 m3. The gauge pressure of the gas is 120 psi ×
1 atm 1.013 × 105 Pa × = 8.269 ×105 Pa, so the absolute pressure of the gas is 14.7 psi 1 atm
8.269 ×105 Pa + 1.013 × 105 Pa = 9.282 × 105 Pa. The temperature of the gas is T = (273 + 20) K = 293 K. The number of moles of the gas in the cylinder is n=
5 −3 3 pV ( 9.282 × 10 Pa )( 2.356 × 10 m ) = = 0.898 mol RT ( 8.31 J mol K )( 293 K )
(a) The number of atoms is
N = nN A = ( 0.898 mol ) ( 6.02 × 1023 mol−1 ) = 5.41× 1023 atoms ≈ 5.4 × 1023 atoms (b) The mass of the helium is
M = nM mol = ( 0.898 mol )( 4 g mol ) = 3.59 g = 3.59 × 10−3 kg ≈ 3.6 × 10−3 kg (c) The number density is
N 5.41 × 1023 atoms = = 2.3 × 1026 atoms m3 V 2.356 × 10−3 m 3 (d) The mass density is
ρ=
M 3.59 × 10−3 kg = = 1.5 kg m3 V 2.356 × 10−3 m3
16.22. Model: The gas is assumed to be ideal and it expands isothermally. Solve: (a) Isothermal expansion means the temperature stays unchanged. That is T2 = T1. (b) The before-and-after relationship of an ideal gas under isothermal conditions is
⎛V ⎞ p p1V1 p2V2 V = ⇒ p2 = p1 1 = p1 ⎜ 1 ⎟ = 1 T1 T1 V2 ⎝ 2V1 ⎠ 2
16.23. Model: The gas is assumed to be ideal. Solve: (a) Isochoric means the volume stays unchanged. That is V2 = V1. (b) The before-and-after relationship of an ideal gas under isochoric conditions is
⎛1 ⎞ ⎜ p1 ⎟ T p1V1 p2V2 p2 = ⇒ T2 = T1 = T1 ⎜ 3 ⎟ = 1 T1 T1 p1 ⎜⎜ p1 ⎟⎟ 3 ⎝ ⎠
16.24. Model: The gas is assumed to be ideal, and since the container is rigid V2 = V1. Solve:
Convert both temperatures to the Kelvin scale. p1V1 p2V2 T ⎛ 283 K ⎞ = ⇒ p2 = p1 2 = 3 atm ⎜ ⎟ = 3.1 atm T1 T1 T1 ⎝ 275 K ⎠
Assess: bit.
On the absolute scale the temperature only went up a little bit, so we expect the pressure to rise a little
16.25. Model: The rigid sphere’s volume does not change, so this is an isochoric process. The air is assumed to be an ideal gas. Solve: (a) When the valve is closed, the air inside is at p1 = 1 atm and T1 = 100°C. The before-and-after relationship of an ideal gas in the closed sphere (constant volume) is
⎛T ⎞ p1V p2V ( 273 + 0 ) K = 0.73 atm = ⇒ p2 = p1 ⎜ 2 ⎟ = (1.0 atm ) T1 T2 ( 273 + 100 ) K ⎝ T1 ⎠ (b) Dry ice is CO2. From Figure 16.4, we can see that the solid-gas transition line gives a temperature of −78°C when p = 1 atm. Cooling the sphere to –78°C gives
⎛T ⎞ ( 273 − 78) K = 0.52 atm p2 = p1 ⎜ 2 ⎟ = (1.0 atm ) 373 K ⎝ T1 ⎠
16.26. Model: Assume the gas is an ideal gas. Visualize: The pressure in the gas must exert exactly enough upward force to counteract the gravitational force on the piston; this is true at both temperatures ( p2 = p1 = p ).
The initial volume of the cylinder is V1 = pr 2 h1 = p ( 0.12m ) ( 0.84m ) = 0.038m3 . The initial temperature is 2
T1 = 303°C + 273 = 576K. Solve: (a)
p=
F mg ( 20kg ) ( 9.80m/s = = 2 A pr 2 p ( 0.12m )
2
) = 4333Pa < 4300Pa
(b) Knowing the initial temperature, pressure, and volume allows us to compute the number of moles of gas (which will stay constant).
n=
( 4333Pa ) ( 0.038m3 ) pV1 = = 0.0344mol < 0.034mol RT1 ( 8.31J/mol ⋅ K )( 576K )
Now apply the n just found and the new temperature (T2 = 15°C + 273 = 288K) to find V2 . V2 =
nRT2 ( 0.0344 mol )( 8.31 J/mol ⋅ K )( 288 K ) = = 0.019 m3 p 4333 Pa
Solve V2 = pr 2 h2 for h2 .
h2 =
V2 0.019m3 = = 0.42m pr 2 p ( 0.12m )2
The new height of the piston is 42 cm. Assess:
Due to a coincidence in the data we could have used the shortcut that T2 = 12 T1 on the absolute scale to
deduce that V2 = 12 V1 1 h2 = 12 h1 .
16.27. Model: In an isochoric process, the volume of the container stays unchanged. Argon gas in the container is assumed to be an ideal gas. Solve: (a) The container has only argon inside with n = 0.1 mol, V1 = 50 cm3 = 50 × 10−6 m3 , and
T1 = 20°C = 293 K. This produces a pressure p1 =
nRT ( 0.1 mol )( 8.31 J mol K )( 293 K ) = = 4.87 × 106 Pa = 4870 kPa ≈ 4900 kPa V1 50 × 10−6 m3
An ideal gas process has p2V2 /T2 = p1V1 / T1. Isochoric heating to a final temperature T2 = 300°C = 573 K has V2 = V1 , so the final pressure is
p2 =
V1 T2 573 p1 = 1 × × 4870 kPa = 9520 kPa ≈ 9500 kPa V2 T1 293
Note that it is essential to express temperatures in kelvins. (b)
Assess:
All isochoric processes will be a straight vertical line on a pV diagram.
16.28. Model: The isobaric heating means that the pressure of the argon gas stays unchanged. Solve:
(a) The container has only argon inside with n = 0.1 mol, V1 = 50 cm3 = 50 × 10−6 m3 , and
T1 = 20°C = 293 K. This produces a pressure p1 =
nRT1 ( 0.1 mol )( 8.31 J mol K )( 293 K ) = = 4.87 × 106 Pa = 4870 kPa ≈ 4900 kPa V1 50 × 10−6 m3
An ideal gas process has p2V2 /T2 = p1V1/T1. Isobaric heating to a final temperature T2 = 300°C = 573 K has p2 = p1 , so the final volume is
V2 =
p1 T2 573 V1 = 1 × × 50 cm3 = 97.8 cm3 ≈ 98 cm3 p2 T1 293
(b)
Assess:
All isobaric processes will be a straight horizontal line on a pV diagram.
16.29. Model: In an isothermal expansion, the temperature stays the same. The argon gas in the container is assumed to be an ideal gas. Solve: (a) The container has only argon inside with n = 0.1 mol,
V1 = 50 cm3 = 50 × 10−6 m3 , and
T1 = 20°C = 293 K. This produces a pressure
p1 =
nRT1 (0.1 mol)(8.31 J/mol K) (293 K) = = 4.87 ×106 Pa = 12.02 atm ≈ 12 atm V1 50 ×10−6 Pa
An ideal-gas process obeys p2V2 /T2 = p1V1/T1. Isothermal expansion to V2 = 200 cm3 gives a final pressure p2 = (b)
T2 V1 200 p1 = 1 × × 12.02 atm = 48 atm T1 V2 50
16.30. Model: Assume the gas to be an ideal gas. Solve: (a) Because the volume stays unchanged, the process is isochoric. (b) The ideal-gas law p1V1 = nRT1 gives
T1 =
5 −6 3 p1V1 ( 3 × 1.013 × 10 Pa )(100 × 10 m ) = = 914 K = 641°C nR ( 0.0040 mol )( 8.31 J mol K )
The final temperature T2 is calculated as follows for an isochoric process:
p1 p2 p ⎛ 1 atm ⎞ ⇒ T2 = T1 2 = ( 914 K ) ⎜ = ⎟ = 305 K = 32°C p1 T1 T2 ⎝ 3 atm ⎠ Assess:
All straight vertical lines on a pV diagram represent isochoric processes.
16.31. Model: Assume that the gas is an ideal gas. Solve: (a) The graph shows that the pressure is inversely proportional to the volume. The process is isothermal. (b) From the ideal-gas law,
T1 =
5 −6 3 p1V1 ( 3 × 1.013 × 10 Pa )(100 × 10 m ) = = 914 K = 641°C nR ( 0.0040 mol )( 8.31 J mol K )
T2 is also 914 K, because the process is isothermal. (c) The before-and-after relationship of an ideal gas under isothermal conditions is p1V1 = p2V2 ⇒ V2 = V1
p1 ⎛ 3 atm ⎞ 3 = (100 cm3 ) ⎜ ⎟ = 300 cm p2 ⎝ 1 atm ⎠
16.32. Model: Assume that the gas is ideal. Solve: (a) Because the process is at a constant pressure, it is isobaric. (b) For an ideal gas at constant pressure,
V2 V1 V 100 cm3 ⇒ T2 = T1 2 = ⎡⎣( 273 + 900 ) K ⎤⎦ = 391 K = 118°C = V1 300 cm3 T2 T1 (c) Using the ideal-gas law p2V2 = nRT2 ,
n=
5 −6 3 p2V2 ( 3 × 1.013 × 10 Pa )(100 × 10 m ) = = 9.35 × 10−3 mol RT2 ( 8.31 J mol K )( 391 K )
Assess: All straight horizontal lines on a pV diagram represent isobaric processes.
16.33. Visualize:
Solve: Suppose we have a 1 m × 1 m × 1 m block of copper of mass M containing N atoms. The atoms are spaced a distance L apart along all three axes of the cube. There are Nx atoms along the x-edge of the cube, Ny atoms along the y-edge, and Nz atoms along the z-edge. The total number of atoms is N = Nx Ny Nz . If L is
expressed in meters, then the number of atoms along the x-edge is N x = (1 m)/L. Thus, 13
N=
⎛ 1 m3 ⎞ 1 m 1 m 1 m 1 m3 × × = 3 ⇒L=⎜ ⎟ L L L L ⎝ N ⎠
This relates the spacing between atoms to the number of atoms in a 1-meter cube. The mass of the large cube of copper is
M = ρ CuV = ( 8920 kg m 3 )(1 m3 ) = 8920 kg But M = mN , where m = 64 u = 64 × (1.661 × 10−27 kg ) is the mass of an individual copper atom. Thus, N=
M 8920 kg = = 8.39 × 1028 atoms m 64 × (1.661 × 10−27 kg ) 13
⎛ 1 m3 ⎞ ⇒ L=⎜ = 2.28 × 10−10 m = 0.228 nm 28 ⎟ 8.39 10 × ⎝ ⎠ Assess:
This is a reasonable interatomic spacing in a crystal lattice.
16.34. Solve: The volume of the cube associated with each atom is Vatom = ( 0.227 × 10−9 m ) = 1.1697 × 10−29 m3 3
The volume of a mole of atoms is
V = Vatom N A = (1.1697 × 10−29 m3 )( 6.02 × 1023 mol−1 ) = 7.0416 × 10−6 m3 mol Thus, the mass of a mole of atoms is
M mol = V ρ = ( 7.0416 m3 mol )( 7950 kg m3 ) = 0.056 kg/mol = 56 g/mol The atomic mass number of the element is 56. Assess: This is likely the most common isotope of iron.
1.0 g . 1.0 cm3 There are 10 protons in each water molecule. 1 mol of water molecules has a mass of 18 g.
16.35. Model: Assume the density of the liquid water is Visualize: Solve:
23 ⎛ 1.0 g ⎞ ⎛ 1 mol ⎞ ⎛ 6.02 × 10 molecules ⎞ ⎛ 10 protons ⎞ 26 1.0L = 1000cm3 ⎜ ⎜ ⎟⎜ ⎜ ⎟ ⎟ ⎟ = 3.3 × 10 protons 3 1 mol ⎝ 1.0 cm ⎠ ⎝ 18 g ⎠⎝ ⎠ ⎝ 1 molecule ⎠
Assess: This is an unimaginably large number, but reasonable considering the data.
16.36. Model: Air is an ideal gas. The pressure at sea level is 1 atm. Solve:
From the ideal gas law pV = NkBT , the number density is
N p 1.013 ×105 Pa = = = 2.5 ×1025 molecules / m3 V kBT (1.38 ×10−23 J/K)(293 K) Assess:
Each cubic meter of air holds an unimaginably large number of molecules.
16.37. Model: Assume the gas in the solar corona is an ideal gas. Solve:
The number density of particles in the solar corona is N V . Using the ideal-gas equation,
pV = NkBT ⇒
N p = V kBT
N ( 0.03 Pa ) = = 1.1 × 1015 particles m3 −23 V (1.38 × 10 J K )( 2 × 106 K ) Assess:
This answer is a lot smaller than the one in the previous problem.
16.38. Model: Assume the gas in the evacuated volume is an ideal gas. Solve:
The number density of particles is N V . Using the ideal-gas equation,
pV = NkBT ⇒
N p = V kBT
The pressure is
p = (1 × 10−10 mm of Hg ) × ⇒
1 atm 1.013 × 105 Pa × = 1.33 × 10−8 Pa 760 mm of Hg 1 atm
N 1.33 × 10−8 Pa = = 3.3 × 1012 m −3 =3.3 × 106 cm −3 V (1.38 × 10−23 J/K)(293 K)
Assess: Even the best vacuum we can achieve in the laboratory contains 3 million molecules per cubic centimeter.
16.39. Model: Assume that the gas in the vacuum chamber is an ideal gas. Solve:
(a) The fraction is
pvacuum chamber 1.0 × 10−10 mm of Hg = = 1.3 × 10−13 patmosphere 760 mm of Hg (b) The volume of the chamber V = π ( 0.20 m ) ( 0.30 m ) = 0.03770 m3 . From the ideal-gas equation 2
pV = NkBT , the number of molecules of gas in the chamber is N=
−13 5 3 pV (1.32 × 10 )(1.013 × 10 Pa )( 0.03770 m ) = = 1.2 × 1011 molecules −23 kBT 1.38 × 10 J K 293 K ) ( )(
16.40 Model: Assume the nebula gas is ideal. Visualize: Use pV = NkBT . We are given N / V = 100 atoms/cm3 = 1× 108 atoms/m3 . Solve:
p=
NkBT = (1.0 × 108 atoms/m3 )(1.38 × 10−23 J/K)(7500 K) = 1.0 ×10−11 Pa = 1.0 × 10−16 atm V
Assess: This is much lower pressure than the best vacuum we can achieve in a laboratory, but it is a higher pressure than in non-nebula space.
16.41. Model: Assume the air is pure N 2 , with a molar mass of 28g/mol. Visualize:
We will use pV = nRT both before and after. Our intermediate goal is n1 − n2 . We are given
T2 = T1 = T = 20°C + 273 = 293 K and V2 = V1 = V = pr 2C = p(0.011 m 2 )(2.0 m) = 7.60 × 1024 m3 . Solve:
We convert the gauge pressures to absolute pressures with
p = pg + 1 atm = pg + 14.7 psi.
p1 = 110 psi + 14.7 psi = 124.7 psi = 860 kPa. p2 = 80 psi + 14.7 psi = 94.7 psi = 653 kPa. n1 − n2 =
p1V1 p2V2 − RT1 RT2
V ( p1 − p2 ) RT 7.60 × 10−4 m3 = (860 kPa − 653 kPa) (8.31 J/mol ⋅ K)(293 K) = 0.0646 mol =
Thus 0.0646 mol of N2 was lost; this is
⎛ 28g ⎞ 0.0646mol ⎜ ⎟ = 1.8g ⎝ 1mol ⎠ Assess:
The result seems to be a reasonable number.
16.42. Model: The carbon dioxide in the cube is an ideal gas. Solve:
Using the ideal gas equation and n = M/M mol ,
pV = nRT ⇒ V =
nRT MRT = p pM mol
The molar mass of CO2 is 44 g/mol or 0.044 kg/mol. Thus, V=
(10,000kg )(8.31 J mol K )( 273 K ) = 5090 m3 (1.013 × 105 Pa ) ( 0.044 kg mol )
The length of the cube is L = (V )1/3 = 17.2 m. Assess: This is sobering when you multiply it by the number of people in the industrialized world. Good thing plants take up CO2 in large quantities.
16.43. Model: Assume the gas is an ideal gas. Solve:
The before-and-after relationship of an ideal gas is 1 p 3V p1V1 p2V2 p V = ⇒ T2 = T1 2 2 = ( 298 K ) 2 1 1 = 447 K = 174°C T1 T2 p1 V1 p1 V1
16.44. Model: Assume the evaporated water is an ideal gas with a molar mass of 18 g/mol. Assume the pressure is 1 atm = 101.3 kPa. Visualize: We are given T = 35°C + 273 = 308 K. n = 1000 g (1 mol/ 18 g ) = 55.6 mol. Solve:
(a) pV = nRT 1V =
nRT ( 55.6 mol )( 8.31 J/mol ⋅ K )( 308 K ) = = 1.4 m3 p 101.3 kPa
(b) In the liquid state r = 1000 kg/m3 .
V=
m 1.0 kg = = 0.0010 m3 r 1000 kg/m3
The factor by which the volume of the evaporated water is larger than the liquid water is 1.4 m3 = 1400 0.0010 m3 Assess:
Gases really do take up a lot more volume than the equivalent mass of a liquid.
16.45. Model: Assume that the steam (as water vapor) is an ideal gas. Solve:
The volume of the liquid water is
V=
m
ρ
=
nM mol
ρ
⎛ pV =⎜ ⎝ RT
5 −6 3 ⎞ M mol 20 (1.013 × 10 Pa )(10,000 × 10 m ) ( 0.018 kg mol ) = ⎟ ⎠ ρ (8.31 J mol K )( 473 K ) (1000 kg m3 )
= 9.28 × 10−5 m 3 = 92.8 cm3 ≈ 93 cm3 Assess:
The liquid takes a lot smaller volume than the same number of atoms as a gas.
16.46. Model: Assume that the steam is an ideal gas. Solve:
V=
(a) The volume of water is
M
ρ
=
nM mol
ρ
=
5 3 pV M mol 50 (1.013 × 10 Pa )( 5.0 m ) ( 0.018 kg mol ) = 0.0815 m3 = 81.5 L ≈ 82 L = RT ρ ( 8.31 J mol K )( 673 K ) (1000 kg m3 )
(b) Using the before-and-after relationship of an ideal gas,
⎛ ( 273 + 150 ) K ⎞ ⎛ 50 atm ⎞ p2V2 p1V1 T p 3 3 3 = ⇒ V2 = 2 1 V1 = ⎜ ⎟⎜ ⎟ ( 5.0 m ) = 78.6 m ≈ 79 m T2 T1 T1 p2 673 K 2.0 atm ⎠ ⎝ ⎠⎝
16.47. Model: We assume that the volume of the tire and that of the air in the tire is constant.
Solve: A gauge pressure of 30 psi corresponds to an absolute pressure of (30 psi) + (14.7 psi) = 44.7 psi. Using the before-and-after relationship of an ideal gas for an isochoric (constant volume) process,
p1 p2 T ⎛ 273 + 45 ⎞ ⇒ p2 = 2 p1 = ⎜ = ⎟ ( 44.7 psi ) = 49.4 psi T1 T1 T2 ⎝ 273 + 15 ⎠ Your tire gauge will read a gauge pressure pg = 49.4 psi − 14.7 psi = 34.7 psi. ≈ 35 psi.
16.48. Model: The air is assumed to be an ideal gas. Solve:
At 20°C and 1 atm pressure, the number of moles in the container is
n1 =
5 −3 3 p1V1 (1.013 × 10 Pa )(10 m ) = = 0.0416 mol RT1 ( 8.31 J mol K )( 293 K )
At 100°C and 1 atm pressure, the number of moles is n2 =
5 −3 3 p2V2 (1.013 × 10 Pa )(10 m ) = = 0.0327 mol RT2 ( 8.31 J mol K )( 373 K )
When heated, the pressure will rise as the number of moles remains n1. When opened, the pressure drops to 1 atm as gas escapes. Thus, the number of moles of air that escape as the container is opened is n1 − n2 = 0.0416 mol − 0.0327 mol = 0.0089 mol.
16.49. Model: The gas’s pressure does not change, so this is an isobaric process. Solve:
The triple point of water is 0.01°C or 273.16 K, so T1 = 273.16 K. Because the pressure is a constant,
V V1 V2 ⎛ 1638 mL ⎞ ⇒ T2 = T1 2 = ( 273.16 K ) ⎜ = ⎟ = 447.44 K = 174.3°C V1 T1 T2 ⎝ 1000 mL ⎠
16.50. Model: Assume the gas in the manometer is an ideal gas. Solve:
In the ice-water mixture the pressure is
p1 = patoms + ρ Hg g (0.120 m) = 1.013 × 105 Pa + (13,600 kg/m3 )(9.8 m/s 2 )(0.120 m) = 1.173 × 105 Pa In the freezer the pressure is
p2 = patm + ρ Hg g (0.030 m) = 1.013 × 105 Pa + (13,600 kg/m3 )(9.8 m/s 2 )(0.030 m) = 1.053 × 105 Pa Assume that a drop in length of 90 mm produces a very small change in gas volume compared with the total volume of the gas cell. This means the volume of the chamber can be considered constant. Hence, ⎛p ⎞ p1 p2 1.053 × 105 Pa = ⇒ T2 = T1 ⎜ 2 ⎟ = ( 273 K ) = 245 K= − 28°C T1 T2 1.173 × 105 Pa ⎝ p1 ⎠ Assess:
This is a reasonable temperature for an industrial freezer.
16.51. Model: The air in the closed section of the U-tube is an ideal gas.
Visualize: The length of the tube is l = 1.0 m and its cross-sectional area is A. Solve: Initially, the pressure of the air in the tube is p1 = patmos and its volume is V1 = Al. After the mercury is poured in, compressing the air, the air-pressure force supports the weight of the mercury. Thus the compressed pressure equals the pressure at the bottom of the column: p2 = patmos + ρ gL. The volume of the compressed air is V2 = A(l − L). Because the mercury is poured in slowly, we will assume that the gas remains in thermal
equilibrium with the surrounding air, so T2 = T1. In an isothermal process, pressure and volume are related by p1V1 = patmos Al = p2V2 = ( patmos + ρ gL) A(l − L)
Canceling the A, multiplying through, and solving for L gives
L=l−
patmos 101,300 Pa = 1.00 m − = 0.240 m = 24.0 cm (13,600 kg/m3 )(9.8 m/s 2 ) ρg
16.52. Model: Assume that the air bubble is always in thermal equilibrium with the surrounding water, and the air in the bubble is an ideal gas. Solve: The pressure inside the bubble matches the pressure of the surrounding water. At 50 m deep, the pressure is
p1 = p0 + ρ water gd = 1.013 × 105 Pa + (1000 kg/m3 )(9.8 m/s 2 )(50 m) = 5.913 × 105 Pa At the lake’s surface, p2 = p0 = 1.013 × 105 Pa. Using the before-and-after relationship of an ideal gas, p2V2 p1V1 p T = ⇒ V2 = V1 1 2 T2 T1 p2 T1 ⇒
5 4π 3 4π 3 ⎛ 5.913 × 10 Pa ⎞ ⎛ 293 K ⎞ r2 = ( 0.005 m ) ⎜ ⎟⎜ ⎟ ⇒ r2 = 0.0091 m 5 3 3 ⎝ 1.013 × 10 Pa ⎠ ⎝ 283 K ⎠
The diameter of the bubble is 2r2 = 0.0182 m = 1.82 cm. Assess: The bubble’s diameter just about doubled; this seems reasonable.
16.53. Model: Assume that the compressed air in the cylinder is an ideal gas. The volume of the air in the cylinder is a constant. Solve: Using the before-and-after relationship of an ideal gas, p2V2 p1V1 T V ⎛ 1223 K ⎞ V1 = ⇒ p2 = p1 2 1 = ( 25 atm ) ⎜ ⎟ = 104 atm T2 T1 T1 V2 ⎝ 293 K ⎠ V1
where we have converted to the Kelvin temperature scale. Because the pressure does not exceed 110 atm, the compressed air cylinder does not blow.
16.54. Model: Assume that the gas is an ideal gas. Solve:
Assess: For the isothermal process, the pressure must be halved as the volume doubles. This is because p1 is proportional to 1/V1 for isothermal processes.
16.55. Model: Assume that the gas is an ideal gas. Solve:
Assess: For the isothermal process, the pressure must double as the volume is halved. This is because p is proportional to 1/V for isothermal processes.
16.56. Model: Assume that the helium gas is an ideal gas. Visualize: Process 1 → 2 is isochoric, process 2 → 3 is isothermal, and process 3 → 1 is isobaric. Solve: The number of moles of helium is
n=
M 8.0 g = = 2.0 mol M mol 4 g mol
Using the ideal-gas equation, V1 =
nRT1 ( 2.0 mol )( 8.31 J mol K ) ⎡⎣( 273 + 37 ) K ⎤⎦ = = 0.0254 m3 ≈ 0.025 m3 p1 2 (1.013 × 105 Pa )
For the isochoric process V2 = V1 , and
p1 p2 T ⎛ 657 + 273 ⎞ = ⇒ p2 = p1 2 = ( 2 atm ) ⎜ ⎟ = 6.0 atm T1 T1 T2 ⎝ 37 + 273 ⎠ For the isothermal process, the equation p3V3 = p2V2 is V3 = V2
p2 ⎛ 6 atm ⎞ 3 3 = ( 0.0254 m3 ) ⎜ ⎟ = 0.0762 m ≈ 0.076 m p3 ⎝ 2 atm ⎠
For the isothermal process, T3 = T2 = 657°C.
16.57. Model: Assume the nitrogen gas is an ideal gas. Solve:
(a) The number of moles of nitrogen is
n=
M 1g ⎛ 1 ⎞ = = ⎜ ⎟ mol M mol 28 g mol ⎝ 28 ⎠
Using the ideal-gas equation,
p1 =
nRT1 (1 28 mol )( 8.31 J mol K )( 298 K ) = = 8.84 × 105 Pa = 884 kPa ≈ 880 kPa V1 (100 × 10−6 m3 )
(b) For the process from state 1 to state 3:
⎛ 1.5 p1 ⎞⎛ 50 cm3 ⎞ p1V1 p3V3 p V = ⇒ T3 = T1 3 3 = ( 298 K ) ⎜ = 223.5 K ≈ −49°C ⎟⎜ 3 ⎟ T1 T3 p1 V1 ⎝ p1 ⎠ ⎝ 100 cm ⎠ For the process from state 3 to state 2: ⎛ p ⎞⎛ V ⎞ ⎛ 2.0 p1 ⎞⎛ 100 cm3 ⎞ p2V2 p3V3 = ⇒ T2 = T3 ⎜ 2 ⎟⎜ 2 ⎟ = ( 223.5 K ) ⎜ = 596 K = 323°C ⎟⎜ 3 ⎟ T2 T3 ⎝ 1.5 p1 ⎠ ⎝ 50 cm ⎠ ⎝ p3 ⎠⎝ V3 ⎠ For the process from state 1 to state 4: ⎛ 1.5 p1 ⎞⎛ 150 cm3 ⎞ p4V4 p1V1 p V = ⇒ T4 = T1 4 4 = ( 298 K ) ⎜ = 670.5 K ≈ 398°C ⎟⎜ 3 ⎟ T4 T1 p1 V1 ⎝ p1 ⎠ ⎝ 100 cm ⎠
16.58. Model: The gas is an ideal gas. Solve:
(a) Using the ideal-gas equation,
T1 =
(1.0 × 105 Pa )( 2.0 m3 ) = 301 K p1V1 = nR ( 80 mol )( 8.31 J mol K )
Because points 1 and 2 lie on the isotherm, T2 = T1 = 301 K. The temperature of the isothermal process is 301 K. (b) The straight-line process 1 → 2 can be represented by the equation
p = (3 − V ) × 105 where V is in m3 and p is in Pa. We can use the ideal gas law to find that the temperature along the line varies as T=
pV 105 = (3V − V 2 ) × nR nR
We can maximize T by setting the derivative dT/dV to zero: dT 105 3 2 = (3 − 2Vmax = 0 ⇒ Vmax = m3 = 1.50 m3 )× nR dV 2 At this volume, the pressure is pmax = 1.5 × 105 Pa and the temperature is Tmax =
pmaxVmax (1.50 × 105 Pa)(1.5 m3 ) = = 338 K nR (80 mol)(8.31 J/mol K)
16.59. Model: Assume the gas is an ideal gas. Solve: units:
(a) We can find the temperatures directly from the ideal-gas law after we convert all quantities to SI
T1 =
3 −6 3 3 p1V1 ( 3.0 atm × 101,300 Pa atm ) (1000 cm × 10 m cm ) = = 366 K = 93°C nR ( 0.10 mol )(8.31 J mol K )
T2 =
3 −6 3 3 p2V2 (1.0 atm × 101,300 Pa atm ) ( 3000 cm × 10 m cm ) = = 366 K = 93°C nR ( 0.10 mol )( 8.31 J mol K )
(b) T2 = T1 , so this is an isothermal process. (c) A constant volume process has V3 = V2 . Because p1 = 3 p2 , restoring the pressure to its original value means
that p3 = 3p2. From the ideal-gas law,
⎛ p ⎞⎛ V ⎞ p3V3 p2V2 = ⇒ T3 = ⎜ 3 ⎟⎜ 3 ⎟ T2 = 3 × 1 × T2 = 3 × 366 K = 1098 K = 825°C T3 T2 ⎝ p2 ⎠⎝ V2 ⎠
16.60. Model: Assume the gas is an ideal gas. Solve:
(a) Using the ideal-gas law,
T1 =
5 −6 3 p1V1 (1.013 × 10 Pa )(100 × 10 m ) = = 244 K = −29°C nR ( 5.0 × 10−3 mol ) ( 8.31 J mol K )
(b) Using the before and after relationship of an ideal gas, 3 p1V1 p2V2 T V ⎛ 2926 K ⎞ ⎛ 100 cm ⎞ ⇒ p2 = p1 2 1 = (1 atm ) ⎜ = = 4.0 atm ⎟⎜ 3 ⎟ T1 V2 T1 T2 ⎝ 244 K ⎠ ⎝ 300 cm ⎠
(c) Using the before and after relationship of an ideal gas,
p3V3 p2V2 p T ⎛ 4 atm ⎞⎛ 2438 K ⎞ 3 3 = ⇒ V3 = 2 3 V2 = ⎜ ⎟⎜ ⎟ ( 300 cm ) = 500 cm T3 T2 p3 T2 ⎝ 2 atm ⎠⎝ 2926 K ⎠
16.61. Model: We assume the oxygen gas is ideal. Visualize: From the figure we glean p2 = 3 p1 and V2 = 3V1 . We are given T1 = 20°C + 273 = 293K. Solve: Use the before-and-after version of the ideal-gas law.
p1V1 p2V2 = T1 T2 T2 =
Assess:
( 3 p1 )( 3V1 ) T = 9T = 9 293K = 2637 K = 2364°C p2V2 T1 = ( ) 1 1 p1V1 p1V1
This is a hot temperature—higher than the melting point of many elements.
16.62. Model: Assume CO2 gas is an ideal gas. Solve:
(a) The molar mass for CO2 is M mol = 44 g/mol, so a 10 g piece of dry ice is 0.2273 mol. This becomes
0.227 mol of gas at 0°C. With V1 = 10,000 cm3 = 0.010 m3 and T1 = 0°C = 273 K, the pressure is p1 =
nRT1 ( 0.2273 mol )( 8.31 J mol K )( 273 K ) = = 5.156 × 104 Pa = 0.509 atm ≈ 0.50 atm V1 0.010 m3
(b) From the isothermal compression,
p2V2 = p1V1 ⇒ V2 = V1
p1 ⎛ 0.509 atm ⎞ −3 3 3 = ( 0.010 m3 ) ⎜ ⎟ = 1.70 × 10 m = 1700 cm p2 ⎝ 3.0 atm ⎠
From the isobaric compression,
T3 = T2 (c)
⎛ 1000 cm 3 ⎞ V3 = ( 273 K ) ⎜ = 161 K = −112°C 3 ⎟ V2 ⎝ 1700 cm ⎠
16.63. Model: The gas in the container is assumed to be an ideal gas. Solve:
(a) The gas starts at pressure p1 = 2.0 atm, temperature T1 = 127°C = (127 + 273) K = 400 K and
volume V1. It is first compressed at a constant temperature T2 = T1 until V2 = 12 V1 and the pressure is p2. It is then further compressed at constant pressure p3 = p2 until V3 = 12 V2 . From the ideal-gas law,
p2V2 p1V1 V T V = ⇒ p2 = p1 1 2 = ( 2.0 atm ) 1 1 × 1 = 4.0 atm V2 T1 V T2 T1 2 1 Note that T2 = T1 = 400 K. Using the ideal-gas law once again, 1 V V p p3V3 p2V2 ⇒ T3 = T2 3 3 = ( 400 K ) 2 2 × 1 = 200 K = −73°C = V2 p2 V2 T3 T2
The final pressure and temperature are 4.0 atm and –73°C. (b)
16.64. Model: Assume that the nitrogen gas is an ideal gas. Solve:
(a) The molar mass of N2 gas is 28 g/mol. The number of moles is n = (5 g)/(28 g / mol) = 0.1786 mol.
The initial conditions are p1 = 3.0 atm and T1 = 293 K. We use the ideal gas law to find the initial volume as follows: V1 =
nRT1 ( 0.1786 mol )( 8.31 J mol K )( 293 K ) = = 1.430 × 10−3 m3 = 1430 cm 3 ≈ 1400 cm3 p1 3.0 atm × 101,300 Pa atm
An isobaric expansion until the volume triples results in V2 = 3V1 = 4290 cm3 . (b) After the expansion,
p2V2 p1V1 p V = ⇒ T2 = 2 2 T1 = 1 × 3 × T1 = 3T1 = 879 K = 606°C T2 T1 p1 V1 (c) A constant volume decrease at V3 = V2 = 4290 cm3 back to T3 = T1 = 13 T2 results in the following:
p3V3 p2V2 T V 1 1 = ⇒ p3 = 3 2 p2 = × 1 × p2 = × 3.0 atm = 1.0 atm T3 T2 T2 V3 3 3 (d) An isothermal compression at T4 = T3 back to the initial volume V4 = V1 = 13 V3 results in the following:
p4V4 p3V3 T V 1 = ⇒ p4 = 4 3 p3 = 1 × 1 × p3 = 3 × 1.0 atm = 3.0 atm T4 T3 T3 V4 3 (e)
16.65. Solve: (a) A gas is compressed isothermally from a volume 300 cm3 at 2 atm to a volume of 100 cm3. What is the final pressure? (b)
(c) The final pressure is p2 = 6 atm.
16.66. Solve: (a) A gas at 400°C and 500 kPa is cooled at a constant volume to a pressure of 200 kPa. What is the final temperature in C°? (b)
(c) The final temperature is T2 = −3.8°C.
16.67. Solve: (a) A gas expands at constant pressure from 200 cm3 at 50°C until the temperature is 400°C. What is the final volume? (b)
(c) The final volume is V2 = 417 cm3 .
16.68. Solve: (a) 0.12 g of neon gas at 2.0 atm and 100 cm3 expands isobarically to twice its initial volume. What is the final temperature of the gas? (b)
(c) The number of moles of neon is
n=
M 0.12 g = = 0.006 mol M mol 20 g
The initial temperature is T1 =
5 −4 3 p1V1 2 (1.013 × 10 Pa )(1.0 × 10 m ) = = 406 K nR ( 0.006 mol )(8.31 J mol K )
The final temperature is T2 =
p2 V2 p 2V1 T1 = ( 406 K ) = 812 K p1 V1 p V1
16.69. Model: Assume that the compressed air is an ideal gas. Solve:
(a) Because the piston is floating in equilibrium,
Fnet = ( p1 − patoms ) A − w = 0 N where the piston’s cross-sectional area A = π (r 2 ) = π (0.050 m) 2 = 7.854 ×10−3 m 2 and the piston’s weight w = (50 kg)(9.8) = 490 N. Thus, p1 =
w 490 N + patmos = + 1.013 × 105 Pa=1.637 × 105 Pa A 7.854 × 10−3 m 2
Using the ideal-gas equation p1V1 = nRT1 ,
(1.637 × 105 Pa)Ah1 = (0.12 mol)(8.31 J/mol K)[(273 + 30) K] With the value of A given above, this equation yields h1 = 0.235 m = 23.5 cm. (b) When the temperature is increased from T1 = 303 K to T2 = (303 + 100) K = 403 K, the volume changes
from V1 = Ah1 to V2 = Ah2 at a constant pressure p2 = p1. From the before-and-after relationship of the ideal gas:
p1 ( Ah1 ) p2 ( Ah2 ) = T1 T2 ⇒ h2 =
p1 T2 ⎛ 403 K ⎞ h1 = (1) ⎜ ⎟ ( 0.235 m ) = 0.313 m = 31.3 cm p2 T1 ⎝ 303 K ⎠
Thus, the piston moves h2 − h1 = 7.8 cm.
16.70. Model: The air in the diving bell is an ideal gas. Visualize:
Solve:
(a) Initially p1 = p0 (atmospheric pressure), V1 = AL, and T1 = 293 K. When the diving bell is
submerged to d = 100 m at the bottom edge, the water comes up height h inside. The volume is V2 = A( L − h) and the temperature is T2 = 283 K. Like a barometer, the pressure at points a and b must be the same. Thus p2 + ρ gh = p0 + ρ gd , or p2 = p0 + ρ g (d − h). Using the before and after relationship of an ideal gas,
p1V1 p2V2 p AL [ p0 + ρ g (d − h)] A( L − h) = ⇒ 0 = T1 T2 293 K 283 K Multiplying this out gives the following quadratic equation for h: ⎛ ⎝
ρ gh 2 − [ p0 + ρ g (d + L)]h + ⎜1 −
283 ⎞ ⎟ p0 L + ρ gLd = 0 293 ⎠
Inserting the known values (using ρ = 1030 kg/m3 for seawater) and dividing by ρ g gives h 2 − 113.04h + 301.03 = 0 ⇒ h = 110 m or 2.7 m
The first solution is not physically meaningful, so the water rises to height h = 2.7 m. (b) To expel all the water, the air pressure inside the bell must be increased to match the water pressure at the bottom edge of the bell, d = 100 m. The necessary pressure is
p = p0 + ρ gd = 101,300 Pa + (1030 kg/m3 )(9.8 m/s 2 )(100 m) = 1111 kPa = 10.96 atm ≈ 11 atm
16.71. Model: Assume the trapped air to be an ideal gas. Visualize:
Initially, as the pipe is touched to the water surface and the gas inside is thus closed off from the air, the pressure p1 = patmos = 1 atm and the volume is V1 = L1 A, where A is the cross-sectional area of the pipe. By pushing the pipe in slowly, the gas temperature in the pipe remains the same as the water temperature. Thus, this is an isothermal compression of the gas with T2 = T1. Solve: From the ideal-gas law,
p2V2 = p1V1 ⇒ p2 L2 A = p1L1 A ⇒ p2 L2 = patmos L1 ⇒ p2 = patmos ( L1 L2 ) As the pipe is pushed down, the increasing water pressure pushes water up into the pipe, compressing the air. In equilibrium, the pressure at points a and b, along a horizontal line, must be equal. (This is like the barometer. If the pressures at a and b weren’t equal, the pressure difference would cause the liquid level in the pipe to move up or down.) The pressure at point a is just the gas pressure inside the pipe: pa = p2 . The pressure at point b is the pressure at depth L2 in water: pb = patmos + ρ gL2 . Equating these gives
p2 = patmos = ρ gL2 Substituting the expression for p2 from the ideal-gas equation above, the pressure equation becomes
patmos L1 = patmos + ρ gL2 ⇒ ρ gL22 + patmos L2 − patmos L1 = 0 L2 This is a quadratic equation for L2 with solutions
L2 =
− patmos ±
( patmos )
2
+ 4 ρ gpatmos L1
2ρ g
Length has to be a positive quantity, so the one physically acceptable solution is L2 =
−101,300 Pa +
(101,300 Pa )
2
+ 4 (1000 kg m3 )( 9.8 m s 2 ) (101,300 Pa )( 3.0 m )
2 (1000 kg m3 )( 9.8 m s 2 )
= 2.4 m
16.72. Model: The gas in the cylinder is assumed to be an ideal gas. Solve:
The gas at T1 = 20°C = 293 K has a pressure p1 = 1 atm and a volume V1 = AL0 . (The pressure has to be
1 atm to balance the external force of the air on the piston.) At a temperature T2 = 100°C = 373 K, its volume becomes V2 = A( L0 + Δx) and its pressure increases to p2 = p1 +
F k Δx = p1 + A A
where F = k Δx is the spring force on the piston. Using the before-and-after relationship of an ideal-gas equation,
p1V1 p2V2 p L A ( p + k Δx A )( L0 + Δx ) A = ⇒ 1 0 = 1 T1 T2 T1 T2 ⇒
T2 ( p1 + k Δx A )( L0 + Δx ) ⎛ k Δx ⎞ ⎛ Δx ⎞ = = ⎜1 + ⎟ ⎟ ⎜1 + T1 p1L0 p1 A ⎠⎝ L0 ⎠ ⎝
L0 can be obtained from the ideal-gas law as follows: L0 =
V1 ⎛ nRT1 ⎞ 1 ( 0.004 mol )( 8.31 J mol K )( 293 K ) =⎜ = 0.0961 m = 9.61 cm ⎟ = A ⎝ p1 ⎠ A (1.013 ×105 Pa )(10 ×10−4 m2 )
Substituting this value of L0 and the values for p1, T1, T2, and k, the above equation can be simplified to 154.02Δx 2 + 25.210Δx − 0.2730 = 0 ⇒ Δx = 0.0102 m = 1.02 cm The spring is compressed by 1.0 cm.
16.73. Model: The gas in containers A and B is assumed to be an ideal gas. Solve:
(a) The number of moles of gas in containers A and B can be expressed as follows:
nA =
5 ( 5.0 ×105 Pa ) ( 4VA ) = 5000 VA pAVA (1.0 × 10 Pa )VA V pV = = 333.3 A nB = B B = RTA R ( 300 K ) R RTB R ( 400 K ) R
where nA and nB are expressed in moles. Let n′A and n′B be the number of moles after the valve is opened. Since the total number of moles is the same, nA + nB = 5333.3
VA = nA′ + nB′ R
The pressure is now the same in both containers:
p′A = ⇒ n′A = nB′
n′A RTA n′ RT = p′B = B B VA VB
TB VA ⎛ 400 K ⎞ VA = n′B ⎜ ⇒ 3n′A = n′B ⎟ TA VB ⎝ 300 K ⎠ 4VA
Solving the above equations, 5333.3
VA V V = nA′ + 3nA′ = 4nA′ ⇒ n′A = 1333.3 A ⇒ n′B = 4000 A R R R
The new pressure is p′A =
n′A RTA 1333.3 ⎛ VA ⎞ 5 = ⎜ ⎟ RTA = 4.0 × 10 Pa VA VA ⎝ R ⎠
p′B =
n′B RTB 4000 ⎛ VA ⎞ 5 = ⎜ ⎟ RTB = 4.0 × 10 Pa VB VB ⎝ R ⎠
Gas flows from B to A until the pressure is 4.0 × 105 Pa. (b) The change is irreversible because opening the valve is like breaking a membrane. It is not a quasi-static process.
16.74 Model: Assume the gas in the left chamber is ideal. Visualize: For the piston to be in equilibrium ( Fx ) net = 0. The initial volume is V0 = L0 A. Solve: (a) The force to the right on the piston by the gas in the left chamber is simply p0 A. . The magnitude of the force to the left on the piston by the spring is k (ΔL). p A ( Fx ) net = p0 A − k (ΔL) ⇒ ΔL = 0 k (b) When the piston is moved x to the right the spring is compressed a total of ΔL + x and the pressure in the left chamber changes to p1 according to the ideal gas law. When the temperature doesn’t change the ideal gas equation reduces to p0V0 = p1V1 , where V0 = L0 A and V1 = ( L0 + x) A . Therefore pL p1 = 0 0 ( L0 + x) Now, similar to part (a) above, ( Fx ) net = p1 A − k (ΔL + x) Substitute in ΔL from part (a) and p1 from above. ( Fx ) net =
p0 L0 ⎛p A ⎞ A− k ⎜ 0 + x⎟ ( L0 + x) ⎝ k ⎠
Now it is time to apply the binomial approximation: ( L0 + x) −1 ≈ 1/ L0 (1 − x / L0 ) for x << L0 .
( Fx ) net = ( p0 L0 ) (1/ L0 (1 − x / L0 ) ) A − p0 A − kx = p0 A − p0 Ax / L0 − p0 A − kx
= − ( p0 A / L0 + k ) x This is clearly a linear restoring force of the form of Hooke’s law where the usual k is replaced with ( p0 A / L0 + k ) . That is, the gas in the left chamber is increasing the k of the system (by p0 A / L0 ) and making it act like a stiffer spring. (c) Since we have a Hooke’s law for which we know the modified k, we simply replace k in the period equation for harmonic oscillators with the modified constant. M T = 2π p0 A / L0 + k Assess: This is simple harmonic motion when the approximation x << L0 is valid.
16-1
17.1. Model: Assume the gas is ideal. The work done on a gas is the negative of the area under the pV curve. Visualize: The gas is compressing, so we expect the work to be positive. Solve: The work done on the gas is
W = − ∫ p dV = − ( area under the pV curve )
(
)
= − − ( 200 cm3 ) ( 200 kPa ) = ( 200 × 10−6 m3 )( 2.0 × 105 Pa ) = 40 J
Assess: The area under the curve is negative because the integration direction is to the left. Thus, the environment does positive work on the gas to compress it.
17.2. Model: Assume the gas is ideal. The work done on a gas is the negative of the area under the pV curve. Visualize: The gas is expanding, so we expect the work to be negative. Solve: The area under the pV curve is the area of the rectangle and triangle. We have
( 200 × 10
−6
m3 )( 200 × 103 Pa ) + 12 ( 200 × 10−6 m3 )( 200 × 103 Pa ) = 60 J
Thus, the work done on the gas is W = −60 J. Assess: The environment does negative work on the gas as it expands.
17.3. Model: Assume the gas is ideal. The work done on a gas is the negative of the area under the pV curve. Solve:
The work done on gas in an isobaric process is
W = − pΔV = − p (Vf − Vi ) Substituting into this equation,
80 J = − ( 200 × 103 Pa ) (V1 − 3V1 ) ⇒ Vi = 2.0 × 10−4 m3 = 200 cm3 Assess:
The work done to compress a gas is positive.
17.4. Model: Helium is an ideal gas that undergoes isobaric and isothermal processes. Solve:
(a) Since the pressure ( pi = pf = p ) is constant the work done is
Won gas = − pΔV = − p (Vf − Vi ) = −
nRTi (Vf − V i ) Vi
= −( 0.10 mol )( 8.31 J mol K )( 573 K )
(1000 cm
3
− 2000 cm3 )
2000 cm3
= 240 J
(b) For compression at a constant temperature,
Won gas = −nRT ln (Vf Vi ) ⎛ 1000 × 10−6 m3 ⎞ = − ( 0.10 mol )( 8.31 J mol K )( 573 K ) ln ⎜ = 330 J −6 3 ⎟ ⎝ 2000 × 10 m ⎠ (c) For the isobaric case,
p=
nRTi = 2.38 × 105 Pa Vi
For the isothermal case, pi = 2.38 × 105 Pa and the final pressure is
pf =
nRTf = 4.76 × 105 Pa Vf
17.5.
Visualize:
Solve:
Because W = − ∫ p dV and this is an isochoric process, W = 0 J. The final point is on a higher isotherm
than the initial point, so Tf > Ti . Heat energy is thus transferred into the gas (Q > 0) and the thermal energy of the gas increases ( Eth f > Eth i ) as the temperature increases.
17.6.
Visualize:
Solve:
Because this is an isobaric process W = − ∫ pdV = − p (Vf − Vi ) . Since Vf is smaller than Vi, W is positive.
That is, the gas is compressed. Since the final point is on a lower isotherm than the initial point, Tf < Ti . In other words, the thermal energy decreases. For this to happen, the heat energy transferred out of the gas must be larger than the work done.
17.7.
Visualize:
Solve:
Because the process is isothermal, ΔEth = Eth f − Eth i = 0 J. According to the first law of
thermodynamics, ΔEth = W + Q. This can only be satisfied if W = −Q. W is positive because the gas is compressing, hence Q is negative. That is, heat energy is removed from the gas.
17.8.
Visualize:
Solve: This is an adiabatic process of gas compression so no heat energy is transferred between the gas and the environment. That is, Q = 0 J. According to the first law of thermodynamics, the work done on a gas in an adiabatic process goes entirely to changing the thermal energy of the gas. The work W is positive because the gas is compressed.
17.9. Solve: The first law of thermodynamics is ΔEth = W + Q ⇒ −200 J = 500 J + Q = Q ⇒ Q = −700 J The negative sign means a transfer of energy from the system to the environment. Assess: Because W > 0 means a transfer of energy into the system, Q must be less than zero and larger in magnitude than W so that Eth f < Eth i .
17.10. Solve: This is an isobaric process. W > 0 because the gas is compressed. This transfers energy into the system. Also, 100 J of heat energy is transferred out of the gas. The first law of thermodynamics is
ΔEth = W + Q = − pΔV + Q = −(4.0 × 105 Pa)(200 − 600) × 10−6 m3 − 100 J = 60 J Thermal energy increases by 60 J.
17.11. Model: The removal of heat from the ice reduces its thermal energy and its temperature. Solve:
The heat needed to change an object’s temperature is Q = McΔT. The mass of the ice cube is
M = ρiceV = (920 kg/m3 )(0.06 × 0.06 × 0.06)m3 = 0.1987 kg The specific heat of ice from Table 17.2 is cice = 2090 J/kg K, so Q = (0.199 kg)(2090 J/kg K)(243 K − 273 K) = −12,460 J ≈ 12,000 J
Thus, the energy removed from the ice block is 12,000 J. Assess: The negative sign with Q means loss of energy.
17.12. Model: The spinning paddle wheel does work and changes the water’s thermal energy and its temperature. Solve: (a) The temperature change is ΔT = Tf − Ti = 25°C − 21°C = 4 K. The mass of the water is
M = (200 × 10−6 m3 )(1000 kg/m3 ) = 0.20 kg The work done is W = ΔEth = Mcwater ΔT = (0.20 kg)(4190 J/kg K)(4 K) = 3350 J ≈ 3400 J (b) Q = 0. No energy is transferred between the system and the environment because of a difference in temperature.
17.13. Model: Heating the mercury at its boiling point changes its thermal energy without a change in temperature. Solve: The mass of the mercury is M = 20 g = 2.0 × 10−2 kg, the specific heat cmercury = 140 J/kg K, the boiling
point Tb = 357°C, and the heat of vaporization Lv = 2.96 × 105 J/kg. The heat required for the mercury to change to the vapor phase is the sum of two steps. The first step is Q1 = Mcmercury ΔT = (2.0 × 10−2 kg)(140 J/kg K)(357°C − 20°C) = 940 J
The second step is Q2 = MLV = (2.0 ×10−2 kg)(2.96 × 105 J/kg) = 5920 J
The total heat needed is 6860 J.
17.14. Model: Heating the mercury changes its thermal energy and its temperature. Solve:
(a) The heat needed to change the mercury’s temperature is
Q = McHg ΔT ⇒ ΔT =
Q 100 J = = 35.7 K ≈ 35°C McHg ( 0.020 kg )(140 J kg K )
(b) The amount of heat required to raise the temperature of the same amount of water by the same number of degrees is
Q = Mcwater ΔT = (0.020 kg)(4190 J/kg K)(35.7 K) = 3000 J Assess: Q is directly proportional to cwater and the specific heat for water is much higher than the specific heat for mercury. This explains why Qwater > Qmercury.
17.15. Model: Changing ethyl alcohol at 20°C to solid ethyl alcohol at its melting point requires two steps:
lowering its temperature from 20°C to −114°C, then changing the ethyl alcohol to its solid phase at −114°C. Solve: The change in temperature is −114°C − 20°C = −134°C = −134 K. The mass is
M = ρV = ( 789 kg/m3 )( 200 × 10−6 m3 ) = 0.1578 kg The heat needed for the two steps is
Q1 = Mcalcohol ΔT = (0.1578 kg)(2400 J/kg K)(−134 K) = −5.07 × 104 J Q2 = − MLf = −(0.1578 kg)(1.09 × 105 J/kg) = −1.72 × 104 J The total heat required is
Q = Q1 + Q2 = −6.79 × 104 J ≈ −6.8 × 104 J Thus, the minimum amount of energy that must be removed is 6.8 × 104 J. Assess: The negative sign with Q indicates that 6.8 × 104 J will be removed from the system.
17.16. Model: Changing solid lead at 20°C to liquid lead at its melting point (Tm = 328°C) requires two steps: raising the temperature to Tm and then melting the solid at Tm to a liquid at Tm. Solve: The equation for the total heat is Q = Q1 + Q2 ⇒ 1000 J = Mclead (Tf − Ti ) + MLf
⇒ 1000 J = M (128 J/kg K)(328 − 20) K + M (0.25 × 105 J/kg) ⇒ M = The maximum mass of lead you can melt with 1000 J of heat is 15.5 g.
1000 J = 15.5 g ( 64,424 J/kg)
17.17. Model: We have a thermal interaction between the copper pellets and the water. Solve:
The conservation of energy equation Qc + Ww = 0 is
Mc cc (Tf − 300°C) + Mw cw (Tf − 20°C) = 0 J Solving this equation for the final temperature Tf gives Tf = =
Mc cc (300°C) + Mw cw (20°C) Mc cc + Mw cw (0.030 kg)(385 J/kg K) (300°C) + (0.10 kg)(4190 J/kg K)(20°C) = 28°C (0.030 kg)(385 J/kg K) + (0.10 kg)(4190 J/kg K)
The final temperature of the water and the copper is 28°C.
17.18. Model: We have a thermal interaction between the copper block and water. Solve:
The conservation of energy equation Qcopper + Qwater = 0 J is
M copper ccopper (Tf − Ti copper ) + M water cwater (Tf − Ti water ) = 0 J Both the copper and the water reach the common final temperature Tf = 25.5°C. Thus M copper (385 J/kg K)(25.5°C − 300°C) + (1.00 × 10−3 m3 )(1000 kg/m3 )(4190 J/kg K)(25.5°C − 20°C) = 0 J ⇒ M copper = 0.218 kg
17.19. Model: We have a thermal interaction between the thermometer and the water. Solve:
The conservation of energy equation Qthermo + Qwater = 0 J is
M thermocthermo (Tf − (Ti ) thermo ) + M water cwater (Tf − (Ti ) water ) = 0 J The thermometer slightly cools the water until both have the same final temperature Tf = 71.2°C. Thus (0.050 kg)(750 J/kg K)(71.2°C − 20.0°C) + (200 × 10−6 m3 )(1000 kg/m3 )(4190 J/kg K)(71.2° C − Ti water ) = 1920 J + 838 (J/K)(71.2°C − Ti water ) = 0 J ⇒ Ti water = 73.5°C Assess:
The thermometer reads 71.2°C for a real temperature of 73.5°C. This is reasonable.
17.20. Model: We have a thermal interaction between the aluminum pan and the water. Solve:
The conservation of energy equation QAl + Qwater = 0 J is
M AlcAl (Tf − Ti A1 ) + M water cwater (Tf − Ti water ) The pan and water reach a common final temperature Tf = 24.0°C
(0.750 kg)(900 J/kg K)(24.0°C − Ti Al ) + (10.0 × 10−3 m3 )(1000 kg/m3 )(4190 J/kg K)(24.0°C − 20.0°C ) = (675.0 J/K)(24.0°C − Ti Al ) + 167,600 J = 0 J ⇒ Ti Al = 272°C = [(272)(9/5) + 32]°F = 522°F
17.21. Model: We have a thermal interaction between the metal sphere and the mercury. Solve:
The conservation of energy equation Qmetal + QHg = 0 J is
M metalcmetal (Tf − Ti metal ) + M Hg cHg (Tf − Ti Hg ) = 0 J The metal and mercury reach a common final temperature Tf = 99.0°C. Thus (0.500 kg)cmetal (99°C − 300°C) + (300 × 10−6 m3 )(13,600 kg/m3 )(140 J/kg K)(99°C − 20°C) = 0 J We find that cmetal = 449 J kg K. The metal is iron.
17.22. Model: Use the models of isochoric and isobaric heating. Note that the change in temperature on the Kelvin scale is the same as the change in temperature on the Celsius scale. Solve: (a) The atomic mass number of argon is 40. That is, M mol = 40 g/mol. The number of moles of argon gas in the container is
n=
M 1.0 g = = 0.025 mol M mol 40 g mol
The amount of heat is Q = nCV ΔT = (0.025 mol)(12.5 J/mol K)(100°C) = 31.25 J ≈ 31 J (b) For the isobaric process Q = nCP ΔT becomes
31.25 J = (0.025 mol)(20.8 J/mol K) ΔT ⇒ ΔT = 60°C
17.23. Model: The heating processes are isobaric and isochoric. O2 is a diatomic ideal gas. Solve:
(a) The number of moles of oxygen is
n=
M 1.0 g = = 0.03125 mol M mol 32 g mol
For the isobaric process, Q = nCp ΔT = (0.03125 mol)(29.2 J/mol K)(100°C) = 91.2 J ≈ 91 J (b) For the isochoric process,
Q = nCV ΔT = 91.2 J = (0.03125 mol)(20.9 J/mol K) ΔT ⇒ ΔT = 140°C
17.24. Model: The heating is an isochoric process. Solve:
The number of moles of helium is
n=
M 2.0 g = = 0.50 mol M mol 4 g mol
For the isochoric processes,
QHe = nCV ΔT = ( 0.50 mol )(12.5 J mol K ) ΔT
⎛ ⎞ M QO2 = nCV ΔT = ⎜ ⎟ ( 20.9 J mol K ) ΔT 32 g mol ⎝ ⎠
Because QHe = QO2 ,
( 0.50 mol )(12.5
⎛ ⎞ M J mol K ) = ⎜ ⎟ ( 20.9 J mol K ) ⇒ M = 9.6 g 32 g mol ⎝ ⎠
17.25. Model: The gas is an ideal gas that is subjected to an adiabatic process. Solve:
(a) For an adiabatic process, γ
γ pf Vfγ = pV i i ⇒
pf ⎛ Vi ⎞ ln(2.5) γ = 1.32 = ⎜ ⎟ ⇒ 2.5 = ( 2.0 ) ⇒ γ = pi ⎝ Vf ⎠ ln(2.0)
(b) Equation 17.39 for an adiabatic process is γ −1 Tf Vfγ −1 = TV ⇒ i i
Tf ⎛ Vi ⎞ =⎜ ⎟ Ti ⎝ Vf ⎠
γ −1
= ( 2.0 )
1.32 −1
= ( 2.0 )
0.32
= 1.25
17.26. Model: We assume the gas is an ideal gas and γ = 1.40 for a diatomic gas. Solve:
Using the ideal-gas law, Vi =
nRTi ( 0.10 mol )( 8.31 J mol K )( 423 K ) = = 1.157 × 10−3 m3 pi ( 3 × 1.013 × 105 Pa )
For an adiabatic process, γ γ pV i i = pf Vf 1γ
⎛ p ⎞ ⇒ Vf = Vi ⎜ i ⎟ ⎝ pf ⎠
1 1.40
⎛ pi ⎞ = (1.157 × 10 m ) ⎜ ⎟ ⎝ 0.5 pi ⎠ −3
3
= 1.9 × 10−3 m3
To find the final temperature, we use the ideal-gas law once again as follows:
Tf = Ti
⎛ 0.5 pi ⎞⎛ 1.90 × 10−3 m3 ⎞ pf Vf = ( 423 K ) ⎜ = 346.9 K ≈ 74°C ⎟⎜ −3 3 ⎟ pi Vi ⎝ pi ⎠ ⎝ 1.157 × 10 m ⎠
17.27. Model: The O2 gas has γ = 1.40 and is an ideal gas. Solve:
(a) For an adiabatic process, pV γ remains a constant. That is, γ
1.40
⎛ Vi ⎞ ⎛ Vi ⎞ pV ⎟ = ( 3.0 atm ) ⎜ ⎟ i i = pf Vf ⇒ pf = pi ⎜ V ⎝ f⎠ ⎝ 2Vi ⎠ γ
γ
1.40
⎛1⎞ = ( 3.0 atm ) ⎜ ⎟ ⎝2⎠
= 1.14 atm ≈ 1.1 atm
(b) Using the ideal-gas law, the final temperature of the gas is calculated as follows:
pV pV p V ⎛ 1.14 atm ⎞ ⎛ 2Vi ⎞ i i = f f ⇒ Tf = Ti f f = ( 423 K ) ⎜ ⎟ = 321.5 K ≈ 48°C ⎟⎜ Ti Tf pi Vi ⎝ 3.0 atm ⎠ ⎝ Vi ⎠
17.28. Visualize: We are asked for the heat-loss rate which is given by Equation 17.48: Q A = k DT Dt L We are given A = 10 m ×14 m = 140 m 2 , L = 0.12 m, and DT = 22°C − 5°C = 17°C = 17 K. We look up the thermal conductivity of concrete in Table 17.5: k = 0.8W/m ⋅ K. Solve:
⎛ 140 m 2 ⎞ Q A = k DT = ( 0.8W/m ⋅ K ) ⎜ ⎟ (17 K ) = 16 kW L Dt ⎝ 0.12 m ⎠ Assess: The answer is in a reasonable range. The heat loss could be reduced with thicker concrete or adding a layer of a different material.
17.29. Visualize: To determine the material we will solve for k in Equation 17.48: Q A = k DT Dt L We are given L = 0.20m and DT = 100 K. We compute A = pr 2 = p ( 0.010m ) = 3.14 × 1024 m 2 . 2
We also convert the heat conduction to watts.
⎛ 1h ⎞ 4.5 ×104 J/h ⎜ ⎟ = 12.5 W ⎝ 3600 s ⎠ Solve:
Solve the equation for k . ⎛ ⎞⎛ 1 ⎞ 0.20m ⎛ Q ⎞L 1 k =⎜ ⎟ = (12.5 W ) ⎜ ⎟ = 80 W/m ⋅ K 24 2 ⎟⎜ 3 14 10 m . × ⎝ Dt ⎠ A DT ⎝ ⎠⎝ 100K ⎠
Look this up in Table 17.5; the value corresponds to iron. Assess: We are grateful that our answer was one of the entries in the table.
17.30. Model: Assume the lead sphere is an ideal radiator with e = 1 . Also assume that the highest temperature the solid lead sphere can have is the melting temperature of lead. Visualize: Use Equation 17.49. First look up the melting temperature of lead in Table 17.3: Tm = 328°C = 601 K. Then compute the surface area of the sphere: A = 4pR 2 = 4p(0.050 m) 2 = 0.0314 m 2 * Solve:
Q = esAT 4 = (1)(5.6731028 W/m 2?K 4 )(0.0314 m 2 )(601 K) 4 = 230 W Dt Assess:
If the sphere were larger it could radiate more power without melting.
17.31. Model: We will ignore the bottom of the head and model it with just the cylindrical sides and top, all
covered in skin. As instructed, assume an emissivity of e = 0.95 . Visualize: The area of the 2 2 A = Aside + Atop = 2prh + pr = 2p(0.10 m)(0.20 m) + p(0.10 m) = 0.157 m 2 .
cylinder
Solve: Use Equation 17.50.
Qnet = esA(T 4 − T04 ) = (0.95)((5.6731028 W/m 2?K 4 )(0.157 m 2 ))((308 K) 4 − (278 K) 4 ) = 26 W Dt Assess:
This is a significant amount of the heat lost by the body. Wearing a hat can help a lot.
is
17.32. Model: There are various steps to the problem. We must (1) raise the the temperature of the ice to
0°C, (2) melt the ice to liquid water at 0°C, (3) raise the water temperature to 100°C, (4) boil the water to produce steam at 100°C , and (5) raise the temperature of the steam to 200°C. Solve: The heat needed for each step is Q1 = MciceDTice = ( 0.0050 kg )( 2090 J/kg?K )( 20 K ) = 209 J < 210 J Q2 = MLf = ( 0.0050 kg ) ( 3.333105 J/kg ) = 1665 J < 1670 J
Q3 = Mcwater DTwater = ( 0.0050 kg )( 4190 J/kg?K )(100 K ) = 2095 J < 2100 J Q4 = MLv = ( 0.0050 kg ) ( 22.63105 J/kg ) = 11,300 J Q5 = Mcsteam DTsteam = ( 0.0050 kg )( 2009 J/kg?K )(100 K ) = 1005 J < 1000 J The total heat is Q = Q1 + Q2 + Q3 + Q4 + Q5 = 16,300 J Assess: It is interesting to note which step required the most heat: The boiling to change the liquid to the gas is by far the largest contributor to the total.
17.33. Solve: The area of the garden pond is A = π ( 2.5 m ) = 19.635 m 2 and its volume is V = A ( 0.30 m ) = 2
5.891 m 3 . The mass of water in the pond is
M = ρV = (1000 kg m3 )(19.635 m3 ) = 5891 kg The water absorbs all the solar power which is
( 400 W
m 2 )(19.635 m 2 ) = 7854 W
This power is used to raise the temperature of the water. That is,
Q = ( 7854 W ) Δt = Mcwater ΔT = ( 5891 kg )( 4190 J kg K )(10 K ) ⇒ Δt = 31,425 s ≈ 8.7 h
17.34. Model: The potential energy of the bowling ball is transferred into the thermal energy of the mixture. We assume the starting temperature of the bowling ball to be 0°C. Solve: The potential energy of the bowling ball is
U g = M ball gh = (11 kg ) ( 9.8 m s 2 ) h = (107.8 kg m s 2 ) h This energy is transferred into the mixture of ice and water and melts 5 g of ice. That is,
(107.8 kg m s2 ) h = ΔEth = M w Lf ⇒ h =
( 0.005 kg ) ( 3.33 × 105
(107.8 kg m s ) 2
J kg )
= 15.4 m
17.35. Model: Heating the water and the kettle raises the temperature of the water to the boiling point and raises the temperature of the kettle to 100°C. Solve: The amount of heat energy from the electric stove’s output in 3 minutes is
Q = ( 2000 J s )( 3 × 60 s ) = 3.6 × 105 J This heat energy heats the kettle and brings the water to a boil. Thus,
Q = M water cwater ΔT + M kettle ckettle ΔT Substituting the given values into this equation, 3.6 × 105 J = M water ( 4190 J kg K )(100°C − 20°C ) + ( 0.750 kg )( 449 J kg K )(100°C − 20°C )
⇒ M water = 0.994 kg The volume of water in the kettle is
V= Assess:
M water
ρ water
=
0.994 kg = 0.994 × 10−3 m3 = 994 cm3 ≈ 990 cm3 1000 kg m3
1 L = 103 cm3, so V ≈ 1 L. This is a reasonable volume of water.
17.36. Model: Each car’s kinetic energy is transformed into thermal energy. Solve:
For each car,
K = 12 Mv 2 = ΔEth = Mccar ΔT ⇒ ΔT =
v2 2ccar
Assume ccar = ciron. The speed of the car is
80 × 1000 m ( 22.22 m s ) = 0.55°C = 22.22 m s ⇒ ΔT = 2 ( 449 J kg K ) 3600 s 2
v = 80 km hr = Assess:
Notice the answer is independent of the car’s mass.
17.37. Model: There are three interacting systems: aluminum, copper, and ethyl alcohol.
Solve: The aluminum, copper, and alcohol form a closed system, so Q = QAl + QCu + Qeth = 0 J. The mass of the alcohol is
M eth = ρV = ( 790 kg m3 )( 50 × 10−6 m3 ) = 0.0395 kg Expressed in terms of specific heats and using the fact that ΔT = Tf – Ti, the Q = 0 J condition is M AlcAl ΔTAl + M Cu cCu ΔTCu + M eth ceth ΔTeth = 0 J Substituting into this expression,
( 0.010 kg )( 900 J kg K )( 298 K − 473 K ) + ( 0.020 kg )( 385 J kg K )( 298 K − T ) + ( 0.0395 kg )( 2400 J kg K )( 298 K − 288 K ) = −1575 J + ( 7.7 J K )( 298 − T ) + 948 J = 0 J ⇒ T = 216.6 K = −56.4°C ≈ −56°C
17.38. Model: There are two interacting systems: aluminum and ice. The system comes to thermal
equilibrium in four steps: (1) the ice temperature increases from −10°C to 0°C, (2) the ice becomes water at 0°C, (3) the water temperature increases from 0°C to 20°C, and (4) the cup temperature decreases from 70°C to 20°C. Solve: The aluminum and ice form a closed system, so Q = Q1 + Q2 + Q3 + Q4 = 0 J. These quantities are
Q1 = M icecice ΔT = ( 0.100 kg )( 2090 J kg K )(10 K ) = 2090 J
Q2 = M ice Lf = ( 0.100 kg ) ( 3.33 × 105 J kg ) = 33,300 J
Q3 = M icecwater ΔT = ( 0.100 kg )( 4190 J kg K )( 20 K ) = 8380 J Q4 = MAl cAl ΔT = MAl ( 900 J kg K )( −50 K ) = − ( 45,000 J kg ) MAl The Q = 0 J equation now becomes 43,770 J – (45,000 J/kg)MAl = 0 J The solution to this is MAl = 0.973 kg.
17.39. Model: There are three interacting systems: metal, aluminum, and water.
Solve: The metal, aluminum container, and water form a closed system, so Qm + QAl + Qw = 0 J, where Qm is the heat transferred to the metal sample. This equation can be written: MmcmΔTm + MAlcAlΔTAl + MwcwΔT = 0 J Substituting in the given values,
( 0.512 kg ) cm ( 351 K − 288 K ) + ( 0.100 kg )( 900 J kg K )( 351 K − 371 K ) + ( 0.325 kg )( 4190 J kg K )( 351 K − 371 K ) = 0 J ⇒ cm = 900 J kg K From Table 17.2, we see that this is the specific heat of aluminum.
17.40. Solve: For a monatomic gas, the molar specific heat at constant volume is CV = 12.5 J mol K . From Equation 17.22,
12.5 J mol K = The gas is therefore neon.
M mol 625 J kg K ⇒ M mol = 20 g mol 1000 g kg
17.41. Model: Heating the water raises its thermal energy and its temperature. Solve: A 5.0 kW heater has power P = 5000 W. That is, it supplies heat energy at the rate 5000 J/s. The heat supplied in time ∆t is Q = 5000∆t J. The temperature increase is ∆TC = (5/9)∆TF = (5/9)(75°) = 41.67°C. Thus
Q = 5000Δt J = Mcw ΔT = (150 kg)(4190 J/kg K)(41.67°C) ⇒ Δt = 5283 s ≈ 87 min Assess:
A time of ≈1.5 hours to heat 40 gallons of water is reasonable.
17.42. Model: Heating the material increases its thermal energy. Visualize: Heat raises the temperature of the substance from –40°C to –20°C, at which temperature a solid to liquid phase change occurs. From –20°C, heat raises the liquid’s temperature up to 40°C. Boiling occurs at 40°C where all of the liquid is converted into the vapor phase. Solve: (a) In the solid phase,
⎛ ΔQ ⎞ 1 ⎛ 20 kJ ⎞ ⎛ 1 ⎞ ΔQ = McΔT ⇒ c = ⎜ ⎟ = 2000 J kg K ⎟ =⎜ ⎟⎜ ⎝ ΔT ⎠ M ⎝ 20 K ⎠ ⎝ 0.50 kg ⎠ (b) In the liquid phase,
c=
1 M
⎛ ΔQ ⎞ ⎛ 1 ⎞ ⎛ 80 kJ ⎞ ⎟⎜ ⎜ ⎟=⎜ ⎟ = 2667 J kg K ⎝ ΔT ⎠ ⎝ 0.50 kg ⎠ ⎝ 60 K ⎠
(c) The melting point Tm = −20°C and the boiling point Tb = +40°C . (d) The heat of fusion is
Lf =
Q 20,000 J = = 4.0 × 104 J kg M 0.50 kg
Lv =
Q 60,000 J = = 1.2 × 105 J kg M 0.50 kg
The heat of vaporization is
17.43. Model: The liquefaction of the nitrogen occurs in two steps: lowering nitrogen’s temperature from
20°C to −196°C, and then liquefying it at −196°C. Assume the cooling occurs at a constant pressure of 1 atm. Solve: The mass of 1.0 L of liquid nitrogen is M = ρV = ( 810 kg m3 )(10−3 m3 ) = 0.810 kg . This mass corres-
ponds to
n=
M 810 g = = 28.9 mols M mol 28 g mol
At constant atmospheric pressure, the heat to be removed from 28.93 mols of nitrogen is Q = MLv + nCP ΔT
= − ( 0.810 kg ) (1.99 × 105 J kg ) + ( 28.9 mols )( 29.1 J mol K )( 77 K − 293 K ) = −3.4 × 105 J
17.44. Model: There are two interacting systems: coffee (i.e., water) and ice. Changing the coffee
temperature from 90°C to 60°C requires four steps: (1) raise the temperature of ice from −20°C to 0°C, (2) change ice at 0°C to water at 0°C, (c) raise the water temperature from 0°C to 60°C, and (4) lower the coffee temperature from 90°C to 60°C. Solve: For the closed coffee-ice system,
Q = Qice + Qcoffee = ( Q1 + Q2 + Q3 ) + ( Q4 ) = 0 J Q1 = M icecice ΔT = M ice ( 2090 J kg K )( 20 K ) = M ice ( 41,800 J kg )
Q2 = M ice Lf = M ice ( 330,000 J kg ) Q3 = M icecwater ΔT = M ice ( 4190 J kg K )( 60 K ) = M ice ( 251,400 J kg ) Q4 = M coffeeccoffee ΔT = ( 300 × 10−6 m3 )(1000 kg m3 ) ( 4190 J kg K )( −30 K ) = −37,000 J The Q = 0 J equation thus becomes M ice ( 41,800 + 330,000 + 251,400 ) J kg − 37,710 J = 0 J ⇒ M ice = 0.061 kg = 61 g Assess:
61 g is the mass of approximately 1 ice cube.
17.45. Model: We have three interacting systems: the aluminum, the air, and the firecracker. The energy released by the firecracker raises the temperature of the aluminum and the air. We will assume that air is predominantly N2. Assume that the surrounding insulation is excellent. Solve: For the closed firecracker + air + aluminum system, energy conservation requires that Q = Qfirecracker + QAl + Qair = 0 J QAl = mAlcAlΔT = (2.0 kg)(900 J/kg K)(3 K) = 5400 J
⎛ pV Qair = nCV ΔT = ⎜ ⎝ RT
=
(1.01×10
5
Pa )( 20 × 10−3 m3 )
(8.31 J/mol K )( 298 K )
⎞ ⎟ CV ΔT ⎠
( 20.8 J/mol K )( 3 K )
= 51 J Thus,
Qfirecracker = −QAl − Qair
= –5450 J That is, ≈ 5500 J of energy are released on explosion of the firecracker. Assess: The negative sign with Qfirecracker means that the firecracker has lost energy.
17.46. Model: There are two interacting systems: the nuclear reactor and the water. The heat generated by the nuclear reactor is used to raise the water temperature. Solve: For the closed reactor-water system, energy conservation per second requires Q = Qreactor + Qwater = 0 J The heat from the reactor in Δt = 1 s is Qreactor = −2000 MJ = −2.0 × 109 J and the heat absorbed by the water is
Qwater = mwater cwater ΔT = mwater ( 4190 J kg K )(12 K ) ⇒ −2.0 × 109 J + mwater ( 4190 J kg K )(12 K ) = 0 J ⇒ mwater = 3.98 × 104 kg Each second, 3.98 × 104 kg of water is needed to remove heat from the nuclear reactor. Thus, the water flow per minute is
3.98 × 104
kg 60 s 1m3 1L × × × = 2.4 × 106 L/min s min 1000 kg 10−3 m3
17.47. Model: We have two interacting systems: the water and the gas. For the closed system comprised of water and gas to come to equilibrium, heat is transferred from one interacting system to the other. Solve: Energy conservation requires that Qair + Qwater = 0 J ⇒ ngasCV (Tf – Ti gas) + mwaterc(Tf – Ti water) = 0 J Using the ideal-gas law,
Ti gas =
pgasVgas ngas R
(10 ×1.013 × 10 Pa )( 4000 × 10 5
=
( 0.40 mol )(8.31 J
−6
mol K )
m3 )
=1219 K
The energy conservation equation with Ti water = 293 K becomes
( 0.40 mol )(12.5
J mol K )(T − 1219 K ) + ( 20 × 10−3 kg ) ( 4190 J kg K )(T − 293 K ) = 0 J ⇒ Tf = 345 K
We can now use the ideal-gas equation to find the final gas pressure.
pV pV T ⎛ 345 K ⎞ i i = f f ⇒ pf = pi f = ⎜ ⎟ (10 atm ) = 2.8 atm Ti Tf Ti ⎝ 1219 K ⎠
17.48. Model: These are isothermal and isobaric ideal-gas processes. Solve:
(a) The work done at constant temperature is
nRT dV = − nRT ( ln Vf − ln Vi ) = − nRT ln (Vf Vi ) V = − ( 2.0 mol )( 8.31 J mol K )( 303 K ) ln ( 13 ) = 5500 J
W = −∫
Vf
Vi
pdV = − ∫
Vf
Vi
(b) The work done at constant pressure is Vf ⎛V ⎞ 2 W = − ∫ p dV = − p (Vf − Vi ) = − p ⎜ i − Vi ⎟ = pVi Vi ⎝3 ⎠ 3 2 2 = nRT = ( 2.0 mol )( 8.31 J mol K )( 303 K ) = 3400 J 3 3
(c) For an isothermal process in which Vf = 13 Vi , the pressure changes to pf = 3 pi = 4.5 atm.
17.49. Model: This is an isothermal process. The work done is positive for a compression. Solve:
For an isothermal process,
W = − nRT ln (Vf Vi ) For the first process,
W = 500 J = −nRT ln ( 12 ) ⇒ nRT = 721.35 J
For the second process, W = − nRT ln ( 101 ) ⇒ W = − ( 721.35 J ) ln ( 101 ) = 1660 J
17.50.
Solve:
Visualize:
(a) The gas exerts a force on the piston of magnitude 2 Fgas on piston = pgas A = ( 3 atm × 101,300 Pa atm ) ⎡π ( 0.080 m ) ⎤ = 6100 N ⎣ ⎦
This force is directed toward the right. (b) The piston is in static equilibrium, so the environment must exert a force on the piston of equal magnitude Fenviron on piston = 6100 N but in the opposite direction, toward the left. (c) The work done by the environment is G G G Wenviron = Fenviron on piston ⋅ Δr = − Fenviron on piston Δx = − ( 6110 N )( 0.10 m ) = −610 J
The work is negative because the force and the displacement are in opposite directions. (d) The work done by the gas is G G Wgas = Fgas on piston ⋅ Δr = + Fgas on piston Δx = ( 6110 N )( 0.10 m ) = 610 J This work is positive because the force and the displacement are in the same direction. (e) The first law of thermodynamics is W + Q = ΔEth where W is Wenviron. So here W = −610 J and we find Q = ΔEth − W = (196 J ) − ( −610 J ) = 806 J
Thus, 806 J of heat is added to the gas.
17.51. Model: This is an isobaric process. Visualize:
Solve: (a) The initial conditions are p1 = 10 atm = 1.013 × 106 Pa, T1 = 50°C = 323 K, V1 = πr2L1 = π(0.050 m)2 (0.20 m) = 1.57 × 10−3 m3. The gas is heated at a constant pressure, so heat and temperature change are related by Q = nCPΔT. From the ideal gas law, the number of moles of gas is
n=
6 −3 3 p1V1 (1.013 × 10 Pa )(1.57 × 10 m ) = = 0.593 mol RT1 ( 8.31 J mol K )( 323 K )
The temperature change due to the addition of Q = 2500 J of heat is thus ΔT =
Q 2500 J = = 203 K nCP ( 0.593 mol )( 20.8 J mol K )
The final temperature is T2 = T1 + ΔT = 526 K = 253°C. (b) Noting that the volume of a cylinder is V = πr2L and that r doesn’t change, the ideal gas relationship for an isobaric process is
V2 V1 L L T 526 K = ⇒ 2 = 1 ⇒ L2 = 2 L1 = ( 20 cm ) = 33 cm T2 T1 T2 T1 T1 323 K
17.52. Model: The process in part (a) is isochoric and the process in part (b) is isobaric. Solve: (a) Initially V1 = (0.20 m)3 = 0.0080 m3 = 8.0 L and T1 = 293 K. Helium has an atomic mass number A = 4, so 3 g of helium is n = M/Mmol = 0.75 mole of helium. We can find the initial pressure from the ideal-gas law: p1 =
nRT1 ( 0.75 mol )( 8.31 J mol K )( 293 K ) = = 228 kPa = 2.25 atm V1 0.0080 m3
Heating the gas will raise its temperature. A constant volume process has Q = nCVΔT, so
ΔT =
Q 1000 J = = 107 K nCV ( 0.75 mol )(12.5 J mol K )
This raises the final temperature to T2 = T1 + ΔT = 400 K. Because the process is isochoric,
p2 p1 T 400 K = ⇒ p2 = 2 p1 = ( 2.25 atm ) = 3.1 atm T2 T1 T1 293 K (b) The initial conditions are the same as part a, but now Q = nCPΔT. Thus, ΔT =
1000 J Q = = 64.1 K nCP ( 0.75 mol )( 20.8 J mol K )
Now the final temperature is T2 = T1 + ΔT = 357 K. Because the process is isobaric,
V2 V1 T 357 K = ⇒ V2 = 2 V1 = ( 0.0080 m3 ) = 0.0097 m3 = 9.7 L T2 T1 T1 293 K
(c)
17.53. Model: Assume the air and the nitrogen are ideal gases. For the piston to be in equilibrium the forces on it must sum to zero. Also assume that the temperature doesn’t change in the second part. 2 Solve: The area is A = pr 2 = p ( 0.040 m ) = 5.031023 m 2 . V1 = Ah1 = ( 5.031023 m 2 ) ( 0.26 m ) = 0.0013 m3 .
Fnet = pin A − mg − pabove A = 0 N
(a)
pin =
2 23 2 mg + pabove A ( 5.1 kg ) ( 9.8 m/s ) + (100 kPa ) ( 5.0310 m ) = = 110 kPa A 5.031023 m 2
(b) Since T1 = T2 , then p1V1 = p2V2 where p1 is pin from above, and p2 is the inside pressure after the new
weight is added. Compute p2 exactly as above replacing 5.1 kg with 5.1 kg + 3.5kg = 8.6kg. p2 =
2 23 2 mg + pabove A ( 8.6 kg ) ( 9.8 m/s ) + (100 kPa ) ( 5.0310 m ) = = 117 kPa A 5.031023 m 2
V2 =
3 p1V1 (110 kPa ) ( 0.0013 m ) = = 1.2231023 m3 p2 117 kPa
h2 =
V2 1.2231023 m 3 = = 0.2447 m < 24 cm A 5.031023 m 2
Assess: The result seems to be a reasonable number; we expected the piston to be a little lower than before due to the increased mass on it.
17.54. Model: This is an isothermal process.
Solve: (a) The final temperature is T2 = T1 because the process is isothermal. (b) The work done on the gas is V2
V2
V1
V1
W = − ∫ pdV = − ∫
nRT1 V dV = − nRT1 ln 2 = −nRT1ln 2 V V1
(c) From the first law of thermodynamics ΔEth = W + Q = 0 J because ΔT = 0 K. Thus, the heat energy transferred to the gas is Q = −W = nRT1 ln 2 .
17.55. Model: The gas is an ideal gas and it goes through an isobaric and an isochoric process.
Solve: (a) The initial conditions are p1 = 3.0 atm = 304,000 Pa and T1 = 293 K. Nitrogen has a molar mass Mmol = 28 g/mol, so 5 g of nitrogen gas has n = M/Mmol = 0.1786 mol. From this, we can find the initial volume:
V1 =
nRT1 ( 0.1786 mol )( 8.31 J mol K )( 293 K ) = = 1.430 × 10−3 m3 ≈ 1400 cm 3 p1 304,000 Pa
The volume triples, so V2 = 3V1 = 4300 cm3. The expansion is isobaric (p 2 = p1 = 3.0 atm), so
V2 V1 V = ⇒ T2 = 2 T1 = ( 3) 293 K = 879 K = 606°C T2 T1 V1 (b) The process is isobaric, so
Q = nCP ΔT = ( 0.1786 mol )( 29.1 J mol K )( 879 K − 293 K ) = 3000 J (c) The pressure is decreased at constant volume (V3 = V2 = 4290 cm3) until the original temperature is reached (T3 = T1 = 293 K). For an isochoric process,
p3 p2 T 293 K = ⇒ p3 = 3 p2 = ( 3.0 atm ) = 1.0 atm T3 T2 T2 879 K (d) The process is isochoric, so
Q = nCV ΔT = ( 0.1786 mol )( 20.8 J mol K )( 293 K − 879 K ) = −2200 J So, 2200 J of heat was removed to decrease the pressure. (e)
17.56. Model: The gas is an ideal gas. Visualize: In the figure, call the upper left corner (on process A) 2 and the lower right corner (on process B) 3. Solve: The change in thermal energy is the same for any gas process that has the same ∆T. Processes A and B have the same ∆T, since they start and end at the same points, so (∆Eth)A = (∆Eth)B. The first law is then (∆Eth)A = QA + WA = (∆Eth)B = QB + WB ⇒ QA – QB = WB – WA In process B, work W = –p∆V = –pi(2Vi – Vi) = –piVi is done during the isobaric process i → 3. No work is done during the isochoric process 3 → f. Thus WB = –piVi. Similarly, no work is done during the isochoric process i → 2 of process A, but W = –p∆V = –2pi(2Vi – Vi) = –2piVi is done during the isobaric process 2→ f. Thus WA = – 2piVi. Combining these, QA – QB = WB – WA = –piVi – (–2piVi) = piVi
17.57. Model: The two processes are isochoric and isobaric. Solve:
Process A is isochoric which means
Tf Ti = pf pi ⇒ Tf = Ti ( pf pi ) = Ti (1 atm 3 atm ) = 13 Ti From the ideal-gas equation,
Ti =
( 3 ×1.013 ×105 Pa )( 2000 × 10−6 m3 ) = 731.4 K ⇒ T = 1 T = 243.8 K pV i i = f 3 i nR ( 0.10 mol )(8.31 J mol K ) ⇒ Tf − Ti = −487.6 K
Thus, the heat required for process A is
QA = nCV ΔT = ( 0.10 mol )( 20.8 J mol K )( −487.6 K ) = −1000 J Process B is isobaric which means
Tf Vf = Ti Vi ⇒ Tf = Ti (Vf Vi ) = Ti ( 3000 cm3 1000 cm3 ) = 3Ti From the ideal-gas equation, Ti =
( 2 × 1.013 × 105 Pa )(1000 × 10−6 m3 ) = 243.8 K pV i i = nR ( 0.10 mol )( 8.31 J mol K ) ⇒ Tf = 3Ti = 731.4 K ⇒ Tf − Ti = 487.6 K
Thus, heat required for process B is
QB = nCP ΔT = ( 0.10 mol )( 29.1 J mol K )( 487.6 K ) = 1400 J Assess:
Heat is transferred out of the gas in process A, but transferred into the gas in process B.
17.58. Model: We have an adiabatic and an isothermal process.
Solve: For the adiabatic process, no heat is added or removed. That is Q = 0 J. Isothermal processes occur at a fixed temperature, so ∆T = 0 K. Thus ∆Eth = 0 J, and the first law of thermodynamics gives
Q = −W = nRT ln (Vf Vi )
The temperature T can be obtained from the ideal-gas equation as follows: pV i i = nRT ⇒ T =
(1.013 ×105 Pa )( 3000 × 10−6 m3 ) = 366 K pV i i = nR ( 0.10 mol )(8.31 J mol K )
Substituting into the equation for Q we get
⎛ 1000 × 10−6 m3 ⎞ Q = ( 0.10 mol )( 8.31 J mol K )( 366 K ) ln ⎜ = −330 J −6 3 ⎟ ⎝ 3000 × 10 m ⎠ That is, 330 J of heat energy is removed from the gas.
17.59. Model: The monatomic gas is an ideal gas which is subject to isobaric and isochoric processes.
Solve: (a) For the isochoric process, V2 = V1 = 800 × 10−6 m3, p1 = 4.0 atm, p2 = 2.0 atm. The temperature T1 of the gas is obtained from the ideal-gas equation as:
T1 =
p1V1 = 390 K nR
where n = 0.10 mol. T2 can be obtained from the ideal-gas equation as follows: p1V1 p2V2 ⎛ 2.0 atm ⎞ = ⇒ T2 = T1 ( p2 p1 ) = ( 390 K ) ⎜ ⎟ = 195 K T1 T2 ⎝ 4.0 atm ⎠
The heat required for the process 1 → 2 is
Q = nCV (T2 − T1 ) = ( 0.10 mol )( 20.8 J/mol K )(195 K − 390 K ) = −406 J ≈ −410 J Because of the negative sign, this is the amount of heat removed from the gas. (b) For this isobaric process, p2 = p3 = 2.0 atm, V2 = 800 × 10−6 m3, and V3 = 1600 × 10−6 m3.
T3 = T2
⎛ 1600 × 10−6 m3 ⎞ V3 = T2 ⎜ = 2T2 = 390 K −6 3 ⎟ V2 ⎝ 800 × 10 m ⎠
Thus, the heat required for the process 2 → 3 is Q = nCP (T3 − T2 ) = ( 0.10 mol )( 29.1 J mol K )(195 K ) = 567 J ≈ 570 J This is heat transferred to the gas. (c) The change in the thermal energy of the gas is ΔEth = ( Q1→ 2 + Q2 →3 ) + (W1→ 2 + W2 →3 ) = −406 J + 567 J + 0 J +W2 → 3 = 162 J − pΔV = 162 J – (2.0 × 1.013 × 105 Pa)(1600 × 10−6 m3 – 800 × 10−6 m3) = 0 J Assess:
This result was expected since T3 = T1.
17.60. Model: Assume that the gas is an ideal gas. Visualize:
The volume of container A is a constant. On the other hand, heating container B causes the volume to change, but the pressure remains the same. Solve: (a) For the heating of the gas in container A, ΔTA = QA / nCV . Similarly, for the gas in container B,
ΔTB = QB / nCP . Because QA = QB and CP > CV, we see that ΔTA > ΔTB . The gases started at the same temperature, so TA > TB.
(b)
(c) The pressure in container B exerted by the gas is equal to the pressure on the gas by the piston. That is,
pB = patmos +
wpiston Apiston
= 1.013 × 105 Pa +
(10 kg ) ( 9.8 m 1.0 × 10
−4
m
s2 ) 2
= 1.08 × 106 Pa
Container A has the same volume, temperature, and number of moles of gas as container B, so PA = PB = 1.08 × 106 Pa. (d) The heating of container B is isobaric, so
Vf Vi T = ⇒ Vf = Vi f Tf Ti Ti We have Ti = 293 K, and Tf can be obtained from
Q = PΔt = nCP (Tf − Ti ) The number of moles of gas is n = PV i i / RTi = 0.355 mol. Thus
( 25 W )(15 s ) = ( 0.355 mol )( 20.8 J
mol K )(Tf − 293 K )
⇒ Tf = 344 K ⇒ Vf = ( 8.0 × 10−4 m3 ) ( 344 K 293 K ) = 9.39 × 10−4 m3 = 939 cm3 ≈ 940 cm3
17.61. Model: Assume that the gas is an ideal gas. A diatomic gas has γ = 1.40. Solve:
(a) For container A,
ViA =
nRTAi ( 0.10 mol )( 8.31 J mol K )( 300 K ) = = 8.20 × 10−4 m3 pAi 3.0 × 1.013 × 105 Pa
For an isothermal process pAfVAf = pAiVAi. This means TAf = TAi = 300 K and
VAf = VAi ( pAi pAf ) = ( 8.20 × 10−4 m3 ) ( 3.0 atm 1.0 atm ) = 2.5 × 10−3 m3 The gas in container B starts with the same initial volume. For an adiabatic process,
⎛p ⎞ pBf VBf = pBiVBi ⇒ VBf = VBi ⎜ Bi ⎟ ⎝ pBf ⎠ γ
γ
1
γ
⎛ 3.0 atm ⎞ = ( 8.20 × 10 m ) ⎜ ⎟ ⎝ 1.0 atm ⎠ −4
3
1 1.40
= 1.8 × 10−3 m3
The final temperature TBf can now be obtained by using the ideal-gas equation: TBf = TiB
(b)
−3 3 pBf VBf ⎛ 1.0 atm ⎞ ⎛ 1.80 × 10 m ⎞ = ( 300 K ) ⎜ = 220 K ⎟⎜ 3 ⎟ −4 pBi VBi ⎝ 3.0 atm ⎠ ⎝ 8.20 × 10 m ⎠
17.62. Model: The gas is assumed to be an ideal gas that is subjected to isobaric and isochoric processes.
Solve: (a) The initial conditions are p1 = 3.0 atm = 304,000 Pa, V1 = 100 cm3 = 1.0 × 10−4 m3, and T1 = 100°C = 373 K. The number of moles of gas is
n=
−4 3 p1V1 ( 304,000 Pa ) (1.0 × 10 m ) = = 9.81 × 10−3 mol RT1 ( 8.31 J mol K ) (373 K)
At point 2 we have p2 = p1 = 3.0 atm and V2 = 300 cm3 = 3V1. This is an isobaric process, so
V2 V1 V = ⇒ T2 = 2 T1 = 3(373 K) = 1119 K T2 T1 V1 The gas is heated to raise the temperature from T1 to T2. The amount of heat required is
Q = nCP ΔT = ( 9.81 × 10−3 mol ) ( 20.8 J mol K )(1119 K − 373 K ) = 152 J This amount of heat is added during process 1 → 2. (b) Point 3 returns to T3 = 100°C = 373 K. This is an isochoric process, so
Q = nCV ΔT = ( 9.81 × 10−3 mol ) (12.5 J mol K )( 373 K − 1119 K ) = −91.5 J This amount of heat is removed during process 2 → 3.
17.63. Assume the gas to be an ideal gas. Solve:
(a) The work done on the gas is the negative of the area under the p-versus-V graph, that is
W = −area under curve = −50.7 J (b) The change in thermal energy is
ΔEth = nCV ΔT = nCV (Tf − Ti )
Using the ideal-gas law to calculate the initial and final temperatures, Ti =
( 4.0 ×1.013 × 105 Pa )(100 × 10−6 m3 ) = 325 K pV i i = nR ( 0.015 mol )(8.31 J mol K )
Tf =
5 −6 3 pfVf (1.013 × 10 Pa )( 300 × 10 m ) = = 244 K nR ( 0.015 mol )( 8.31 J mol K )
⇒ ΔEth = ( 0.015 mol )(12.5 J mol K )( 244 K − 325 K ) = −15.2 J ≈ −15 J (c) From the first law of thermodynamics, ΔEth = Q + W ⇒ Q = ΔEth − W = −15.2 J − ( −50.7 J ) = 35.5 J ≈ 36 J
That is, 36 J of heat energy is transferred to the gas.
17.64. Model: Assume that the gas is an ideal gas and that the work, heat, and thermal energy are connected by the first law of thermodynamics. Solve: (a) For point 1, V1 = 1000 cm3 = 1.0 × 10−3 m3, T1 = 133°C = 406 K, and the number of moles is
n=
⎛ 120 × 10−3 g ⎞ M =⎜ ⎟ = 0.030 mol M mol ⎝ 4 g/mol ⎠
Thus, the pressure p1 is
p1 =
nRT1 = 1.012 × 105 Pa = 1.0 atm V1
The process 1 → 2 is isochoric (V2 = V1) and p2 = 5p1 = 5.0 atm. Thus,
T2 = T1 ( p2 p1 ) = ( 406 K )( 5 ) = 2030 K = 1757°C The process 2 → 3 is isothermal (T2 = T3), so V3 = V2(p2/p3) = V2(p2/p1) = 5V2 = 5000 cm3 p (atm)
T (°C)
V (cm3)
1.0 5.0 1.0
133 1757 1757
1000 1000 5000
Point 1 Point 2 Point 3
(b) The work W1→ 2 = 0 J because it is an isochoric process. The work in process 2 → 3 can be found using Equation 17.16 as follows:
W2 →3 = − nRT2 ln (V3 V2 ) = − ( 0.030 mol )( 8.31 J mol K )( 2030 K ) ln ( 5 ) = −815 J The work in the isobaric process 3 → 1 is
W3→1 = − p (Vf − Vi ) = − (1.012 × 105 Pa )(1.0 × 10−3 m3 − 5.0 × 10−3 m3 ) = 405 J (c) The heat transferred in process 1 → 2 is
Q1→ 2 = nCV ΔT = ( 0.030 mol )(12.5 J mol K )( 2030 K − 406 K ) = 609 J
The heat transferred in the isothermal process 2 → 3 is Q2 →3 = −W2 →3 = 815 J . The heat transferred in the isobaric process 3 → 1 is Q3→1 = nCP ΔT = ( 0.030 mol )( 20.8 J mol K )( 406 K − 2030 K ) = −1013 J
17.65. Model: The air is assumed to be an ideal diatomic gas that is subjected to an adiabatic process.
Solve: The air admitted into the cylinder at T0 = 30°C = 303 K and p0 = 1 atm = 1.013 × 105 Pa has a volume V0 = 600 × 10−6 m3 and contains pV n = 0 0 = 0.024 mol RT0
Using Equation 17.36 and the fact that Q = 0 J for an adiabatic process, ΔEth = Q + W = nCV ΔT ⇒ W = nCVT
⇒ 400 J = (0.024 mol)(20.8 J/mol K)(Tf – 303 K) ⇒ Tf = 1100 K For an adiabatic process Equation 17.40 is 1
TfVf Assess:
γ −1
γ −1
= T0V0
1
⎛ T ⎞ γ −1 ⎛ 303 K ⎞1.4 −1 ⇒ Vf = V0 ⎜ 0 ⎟ = ( 6.0 × 10−4 m3 ) ⎜ = 2.39 × 10−5 m3 ≈ 24 cm3 ⎟ T ⎝ 1100 K ⎠ ⎝ f⎠
Note that W is positive because the environment does work on the gas.
17.66. Model: γ is 1.40 for a diatomic gas and 1.67 for a monoatomic gas. Solve:
(a) We will assume that air is a diatomic gas. For an adiabatic process, γ −1 Tf Vfγ −1 = TV i i
Thus 1
1
⎛ Vi ⎞ ⎛ Tf ⎞ γ −1 ⎛ 1123 K ⎞1.40 −1 = 26.4 ⎜ ⎟=⎜ ⎟ =⎜ ⎟ ⎝ 303 K ⎠ ⎝ Vf ⎠ ⎝ Ti ⎠ (b) For argon, a monatomic gas, 1
⎛ Vi ⎞ ⎛ 1123 K ⎞1.67 −1 = 7.07 ⎜ ⎟=⎜ ⎟ ⎝ Vf ⎠ ⎝ 303 K ⎠
17.67. Model: Air is assumed to be an ideal diatomic gas that is subjected to an adiabatic process. Solve:
(a) Equation 17.40 for an adiabatic process is 1
γ −1
TfVf
= TV i i
γ −1
V ⎛ T ⎞ γ −1 ⇒ f =⎜ i ⎟ Vi ⎝ Tf ⎠
For the temperature to increase from Ti = 20°C = 293 K to Tf = 1000°C = 1273 K, the compression ratio will be 1
V Vf ⎛ 293 K ⎞1.4 −1 1 = 39.3 =⎜ = 0.02542 ⇒ max = ⎟ Vmin 0.02542 Vi ⎝ 1273 K ⎠ (b) From the Equation 17.39, γ
pf ⎛ Vi ⎞ 1.4 pfVf = pV = ⎜ ⎟ = ( 39.3) = 171 i i ⇒ pi ⎝ Vf ⎠ γ
γ
17.68. Model: The helium gas is assumed to be an ideal gas that is subjected to an isobaric process. Solve:
(a) The number of moles in 2.0 g of helium is
n=
M 2.0 g = = 0.50 mol M mol 4.0 g mol
At Ti = 100°C = 373 K and pi = 1.0 atm = 1.013 × 105 Pa, the gas has a volume Vi =
nRTi = 0.0153 m3 = 15.3 L pi
For an isobaric process (pf = pi) that doubles the volume Vf = 2Vi,
(Tf
Ti ) = (Vf Vi ) = 2 ⇒ Tf = 2Ti = 2 ( 373 K ) = 746 K = 473°C
(b) The work done by the environment on the gas is 5 3 W = − pi (Vf − Vi ) = − pV i i ( 2 − 1) = − (1.013 × 10 Pa )( 0.0153 m ) = −1550 J
(c) The heat input to the gas is
Q = nCP (Tf − Ti ) = ( 0.50 mol )( 20.8 J mol K )( 746 K − 373 K ) = 3880 J ≈ 3900 J (d) The change in the thermal energy of the gas is
ΔEth = Q + W = 3880 J − 1550 J = 2330 J ≈ 2300 J
(e)
Assess:
The internal energy can also be calculated as follows:
ΔEth = nCV ΔT = ( 0.5 mol )(12.5 J mol K )( 746 K − 373 K ) = 2330 J This is the same result as we got in part (d).
17.69. Model: The helium gas is assumed to be an ideal gas that is subjected to an isothermal process. Solve:
(a) The number of moles in 2.0 g of helium gas is
n=
M 2.0 g = = 0.50 mol M mol 4.0 g mol
At Ti = 100°C = 373 K and pi = 1.0 atm = 1.013 × 105 Pa, the gas has a volume Vi =
nRTi = 0.0153 m3 = 15.3 L pi
For an isothermal process (Tf = Ti) that doubles the volume Vf = 2Vi, 1 pf Vf = pV i i ⇒ pf = pi (Vi Vf ) = (1.0 atm ) ( 2 ) = 0.50 atm
(b) The work done by the environment on the gas is
W = − nRTi ln (Vf Vi ) = − ( 0.50 mol )( 8.31 J mol K )( 373 K ) ln ( 2 ) = −1074 J ≈ −1070 J (c) Because ΔEth = Q + W = 0 J for an isothermal process, the heat input to the gas is Q = −W = 1074 J ≈ 1070 J . (d) The change in internal energy ΔEth = 0 J.
(e)
17.70. Model: The nitrogen gas is assumed to be an ideal gas that is subjected to an adiabatic process. Solve:
(a) The number of moles in 14.0 g of N2 gas is
n=
M 14.0 g = = 0.50 mol M mol 28 g mol
At Ti = 273 K and pi = 1.0 atm = 1.013 × 105 Pa, the gas has a volume Vi =
nRTi = 0.0112 m3 = 11.2 L pi
For an adiabatic process that compresses to a pressure pf = 20 atm, we can use Equation 17.39 and Equation 17.40 as follows: TfVf
γ −1
= TV i i
γ −1
⎛T ⎞ ⎛V ⎞ ⇒⎜ f ⎟=⎜ i ⎟ ⎝ Ti ⎠ ⎝ Vf ⎠
γ −1
1
⎛ p ⎞γ V pf Vf = piVi ⇒ ⎜ f ⎟ = i Vf ⎝ pi ⎠ γ
γ
Combining the above two equations yields
(Tf
Ti ) = ( pf pi )
γ −1 γ
⇒ Tf = Ti ( 20 )
1.4 −1.0 1.4
= Ti ( 2.3535 ) = 643 K
(b) The work done on the gas is
W = ΔEth = nCV (Tf − Ti ) = ( 0.50 mol )( 20.8 J mol K )( 643 K − 273 K ) = 3850 J ≈ 3800 J (c) The heat input to the gas is Q = 0 J. (d) From the above equation, 1
1 Vi ⎛ pf ⎞ γ V = ⎜ ⎟ = ( 20 )1.4 = 8.5 = max Vf ⎝ pi ⎠ Vmin
(e)
17.71. Model: The gas is assumed to be an ideal gas that is subjected to an isochoric process. Solve:
(a) The number of moles in 14.0 g of N2 gas is
n=
M 14.0 g = = 0.50 mol M mol 28 g mol
At Ti = 273 K and pi = 1.0 atm = 1.013 × 105 Pa, the gas has a volume Vi =
nRTi = 0.0112 m3 = 11.2 L pi
For an isochoric process (Vi = Vf), Tf pf 20 atm = = = 20 ⇒ Tf = 20 ( 273 K ) = 5460 K ≈ 5500 K Ti pi 1 atm (b) The work done on the gas is W = − pΔV = 0 J. (c) The heat input to the gas is Q = nCV (Tf − Ti ) = ( 0.50 mol )( 20.8 J mol K )( 5460 K − 273 K ) = 5.4 × 10 4 J
(d) The pressure ratio is
pmax pf 20 atm = = = 20 pmin pi 1 atm
(e)
17.72. Model: The air is assumed to be an ideal gas. Because the air is compressed without time to exchange heat with its surroundings, the compression is an adiabatic process. Solve: The initial pressure of air in the mountains behind Los Angeles is pi = 60 × 103 Pa at Ti = 273 K. The pressure of this air when it is carried down to the elevation near sea level is pf = 100 × 103 Pa. The adiabatic compression of a gas leads to an increase in temperature according to Equation 17.39 and Equation 17.40, which are γ −1
TfVf
= TV i i
γ −1
⎛T ⇒⎜ f ⎝ Ti
γ −1
⎞ ⎛ Vi ⎞ ⎟=⎜ ⎟ ⎠ ⎝ Vf ⎠
1
⎛ p ⎞γ V pfVf = piVi ⇒ ⎜ f ⎟ = i Vf ⎝ pi ⎠ γ
γ
Combining these two equations,
(Tf
Ti ) = ( pf pi )
γ −1 γ
⎛ 100 × 103 Pa ⎞ ⇒ Tf = Ti ⎜ ⎟ 3 ⎝ 60 × 10 Pa ⎠
1.4 −1 1.4
⎛5⎞ = ( 273 K ) ⎜ ⎟ ⎝3⎠
0.286
= 316 K = 43ºC = 109ºF
17.73. Model: Assume the collector has emissivity e = 1.0 . We will use Equation 17.50 and divide both sides by A to get intensity in W/m 2 . At the equilibrium temperature the collector will absorb 800 W/m 2 on the sun side and must radiate the same amount from the same side. T0 = 20°C = 293 K . Solve: Q/ Dt = es (T 4 − T04 ) A Solve this for T . Q/ Dt 1 = T 4 − T04 A es
T=
4
Q/ Dt 1 + T04 = A es
4
800 W/m 2 4 + ( 293K ) = 383K = 110°C 28 2 4 1 0 5 67 10 W/m K . . × ⋅ ( )( )
Assess: 110°C is hot enough to boil water, so this seems feasible as a hot water heater, at least during the 5 hours a day the collector gets 800 W/m 2 . You can store the hot water in an insulated tank for use at other times of the day.
17.74. Model: This is a problem about conduction of the heat from inside the box to outside. 100 W of heat is generated by the light bulb inside the box, so in equilibrium that is how much must be conducted away through the sides of the box. Visualize: We are given L = 0.012 m , Q/ Dt = 100 W and the thermal conductivity of concrete k = 0.8 W/m ⋅ K . We compute A = 6 ( 0.20m × 0.20 m ) = 0.24m 2 . TC = 20°C = 293K .
Solve:
Use Equation 17.48 for the rate of heat transfer by conduction. Q A = k (TH − TC ) Dt L
Solve for TH . ⎛Q⎞ L = (TH − TC ) ⎜ ⎟ ⎝ Dt ⎠ kA
0.012 m ⎛Q⎞ L TH = ⎜ ⎟ + TC = (100 W ) + 293 K = 299 K = 26°C D t kA 0 8 W/m . ?K ) ( 0.24 m 2 ) ⎝ ⎠ ( Assess:
We expected TH to be a few degrees hotter than TC , and so it is.
17.75. Model: Refer to Example 17.11. Assume the surface of the earth is an ideal radiator with e = 1 . Visualize: Only half the earth faces the sun at a time, so the earth intercepts 1370 W/m 2 over a cross section of pRe2 = p ( 6.37 × 106 m ) = 1.275 × 1014 m 2 . If the earth’s surface absorbs 70% of the incident power then the total 2
power absorbed is
( 0.70 ) (1370 W/m 2 )(1.275 × 1014 m 2 ) = 1.222 × 1017 W
. This much power must also be
radiated away from the earth in equilibrium. Solve: The power is radiated from the whole spherical surface of the earth: A = 4pRe2 . Use Equation 17.49 to find the temperature of the earth.
⎡ Q/ Dt ⎤ ⎥ T =⎢ 2 ⎢⎣ es ( 4pRe ) ⎥⎦
1/ 4
⎡ ⎤ 1.22231017 W ⎥ =⎢ 2 ⎢ (1) ( 5.6731028 W/m 2?K ) 4p ( 6.373106 m ) ⎥ ⎣ ⎦
1/ 4
= 255 K = 218°C
Assess: Without a moderate greenhouse effect the earth would be too cold for mammal life. On the other hand, when the greenhouse effect gets out of hand, such as on Venus, it can be too hot for life.
17.76. Solve: (a) 50 J of work are done on a gas to compress it to one-third of its original volume at a constant temperature of 77°C. How many moles of the gas are in the sample? (b) The number of moles is
n=
50 J = 0.0156 mol − ( 8.31 J mol K )( 350 K ) ( ln 13 )
17.77. Solve: (a) A heated 500 g iron slug is dropped into a 200 cm3 pool of mercury at 15°C. If the mercury temperature rises to 90°C, what was the initial temperature of the iron slug? (b) The initial temperature was 217°C .
17.78. Solve: (a) A diatomic gas is adiabatically compressed from 1 atm pressure to 10 atm pressure. What is the compression ratio Vmax/Vmin? 1
(b) The ratio is Vmax Vmin = 101.4 = 5.18 .
17.79. Model: Assume the helium gas to be an ideal gas. The gas is subjected to isothermal, isochoric, and adiabatic processes. Visualize: Please refer to Figure CP17.79. The gas at point 1 has volume V1 = 1000 cm3 = 1.0 × 10−3 m3 and pressure p1 = 3.0 atm. At point 2, V2 = 3000 cm3 = 3.0 × 10−3 m3 and p2 = 1.0 atm. These values mean that T2 = T1, so process 1 → 2 is an isothermal process. The process 2 → 3 occurs at constant volume and is thus an isochoric process. Finally, because temperature T3 is lower than T1 or T2, the process 3 → 1 is an adiabatic process. Solve: (a) The number of moles of gas is n=
0.120 g = 0.030 mols 4 g mol
The temperature T1 can be calculated to be T1 =
5 −3 3 p1V1 ( 3.0 × 1.013 × 10 Pa )(1.0 × 10 m ) = = 1219 K = 946°C nR ( 0.030 mol )(8.31 J mol K )
For the isothermal process 1 → 2, T2 = 1219 K. For the adiabatic process 3 → 1, p3 = p1 (V1 V3 )
γ
1.67
⎛ 1000 cm3 ⎞ = ( 3.0 atm ) ⎜ 3 ⎟ ⎝ 3000 cm ⎠ Point
= 0.48 atm
p (atm)
T3 = T1 (V1 V3 )
γ −1
T (°C)
= (1219 K ) ( 13 )
0.67
= 583 K=310°C
V (cm3)
1 3.0 946 1000 2 1.0 946 3000 3 0.48 310 3000 Note that the values obtained above are consistent with the isochoric process 2 → 3, for which p2 p3 = ⇒ p2 = (T2 T3 ) p3 = (1219 K 583 K )( 0.48 atm ) = 1 atm T2 T3 (b) From Equation 17.15,
⎛V ⎞ W1→ 2 = − nRT1 ln ⎜ 2 ⎟ = − ( 0.030 mol )( 8.31 J mol K )(1219 K ) ln ( 3) = −334 J ⎝ V1 ⎠
The work done in the ischochoric process is W2 →3 = 0 J . The work done in the adiabatic process is
W3→1 = nCV (T1 − T3 ) = ( 0.030 mol )(12.5 J mol K )(1219 K − 583 K ) = 239 J (c) For the process 1 → 2, ΔT = 0 K ⇒ ΔEth = 0 J ⇒ Q = −W = 334 J
For the process 2 → 3, W = 0 J,
ΔEth = Q = nCV (T3 − T2 ) = −239 J For the process 3 → 1, Q = 0 J.
17.80. Model: The gas is an ideal gas, and its thermal energy is the total kinetic energy of the moving molecules. Visualize: Please refer to Figure P17.80. Solve: (a) The piston is floating in static equilibrium, so the downward force of gravity on the piston’s mass must exactly balance the upward force of the gas, Fgas = pA where A = πr2 is the area of the face of the piston. Since the upper part of the cylinder is evacuated, there is no gas pressure force pushing downward. Thus,
M piston g = pA ⇒ p =
M piston g A
=
ρCuVpiston g A
= ρ Cu gh = ( 8920 kg m3 )( 9.80 m s 2 ) ( 0.040 m ) = 3500 Pa
(b) The gas volume is V1 = π r 2 L = π (0.030) 2 (0.20 m) = 5.65 × 10−4 m3 . The number of moles is
n=
−4 3 p1V1 ( 3500 Pa ) ( 5.65 × 10 m ) = = 8.12 × 10−4 mol RT1 ( 8.31 J mol K )( 293 K )
The number of molecules is N = nNA = (8.12 × 10−4 mol)(6.02 × 1023 mol−1) = 4.9 × 1020 (c) The pressure in the gas is determined simply by the weight of the piston. That will not change as heat is added, so the heating takes place at constant pressure with Q = nCPΔT. The temperature increase is
ΔT =
2.0 J Q = = 85 K nCP ( 8.12 × 10−4 mol ) ( 29.1 J mol K )
This raises the gas temperature to T2 = T1 + ΔT = 378 K = 105°C. (d) Noting that the volume of a cylinder is V = πr2L and that r doesn’t change, the ideal-gas relationship for an isobaric process is V2 V1 L L T ⎛ 378 K ⎞ = ⇒ 2 = 1 ⇒ L2 = 2 L1 = ⎜ ⎟ 20 cm = 25.8 cm T2 T1 T2 T1 T1 ⎝ 293 K ⎠ (e) The work done by the gas is Wgas = Fgas Δy . The force exerted on the piston by the gas is
Fgas = pA = pπ r 2 = 9.90 N This force is applied through Δy = 5.8 cm = 0.058 m, so the work done is Wgas = (9.90 N)(0.058 m) = 0.574 J ≈ 0.57 J
Thus, 0.57 J is the work done by the gas on the piston. The work done on the gas is 0.57 J.
17.81. Model: There is a thermal interaction between the iron, assumed to be initially at room temperature (20°C), and the liquid nitrogen. The boiling point of liquid nitrogen is –196°C = 77 K. Solve: The piece of iron has mass Miron = 197 g = 0.197 kg and volume Viron = Miron/ρiron = (0.197 kg)/(7870 kg/m3) = 25 × 10−6 m3 = 25 cm3 = 25 mL. The heat lost by the iron is Qiron = Mironciron∆T = (0.197 kg)(449 J/kg K)(77 K – 293 K) = –1.911 × 104 J This heat causes mass M of the liquid nitrogen to boil. Energy conservation requires Qiron + QN2 = Qiron + MLf = 0 ⇒ M = −
Qiron 1.911 × 104 J = = 0.0960 kg Lf 1.99 × 105 J/kg
The volume of liquid nitrogen boiled away is thus
Vboil =
M
ρ N2
=
0.0960 kg = 1.19 × 10−4 m3 = 119 mL 810 kg/m3
Now the volume of nitrogen gas (at 77 K) is 1500 mL before the iron is dropped in. The volume of the piece of iron excludes 25 mL of gas, so the initial gas volume, when the lid is sealed and the liquid starts to boil, is V1 = 1475 mL. The pressure is p1 = 1.0 atm and the temperature is T1 = 77 K. Thus the number of moles of nitrogen gas is n1 =
p1V1 (101,300 Pa)(1.475 × 10−3 m3 ) = = 0.234 mol RT1 (8.31 J/mol K)(77 K)
119 mL of liquid boils away, so the gas volume increases to V2 = V1 + Vboil = 1475 mL + 119 mL = 1594 mL. The temperature is still T2 = 77 K, but the number of moles of gas has been increased by the liquid that boiled. The number of moles that boiled away is 96.0 g nboil = = 3.429 mol 28 mol/g
Thus the number of moles of nitrogen gas increases to n2 = n1 + nboil = 0.234 mol + 3.429 mol = 3.663 mol. Consequently, the gas pressure increases to p2 =
n2 RT2 (3.663 mol)(8.31 J/mol K)(77 K) = = 1.470 × 106 Pa = 14.5 atm ≈ 15 atm V2 1.594 × 10−3 m3
Assess: Don’t try this! The large pressure increase could cause a flask of liquid nitrogen to explode, leading to serious injuries.
17.82.
Model: Ignore any radiation and assume all of the heat is transferred by conduction through the compound rod from the hot end to the cold end. Visualize: Look up the thermal conductivities of the two metals in Table 17.5: kCu = 400 W/m? k and
kFe = 80 W/m?K . TH = 373 K and TC = 273 K. Equation 17.48 will be applied to each rod and all of the heat must go through both rods at the same rate. ⎛Q⎞ ⎛Q⎞ ⎜ ⎟ =⎜ ⎟ D t ⎝ ⎠Cu ⎝ Dt ⎠ Fe
Solve:
Since A and L are the same for the two rods, then kCu DTCu = kFeDTFe
Label the temperature at the joining point in the middle with a subscript M.
kCu (TH − TM ) = kFe (TM − TC ) Solve for TM . kCuTH − kCuTM = kFeTM − k FeTC kCuTH + kFeTC = kFeTM + kCuTM
( kFe + kCu )TM = kCuTH + kFeTC TM = TM =
kCuTH + kFeTC kCu + kFe
(400 W/m ⋅ K)(373 K) + (80 W/m ⋅ K)(273 K) = 356 K = 83°C 400 W/m ⋅ K + 80 W/m ⋅ K
Assess: Had we not arrived at an answer between 100°C and 0°C we would have been very worried. Furthermore, because the conductivity of copper is greater than the conductivity of iron we expected the answer to be above 50°C.
17.83. Model: Assume the gas is ideal. The process is not adiabatic, despite Equation 17.38, because the exponent
given in the problem is not equal to g . The gas is identified as diatomic which means that g = 1.40 , but the exponent is
explicitly given as 2, so while we are given pV 2 = constant, it is not true that pV g = constant. We will use the fact that the gas is diatomic to deduce CV = 52 R = 20.8J/mol ⋅ k , although we won’t specifically need g = 1.40 . We are given n = 0.020mol , Ti = 293K , Vi = 1500cm3 , and Vf = 500cm3 .
Visualize: Solve:
(a) If we use the ideal-gas-law expression p = nRT/V in pV 2 = constant we get TV = constant .
Tf =
TV (293 K)(1500 cm3 ) i i = = 879 K = 606°C Vf 500 cm3
(b) The strategy to find Q will be to use the first law Q = DEth − W . We will use Eth = nCV DT and W = 2∫ pdV.
DEth = nCV DT = ( 0.020 mol )( 20.8 J/mol ⋅ k )( 879 K − 293 K ) = 243.8 J To do the W = 2∫ pdV integral we need to know what the constant is in pV 2 = constant . Use the ideal gas law to compute pi . pi =
nRTi ( 0.020 mol )( 8.31 J/mol ⋅ K )( 293 K ) = = 32,460 Pa Vi 1.5 ×1023 m3
23 2 constant = pV m 3 ) = 0.0730 Pa ⋅ m 6 i i = ( 32,460Pa ) (1.5 × 10 2
Vf
W = − ∫ pdV = 2Ñ
Vi
constant V2
dV = 2( 0.0730 Pa ⋅ m
6
)Ñ
0.00050 m3
0.0015 m3
0.00050 m3
dv ⎡1⎤ = ( 0.0730 Pa ⋅ m 6 ) ⎢ ⎥ = 97.4 J 2 V ⎣V ⎦ 0.0015 m3
Q = DEth − W = 243.8 J − 97.4 J = 146.4 J < 150 J (c) We also need pf to complete the pV diagram. We could use the ideal-gas law again to compute pf , but let’s do it another way. 2 2 pV i i = pf Vf 2
2
⎛V ⎞ ⎛ 1500 cm3 ⎞ pf = pi ⎜ i ⎟ = pi ⎜ = 9 pi = 292,200 Pa 3 ⎟ ⎝ 500 cm ⎠ ⎝ Vf ⎠
Assess: Both Q and W are positive because heat was added to the system and work was done on the system.
17-1
18.1. Solve: We can use the ideal-gas law in the form pV = NkBT to determine the Loschmidt number (N/V):
(1.013 × 105 Pa ) = 2.69 × 1025 m−3 N p = = V kBT (1.38 × 10−23 J K ) ( 273 K )
18.2. Solve: The volume of the nitrogen gas is 1.0 m3 and its temperature is 20°C or 293 K. The number of gas molecules can be found as N = nN A =
(1.013 × 105 Pa )(1.0 m3 ) 6.02 × 1023 mol−1 = 2.5 × 1025 pV NA = ( ) RT ( 8.31 J mol K )( 293 K )
According to Figure 18.2, 12% of the molecules have a speed between 700 and 800 m/s, 7% between 800 and 900 m/s, and 3% between 900 and 1000 m/s. Thus, the number of molecules in the cube with a speed between 700 m/s and 1000 m/s is (0.22)(2.51 × 1025) = 5.5 × 1024.
18.3. Solve: Nitrogen is a diatomic molecule, so r ≈ 1.0 × 10−10 m. We can use the ideal-gas law in the form pV = NkBT and Equation 18.3 for the mean free path to obtain p:
λ=
(1.38 ×10−23 J k ) ( 293 K ) = 0.023 Pa kBT 1 kBT = ⇒ p = = 2 4 2π ( N V ) r 2 4 2π pr 2 4 2πλ r 2 4 2π (1.0 m ) (1.0 × 10−10 m )
Assess: In Example 18.1 λ = 225 nm at STP for nitrogen. λ = 1.0 m must therefore require a very small pressure.
18.4. Solve: (a) Air is primarily comprised of diatomic molecules, so r ≈ 1.0 × 10−10 m. Using the ideal-gas
law in the form pV = NkBT, we get
N p = = V kBT
1.013 × 105 Pa 760 mm of Hg = 3.30 × 1012 m −3 J K ) ( 293 K )
1.0 × 10−10 mm of Hg ×
(1.38 ×10
−23
(b) The mean free path is
λ=
1 1 = = 1.71 × 106 m 2 4 2π ( N V ) ( r 2 ) 4 2π ( 3.30 × 1012 m −3 )(1.0 × 10−10 m )
Assess: The pressure p in the vacuum chamber is 1.33 × 10−8 Pa = 1.32 × 10−13 atm. A mean free path of 1.71 × 106 m is large but not unreasonable.
18.5. Solve: (a) The mean free path of a molecule in a gas at temperature T1, volume V1, and pressure p1 is λ1 = 300 nm. We also know that
λ=
1 ⇒ λ ∝V 4 2π ( N / V )r 2
Although T2 = 2T1, constant volume (V2 = V1) means that λ2 = λ1 = 300 nm. (b) For T2 = 2 T1 and p2 = p1, the ideal gas equation gives
p1V1 pV p1V2 ⇒ V2 = 2V1 = 2 2 = NkBT1 NkBT2 NkB ( 2T1 )
Because λ ∝ V , λ2 = 2λ1 = 2(300 nm) = 600 nm.
18.6. Solve: Neon is a monatomic gas and has a radius r ≈ 5.0 × 10−11 m. Using the ideal-gas equation,
(150 ) (1.013 × 105 Pa ) N p = = = 3.695 × 1027 m −3 V kBT (1.38 × 10−23 J/K ) ( 298 K ) Thus, the mean free path of a neon atom is
λ=
1 1 = = 6.09 × 10−9 m 2 2 27 4 2π ( N V ) r 4 2π ( 3.695 × 10 m −3 )( 5.0 × 10−11 m )
Since the atomic diameter of neon is 2 × (5.0 × 10−11 m) = 1.0 × 10–10 m,
λ=
6.09 × 10−9 m = 61 atomic diameters 1.0 × 10−10 m
18.7. Solve: The number density of the Ping-Pong balls inside the box is 2000 N = = 2000 m −3 V 1.0 m3 With r = (3.0 cm)/2 = 1.5 cm, the mean free path of the balls is
λ=
1 = 0.125 m = 12.5 cm 4 2π ( N V ) ( r 2 )
18.8. Solve: (a) The average speed is vavg =
1 ⎛ n = 25 ⎞ 220 m/s = 20.0 m/s ⎜ ∑ n ⎟ m/s = 11 ⎝ n =15 ⎠ 11
(b) The root-mean-square speed is vrms =
(v2 )
1
avg
1
⎛ 1 n = 25 ⎞ 2 ⎛ 4510 ⎞ 2 = ⎜ ∑ n 2 ⎟ m/s = ⎜ ⎟ = 20.2 m/s ⎝ 11 ⎠ ⎝ 11 n =15 ⎠
18.9. Solve: (a) In tabular form we have Particle
vx (m/s)
vy (m/s)
1 20 30 2 70 −40 3 −80 −10 4 60 −20 5 0 −50 6 40 −20 Average 0 0 G G G ˆ The average velocity is vavg = 0 i + 0 ˆj .
vx2 (m/s)2
v y2 (m/s)2
v2 (m/s)2
v (m/s)
400 1600 6400 3600 0 1600
900 4900 100 400 2500 400
1300 6500 6500 4000 2500 2000 3800
36.06 80.62 80.62 63.25 50.00 44.72 59.20
(b) The average speed is vavg = 59 m/s . (c) The root-mean-square speed is vrms =
(v ) 2
avg
= 3800 m 2 / s 2 = 62 m/s .
18.10. Solve: (a) The most probable speed is 4.0 m/s. (b) The average speed is vavg =
2 × 2 m/s + 4 × 4 m/s + 3 × 6 m/s + 1 × 8 m/s = 4.6 m/s 2 + 4 + 3 +1
(c) The root-mean-square speed is
2 × ( 2 m/s ) + 4 × ( 4 m/s ) + 3 × ( 6 m/s ) + 1 × ( 8 m/s ) = 4.9 m/s 2 + 4 + 3 +1 2
vrms =
2
2
2
18.11. Solve: (a) The atomic mass number of argon is 40. This means the mass of an argon atom is m = 40 u = 40(1.661 × 10−27 kg) = 6.64 × 10−26 kg The pressure of the gas is 2 ⎛N⎞ 2 p = 13 ⎜ ⎟ mvrms = 13 ( 2.00 × 1025 m −3 )( 6.64 × 10−26 kg ) ( 455 m/s ) = 9.16 × 104 Pa ⎝V ⎠
(b) The temperature of the gas in the container can be obtained from the ideal-gas equation in the form pV = NkBT:
T=
pV 9.16 × 104 Pa = = 332 K 25 NkB ( 2.00 × 10 m −3 )(1.38 × 10−23 J/K )
18.12. Model: Pressure is due to random collisions of gas molecules with the walls. Solve:
According to Equation 18.8, the collision rate with one wall is
rate of collisions =
N coll F pA = net = Δtcoll 2mvx 2mvx
where Fnet = pA is the force exerted on area A by the gas pressure. However, this equation assumed that all molecules are moving in the x-direction with constant speed. The rms speed vrms is for motion in three dimensions at varying speeds. Consequently, we need to replace vx not with (vx)avg, which is zero, but with vx → (vx2)avg =
2 vrms v = rms 3 3
With this change, rate of collisions =
3 pA 3(2 × 101,300 Pa)(0.10 m × 0.10 m) = = 6.5 × 1025 s −1 2mvrms 2(28 × 1.661× 10−27 kg)(576 m/s)
This collision rate can also be found by using the expression in Eq. 18.10, making the same change in vx, and using the ideal-gas law to determine N/V.
18.13. Visualize: We will use Equation 18.18 to find m of the atoms. p=
1N 2 mvrms 3V
We are given p = 2.0 atm = 202,650 Pa , N /V = 4.2 × 1025 m −3 , and vrms = 660 m/s. Solve: Solve for m.
m=
3 p V 3(202,650 Pa) 1 = = 3.32 × 10−26 kg = 20 u 2 vrms N (660 m/s) 2 4.2 ×1025 m −3
With an atomic mass of 20 the gas is likely neon. Assess: Neon is gaseous at room temperature, so is a likely choice. As a noble gas, it also doesn’t form many molecules.
18.14. Visualize: Equation 18.26 will give us vrms from the temperature and mass of the particles. vrms =
3kBT m
We are given T = 1100°C = 1373 K . The atomic mass of neon atoms is 20 u = 3.321 × 10−26 kg; the molecular
mass of oxygen molecules is 32 u = 5.31 × 10−26 kg. Solve: (a) vrms =
3(1.38 × 10−23 J/K)(1373 K) = 1310 m/s 3.321 × 10−26 kg
vrms =
3(1.38 × 10−23 J/K)(1373 K) = 1030 m/s 5.31 × 10−26 kg
(b)
Assess: These are fast speeds, but the temperature is high.
18.15. Visualize: Use Equation 18.26. vrms =
3kBT m
We are given vrms = 1.5 m/s and m = 28 u = 4.650 × 10−26 kg. Solve: Solve the equation for T . T=
Assess:
2 vrms m (1.5 m/s) 2 (4.650 × 10−26 kg) = = 2.5 mK 3kB 3(1.38 × 10−23 J/K)
2.5 mK is close to absolute zero, but that’s how cold it would have to be for vrms to be 1.5 m/s.
18.16.
Solve: Because the neon and argon atoms in the mixture are in thermal equilibrium, the temperature of each gas in the mixture must be the same. That is, using Equation 18.26, 2 2 mAr vrms Ar = mNe vrms Ne
vrms Ar = vrms Ne
20 u mNe = ( 400 m/s ) = 283 m/s 40 u mAr
18.17. Solve: The average translational kinetic energy per molecule is 3kBT 1 2 3 Pavg = mvrms = kBT ⇒ vrms = 2 2 m Since we want the vrms for H2 and N2 to be equal,
3kBTH2 mH 2
=
mH 3kBT ⎛ 2u ⎞ ⇒ TH2 = 2 TN2 = ⎜ ⎟ ( 373 K ) = 27 K = −246°C mN2 mN 2 ⎝ 28 u ⎠
18.18. Solve: The formula for the root-means-square speed as a function of temperature is
( vrms )T =
3kBT m
(a) For ( vrms )T = 12 ( vrms )STP ,
3kBT 1 3kB ( 273 K ) 1 =2 ⇒ T = (273 K) = 68 K= − 205°C m m 4 (b) For ( vrms )T = 2 ( vrms )STP ,
3kB ( 273 K ) 3kBT =2 ⇒ T = 4 ( 273 K ) = 1090 K = 817°C m m
18.19.
Solve:
Solve Equation 18.18 for vrms to see that if both p and V are doubled then vrms is also doubled.
vrms =
3 pV Nm
So the new rms speed will be 800 m/s. Assess: Think microscopically; for the pressure to double when the volume is doubled the particles will have to be going a lot faster and hit the walls more often.
18.20. Solve: The formula for the root-means-square speed as a function of temperature is
( vrms )T =
3kBT m
The ratio at 20°C and at 100°C is
( vrms )100 ( vrms )20
=
373 K = 1.13 293 K
18.21. Solve: molecule is
Assuming ideal-gas behavior and ignoring relativistic effects, the root-mean-square speed of a
vrms =
3kBT m
The temperature where vrms is the speed of light (c) for a hydrogen molecule
mv 2 ( 2 u ) c 2 = 2 (1.66 ×10−27 kg )( 3.0 ×108 m/s ) = 7.22 ×1012 K T = rms = 3kB 3kB 3 (1.38 × 10−23 J/K ) 2
18.22. Solve: (a) The average translational kinetic energy per molecule is 1 2 3 Pavg = mvrms = kBT 2 2
This means Pavg doubles if the temperature T doubles. (b) The root-mean-square speed vrms increases by a factor of (c) The mean free path is
λ=
2 as the temperature doubles.
1 4 2π ( N V ) r 2
Because N/V and r do not depend on T, doubling temperature has no effect on λ.
18.23. Solve: (a) The total translational kinetic energy of a gas is K micro = 32 N A kBT = 32 nRT . For H2 gas at STP, K micro =
3 (1.0 mol )(8.31 J/mol K )( 273 K ) = 3400 J 2
K micro =
3 (1.0 mol )(8.31 J/mol K )( 273 K ) = 3400 J 2
(b) For He gas at STP,
(c) For O2 gas at STP, K micro = 3400 J. Assess: The translational kinetic energy of a gas depends on the temperature and the number of molecules but not on the molecule’s mass.
18.24. Solve: (a) The mean free path is λ=
1 4 2π ( N V ) r 2
where r ≈ 0.5 × 10−10 m is the atomic radius for helium and N/V is the gas number density. From the ideal-gas law,
N p 0.10 atm × 101,300 Pa/atm = = = 7.34 × 1025 m −3 V kT (1.38 ×10−23 J/K ) (10 K ) ⇒λ =
1
4 2π ( 7.34 × 10 m 25
−3
)( 0.5 × 10
−10
m)
2
= 3.1 × 10−7 m = 310 nm
(b) The root-mean-square speed is
vrms =
3 (1.38 × 10−23 J/K ) (10 K ) 3kBT = = 250 m/s m 4 × (1.661 × 10−27 kg )
where we used A = 4 u as the atomic mass of helium. (c) The average energy per atom is ε avg = 32 kBT = 32 (1.38 × 10−23 J/K ) (10 K ) = 2.1 × 10−22 J .
18.25. Solve: (a) The average kinetic energy of a proton at the center of the sun is 3 Pavg = kBT ≈ 2
3 2
(1.38 × 10
−23
J/K )( 2.0 × 107 K ) = 4.1× 10−16 J
(b) The root-mean-square speed of the proton is
vrms =
3 (1.38 × 10−23 J/K )( 2.0 × 107 K ) 3kBT ≈ = 7.0 × 105 m/s m 1.67 × 10−27 kg
18.26. Solve: (a) Since the hydrogen in the sun’s atmosphere is monatomic, the average translational kinetic energy per atom is 3 Pavg = kBT = 32 (1.38 × 10−23 J/K ) ( 6000 K ) = 1.24 × 10−19 J 2 (b) The root-mean-square speed is
vrms =
3 (1.38 × 10−23 J/K ) ( 6000 K ) 3kBT = = 1.22 × 104 m/s mH 1.67 × 10−27 kg
18.27. Model: For a gas, the thermal energy is the total kinetic energy of the moving molecules. That is, Eth = Kmicro. Solve: The number of atoms is
N=
M 0.0020 kg = = 3.01 × 1023 m 6.64 × 10−27 kg
Because helium atoms have an atomic mass number A = 4, the mass of each helium atom is m = 4 u = 4 (1.661 × 10−27 kg ) = 6.64 × 10−27 kg The average kinetic energy of each atom is 2 K avg = 12 mvavg = 12 ( 6.64 × 10−27 kg ) ( 700 m s ) = 1.63 × 10−21 J 2
Thus the thermal energy of the gas is Eth = K micro = NK avg = ( 3.01× 1023 )(1.63 × 10−21 J ) = 490 J
18.28. Model: For a gas, the thermal energy is the total kinetic energy of the moving molecules. Solve:
Neon atoms have an atomic mass number A = 14, so the mass of each molecule is m = 28 u = 28(1.661 × 10−27 kg) = 4.51 × 10−26 kg
The number of molecules in the gas is
N=
M 0.010 kg = = 2.218 × 1023 m 4.51 × 10−26 kg
The thermal energy is 2 Eth = NK avg = N ( 12 mvavg ) ⇒ vavg =
2 Eth = Nm
2 (1700 J )
( 2.218 ×10 )( 4.51×10 23
−26
kg )
= 580 m/s
18.29. Solve: The volume of the air is V = 6.0 m × 8.0 m × 3.0 m = 144.0 m3, the pressure p = 1 atm = 1.013
× 105 Pa, and the temperature T = 20°C = 293 K. The number of moles of the gas is pV n= = 5991 mol RT This means the number of molecules is
N = nN A = ( 5991 mols ) ( 6.022 × 1023 mol−1 ) = 3.61 × 1027 molecules
Since air is a diatomic gas, the room’s thermal energy is Eth = N Pavg = N ( 52 kBT ) = 3.6 × 107 J Assess:
The room’s thermal energy can also be obtained as follows:
Eth = nCVT = (5991 mol)(20.8 J/mol K)(293 K) = 3.6 × 107 J
18.30. Solve: The thermal energy of a solid is Eth = 3 NkBT = 3nRT The volume of lead V = 100 cm3 = 10−4 m3, which means the mass is M = ρV = (11,300 kg/m3)(10−4 m3) = 1.13 kg
Because the atomic mass number of Pb is 207, the number of moles is n=
M 1.13 kg = = 5.459 mol M mol 0.207 kg/mol
⇒ Eth = 3( 5.459 mols )( 8.31 J/mol K )( 293 K ) = 3.99 × 104 J
18.31. Solve: (a) For a monatomic gas, ΔEth = nCV ΔT = 1.0 J = (1.0 mol )(12.5 J/mol K ) ΔT ⇒ ΔT = 0.080°C or 0.080 K (b) For a diatomic gas,
1.0 J = (1.0 mol)(20.8 J/mol K)ΔT ⇒ ΔT = 0.048°C or K (c) For a solid,
1.0 J = (1.0 mol)(25.0 J/mol K) ΔT ⇒ ΔT = 0.040°C or K
18.32. Solve: The conservation of energy equation (ΔEth)gas + (ΔEth)solid = 0 J is ngas ( CV )gas (Tf − Ti )gas + nsolid ( CV )solid (Tf − Ti )solid = 0 J
⇒ (1.0 mol)(12.5 J/mol K)(−50 K) + (1.0 mol)(25.0 J/mol K)(ΔT)solid = 0 ⇒ (ΔT)solid = 25°C The temperature of the solid increases by 25°C.
Refer to Figure 18.13. At low temperatures, CV = 32 R = 12.5 J/mol K. At room temperature and modestly hot temperatures, CV = 52 R = 20.8 J/mol K. At very hot temperatures, CV = 72 R = 29.1 J/mol K. Solve: (a) The number of moles of diatomic hydrogen gas in the rigid container is
18.33. Visualize:
0.20 g = 0.10 mol 2 g/mol
The heat needed to change the temperature of the gas from 50 K to 100 K at constant volume is Q = ΔEth = nCV ΔT = (0.10 mol)(12.5 J/mol K)(100 K – 50 K) = 62 J (b) To raise the temperature from 250 K to 300 K, Q = ΔEth = (0.10 mol)(20.8 J/mol K)(300 K – 250 K) = 100 J (c) To raise the temperature from 550 K to 600 K, Q = 100 J. (d) To raise the temperature from 2250 K to 2300 K, Q = ΔEth = nCVΔT = (0.10 mol)(29.1 J/mol K)(50 K) = 150 J.
18.34. Model: The potential energy of the bowling ball is transferred into the thermal energy of the mixture. We assume the starting temperature of the bowling ball to be 0°C. Solve: The potential energy of the bowling ball is
U g = M ball gh = (11 kg)(9.8 m/s 2 ) h = (107.8 kg m/s 2 )h This energy is transferred into the mixture of ice and water and melts 5 g of ice. That is, (107.8 kg m/s 2 )h = ΔEth = M w Lf ⇒ h =
(0.005 kg)(3.33 × 105 J/kg) = 15.4 m (107.8 kg m/s 2 )
18.35. Model: Heating the water and the kettle raises the temperature of the water to the boiling point and raises the temperature of the kettle to 100°C. Solve: The amount of heat energy from the electric stove’s output in 3 minutes is
Q = ( 2000 J s )( 3 × 60 s ) = 3.6 × 105 J This heat energy heats the kettle and brings the water to a boil. Thus,
Q = M water cwater ΔT + M kettle ckettle ΔT Substituting the given values into this equation, 3.6 × 105 J = M water ( 4190 J kg K )(100°C − 20°C ) + ( 0.750 kg )( 449 J kg K )(100°C − 20°C )
⇒ M water = 0.994 kg The volume of water in the kettle is
V= Assess:
3
3
M water
ρ water
=
0.994 kg = 0.994 × 10−3 m3 = 994 cm3 ≈ 990 cm3 1000 kg m3
1 L = 10 cm , so V ≈ 1 L. This is a reasonable volume of water.
18.36. Solve: The equilibrium condition is (εA)avg = (εB)avg = (εtot)avg ⇒ EAf = EBf = Etot nA nB nA + nB Thus the final thermal energies are
⎛ nA ⎞ 4.0 mols ⎛ ⎞ EAf = ⎜ ⎟ Etot = ⎜ ⎟ ( 9000 J + 5000 J ) = 8000 J 4.0 mols 3.0 mols n n + + ⎝ ⎠ B ⎠ ⎝ A ⎛ nB ⎞ ⎛ 3.0 mols ⎞ EBf = ⎜ ⎟ Etot = ⎜ ⎟ (14,000 J ) = 6000 J n n + ⎝ 7.0 mols ⎠ B ⎠ ⎝ A Because EAi = 9000 J and EAf = 8000 J, 1000 J of heat energy is transferred from gas A to gas B.
18.37. Solve: The mean free path for a monatomic gas is λ=
1 V ⇒ = 4π 2λ r 2 N 4π 2 ( N V ) r 2
For λ = 20r, meaning that the mean free path equals the atomic diameter, V V /N ⎛ 4π 3 ⎞ = 4π 2 ( 20r ) ( r 2 ) = ⎜ r ⎟ 60 2 ⇒ 4π 3 = 60 2 = 84.8 N r ⎝ 3 ⎠ 3
18.38. Visualize:
Solve:
The average energy of an oxygen molecule at 300 K is
ε avg =
Eth 5 = kBT = N 2
5 2
(1.38 × 10
−23
J/K ) ( 300 K ) = 1.035 × 10−20 J
The energy conservation equation Ugf + Kf = Ugi + Ki with K f = ε avg is
mgyf + ε avg = mgyi + 0 J With yi = h and yf = 0 m, we have
mgh = 1.035 × 10−20 J ⇒ h =
1.035 × 10−20 J = 1.99 × 104 m ( 32 ×1.66 ×10−27 kg )(9.8 m/s 2 )
18.39. Solve: (a) To identify the gas, we need to determine its atomic mass number A or, equivalently, the mass m of each atom or molecule. The mass density ρ and the number density (N/V ) are related by ρ = m(N/V ) , so the mass is m = ρ (V/N ) . From the ideal-gas law, the number density is 50,000 Pa N p = = = 1.208 × 1025 m −3 V kT (1.38 × 10−23 J/K ) ( 300 K ) Thus, the mass of an atom is m=ρ
V 8.02 × 10−2 kg/m3 = = 6.64 × 10−27 kg N 1.208 × 1025 m −3
Converting to atomic mass units,
A = 6.64 × 10−27 kg ×
1u = 4.00 u 1.661 × 10−27 kg
This is the atomic mass of helium. (b) Knowing the mass, we find vrms to be vrms =
3 (1.38 × 10−23 J/K ) ( 300 K ) 3kBT = = 1370 m/s m 6.64 × 10−27 kg
(c) A typical atomic radius is r ≈ 0.5 × 10−10 m. The mean free path is thus
λ=
1 1 = = 1.86 × 10−6 m = 1.86 μ m 2 4 2π ( N V ) r 2 4 2π (1.208 × 1025 m −3 )( 0.5 × 10−10 m )
18.40. Solve: (a) The number density is N / V = 1 cm −3 = 106 m −3 . Using the ideal-gas equation, N kBT ≈ (1 × 106 m −3 )(1.38 × 10−23 J/K ) ( 3 K ) V 1 atm = 4 × 10−17 Pa × = 4 × 10−22 atm 1.013 × 105 Pa
p=
(b) For a monatomic gas, 3 (1.38 × 10−23 J/K ) ( 3 K ) 3kBT = 270 m/s = m 1.67 × 10−27 kg
vrms =
(c) The thermal energy is Eth = 32 NkBT , where N = (106 m −3 )V . Thus
Eth = 1.0 J =
3 2
(10
6
m −3 )V (1.38 × 10−23 J/K ) ( 3 K ) ⇒ V = 1.6 × 1016 m3 = L3 ⇒ L = 2.5 × 105 m
18.41. Solve: Mass m of a dust particle is 3 m = ρV = ρ ( 43 π r 3 ) = ( 2500 kg/m3 ) ⎡ 43 π ( 5 × 10−6 m ) ⎤ = 1.3 × 10−12 kg ⎥⎦ ⎣⎢
The root-mean-square speed of the dust particles at 20°C is vrms =
3 (1.38 × 10−23 J/K ) ( 293 K ) 3kBT = = 9.6 × 10−5 m/s m 1.3 × 10−12 kg
18.42. Solve: Fluorine has atomic mass number A = 19. Thus the root-mean-square speed of 238UF6 is vrms ( 238 UF6 ) =
3kBT 3kBT = m 238 u + 6 × 19 u
The ratio of the root-mean-square speed for the molecules of this isotope and the 235UF6 molecules is
vrms ( 235 UF6 ) vrms (
238
UF6 )
=
( 238 + 6 × 19 ) u ( 235 + 6 × 19 ) u
=
352 = 1.0043 349
18.43. Solve: (a) If the electron can be thought of as a point particle with zero radius, then it will collide with any gas particle that is within r of its path. Hence, the number of collisions Ncoll is equal to the number of gas particles in a cylindrical volume of length L. The volume of a cylinder is Vcyl = AL = (π r 2 ) L . If the number density of the gas is (N/V) particles per m3, then the number of collisions along a trajectory of length L is
N coll = Introducing a factor of
N N L 1 Vcyl = (π r 2 L ) ⇒ λelectron = = V V N coll π ( N / V ) r 2
2 to account for the motion of all particles,
λelectron =
L 1 = N coll 2π ( N / V ) r 2
(b) Assuming that most of the molecules in the accelerator are diatomic,
5.0 × 104 m =
1
2π ( N / V ) (1.0 × 10
−10
m)
2
⇒ N / V = 4.50 × 1014 m −3
From the ideal-gas equation, p=
N kBT = ( 4.50 × 1014 m −3 )(1.38 × 10−23 J/K ) ( 293 K ) = 1.82 × 10−6 Pa = 1.80 × 10−11 atm V
18.44. Solve: The pressure on the wall with area A = 10 cm2 = 10 × 10−4 m2 is p=
F Δ ( mv ) N = A AΔt
where N Δt is the number of N2 molecules colliding with the wall every second and Δ(mv) is the change in momentum for one collision. The mass of the nitrogen molecule is m = 28 u = 28 (1.66 × 10−27 kg) = 4.648 × 10−26 kg and Δv = 400 m/s − (−400 m/s) = 800 m/s . Thus,
( 4.648 ×10 p=
−26
kg ) ( 800 m/s ) ( 5.0 × 1023 s −1 ) 1.0 × 10−3 m 2
= 1.9 × 104 Pa
18.45. Solve: (a) The cylinder volume is V = πr2L = 1.571 × 10–3 m3. Thus the number density is N 2.0 × 1022 = = 1.273 × 1025 m −3 ≈ 1.3 × 1025 m −3 V 1.571 × 10−3 m3 (b) The mass of an argon atom is m = 40 u = 40(1.661 × 10−27 kg) = 6.64 × 10−26 kg ⇒ vrms =
3 (1.38 × 10−23 J/K ) ( 323 K ) 3kBT = = 449 m/s ≈ 450 m/s m 6.64 × 10−26 kg
(c) vrms is the square root of the average of v2. That is, 2 vrms = ( v2 )
avg
= ( vx2 )
avg
+ ( v y2 )
avg
+ ( vz2 )
avg
An atom is equally likely to move in the x, y, or z direction, so on average ( vx2 ) 2 vrms = 3 ( vx2 )
avg
⇒ ( vx )rms =
(v )
2 x avg
=
avg
= ( v y2 )
avg
= ( vz2 )
avg
. Hence,
vrms = 259 m/s ≈ 260 m/s 3
(d) When we considered all the atoms to have the same velocity, we found the collision rate to be 12 ( N/V ) Avx (see Equation 18.10). Because the atoms move with different speeds, we need to replace vx with (vx)rms. The end of the cylinder has area A = πr2 = 7.85 × 10–3 m2. Therefore, the number of collisions per second is 1 2
( N / V ) A ( vx )rms = 12 (1.273 × 1025 m−3 )( 7.85 × 10−3 m2 ) ( 259 m/s ) = 1.3 × 1025 s−1
(e) From kinetic theory, the pressure is 2 ⎛N⎞ ⎛N⎞ 2 p = 13 ⎜ ⎟ m ( v 2 ) = 13 ⎜ ⎟ mvrms = 13 (1.273 × 1025 m −3 )( 6.64 × 10−26 kg ) ( 449 m/s ) = 56,800 Pa ≈ 57,000 Pa avg V V ⎝ ⎠ ⎝ ⎠ (f) From the ideal-gas law, the pressure is
22 −23 NkBT ( 2.0 × 10 )(1.38 × 10 J/K ) ( 323 K ) = = 56,700 Pa ≈ 57,000 Pa V 1.571 × 10−3 m3 Assess: The very slight difference between parts (e) and (f) is due to rounding errors; to two significant figures they are the same.
p=
18.46. Model: Pressure is due to random collisions of gas molecules with the walls. Solve:
According to Equation 18.9, the collision rate with one wall is N 1N rate of collisions = coll = Avx Δtcoll 2 V
However, this equation assumed that all molecules are moving in the x-direction with constant speed. The rms speed vrms is for motion in three dimensions at varying speeds. Consequently, we need to replace vx not with (vx)avg, which is zero, but with vx → (vx2)avg =
2 vrms v = rms 3 3
With this change,
rate of collisions =
1 N Avrms 2 3V
The molecular mass of nitrogen is A = 28 u, thus the rms speed of the molecules at 20°C is vrms =
3kBT 3(1.38 × 10 −23 J/K)(293 K) = = 510 m/s m 28(1.661 × 10 –27 kg)
With N = 0.10NA = 6.02 × 1022 molecules, the number density is N 6.02 × 1022 = = 6.02 × 1025 m −3 V 0.10 m × 0.10 m × 0.10 m Thus 1 rate of collisions = (6.02 × 1025 m −3 )(0.10 m × 0.10 m)(510 m/s)=8.9 × 1025 s −1 2 3
18.47.
Model:
Assume the gas is ideal so that Equation 18.30 will apply.
ΔEth = nCV ΔT Visualize: Solve:
We see from the graph that ΔT = 200 K and ΔEth = 800 J. We are also given n = 0.14 mol.
Solve for CV .
CV = Assess:
This is about
7 2
R.
ΔEth 800 J = = 29 J/mol ⋅ K nΔT (0.14 mol)(200 K)
18.48. Solve: (a) The number of molecules of helium is 5 −6 3 pV ( 2.0 × 1.013 × 10 Pa )(100 × 10 m ) = = 3.936 × 1021 kBT (1.38 ×10−23 J/K ) ( 373 K )
N helium =
⇒ nhelium =
3.936 × 1021 = 6.536 × 10−3 mol 6.022 × 1023 mol−1
The initial internal energy of helium is
Ehelium i = 32 N helium kBT = 30.4 J ≈ 30 J The number of molecules of argon is N argon =
5 −6 3 pV ( 4.0 × 1.013 × 10 Pa )( 200 × 10 m ) = = 8.726 × 1021 kBT (1.38 ×10−23 J/K ) ( 673 K )
⇒ nargon =
8.726 × 1021 = 1.449 × 10−2 mol 6.022 × 1023 mol−1
The initial thermal energy of argon is Eargon i = 32 N argon kBT = 121.6 J ≈ 122 J (b) The equilibrium condition for monatomic gases is
(εhelium f)avg = (εargon f)avg = (εtotal)avg ⇒
Ehelium f nhelium
=
Eargon f nargon
=
Etot ( 30.4 + 121.6 ) J = = 7228 J/mol ntot ( 6.54 × 10−3 + 1.449 × 10−2 ) mol
⇒ Ehelium f = ( 7228 J/mol ) nhelium = ( 7228 J/mol ) ( 6.54 × 10−3 mol ) = 47.3 J ≈ 47 J Eargon f = ( 7228 J/mol ) nargon = ( 7228 J/mol ) (1.449 × 10−2 mol ) = 104.7 J ≈ 105 J (c) The amount of heat transferred is
Ehelium f − Ehelium i = 47.3 J − 30.4 J = 16.9 J
Eargon f − Eargon i = 104.7 J − 121.6 J = −16.9 J
The helium gains 16.9 J of heat energy and the argon loses 16.9 J. Thus approximately 17 J are transferred from the argon to the helium. (d) The equilibrium condition for monatomic gases is
( ε helium )avg = (ε argon )avg
⇒
Ehelium f N helium
=
Eargon f N argon
= 32 kBTf
Substituting the above values, 47.3 J 104.7 J = = 3.936 × 1021 8.726 × 1021
3 2
(1.38 × 10
−23
J/K ) Tf ⇒ TF = 580 K = 307°C
(e) The final pressure of the helium and argon are
phelium f =
pargon f =
21 −23 N helium kBT ( 3.936 × 10 )(1.38 × 10 J/K ) ( 580 K ) = = 3.15 × 105 Pa ≈ 3.1 atm Vhelium 100 × 10−6 m 3
N argon kBT Vargon
(8.726 ×10 )(1.38 ×10 21
=
200 × 10
−6
J/K ) ( 580 K )
−23
m
3
= 3.49 × 105 Pa ≈ 3.4 atm
18.49. Solve: (a) The number of moles of helium and oxygen are nhelium =
2.0 g = 0.50 mol 4.0 g/mol
noxygen =
8.0 g = 0.25 mol 32.0 g/mol
Since helium is a monoatomic gas, the initial thermal energy is Ehelium i = nhelium ( 32 RThelium ) = ( 0.50 mol ) ( 32 ) ( 8.31 J/mol K )( 300 K ) = 1870 J ≈ 1900 J Since oxygen is a diatomic gas, the initial thermal energy is
Eoxygen i = noxygen ( 52 RToxygen ) = ( 0.25 mol ) ( 52 ) ( 8.31 J/mol K )( 600 K ) = 3116 J ≈ 3100 J (b) The total initial thermal energy is Etot = Ehelium i + Eoxygen i = 4986 J
As the gases interact, they come to equilibrium at a common temperature Tf. This means 4986 J = nhelium ( 32 RTf ) + noxygen ( 52 RTf )
⇒ Tf =
(
1 2
4986 J = R ) ( 3nhelium + 5noxygen )
1 2
4986 J = 436.4 K = 436 K ( 8.31 J/mol K )( 3 × 0.50 mol + 5 × 0.25 mol )
The thermal energies at the final temperature Tf are Ehelium f = nhelium ( 32 RTf ) = ( 32 ) ( 0.50 mol )( 8.31 J/mol K )( 436.4 K ) = 2700 J Eoxygen f = noxygen ( 52 RTf ) = ( 52 ) ( 0.25 mol )( 8.31 J/mol K )( 436.4 K ) = 2300 J (c) The change in the thermal energies are
Ehelium f − Ehelium i = 2720 J − 1870 J = 850 J
Eoxygen f − Eoxygen i = 2266 J − 3116 J = −850 J
The helium gains energy and the oxygen loses energy. (d) The final temperature can also be calculated as follows:
Ehelium f = ( nhelium ) 32 RTf ⇒ 2720 J = (0.50 mol)(1.5)(8.31 J/mol K)Tf ⇒ Tf = 436.4 K ≈ 436 K
18.50. Solve: The thermal energy of any ideal gas is related to the molar specific heat at constant volume by Eth = nCVT Since CP = CV + R ,
20.8 J/mol K = CV + R ⇒ CV = 12.5 J/mol K The number of moles of the gas is n=
1.0 × 1020 = 1.66 × 10−4 mol 6.02 × 1023 mol −1
Thus T=
(1.66 ×10
−4
(1.0 J ) = 482 K mols ) (12.5 J/mol K )
18.51. Solve: For a system with n degrees of freedom, the molar specific heat is CV = nR/2. The specific heat
ratio is
γ=
CP CV + R R R R ⇒ = γ − 1 = 1.29 − 1 = 0.29 ⇒ CV = = =1+ = 3.5 R = 72 R 0.29 CV CV CV CV
Thus, the system has 7 degrees of freedom.
18.52. Solve: As the volume V1 of a gas increases to V2 = 2V1 at a constant pressure p1 = p2, the temperature of the gas changes from T1 to T2 as follows: V p p2V2 p1V1 ⇒ T2 = T1 2 2 = 2T1 = V1 p1 T2 T1
Since the process occurs at constant pressure the heat transferred is
Q = nCP ΔT = nCP (T2 − T1 ) = nCP ( 2T1 − T1 ) = nCPT1 For a monatomic gas, CP = CV + R = 32 R + R = 52 R
For the diatomic gas, CP = CV + R = 52 R + R = 72 R
Thus
n ( 7 R ) T1 Qdiatomic = 52 = 1.40 Qmonatomic n ( 2 R ) T1
18.53. Solve: Assuming that the systems are thermally isolated except from each other, the total energy for the two thermally interacting systems must remain the same. That is, E1i + E2i = E1f + E2f ⇒ E1f − E1i = E2i − E2f = − ( E2f − E1f ) ⇒ ΔE1 = −ΔE2 No work is done on either system, so from the first law ΔE = Q. Thus
Q1 = −Q2 That is, the heat lost by one system is gained by the other system.
18.54. Solve: (a) From Equation 18.26 vrms = 3kBT m . For an adiabatic process γ −1
γ −1
TfVf
= TV i i
γ −1
⎛V ⎞ ⇒ Tf = Ti ⎜ i ⎟ ⎝ Vf ⎠
⇒ Tf = Ti ( 8 ) 3 = 4Ti 5 −1
The root-mean-square speed increases by a factor of 2 with an increase in temperature. (b) From Equation 18.3 λ = [4 2π ( N / V )r 2 ]−1. A decrease in volume decreases the mean free path by a factor of 1/8. (c) For an adiabatic process, γ −1
γ −1
Tf Vf
= TV i i
γ −1
⎛V ⎞ ⇒ Tf = Ti ⎜ i ⎟ ⎝ Vf ⎠
= Ti ( 8 ) 3 = 4Ti 5 −1
Because the decrease in volume increases Tf , the thermal energy increases by a factor of 4. (d) The molar specific heat at constant volume is CV = 32 R, a constant. It does not change.
18.55. Model: Assume the gas is monatomic. Visualize: From the equipartition theorem there is 12 nRT of energy for each degree of freedom. For a twodimensional monatomic gas there are only two degrees of freedom. 2 Solve: (a) CV = R = R 2 (b) Equation 17.34 gives Cp = CV + R , so
Cp = R + R = 2 R Assess:
It would be nice to measure the values and compare with these predictions.
18.56. Solve: (a) The thermal energy of a monatomic gas of N molecules is Eth = N Pavg , where Pavg = 32 kBT . A monatomic gas molecule has 3 degrees of freedom. However, a two-dimensional monatomic gas molecule has only 2 degrees of freedom. Thus, ⎛2 ⎞ Eth = N ⎜ kBT ⎟ = NkBT = nRT 2 ⎝ ⎠ If the temperature changes by ΔT, then the thermal energy changes by ΔEth = nRΔT . Comparing this form with ΔEth = nCV ΔT , we have CV = R = 8.31 J/mol K. (b) A two-dimensional solid has 2 degrees of freedom associated with the kinetic energy and 2 degrees of freedom associated with the potential energy (or 2 spring directions), giving a total of 4 degrees of freedom. Thus, Eth = N Pavg and Pavg = 42 kBT . Or Eth = 2NkBT = 2nRT. For a temperature change of ΔT, ΔEth = 2nRΔT = nCVΔT ⇒ CV = 2 R = 16.6 J/mol K.
18.57. Solve: (a) The rms speed is vrms =
v 3kBT 32 u ⇒ rms hydrogen = =4 vrms oxygen 2u m
(b) The average translational energy is P = 32 kBT . Thus
Pavg hydrogen Pavg oxygen
=
Thydrogen Toxygen
=1
(c) The thermal energy is
Eth = 52 nRT
⇒
Eth hydrogen Eth oxygen
=
nhydrogen noxygen
=
mhydrogen 32.0 g/mol = 16 2.0 g/mol moxygen
18.58. Visualize: We are given vrms = 1800 m/s. For 1.0 g of molecular hydrogen gas ⎛ 1 mol ⎞ ⎛ 23 molecules ⎞ 23 N = 1.0 g ⎜ ⎟ = 0.5 mol ⎜ 6.02 × 10 ⎟ = 3.01× 10 molecules mol ⎠ ⎝ ⎝ 2.0 g ⎠ −27 The mass of one molecule is 2(1.661× 10 kg) = 3.322 ×10−27 kg. Solve: (a) 2 ⎛1⎞ 2 ⎛1⎞ ∈total = N ⎜ ⎟ mvrms = ( 3.01× 1023 ) ⎜ ⎟ ( 3.322 × 10−27 kg ) (1800 m/s ) = 1.6 kJ ⎝2⎠ ⎝2⎠ (b) The temperature is given by Equation 18.25. 2 2 ⎛1⎞ 2 3.322 ×10−27 kg 2 T= ∈ave = (1800 m/s ) = 260 K ⎜ ⎟ mvrms = −23 3kB 3kB ⎝ 2 ⎠ 3 (1.38 × 10 J/K )
For diatomic gases Equation 18.37 gives the thermal energy. 5 5 Eth = nRT = ( 0.50 mol )( 8.31 J/mol ⋅ K )( 260 K ) = 2.7 kJ 2 2 (c) There is a net loss of 700 J of energy from the system, so the new Eth = 2000 J. Solve Equation 18.37 for T. 2 2 T= Eth = ( 2000 J ) = 192.5 K 5nR 5 ( 0.50 mol )( 8.31 J/mol ⋅ K ) Now Equation 18.26 gives the new rms speed.
3 (1.38 ×10−23 J/K ) (192.5 K ) 3kBT = = 1549 m/s ≈ 1500 m/s vrms = m 3.322 × 10−27 kg Assess: With the net loss of energy we expected vrms to decrease as the temperature decreased.
18.59. Solve: (a) The escape speed is the speed with which a mass m can leave the earth’s surface and
escape to infinity (rf = ∞) with no left over speed (vf = 0). The conservation of energy equation Kf + Uf = Ki + Ui is 2 0 + 0 = 12 mvesc −
GM e m 2GM e ⇒ vesc = Re Re
The rms speed of a gas molecule is vrms = (3kBT/m)1/2. Equating vesc and vrms, and squaring both sides, the temperature at which the rms speed equals the escape speed is
⎛ 2GM e ⎞ 3kBT 2GM e = ⇒ T = m⎜ ⎟ m Re ⎝ 3kB Re ⎠ For a nitrogen molecule, with m = 28 u, the temperature is
⎛ 2(6.67 × 10−11 N m 2 / kg 2 )(5.98 × 1024 kg) ⎞ T = (28 × 1.661 × 10−27 kg) ⎜ ⎟ = 141,000 K 3(1.38 × 10−23 J/K)(6.37 × 106 m) ⎝ ⎠ (b) For a hydrogen molecule, with m = 2 u, the temperature is less by a factor of 14, or T = 10,100 K. (c) The average translational kinetic energy of a molecule is ε avg = 32 kBT = 6.1 × 10–21 J at a typical atmosphere 2 temperature of 20°C. The kinetic energy needed to escape is K esc = 12 mvesc . For nitrogen molecules, Kesc =
2.9 × 10–18 J. Thus εavg/Kesc = 0.002 = 0.2%. Earth will retain nitrogen in its atmosphere because the molecules are moving too slowly to escape. But for hydrogen molecules, with Kesc = 2.1 × 10–19 J, the ratio is εavg/Kesc = 0.03 = 3%. Thus a large enough fraction of hydrogen molecules are moving at escape speed, or faster, to allow hydrogen to leak out of the atmosphere into space. Consequently, earth’s atmosphere does not contain hydrogen.
18.60. Solve: (a) The thermal energy of a monatomic gas of n1 moles is E1 = 32 n1RT . The thermal energy of a diatomic gas of n2 moles is E2 = 52 n2 RT . The total thermal energy of the mixture is Eth = 12 ( 3n1 + 5n2 ) RT ⇒ ΔEth = 12 ( 3n1 + 5n2 ) RΔT Comparing this expression with
ΔEth = ntotal CV ΔT = ( n1 + n2 ) CV ΔT
we get CV =
( 3n1 + 5n2 ) R 2 ( n1 + n2 )
(b) For a diatomic gas, n1 → 0, and CV = 52 R . For a monotomic gas, n2 → 0, and CV = 32 R .
18.61. Solve: (a) The thermal energy is
Eth = ( Eth ) N + ( Eth )O = 52 N N2 kBT + 52 N O2 kBT = 52 N total kBT 2
2
where Ntotal is the total number of molecules. The identity of the molecules makes no difference since both are diatomic. The number of molecules in the room is pV (101,300 Pa )( 2 m × 2 m × 2 m ) N total = = = 2.15 × 10 26 −23 kBT 1.38 × 10 J/K 273 K ) ( )( The thermal energy is Eth =
5 2
( 2.15 × 10 )(1.38 × 10 26
−23
J/K ) ( 273 K ) = 2.03 × 106 J ≈ 2.0 × 106 J
(b) A 1 kg ball at height y = 1 m has a potential energy U = mgy = 9.8 J. The ball would need 9.8 J of initial kinetic energy to reach this height. The fraction of thermal energy that would have to be conveyed to the ball is 9.8 J = 4.8 × 10−6 2.03 × 106 J (c) A temperature change ΔT corresponds to a thermal energy change ΔEth = 52 N total kBΔT . But Using this, we can write E ΔEth −9.8 J ΔEth = th ΔT ⇒ ΔT = T= 273 K = −0.0013 K T Eth 2.03 × 106 J
5 2
N total kB = Eth T .
The room temperature would decrease by 0.0013 K or 0.0013°C. (d) The situation with the ball at rest on the floor and in thermal equilibrium with the air is a very probable distribution of energy and thus a state with high entropy. Although energy would be conserved by removing energy from the air and transferring it to the ball, this would be a very improbable distribution of energy and thus a state of low entropy. The ball will not be spontaneously launched from the ground because this would require a decrease in entropy, in violation of the second law of thermodynamics. As another way of thinking about the situation, the ball and the air are initially at the same temperature. Once even the slightest amount of energy is transferred from the air to the ball, the air’s temperature will be less than that of the ball. Any further flow of energy from the air to the ball would be a situation in which heat energy is flowing from a colder object to a hotter object. This cannot happen because it would violate the second law of thermodynamics.
18.62. Solve: This is not a wise investment. Although an invention to move energy from the hot brakes to the forward motion of the car would not violate energy conservation, it would violate the second law of thermodynamics. The forward motion of the car is a very improbable distribution of energy. It happens only when a force accelerates the car and then sustains the motion against the retarding forces of friction and air resistance. The moving car is a state of low entropy. By contrast, the random motion of many atoms in the hot brakes is a state of high probability and high entropy. To convert the random motion of the atoms in the brakes back into the forward motion of the car would require a decrease of entropy and thus would violate the second law of thermodynamics. In other words, the increasing temperature of the brakes as the car stops is an irreversible process that cannot be undone.
18.63. Solve: (a) We are given that
( vrms i ) =
3kBTi m
( vrms f ) =
3kBTf = 2 ( vrms i ) m
This means that Tf = 4Ti. Using the ideal-gas law, it also means that pfVf = 4piVi. Since the pressure is directly proportional to the volume during the process, we have pi / V i = pf /Vf. Combining these two equations gives pf = 2pi and Vf = 2Vi.
(b) The change in thermal energy for any ideal gas process is related to the molar specific heat at constant volume by ΔEth = nCV (Tf − Ti ) . The work done on the gas is
W = − ∫ pdV = − (area under the p-versus-V graph) = − 32 pV i i The first law of thermodynamics ΔEth = Q + W can be written 3 Q = ΔEth − W = nCV (Tf − Ti ) + 32 pV i i = 3nCVTi + 2 pV i i 15 3 = 3n ( 52 R ) Ti + 32 pV i i = 2 piVi + 2 pV i i = 9 pV i i
18.64. Solve: The thermal energy of a monatomic gas of n1 moles is E1 = 32 n1RT . The thermal energy of a diatomic gas of n2 moles is E2 = 52 n2 RT . The total thermal energy of the mixture is
Eth = E1 + E2 = Comparing this expression with
1 2
( 3n1 + 5n2 ) RT
⇒ ΔEth = 12 ( 3n1 + 5n2 ) RΔT
ΔEth = ntotal CV ΔT = ( n1 + n2 ) CV ΔT
we get
( n1 + n2 ) CV =
( 3n1 + 5n2 ) R 2
The requirement that the ratio of specific heats is 1.50 means
γ = 1.50 = The above equation is then
( n1 + n2 )( 2 R ) =
CP CV + R R = =1+ ⇒ CV = 2 R CV CV CV
( 3n1 + 5n2 ) R ⇒ 4n 2
1
+ 4n2 = 3n1 + 5n2 ⇒ n1 = n2
Thus, monatomic and diatomic molecules need to be mixed in the ratio 1:1. Or the fraction of the molecules that are monatomic needs to be 12 .
18.65.
Solve: (a) The thermal energy of a monatomic gas of n1 moles at an initial temperature T1i is E1i = 32 n1RT1i . The thermal energy of a diatomic gas of n2 moles at an initial temperature T2i is E2i = 52 n2 RT2i . Consequently, the total initial energy is
Etot = E1i + E2i =
3n1RT1i + 5n2 RT2i ( 3n1T1i + 5n2T2i ) R = 2 2
After the gases interact, they come to equilibrium with T1f = T2f = Tf. Then their total energy is Etot =
( 3n1 + 5n2 ) RTf 2
No energy is lost, so these two expressions for Etot must be equal. Thus,
( 3n1T1i + 5n2T2i ) R = ( 3n1 + 5n2 ) RTf 2
⇒ Tf =
2
3n1T1i + 5n2T2i 2 Etot 2 ( E1i + E2i ) = = 3n1 + 5n2 R ( 3n1 + 5n2 ) R ( 3n1 + 5n2 )
The thermal energies at the final temperature Tf are E1f = 32 n1RTf = 32 n1R
⎛ 3n1 ⎞ 2 E1i + E2i =⎜ ⎟ ( E1i + E2i ) R ( 3n1 + 5n2 ) ⎝ 3n1 + 5n2 ⎠
E2f = 52 n2 RTf = 52 n2 R
2 E1i + E2i ⎛ 5n2 ⎞ =⎜ ⎟ ( E1i + E2i ) R 3n1 + 5n2 ⎝ 3n1 + 5n2 ⎠
(b) In part (a) we found that
Tf =
3n1T1i + 5n2T2i 3n1 + 5n2
(c) 2 g of He at T1i = 300 K are n1 = (2 g)/(4 g/mol) = 0.50 mol. Oxygen has an atomic mass of 16, so the molecular mass of oxygen gas (O2) is A = 32 g/mol. 8 g of O2 at T2i = 600 K are n2 = (8 g)/(32 g/mol) = 0.25 mol. The final temperature is
Tf =
3 ( 0.50 mol )( 300 K ) + 5 ( 0.25 mol )( 600 K ) = 436 K 3 ( 0.50 mol ) + 5 ( 0.25 mol )
The heat flows to the two gases are found from Q = nCVΔT. For helium,
Q = n1CV ΔT = ( 0.50 mol )(12.5 J/mol K )( 436 K − 300 K ) = 850 J For O2,
Q = n2CV ΔT = ( 0.25 mol )( 20.8 J/mol K )( 436 K − 600 K ) = −850 J
So 850 J of heat is transferred from the oxygen to the helium.
Solve: (a) The engine has a thermal efficiency of η = 40% = 0.40 and a work output of 100 J per cycle. The heat input is calculated as follows:
19.1.
η=
Wout 100 J ⇒ 0.40 = ⇒ QH = 250 J QH QH
(b) Because Wout = QH − QC , the heat exhausted is
QC = QH − Wout = 250 J − 100 J = 150 J
19.2.
Solve: During each cycle, the work done by the engine is Wout = 20 J and the engine exhausts
QC = 30 J of heat energy. Because Wout = QH − QC , QH = Wout + QC = 20 J + 30 J = 50 J Thus, the efficiency of the engine is
η =1−
QC 30 J =1− = 0.40 QH 50 J
19.3. Solve: (a) During each cycle, the heat transferred into the engine is QH = 55 kJ , and the heat exhausted is QC = 40 kJ . The thermal efficiency of the heat engine is
η = 1−
QC 40 kJ =1− = 0.27 = 27% QH 55 kJ
(b) The work done by the engine per cycle is
Wout = QH − QC = 55 kJ − 40 kJ = 15 kJ
19.4. Solve: The coefficient of performance of the refrigerator is K=
QC QH − Win 50 J − 20 J = = = 1.5 Win Win 20 J
19.5. Solve: (a) The heat extracted from the cold reservoir is calculated as follows: K=
QC Q ⇒ 4.0 = C ⇒ QC = 200 J Win 50 J
(b) The heat exhausted to the hot reservoir is
QH = QC + Win = 200 J + 50 J = 250 J
19.6. Solve:
Model: Assume that the car engine follows a closed cycle. (a) Since 2400 rpm is 40 cycles per second, the work output of the car engine per cycle is
Wout = 500
kJ 1s kJ × = 12.5 s 40 cycles cycle
(b) The heat input per cycle is calculated as follows:
η=
Wout 12.5 kJ ⇒ QH = = 62.5 kJ QH 0.20
The heat exhausted per cycle is QC = QH − Win = 62.5 kJ − 12.5 kJ = 50 kJ
19.7.
Solve: The amount of heat discharged per second is calculated as follows:
η=
⎛1 ⎞ Wout Wout ⎛ 1 ⎞ = ⇒ QC = Wout ⎜ − 1⎟ = ( 900 MW ) ⎜ − 1⎟ = 1.913 × 109 W η 0.32 QH QC + Wout ⎝ ⎠ ⎝ ⎠
That is, each second the electric power plant discharges 1.913 × 109 J of energy into the ocean. Since a typical American house needs 2.0 × 104 J of energy per second for heating, the number of houses that could be heated with the waste heat is (1.913 × 109 J ) ( 2.0 × 104 J ) = 96,000 .
19.8. Solve: The amount of heat removed from the water in cooling it down in 1 hour is QC = mwater cwater ΔT . The mass of the water is
mwater = ρ waterVwater = (1000 kg/m3 ) (1 L ) = (100 kg/m3 )(10−3 m3 ) = 1.0 kg ⇒ QC = (1.0 kg )( 4190 J/kg K )( 20°C − 5°C ) = 6.285 × 10 4 J The rate of heat removal from the refrigerator is 6.285 × 104 J = 17.46 J/s 3600 s The refrigerator does work W = 8.0. W = 8.0 J/s to remove this heat. Thus the performance coefficient of the refrigerator is QC =
K=
17.46 J/s = 2.2 8.0 J/s
19.9.
Model: Process A is isochoric, process B is isothermal, process C is adiabatic, and process D is isobaric. Solve: Process A is isochoric, so the increase in pressure increases the temperature and hence the thermal energy. Because ΔEth = Q − Ws and Ws = 0 J , Q increases for process A. Process B is isothermal, so T is constant
and hence ΔEth = 0 J. The work done Ws is positive because the gas expands. Because Q = Ws + ΔEth , Q is positive for process B. Process C is adiabatic, so Q = 0 J. Ws is positive because of the increase in volume. Since Q = 0 J = Ws + ΔEth , ΔEth is negative for process C. Process D is isobaric, so the decrease in volume leads to a decrease in temperature and hence a decrease in the thermal energy. Due to the decrease in volume, Ws is negative. Because Q = Ws + ΔEth , Q also decreases for process D. A B C D
ΔEth + 0 − −
Ws 0 + + −
Q + + 0 −
19.10. Model: Process A is adiabatic, process B is isothermal, and process C is isochoric.
Solve: Process A is adiabatic, so Q = 0 J. Work Ws is positive as the gas expands. Since Q = Ws + ∆Eth = 0 J, ∆Eth must be negative. The temperature falls during an adiabatic expansion. Process B is isothermal so ∆T = 0 and ∆Eth = 0 J. The gas is compressed, so Ws is negative. Q = Ws for an isothermal process, so Q is negative. Heat energy is withdrawn during the compression to keep the temperature constant. Process C is isochoric. No work is done (Ws = 0 J), and Q is positive as heat energy is added to raise the temperature (∆Eth positive). ∆Eth Ws Q A – 0 + B 0 – – C 0 + +
19.11. Solve: The work done by the gas per cycle is the area inside the closed p-versus-V curve. The area inside the triangle is Wout =
1 2
( 3 atm − 1 atm ) ( 600 × 10−6 m3 − 200 × 10−6 m3 )
⎛ 1.013 × 105 Pa ⎞ −6 3 = 12 ⎜ 2 atm × ⎟ ( 400 × 10 m ) = 40.5 J 1 atm ⎝ ⎠
19.12.
Solve:
The work done by the gas per cycle is the area enclosed within the pV curve. We have
60 J = 12 ( pmax − 100 kPa ) ( 800 cm3 − 200 cm3 ) ⇒
2 ( 60 J ) = pmax − 1.0 × 105 Pa 600 × 10−6 m3
⇒ pmax = 3.0 × 105 Pa = 300 kPa
19.13.
Model: The heat engine follows a closed cycle, starting and ending in the original state. The cycle consists of three individual processes. Solve: (a) The work done by the heat engine per cycle is the area enclosed by the p-versus-V graph. We get
Wout =
1 2
( 200 kPa ) (100 × 10−6 m3 ) = 10 J
The heat energy transferred into the engine is QH = 120 J. Because Wout = QH − QC , the heat energy exhausted is QC = QH − Wout = 120 J − 10 J = 110 J (b) The thermal efficiency of the engine is
η= Assess:
Wout 10 J = = 0.0833 QH 120 J
Practical engines have thermal efficiencies in the range η ≈ 0.1 − 0.4 .
19.14. Solve:
Model: The heat engine follows a closed cycle, which consists of four individual processes. (a) The work done by the heat engine per cycle is the area enclosed by the p-versus-V graph. We get
Wout = ( 400 kPa − 100 kPa ) (100 × 10−6 m3 ) = 30 J The heat energy leaving the engine is QC = 90 J + 25 J = 115 J . The heat input is calculated as follows:
Wout = QH − QC ⇒ QH = QC + Wout = 115 J + 30 J = 145 J (b) The thermal efficiency of the engine is
η= Assess:
Wout 30 J = = 0.207 QH 145 J
Practical engines have thermal efficiencies in the range η ≈ 0.1 − 0.4 .
19.15. Solve:
Model: The heat engine follows a closed cycle. The work done by the gas per cycle is the area inside the closed p-versus-V curve. We get
Wout =
1 2
( 300 kPa − 100 kPa ) ( 600 cm3 − 300 cm3 ) = 12 ( 200 × 103 Pa )( 300 × 10−6 m3 )
Because Wout = QH − QC , the heat exhausted is QC = QH − Wout = (225 J + 90 J) − 30 J = 315 J − 30 J = 285 J
= 30 J
19.16. Solve:
Model: The heat engine follows a closed cycle. (a) The work done by the gas per cycle is the area inside the closed p-versus-V curve. We get
Wout =
1 2
( 300 kPa − 100 kPa ) ( 600 cm3 − 200 cm3 ) = 12 ( 200 × 103 Pa )( 400 × 10−6 m3 )
The heat exhausted is QC = 180 J + 100 J = 280 J . The thermal efficiency of the engine is
η=
Wout Wout 40 J = = = 0.125 QH QC + Wout 280 J + 40 J
(b) The heat extracted from the hot reservoir is QH = QC + Wout = 320 J.
= 40 J
19.17. Solve:
Model: The Brayton cycle involves two adiabatic processes and two isobaric processes. From Equation 19.21, the efficiency of a Brayton cycle is ηB = 1 − rp(1−γ ) γ , where rp is the pressure ratio
pmax/pmin. The specific heat ratio for a diatomic gas is
γ=
CP = CV
7 2 5 2
R = 1.4 R
Solving the above equation for rp,
(1 − ηB ) = rp(1−γ ) / γ
γ
−1.4
⇒ rp = (1 − ηB )1−γ = (1 − 0.60 ) 0.4 = 25
19.18.
Model: The Brayton cycle involves two adiabatic processes and two isobaric processes. The adiabatic processes involve compression and expansion through the turbine. Solve: The thermal efficiency for the Brayton cycle is ηB = 1 − rp(1−γ ) / γ , where γ = CP / CV and rp is the pressure
ratio. For a diatomic gas γ = 1.4. For an adiabatic process, p1V1γ = p2V2γ ⇒ p2 / p1 = (V1 / V2 )
γ
Because the volume is halved, V2 = 12 V1 and hence rp = p2 / p1 = ( 2 ) = 21.4 = 2.639 γ
The efficiency is −0.4
ηB = 1 − ( 2.639 ) 1.4 = 0.242
19.19. Model: The efficiency of a Carnot engine (ηCarnot) depends only on the temperatures of the hot and cold reservoirs. On the other hand, the thermal efficiency (η) of a heat engine depends on the heats QH and QC. Solve:
(a) According to the first law of thermodynamics, QH = Wout + QC . For engine (a), QH = 50 J, QC = 20 J
and Wout = 30 J, so the first law of thermodynamics is obeyed. For engine (b), QH = 10 J, QC = 7 J and Wout = 4 J, so the first law is violated. For engine (c) the first law of thermodynamics is obeyed. (b) For the three heat engines, the maximum or Carnot efficiency is
ηCarnot = 1 −
TC 300 K =1− = 0.50 TH 600 K
Engine (a) has
η = 1−
QC Wout 30 J = = = 0.60 QH QH 50 J
This is larger than ηCarnot, thus violating the second law of thermodynamics. For engine (b),
η=
Wout 4 J = = 0.40 < ηCarnot QH 10 J
so the second law is obeyed. Engine (c) has a thermal efficiency that is 10 J η= = 0.333 < ηCarnot 30 J so the second law of thermodynamics is obeyed.
19.20. Model: For a refrigerator QH = QC + Win, and the coefficient of performance and the Carnot coefficient of performance are
K=
QC Win
K Carnot =
TC TH − TC
Solve: (a) For refrigerator (a) QH = QC + Win (60 J = 40 J + 20 J) , so the first law of thermodynamics is obeyed. For refrigerator (b) 50 J = 40 J + 10 J , so the first law of thermodynamics is obeyed. For the refrigerator (c) 40 J ≠ 30 J + 20 J , so the first law of thermodynamics is violated. (b) For the three refrigerators, the maximum coefficient of performance is
K Carnot =
TC 300 K = =3 TH − TC 400 K − 300 K
For refrigerator (a), K=
QC 40 J = = 2 < K Carnot Win 20 J
so the second law of thermodynamics is obeyed. For refrigerator (b), K=
QC 40 J = = 4 > K Carnot Win 10 J
so the second law of thermodynamics is violated. For refrigerator (c), K= so the second law is obeyed.
30 J = 1.5 < K Carnot 20 J
19.21. Model: The efficiency of a Carnot engine depends only on the absolute temperatures of the hot and cold reservoirs. Solve: The efficiency of a Carnot engine is
ηCarnot = 1 −
TC TC ⇒ TC = 280 K = 7°C ⇒ 0.60 = 1 − TH ( 427 + 273) K
Assess: A “real” engine would need a lower temperature than 7oC to provide 60% efficiency because no real engine can match the Carnot efficiency.
19.22. Model: Assume that the heat engine follows a closed cycle. Solve:
(a) The engine’s efficiency is
η=
Wout Wout 200 J = = = 0.25 = 25% QH QC + Wout 600 J + 200 J
(b) The thermal efficiency of a Carnot engine is ηCarnot = 1 − TC TH . For this to be 25%, 0.25 = 1 −
TC
( 400 + 273) K
⇒ TC = 504.8 K = 232°C
19.23. Model: The efficiency of an ideal engine (or Carnot engine) depends only on the temperatures of the hot and cold reservoirs. Solve: (a) The engine’s thermal efficiency is
η=
Wout Wout 10 J = = = 0.40 = 40% QH QC + Wout 15 J + 10 J
(b) The efficiency of a Carnot engine is ηCarnot = 1 − TC TH . The minimum temperature in the hot reservoir is found as follows:
0.40 = 1 −
293 K ⇒ TH = 488 K = 215°C TH
This is the minimum possible temperature. In a real engine, the hot-reservoir temperature would be higher than 215°C because no real engine can match the Carnot efficiency.
19.24. Solve: (a) The efficiency of the Carnot engine is ηCarnot = 1 −
TC 300 K =1− = 0.40 = 40% TH 500 K
(b) An engine with power output of 1000 W does Wout = 1000 J of work during each Δt = 1 s. A Carnot engine has a heat input that is Qin =
Wout
ηCarnot
=
1000 J = 2500 J 0.40
during each Δt = 1 s. The rate of heat input is 2500 J/s = 2500 W. (c) Wout = Qin − Qout , so the heat output during Δt = 1 s is Qout = Qin − Wout = 1500 J . The rate of heat output is
thus 1500 J/s = 1500 W.
19.25.
Visualize:
We will use Equation 19.29 for the efficiency of a Carnot engine.
ηCarnot = 1 −
TC TH
We are given TH = 673K and the original efficiency ηCarnot = 0.40 . Solve:
First solve for TC . TC = TH (1 − ηCarnot ) = (673K)(1 − 0.40) = 404K
′ Solve for TC′ again with ηCarnot = 0.60. ′ ) = (673K)(1 − 0.60) = 269K TC′ = TH (1 − ηCarnot The difference of these TC ’s is 135 K, so the temperature of the cold reservoir should be decreased by 135 degrees to raise the efficiency from 40% to 60%. Assess: We expected to have to lower TC by quite a bit to get the better efficiency.
19.26. Model: The maximum possible efficiency for a heat engine is provided by the Carnot engine. Solve:
The maximum efficiency is
ηmax = ηCarnot = 1 −
TC ( 273 + 20 ) K = 0.6644 =1− TH ( 273 + 600 ) K
Because the heat engine is running at only 30% of the maximum efficiency, η = ( 0.30 )ηmax = 0.1993 . The
amount of heat that must be extracted is QH =
Wout
η
=
1000 J = 5.0 kJ 0.1993
19.27. Model: The coefficient of performance of a Carnot refrigerator depends only on the temperatures of the cold and hot reservoirs. Solve: (a) The Carnot performance coefficient of a refrigerator is
K Carnot =
TC ( −20 + 273) K = = 6.33 TH − TC ( 20 + 273) K − ( −20 + 273) K
(b) The rate at which work is done on the refrigerator is found as follows:
K=
QC Q 200 J/s ⇒ Win = C = = 32 J/s = 32 W Win K 6.325
(c) The heat exhausted to the hot side per second is
QH = QC + Win = 200 J/s + 32 J/s = 232 J/s = 232 W
19.28. Visualize: We are given TH = 773K and TC = 273K , therefore (by Equation 19.29) the Carnot 273 K efficiency is nCarnot = 1 − 773 K = 0.647
We are also given η = 0.60(ηCarnot ) . Solve:
Rearrange Equation 19.7: QH = Wout (1 − η ) . Wout is the same for both engines, so it cancels.
QH Wout (1 − η ) 1 − 0.60(ηCarnot ) 1 − 0.60(0.647) = = = = 1.73 (QH )Carnot Wout (1 − ηCarnot ) 1 − ηCarnot 1 − 0.647 Assess: This engine requires 1.73 times as much heat energy in during each cycle as a Carnot engine to do the same amount of work.
19.29. Model: The minimum possible value of TC occurs with a Carnot refrigerator.
Solve:
(a) For the refrigerator, the coefficient of performance is
K=
QC ⇒ QC = KWin = ( 5.0 )(10 J ) = 50 J Win
The heat energy exhausted per cycle is QH = QC + Win = 50 J + 10 J = 60 J (b) If the hot-reservoir temperature is 27°C = 300 K, the lowest possible temperature of the cold reservoir can be obtained as follows:
K Carnot =
TC TC ⇒ 5.0 = ⇒ TC = 250 K = −23°C TH − TC 300 K − TC
19.30. Model: The coefficient of performance of a Carnot refrigerator depends only on the temperatures of the cold and hot reservoirs. Solve: Using the definition of the coefficient of performance for a Carnot refrigerator,
K Carnot =
TC ( −13 + 273) K = 260 K ⇒ T = 312 K ⇒ 5.0 = H TH − TC TH − ( −13 + 273) K TH − 260 K
To increase the coefficient of performance to 10.0, the hot-reservoir temperature changes to TH′ . Thus, K Carnot = 10.0 =
260 K ⇒ TH′ = 286 K TH′ − 260 K
Since TH′ is less than TH, the hot-reservoir temperature must be decreased by 26 K or 26 ° C.
19.31.
Solve:
The work done by the engine is equal to the change in the gravitational potential energy.
Thus, Wout = ΔUgrav = mgh = (2000 kg)(9.8 m/s2)(30 m) = 588,000 J The efficiency of this engine is
⎛
η = 0.40 ηCarnot = 0.40 ⎜1 − ⎝
⎛ ( 273 + 20 ) K ⎞ = 0.3484 TC ⎞ ⎟⎟ ⎟ = 0.40 ⎜⎜1 − TH ⎠ ⎝ ( 273 + 2000 ) K ⎠
The amount of heat energy transferred is calculated as follows:
η=
Wout W 588,000 J ⇒ QH = out = = 1.7 × 106 J η QH 0.3484
19.32.
Solve:
The mass of the water is
100 × 10−3 L ×
10−3 m3 1000 kg × = 0.100 kg 1L m3
The heat energy is removed from the water in three steps: (1) cooling from +15°C to 0°C, (2) freezing at 0°C, and (3) cooling from 0°C to −15°C . The three heat energies are Q1 = mcΔT = ( 0.100 kg )( 4186 J/kg K )(15 K ) = 6279 J
Q2 = mLf = ( 0.100 kg ) ( 3.33 × 105 J/kg ) = 33,300 J
Q3 = mcΔT = ( 0.100 kg )( 2090 J/kg K )(15 K ) = 3135 J ⇒ QC = Q1 + Q2 + Q3 = 42,714 J
Using the performance coefficient, Q 42,714 J 42,714 J K = C ⇒ 4.0 = ⇒ Win = = 10,679 J Win Win 4.0
The heat exhausted into the room is thus QH = QC + Win = 42,714 J + 10,679 J = 5.3 × 104 J
19.33. Solve: An adiabatic process has Q = 0 and thus, from the first law, Ws = –∆Eth. For any ideal-gas process, ∆Eth = nCV∆T, so Ws = –nCV∆T. We can use the ideal-gas law to find T=
pV Δ ( pV ) ( pV )f − ( pV )i pfVf − pV i i ⇒ ΔT = = = nR nR nR nR
Consequently, the work is
CV ⎛ p V − pV i i ⎞ Ws = −nCV ΔT = − nCV ⎜ f f ( pfVf − pV i i) ⎟=− nR R ⎝ ⎠ Because CP = CV + R, we can use the specific heat ratio γ to find
γ=
C P CV + R C V / R + 1 = = ⇒ CV CV CV / R
CV 1 = R γ −1
With this, the work done in an adiabatic process is Ws = −
CV 1 ( pfVf − pV ( pf Vf − pV i i) = − i i ) = ( pf Vf − pV i i )/(1 − γ ) R γ −1
19.34. Visualize: We
are
QH = msteam Lv = (10kg)(22.6 × 105 J/kg) = 22.6 × 106 J
given
QC = mice Lf = (55kg) (3.33 × 10 J/kg) = 18.3 × 10 J . 5
Solve:
Equation 19.6 gives Wout = QH − QC .
P= Assess:
6
Wout QH − QC 22.6 × 106 J − 18.3 × 106 J ⎛ 1h ⎞ = = ⎜ ⎟ = 1200 W 1h Δt Δt ⎝ 3600s ⎠
This is a reasonable answer.
and
Solve: For any heat engine, η = 1 – QC/QH. For a Carnot heat engine, ηCarnot = 1 – TC/TH. Thus a property of the Carnot cycle is that QC/QH = TC/TH. Consequently, the coefficient of performance of a Carnot refrigerator is
19.35.
K Carnot =
QC QC QC /QH T C / TH TC = = = = Win QH − QC 1 − QC /QH 1 − T C / TH TH − TC
19.36. Visualize: Equation 19.29 gives ηCarnot = 1 − TT . We are given ηCarnot = 1/ 3. C
H
Solve:
TC 1 T 2 3 = ⇒ C = ⇒ TH = TC TH 3 TH 3 2 Equation 19.30 gives the coefficient of performance for the Carnot refrigerator.
ηCarnot = 1 −
K Carnot =
Assess:
1 TC T =3 C =3 =2 TH − TC 2 TC − TC 2 − 1
This result is in the ballpark for coefficients of performance.
19.37. Visualize: We are given TH = 298 K and TC = 273 K . See Figure 19.11. Solve:
QC = mLf = (10kg)(3.33 × 105 J/kg) = 3.33 × 106 J .
(a) For a Carnot cycle ηCarnot = 1 − THC but that must also equal η = 1 − QHC , so T
QH = QC
Q
QC QH
T
= THC .
⎛ 298K ⎞ TH 6 = (3.33 × 106 J) ⎜ ⎟ = 3.63 × 10 J TC ⎝ 273K ⎠
(b)
Win = QH − QC = 3.63 × 106 J − 3.33 × 106 J = 0.30 × 106 J = 3.0 × 105 J Assess:
This is a reasonable amount of work to freeze 10 kg of water.
19.38. Model: We will use the Carnot engine to find the maximum possible efficiency of a floating power plant. Solve:
The efficiency of a Carnot engine is
ηmax = ηCarnot = 1 −
TC ( 273 + 5) K = 0.0825 ≈ 8% =1− TH ( 273 + 30 ) K
19.39.
Model: The ideal gas in the Carnot engine follows a closed cycle in four steps. During the isothermal expansion at temperature TH, heat QH is transferred from the hot reservoir into the gas. During the isothermal compression at TC, heat QC is removed from the gas. No heat is transferred during the remaining two adiabatic steps. Solve: The thermal efficiency of the Carnot engine is
ηCarnot = 1 −
TC Wout 323 K Wout = ⇒ Wout = 436 J ⇒1− = TH QH 573 K 1000 J
Using QH = QC + Wout , we obtain Qisothermal = QC = QH − Wout = 1000 J − 436 J ≈ 560 J
19.40.
Solve:
Substituting into the formula for the efficiency of a Carnot engine,
ηCarnot = 1 −
TC TC ⇒ 0.25 = 1 − ⇒ TC = 240 K = −33°C TH TC + 80 K
The hot-reservoir temperature is TH = TC + 80 K = 320 K = 47 ° C.
19.41. Model: Assume the gas is monatomic. Visualize:
We are given TC = 273 K and QC = 15 J per cycle.
For 98 cycles Wout = mgh = (10kg)(9.8kg/m 2 )(10m) = 9800 J ; for one cycle this would be Wout = 100 J. Solve: Do the computation for one cycle. QH = Wout + QC = 100J + 15J = 115J For a Carnot cycle ηCarnot = 1 − THC but that must also equal η = 1 − QHC , so T
Q
TH = TC Assess:
QC QH
T
= THC .
⎛ 115J ⎞ QH = (273K) ⎜ ⎟ = 2093K ≈ 2100K QC ⎝ 15J ⎠
TH must be this high to give the 87% efficiency.
19.42.
Solve:
If QC =
2 3
QH, then Wout = QH – QC =
η=
1 3
QC. Thus the efficiency is
Wout 13 QH 1 = = QH QH 3
The efficiency of a Carnot engine is η = 1 – TC/TH. Thus T 1 TC 2 ⇒ = 1− C = TH 3 TH 3
19.43.
Solve:
(a) Q1 is given as 1000 J. Using the energy transfer equation for the heat engine,
QH = QC + Wout ⇒ Q1 = Q2 + Wout ⇒ Q2 = Q1 − Wout The thermal efficiency of a Carnot engine is
η = 1−
TC W 300 K =1− = 0.50 = out TH 600 K Q1
⇒ Q2 = Q1 − ηQ1 = Q1 (1 − η ) = (1000 J )(1 − 0.50 ) = 500 J To determine Q3 and Q4, we turn our attention to the Carnot refrigerator, which is driven by the output of the heat engine with Win = Wout. The coefficient of performance is K=
TC 400 K Q Q Q = = 4.0 = C = 4 = 4 TH − TC 500 K − 400 K Win Wout ηQ1
⇒ Q4 = KηQ1 = ( 4.0 )( 0.50 )(1000 J ) = 2000 J Using now the energy transfer equation Win + Q4 = Q3 , we have
Q3 = Wout + Q4 = ηQ1 + Q4 = ( 0.50 )(1000 J ) + 2000 J = 2500 J (b) From part (a) Q3 = 2500 J and Q1 = 1000 J , so Q3 > Q1 . (c) Although Q1 = 1000 J and Q3 = 2500 J, the two devices together do not violate the second law of thermodynamics. This is because the hot and cold reservoirs are different for the heat engine and the refrigerator.
19.44.
Model: A heat pump is a refrigerator that is cooling the already cold outdoors and warming the indoors with its exhaust heat. Solve: (a) The coefficient of performance for this heat pump is K = 5.0 = QC Win , where QC is the amount of
heat removed from the cold reservoir. QH is the amount of heat exhausted into the hot reservoir. QH = QC + Win, where Win is the amount of work done on the heat pump. We have QC = 5.0Win ⇒ QH = 5.0Win + Win = 6.0Win
If the heat pump is to deliver 15 kJ of heat per second to the house, then 15 kJ = 2.5 kJ 6.0 In other words, 2.5 kW of electric power is used by the heat pump to deliver 15 kJ/s of heat energy to the house. (b) The monthly heating cost in the house using an electric heater is QH = 15 kJ = 6.0Win ⇒ Win =
15 kJ 3600 s 1$ × 200 h × × = $270 s 1h 40 MJ
The monthly heating cost in the house using a heat pump is 2.5 kJ 3600 s 1$ × 200 h × × = $45 s 1h 40 MJ
19.45. Visualize: We are given TC = 275 K , TH = 295 K. We are also given that in one second Win = 100 J and QC = (1 s)(100 kJ/min)(1 min/60 s) = 1667 J. Solve: The coefficient of performance of a refrigerator is given in Equation 19.10.
K=
QC 1667 J = = 16.67 Win 100 J
However the coefficient of performance of a Carnot refrigerator is given in Equation 19.30. K Carnot =
TC 275 K = = 13.75 TH − TC 20 K
However, informal statement #8 of the second law says that the coefficient of performance cannot exceed the Carnot coefficient of performance, so the salesman is making false claims. You should not buy the DreamFridge. Assess: The second law imposes real-world restrictions.
19.46. Solve: The maximum possible efficiency of the heat engine is ηmax = 1 −
TC 300 K =1− = 0.40 TH 500 K
The efficiency of the engine designed by the first student is Wout 110 J = = 0.44 QH 250 J Because η1 > ηmax, the first student has proposed an engine that would violate the second law of thermodynamics. His or her design will not work. The efficiency of the engine designed by the second student is η2 = 90 J 250 J = 0.36 < ηmax in agreement with the second law of thermodynamics. Applying the first law of thermodynamics,
η1 =
QH = QC + Wout ⇒ 250 J = 170 J + 90 J we see the first law is violated. This design will not work as claimed. The design by the third student satisfies the first law of thermodynamics because QH = QC + Wout = 250 J . The thermal efficiency of this engine is η3 = 0.36 < ηmax , which satisfies the second law of thermodynamics. The data presented by students 1 and 2 are faulty. Only student 3 has an acceptable design.
19.47. Model: The power plant is to be treated as a heat engine. Solve:
(a) Every hour 300 metric tons or 3 × 105 kg of coal is burnt. The volume of coal is 3 ×105 kg m3 × × 24 h = 4800 m3 1h 1500 kg
The height of the room will be 48 m. (b) The thermal efficiency of the power plant is
η=
Assess:
7.50 ×108 J/s 7.50 ×108 J Wout = = = 0.32 = 32% 5 6 1h QH 3 ×10 kg 28 ×10 J 2.333 ×109 J × × 1h kg 3600 s
An efficiency of 32% is typical of power plants.
19.48. Model: The power plant is treated as a heat engine. Solve:
(a) The maximum possible thermal efficiency of the power plant is
ηmax = 1 −
TC 303 K =1− = 0.47 = 47% TH 573 K
(b) The plant’s actual efficiency is
η=
Wout 700 × 106 J/s = 0.35 = 35% QH 2000 × 106 J/s
(c) Because QH = QC + Wout,
QC = QH − Wout ⇒ QC = 2.0 × 109 J/s − 0.7 × 109 J/s = 1.3 × 109 J/s The mass of water that flows per second through the condenser is m = 1.2 ×108
L 1 hr 10−3 m3 1000 kg × × × = 3.333 × 104 kg h 3600 s 1L m3
The change in the temperature as QC = 1.3 × 109 J of heat is transferred to m = 3.333 × 104 kg of water is calculated as follows: QC = mcΔT ⇒ 1.3 × 109 J = ( 3.333 × 104 kg ) ( 4186 J/kg K ) ΔT ⇒ ΔT = 9°C The exit temperature is18°C + 9°C = 27°C .
19.49. Model: The power plant is treated as a heat engine. Solve:
The mass of water per second that flows through the plant every second is
m = 1.0 ×108
L 1 hr 10−3 m3 1000 kg × × × = 2.778 × 104 kg/s h 3600 s 1L m3
The amount of heat transferred per second to the cooling water is thus
QC = mcΔT = ( 2.778 × 104 kg/s ) ( 4186 J/kg K )( 27°C − 16°C ) = 1.279 × 109 J/s The amount of heat per second into the power plant is
QH = Wout + QC = 0.750 × 109 J/s + 1.279 × 109 J/s = 2.029 × 109 J/s Finally, the power plant’s thermal efficiency is
η=
Wout 0.750 × 109 J/s = = 0.37 = 37% QH 2.029 × 109 J/s
19.50. Solve: (a) The energy supplied in one day is J 3600 s 24 h Wout = 1.0 ×109 × × = 8.64 × 1013 J s 1h 1d (b) The volume of water is V = 1 km3 = 109 m3 . The amount of energy is
⎛ 1000 kg ⎞ 15 QH = mcΔT = (109 m3 ) ⎜ ⎟ ( 4190 J/kg K )(1.0 K ) = 4.19 × 10 J 3 ⎝ m ⎠ (c) While it’s true that the ocean contains vast amounts of thermal energy, that energy can be extracted to do useful work only if there is a cold reservoir at a lower temperature. That is, the ocean has to be the hot reservoir of a heat engine. But there’s no readily available cold reservoir, so the ocean’s energy cannot readily be tapped. There have been proposals for using the colder water near the bottom of the ocean as a cold reservoir, pumping it up to the surface where the heat engine is. Although possible, the very small temperature difference between the surface and the ocean depths implies that the maximum possible efficiency (the Carnot efficiency) is only a few percent, and the efficiency of any real ocean-driven heat engine would likely be less than 1%—perhaps much less. Thus the second law of thermodynamics prevents us from using the thermal energy of the ocean. Save your money. Don’t invest.
19.51. Visualize: If we do this problem on a “per second” basis then in one second QC = (1 s)(5.0 ×105 J/min) (1 min/60 s) = 8.33 ×103 J. QH = (1 s)(8.0 ×105 J/min)(1 min / 60 s) = 13.33 × 103 J. Solve: (a) Again, in one second Win = QH − QC = 13.33 × 103 J − 8.33 × 103 J = 5.0 × 103 J Since this is per second then the power required by the compressor is P = 5.0 kW. (b) Assess:
K= The result is typical for air conditioners.
QC 8.33 × 103 J = = 1.7 Win 5.0 × 103 J
19.52. Model: The heat engine follows a closed cycle with process 1 → 2 and process 3 → 4 being isochoric and process 2 → 3 and process 4 → 1 being isobaric. For a monatomic gas, CV = 32 R and CP = 52 R. Visualize: Please refer to Figure P19.52. Solve: (a) The first law of thermodynamics is Q = ΔEth + WS . For the isochoric process 1 → 2, WS 1 → 2 = 0 J. Thus, Q1→ 2 = 3750 J = ΔEth = nCV ΔT ⇒ ΔT =
3750 J 3750 J 3750 J = = = 301 K nCV (1.0 mol ) ( 32 R ) (1.0 mol ) ( 32 ) (8.31 J/mol K ) ⇒ T2 − T1 = 300.8 K ⇒ T2 = 300.8 K + 300 K = 601 K
To find volume V2, V2 = V1 =
nRT1 (1.0 mol )( 8.31 J/mol K )( 300 K ) = = 8.31× 10−3 m3 p1 3.0 × 105 Pa
The pressure p2 can be obtained from the isochoric condition as follows: p2 p1 T ⎛ 601 K ⎞ 5 5 = ⇒ p2 = 2 p1 = ⎜ ⎟ ( 3.00 × 10 Pa ) = 6.01 × 10 Pa T2 T1 T1 300 K ⎝ ⎠ With the above values of p2, V2 and T2, we can now obtain p3, V3 and T3. We have V3 = 2V2 = 1.662 × 10−2 m3
p3 = p2 = 6.01× 105 Pa
T3 T2 V = ⇒ T3 = 3 T2 = 2T2 = 1202 K V3 V2 V2 For the isobaric process 2 → 3,
Q2 →3 = nCP ΔT = (1.0 mol ) ( 52 R ) (T3 − T2 ) = (1.0 mol ) ( 52 ) ( 8.31 J/mol K )( 601 K ) = 12,480 J WS 2 →3 = p3 (V3 − V2 ) = ( 6.01× 105 Pa )( 8.31× 10−3 m3 ) = 4990 J ΔEth = Q2 →3 − WS 2 →3 = 12,480 J − 4990 J = 7490 J We are now able to obtain p4, V4 and T4. We have V4 = V3 = 1.662 × 10−2 m3
p4 = p1 = 3.00 × 105 Pa
⎛ 3.00 × 105 Pa ⎞ T4 T3 p = ⇒ T4 = 4 T3 = ⎜ ⎟ (1202 K ) = 600 K 5 p4 p3 p3 ⎝ 6.01 × 10 Pa ⎠ For isochoric process 3 → 4,
Q3→ 4 = nCV ΔT = (1.0 mol ) ( 32 R ) (T4 − T3 ) = (1.0 mol ) ( 32 ) ( 8.31 J/mol K )( −602 ) = −7500 J WS 3→ 4 = 0 J ⇒ ΔEth = Q3→ 4 − WS 3→ 4 = −7500 J For isobaric process 4 → 1, Q4 →1 = nCP ΔT = (1.0 mol ) 52 ( 8.31 J/mol K )( 300 K − 600 K ) = −6230 J
WS 4 →1 = p4 (V1 − V4 ) = ( 3.00 × 105 Pa ) × (8.31 × 10−3 m3 − 1.662 × 10−2 m3 ) = −2490 J ΔEth = Q4 →1 − WS 4 →1 = −6230 J − ( −2490 J ) = −3740 J WS (J)
Q (J)
ΔEth (J)
1→2
0
3750
3750
2→3
4990
12,480
7490
3→4
0
–7500
–7500
4→1 Net
–2490
–6230
–3740
2500
2500
0
(b) The thermal efficiency of this heat engine is
η= Assess:
Wout Wout 2500 J = = = 0.15 = 15% QH Q1→ 2 + Q2 →3 3750 J + 12,480 J
For a closed cycle, as expected, (Ws)net = Qnet and (ΔEth)net = 0 J
19.53. Model: The heat engine follows a closed cycle. For a diatomic gas, CV = 52 R and CP = 72 R. Visualize: Please refer to Figure P19.53. Solve: (a) Since T1 = 293 K, the number of moles of the gas is n=
5 −6 3 p1V1 ( 0.5 × 1.013 × 10 Pa )(10 × 10 m ) = = 2.08 × 10−4 mol RT1 ( 8.31 J/mol K )( 293 K )
At point 2, V2 = 4V1 and p2 = 3 p1 . The temperature is calculated as follows:
p1V1 p2V2 p V = ⇒ T2 = 2 2 T1 = ( 3)( 4 )( 293 K ) = 3516 K T1 T2 p1 V1 At point 3, V3 = V2 = 4V1 and p3 = p1 . The temperature is calculated as before: T3 =
p3 V3 T1 = (1)( 4 )( 293 K ) = 1172 K p1 V1
For process 1 → 2, the work done is the area under the p-versus-V curve. That is, Ws = ( 0.5 atm ) ( 40 cm3 − 10 cm3 ) + 12 (1.5 atm − 0.5 atm ) ( 40 cm3 − 10 cm3 ) ⎛ 1.013 × 105 Pa ⎞ = ( 30 × 10−6 m3 ) (1 atm ) ⎜ ⎟ = 3.04 J 1 atm ⎝ ⎠ The change in the thermal energy is
ΔEth = nCV ΔT = ( 2.08 × 10−4 mol ) 52 ( 8.31 J/mol K )( 3516 K − 293 K ) = 13.93 J The heat is Q = Ws + ΔEth = 16.97 J . For process 2 → 3, the work done is Ws = 0 J and Q = ΔEth = nCV ΔT = n ( 52 R ) (T3 − T2 )
= ( 2.08 × 10−4 mol ) 52 ( 8.31 J/mol K )(1172 K − 3516 K ) = −10.13 J
For process 3 → 1,
Ws = ( 0.5 atm ) (10 cm3 − 40 cm3 ) = ( 0.5 × 1.013 × 105 Pa )( −30 × 10−6 m3 ) = −1.52 J ΔEth = nCV ΔT = ( 2.08 × 10−4 mol ) 52 ( 8.31 J/mol K )( 293 K − 1172 K ) = −3.80 J The heat is Q = ΔEth + Ws = −5.32 J . 1→2 2→3 3→1 Net
Ws (J)
Q (J)
ΔEth
3.04 0 −1.52 1.52
16.97 −10.13 −5.32 1.52
13.93 −10.13 −3.80 0
(b) The efficiency of the engine is
η=
Wnet 1.52 J = = 0.090 = 9.0% QH 16.97 J
(c) The power output of the engine is 500
Assess:
revolutions 1 min Wnet 500 × × = × 1.52 J/s = 13 W min 60 s revolution 60
For a closed cycle, as expected, (Ws)net = Qnet and (ΔEth)net = 0 J.
19.54. Model: For the closed cycle, process 1 → 2 is isothermal, process 2 → 3 is isobaric, and process 3 → 1 is isochoric. Solve: (a) We first need to find the conditions at points 1, 2, and 3. We can then use that information to find WS and Q for each of the three processes that make up this cycle. Using the ideal-gas equation the number of moles of the gas is n=
5 −6 3 p1V1 (1.013 × 10 Pa )( 600 × 10 m ) = = 0.0244 mol RT1 (8.31 J/mol K )( 300 K )
We are given that γ = 1.25, which means this is not a monatomic or a diatomic gas. The specific heats are
CV =
R = 4R 1−γ
C P = CV + R = 5 R
At point 2, process 1 → 2 is isothermal, so we can find the pressure p2 as follows: p1V1 = p2V2 ⇒ p2 =
V1 6.00 × 10−4 m3 p1 = p1 = 3 p1 = 3 atm = 3.039 × 105 Pa V2 2.00 × 10−4 m3
At point 3, process 2 → 3 is isobaric, so we can find the temperature T3 as follows: V2 V3 V 6.00 × 10−4 m 3 T2 = 3T2 = 900 K = ⇒ T3 = 3 T2 = T2 T3 V2 2.00 × 10−4 m3 Point
V (m3 )
P (Pa)
6.00 × 10 2.00 × 10−4 6.00 × 10−4
1.0 atm = 1.013 × 10 3.0 atm = 3.039 × 105 3.0 atm = 3.039 × 105
1 2 3
T (K) −4
5
300 300 900
Process 1 → 2 is isothermal:
(WS )12 = p1V1 ln (V2
V1 ) = −66.8 J
Q12 = (WS )12 = −66.8 J
Process 2 → 3 is isobaric:
(WS )23 = p2ΔV = p2 (V3 − V2 ) = 121.6 J
Q23 = nCP ΔT = nCP (T3 − T2 ) = 608.3 J
Process 3 → 1 is isochoric:
(WS )31 = 0 J
Q31 = nCV ΔT = nCV (T1 − T3 ) = −486.7 J
We find that
(WS )cycle = −66.8 J + 121.6 J + 0 J = 54.8 J
Qcycle = −66.8 J + 608.3 J − 486.7 J = 54.8 J
These are equal, as they should be. Knowing that the work done is Wout = (WS)cycle = 54.8 J/cycle, an engine operating at 20 cycles/s has power output Pout =
54.8 J 20 cycle J × = 1096 = 1096 W ≈ 1.10 kW cycle s s
(b) Only Q23 is positive, so Qin = Q23 = 608 J. Thus, the thermal efficiency is
η=
Wout 54.8 J = = 0.090 = 9.0% Qin 608.3 J
19.55. Model: For the closed cycle of the heat engine, process 1 → 2 is isochoric, process 2 → 3 is adiabatic, and process 3 → 1 is isobaric. CV = 32 R and CP = 52 R for a monatomic gas. Visualize: Please refer to Figure P19.55. Solve: (a) Using the ideal-gas equation, the number of moles is n=
5 −6 3 p1V1 (1.0 × 10 Pa ) (100 × 10 m ) = = 0.00401 mol RT1 (8.31 J/mol K)(300 K)
We can use the adiabat 2 → 3 to calculate p2 as follows: γ
γ
⎛V ⎞ ⎛V ⎞ ⎛ 600 cm3 ⎞ p3V3 = p2V2 ⇒ p2 = p3 ⎜ 3 ⎟ = p1 ⎜ 3 ⎟ = (1.0 × 105 Pa ) ⎜ 3 ⎟ ⎝ 100 cm ⎠ ⎝ V2 ⎠ ⎝ V2 ⎠ γ
5/3
γ
= 1.981 × 106 Pa
T2 can be determined from the ideal-gas equation and is
T2 =
6 −6 3 p2V2 (1.981 × 10 Pa )(100 × 10 m ) = = 5943 K nR (8.31 J/mol K)(0.00401 mol)
At point 3, p3 = p1 and T3 T1 V 600 × 10−6 m3 = ⇒ T3 = 3 T1 = ( 300 K ) = 1800 K V3 V1 V1 100 × 10−6 m3 Now we can calculate Ws, Q, and ΔEth for the three processes involved in the cycle. For process 1 → 2, Ws 1→2 = 0 J and ΔEth = Q1→ 2 = nCV (T2 − T1 ) = n ( 32 R ) (T2 − T1 ) = 282.2 J
For process 2 → 3, Q2→3 = 0 J and ΔEth = nCV (T3 − T2 ) = n ( 32 R ) (T3 − T2 ) = –207.2 J
Because, ΔEth = W + Q, W = −Ws, so Ws 2→3 = −ΔEth = +207.2 J for process 2 → 3. For process 3 → 1, ΔEth = nCV (T1 − T3 ) = n ( 32 R ) (T1 − T3 ) = −75.0 J Q3→1 = nCP (T1 − T3 ) = n ( 52 R ) (T1 − T3 ) = −125.0 J
The work done Ws 3→1 is the area under the p-versus-V graph. We have
Ws 3→1 = (100 × 103 Pa )(100 × 10−6 m3 − 600 × 10−6 m3 ) = −50.0 J
1→2 2→3 3→1 Net
Ws (J)
Q (J)
ΔEth (J)
0 207.2 −50.0 157.2
282.2 0 −125.0 157.2
282.2 −207.2 −75.0 0
(b) The thermal efficiency of the engine is
Wout 157.2 J = = 0.52 = 52% QH 282.2 J As expected for a closed cycle, (Ws)net = Qnet and (ΔEth)net = 0 J.
η=
Assess:
19.56. Model: For the closed cycle of the heat engine, process 1 → 2 is isochoric, process 2 → 3 is adiabatic, and process 3 → 1 is isothermal. For a diatomic gas CV = 52 R and γ = 75 . Solve:
(a) From the graph V2 = 1000 cm3 .
The pressure p2 lies on the adiabat from 2 → 3. We can find the pressure as follows: γ
⎛V ⎞ ⎛ 4000 cm3 ⎞ p2V2γ = p3V3γ ⇒ p2 = p3 ⎜ 3 ⎟ = (1.0 ×105 Pa ) ⎜ 3 ⎟ ⎝ 1000 cm ⎠ ⎝ V2 ⎠
7/5
= 6.964 ×105 Pa ≈ 700 kPa
The temperature T2 can be obtained from the ideal-gas equation relating points 1 and 2:
⎛ 6.964 × 105 Pa ⎞ p1V1 p2V2 p V = ⇒ T2 = T1 2 2 = ( 300 K ) ⎜ ⎟ (1) = 522.3 K ≈ 522 K 5 T1 T2 p1 V1 ⎝ 4.0 × 10 Pa ⎠ (b) The number of moles of the gas is
R=
5 −3 3 p1V1 ( 4.0 × 10 Pa )(1.0 × 10 m ) = = 0.1604 mol RT1 (8.31 J/mol K)(300 K)
For isochoric process 1 → 2 , Ws = 0 J and Q = ΔEth = nCV ΔT = n ( 52 R ) ΔT = 741.1 J For adiabatic process 2 → 3, Q = 0 J and ΔEth = nCV ΔT = n ( 52 R ) (T3 − T2 ) = −741.1 J Using the first law of thermodynamics, ΔEth = ΔWs + Q, which means Ws = −ΔEth = +741.1 J. Ws can also be determined from
Ws =
p3V3 − p2V2 nR (T3 − T2 ) ( 34 J/K ) ( 300 K − 522.3 K ) = = = 741.1 J 1−γ 1−γ ( − 52 )
For isothermal process 3 → 1, ΔEth = 0 J and
Ws = nRT1 ln
V1 = −554.5 J V3
Using the first law of thermodynamics, ΔEth = −Ws + Q , Q = Ws = −554.5 J . 1→2 2→3 3→1 Net
ΔEth (J)
Ws (J)
Q (J)
741.1 −741.1 0 0
0 741.1 −554.5 186.6
741.1 0 −554.5 186.6
(c) The work per cycle is 186.6 J and the thermal efficiency is
η=
Ws 186.6 J = = 0.25 = 25% QH 741.1 J
19.57. Model: For the closed cycle of the heat engine, process 1 → 2 is adiabatic, process 2 → 3 is isothermal, and process 3 → 1 is isochoric. For a diatomic gas CV = 52 R and γ = 75 . Solve: (a) From the graph V1 = 4000 cm3 . The number of moles of gas is n=
5 −6 3 p2V2 ( 4.0 × 10 Pa )(1000 × 10 m ) = = 0.1203 mol RT2 (8.31 J/mol K)(400 K)
Using p1V1γ = p2V2γ , and reading V1 from the graph, γ
⎛V ⎞ 7/5 p1 = p2 ⎜ 2 ⎟ = ( 4.0 × 105 Pa ) ( 14 ) = 5.743 × 104 Pa ≈ 5.7 kPa ⎝ V1 ⎠ With p1, V1, and nR having been determined, we can find T1 using the ideal-gas equation: T1 =
4 −6 3 p1V1 ( 5.743 × 10 )( 4000 × 10 m ) = = 229.7 K ≈ 230 K nR (8.31 J/mol K)(0.1203 mol)
(b) For adiabatic process 1 → 2, Q = 0 J and
Ws =
5 −3 3 4 −3 3 p2V2 − p1V1 ( 4.0 × 10 Pa )(1.00 × 10 m ) − ( 5.743 × 10 Pa )( 4.00 × 10 m ) = = −425.7 K 1−γ 1 − 1.4
Because ΔEth = −Ws + Q , ΔEth = −Ws = 425.7 K . For isothermal process 2 → 3, ΔEth = 0 J . From Equation 17.15,
Ws = nRT2 ln
V3 = 554.5 J V2
From the first law of thermodynamics, Q = Ws = 554.5 J for process 2 → 3. For isochoric process 3 → 1, Ws = 0 J and Q = ΔEth = nCV ΔT = n ( 52 R ) (T1 − T3 ) = −425.7 J
1→2 2→3 3→1 Net
ΔEth (J)
Ws (J)
Q (J)
425.7 0 −425.7 0
−425.7 554.5 0 128.8
0 554.5 −425.7 128.8
(c) The work per cycle is 128.8 J and the thermal efficiency of the engine is
η= Assess:
Ws 128.8 J = = 0.23 = 23% QH 554.5 J
As expected, for a closed cycle (Ws )net = Qnet and ( ΔEth )net = 0 J.
19.58. Model: In the Brayton cycle, process 1 → 2 and process 3 → 4 are adiabatic, and process 2 → 3 and process 4 → 2 are isobaric. We will assume the gas to be diatomic, with CP = 72 R and γ = 1.40. Visualize: Please refer to Figure P19.58. Solve: The number of moles of gas is n=
5 3 p1V1 (1.013 × 10 Pa )( 2.0 m ) = = 81.27 mol RT1 (8.31 J/mol K)(300 K)
We can find the volume at 2 from the adiabatic equation p1V1γ = p2V2γ , 1
1
⎛ p ⎞γ ⎛ 1 ⎞1.4 V2 = V1 ⎜ 1 ⎟ = ( 2.0 m3 ) ⎜ ⎟ = 0.3861 m3 ⎝ 10 ⎠ ⎝ p2 ⎠ The temperature T2 is determined from the ideal-gas equation as follows: 2 p1V1 p2V2 p V ⎛ 10 atm ⎞ ⎛ 0.3861 m ⎞ = ⇒ T2 = T1 2 2 = ( 300 K ) ⎜ ⎟ = 579.2 K ⎟⎜ 3 T1 T2 p1 V1 ⎝ 1 atm ⎠ ⎝ 2.0 m ⎠
To find T3 we use the heat input: QH = 2.0 × 106 J = nCP ΔT = 72 n RΔT =
7 2
( 675.33 J/K ) ΔT
⇒ ΔT = 846.1 K ⇒ T3 = T2 + ΔT = 579.2 K + 846.1 K = 1425.3 K We are now able to find V3 using the ideal-gas law. We have V3 V2 T ⎛ 1425.3 K ⎞ 3 = ⇒ V3 = V2 3 = ( 0.3861 m3 ) ⎜ ⎟ = 0.9501 m T3 T2 T2 ⎝ 579.2 K ⎠
The volume at 4 is found by again using the adiabatic equation p3V3γ = p4V4γ , 1
1 ⎛ p ⎞γ V4 = V3 ⎜ 3 ⎟ = ( 0.9501 m 3 ) (10 )1.4 = 4.921 m3 ⎝ p4 ⎠
T4 is now obtained from T4 =
5 3 p4V4 (1.013 × 10 Pa )( 4.921 m ) = = 738.2 K nR (81.27 mol)(8.31 J/mol K)
To find the engine’s efficiency, we must first find the total work done per cycle which is Wnet = W1→ 2 + W2 →3 + W3→ 4 + W4 →1 The four contributions are W1→ 2 =
p2V2 − p1V1 nR (T2 − T1 ) ( 81.27 mol )( 8.31 J/mol K )( 579.2 K − 300 K ) = = = −4.714 × 105 J 1−γ 1−γ 1 − 1.4
W2 →3 = p2 ΔV = (1.013 × 106 Pa )( 0.9501 m3 − 0.3861 m3 ) = 5.713 × 105 J W3→ 4 =
nR (T4 − T3 ) ( 81.27 mol )( 8.31 J/mol K )( 738.2 K − 1425.3 K ) = = +1.160 × 106 J 1−γ 1 − 1.4
W4 →1 = p4 ΔV = (1.013 × 105 Pa )( 2.0 m3 − 4.921 m3 ) = −2.959 × 105 J Thus, Wnet = 9.64 × 105 J . The thermal efficiency is
η=
Wnet 9.64 × 105 J ⇒η = = 0.482 = 48.2% QH 2.0 × 106 J
As a comparison, we can calculate η from Equation 19.21 which is
η =1−
1
(r ) p
where rp = pmax pmin = 10 .
( ) γ −1 γ
=1−
1
(10 )
0.4 1.4
= 48.2%
19.59. Model: Process 1 → 2 of the cycle is isochoric, process 2 → 3 is isothermal, and process 3 → 1 is isobaric. For a monatomic gas, CV = 32 R and CP = 52 R. Visualize: Please refer to Figure P19.59. Solve: (a) At point 1: The pressure p1 = 1.0 atm = 1.013 × 105 Pa and the volume V1 = 1000 ×10−6 m 3 = 1.0 × 10−3 m3 . The number of moles is
n=
0.120 g = 0.030 mol 4 g/mol
Using the ideal-gas law, T1 =
5 −3 3 p1V1 (1.013 × 10 Pa )(1.0 × 10 m ) = = 406 K nR ( 0.030 mol )(8.31 J/mol K )
At point 2: The pressure p2 = 5.0 atm = 5.06 × 105 Pa and V2 = 1.0 × 10−3 m3 . The temperature is T2 =
5 −3 3 p2V2 ( 5.06 × 10 Pa )(1.0 × 10 m ) = = 2030 K nR ( 0.030 mol )(8.31 J/mol K )
At point 3: The pressure is p3 = 1.0 atm = 1.013 × 105 Pa and the temperature is T3 = T2 = 2030 K . The volume is V3 = V2
p2 ⎛ 5 atm ⎞ −3 3 = (1.0 × 10−3 m3 ) ⎜ ⎟ = 5.0 × 10 m p3 1 atm ⎝ ⎠
(b) For isochoric process 1 → 2, W1→2 = 0 J and Q1→ 2 = nCV ΔT = ( 0.030 mol ) ( 32 R ) ( 2030 K − 406 K ) = 607 J
For isothermal process 2 → 3, ΔEth 2 →3 = 0 J and
Q2 →3 = W2 →3 = nRT2 ln
⎛ 5.0 × 10−3 m3 ⎞ V3 = ( 0.030 mol )(8.31 J/mol K )( 2030 K ) ln ⎜ = 815 J −3 3 ⎟ V2 ⎝ 1.0 × 10 m ⎠
For isobaric process 3 → 1,
W3→1 = p3ΔV = (1.013 × 105 Pa )(1.0 × 10−3 m3 − 5.0 × 10−3 m3 ) = −405 J Q3→1 = nCP ΔT = ( 0.030 mol ) ( 52 ) ( 8.31 J/mol K )( 406 K − 2030 K ) = −1012 J The total work done is Wnet = W1→ 2 + W2 →3 + W3→1 = 410 J. The total heat input is QH = Q1→ 2 + Q2 →3 = 1422 J . The efficiency of the engine is
η=
Wnet 410 J = = 20% QH 1422 J
(c) The maximum possible efficiency of a heat engine that operates between Tmax and Tmin is
ηmax = 1 − Assess:
Tmin 406 K =1− = 80% Tmax 2030 K
The actual efficiency of an engine is less than the maximum possible efficiency.
19.60. Model: The process 2 → 3 of the heat engine cycle is isochoric and the process 3 → 1 is isobaric. For a monatomic gas CV = 32 R and CP = 52 R . Solve: (a) The three temperatures are T1 =
5 3 p1V1 ( 4.0 × 10 Pa )( 0.025 m ) = = 601.7 K ≈ 602 K nR ( 2.0 mol )( 8.31 J/mol K )
T2 =
5 3 p2V2 ( 6.0 × 10 Pa )( 0.050 m ) = = 1805.1 K ≈ 1805 K nR ( 2.0 mol )( 8.31 J/mol K )
T3 =
5 3 p3V3 ( 4.0 × 10 Pa )( 0.050 m ) = = 1203.4 K ≈ 1203 K nR ( 2.0 mol )( 8.31 J/mol K )
(b) For process 1 → 2, the work done is the area under the p-versus-V graph. The work and the change in internal energy are
Ws = 12 ( 6.0 × 105 Pa − 4.0 × 105 Pa )( 0.050 m3 − 0.025 m3 ) + ( 4.0 × 105 Pa )( 0.050 m3 − 0.025 m3 ) = 1.25 × 104 J ΔEth = nCV ΔT = ( 2.0 mol ) ( 32 R ) (T2 − T1 ) = ( 2.0 mol ) ( 32 ) ( 8.31 J/mol K )(1805.1 K − 601.7 K ) = 3.00 × 104 J The heat input is Q = Ws + ΔEth = 4.25 × 104 J . For isochoric process 2 → 3, Ws = 0 J and Q = ΔEth = nCV ΔT = ( 2.0 mol ) 32 ( 8.31 J/mol K )(1203.4 K − 1805.1 K ) = −1.50 × 104 J For isobaric process 3 → 1, the work done is the area under the p-versus-V curve. Hence,
Ws = ( 4.0 × 105 Pa )( 0.025 m3 − 0.050 m3 ) = −1.0 × 104 J ΔEth = nCV ΔT = n ( 32 R ) (T1 − T3 ) = ( 2.0 mol ) 32 ( 8.31 J/mol K )( 601.7 K − 1203.4 K ) = −1.5 × 104 J The heat input is Q = WS + ΔEth = −2.50 × 104 J .
ΔEth (J) 1→2 2→3 3→1 Net
Ws (J) 4
3.0 × 10 −1.5 × 104 −1.5 × 104 0
1.25 × 10 0 −1.0 ×104 2.5 × 103
(c) The thermal efficiency is
η=
Q (J) 4
Wnet 2.5 × 103 J = = 5.9% QH 4.25 × 104 J
4.25 × 104 −1.50 × 104 −2.50 × 104 2.5 × 103
19.61. Model: The closed cycle in this heat engine includes adiabatic process 1 → 2, isobaric process 2 → 3, and isochoric process 3 → 1. For a diatomic gas, CV = 52 R, CP = 72 R, and γ = 75 =1.4. Visualize: Please refer to Figure P19.61. Solve: (a) We can find the temperature T2 from the ideal-gas equation as follows: T2 =
5 −3 3 p2V2 ( 4.0 × 10 Pa )(1.0 × 10 m ) = = 2407 K nR ( 0.020 mol )( 8.31 J/mol K )
We can use the equation p2V2γ = p1V1γ to find V1, 1/ γ
⎛p ⎞ V1 = V2 ⎜ 2 ⎟ ⎝ p1 ⎠
1/1.4
⎛ 4.0 × 105 Pa ⎞ = (1.0 × 10−3 m3 ) ⎜ ⎟ 5 ⎝ 1.0 × 10 Pa ⎠
= 2.692 × 10−3 m3
The ideal-gas equation can now be used to find T1,
T1 =
5 −3 3 p1V1 (1.0 × 10 Pa )( 2.692 × 10 m ) = = 1620 K nR ( 0.020 mol )(8.31 J/mol K )
At point 3, V3 = V1 so we have
T3 =
5 −3 3 p3V3 ( 4 × 10 Pa )( 2.692 × 10 m ) = = 6479 K nR ( 0.020 mol )( 8.31 J/mol K )
(b) For adiabatic process 1 → 2, Q = 0 J, ΔEth = −Ws , and
Ws =
p2V2 − p1V1 nR (T2 − T1 ) ( 0.020 mol )( 8.31 J/mol K )( 2407 K − 1620 K ) = = = −327.0 J 1−γ 1−γ (1 − 1.4 )
For isobaric process 2 → 3, Q = nCP ΔT = n ( 72 R ) ( ΔT ) = ( 0.020 mol ) 72 ( 8.31 J/mol K )( 6479 K − 2407 K ) = 2369 J ΔEth = nCV ΔT = n ( 52 R ) ΔT = 1692 J The work done is the area under the p-versus-V graph. Hence,
Ws = ( 4.0 × 105 Pa )( 2.692 × 10−3 m3 − 1.0 × 10−3 m3 ) = 677 J For isochoric process 3 → 1, Ws = 0 J and ΔEth = Q = nCV ΔT = ( 0.020 mol ) ( 52 ) ( 8.31 J/mol K )(1620 K − 6479 K ) = −2019 J 1→2 2→3 3→1 Net
ΔEth (J)
Ws (J)
Q (J)
327 1692 −2019 0
−327 677 0 350
0 2369 −2019 350
(c) The engine’s thermal efficiency is
η=
350 J = 0.15 = 15% 2369 J
19.62. Model: The closed cycle of the heat engine involves the following processes: one isothermal compression, one isobaric expansion, and one isochoric cooling.
Visualize:
Solve:
The number of moles of helium is
n=
2.0 g = 0.50 mol 4.0 g/mol
The total work done by the engine per cycle is Wnet = W1→ 2 + W2 →3 + W3→1 . For the isothermal process, W1→ 2 = nRT1 ln
V2 = ( 0.50 mol )( 8.31 J/mol K )( 273 K ) ln ( 0.5) = −786.2 J V1
For the isobaric process, ⎛V ⎞ W2 →3 = p2 ( ΔV ) = ( 2 p1 ) ⎜ 1 ⎟ = p1V1 = nRT1 = ( 0.50 mol )( 8.31 J/mol K )( 273 K ) = 1134.3 J ⎝2⎠ For the isochoric process, W3→1 = 0 J. Thus, Wnet = 348.1 J ≈ 350 J . For calculating heat transfers, note that T2 = T1 = 273 K and, since process 3 → 1 is isochoric, T3 = (2 atm/1 atm) T1 = 546 K. For the isothermal process, Q1→ 2 = W1→ 2 = −786.2 J because ΔEth 1→ 2 = 0 J. For the isobaric process, Q2 →3 = nCP ΔT = n ( 52 R ) (T3 − T2 ) =
5 2
( 0.50 mol )(8.31 J/mol K )( 273 K ) = 2835.8 J
For the isochoric process, Q3→1 = nCv ΔT = n ( 32 R ) (T1 − T3 ) = ( 32 ) ( 0.50 mol )( 8.31 J/mol K )( –273 K ) = −1701.5 J From Q1→2, Q2→3, and Q3→1, we see that QH = 2835.8 J. Thus, the thermal efficiency is
η=
348.1 J = 0.12 = 12% 2835.8 J
19.63. Model: The closed cycle of the heat engine involves the following four processes: isothermal expansion, isochoric cooling, isothermal compression, and isochoric heating. For a monatomic gas CV = 32 R .
Visualize:
Solve:
Using the ideal-gas law,
p1 =
nRT1 ( 0.20 mol )( 8.31 J/mol K )( 600 K ) = = 4.986 × 105 Pa V1 2.0 × 10−3 m3
At point 2, because of the isothermal conditions, T2 = T1 = 600 K and
p2 = p1
⎛ 2.0 × 10−3 m3 ⎞ V1 = ( 4.986 × 105 Pa ) ⎜ = 2.493 × 105 Pa −3 3 ⎟ V2 ⎝ 4.0 × 10 m ⎠
At point 3, because it is an isochoric process, V3 = V2 = 4000 cm3 and p3 = p2
T3 ⎛ 300 K ⎞ 5 = ( 2.493 × 105 Pa ) ⎜ ⎟ = 1.247 × 10 Pa T2 ⎝ 600 K ⎠
Likewise at point 4, T4 = T3 = 300 K and p4 = p3
⎛ 4.0 × 10−3 m3 ⎞ V3 = (1.247 × 105 Pa ) ⎜ = 2.493 × 105 Pa 3 ⎟ −3 V4 2.0 10 m × ⎝ ⎠
Let us now calculate Wnet = W1→2 + W2→3 + W3→4 + W4→1. For the isothermal processes,
W1→ 2 = nRT1 ln
W3→ 4 = nRT3 ln
V2 = ( 0.20 mol )( 8.31 J/mol K )( 600 K ) ln ( 2 ) = 691.2 J V1
V4 = ( 0.20 mol )( 8.31 J/mol K )( 300 K ) ln ( 12 ) = −345.6 J V3
For the isochoric processes, W2→3 = W4→1 = 0 J. Thus, Wnet = 345.6 J ≈ 350 J . Because Q = Ws + ΔEth, Q1→ 2 = W1→ 2 + ( ΔEth )1→ 2 = 691.2 J + 0 J = 691.2 J For the first isochoric process, Q2 →3 = nCV ΔT = ( 0.20 mol ) ( 32 R ) (T3 − T2 ) = ( 0.20 mol ) 32 ( 8.31 J/mol K )( 300 K − 600 K ) = −747.9 K For the second isothermal process Q3→ 4 = W3→ 4 + ( ΔEth )3→ 4 = −345.6 J + 0 J = −345.6 J
For the second isochoric process, Q4 →1 = nCV ΔT = n ( 32 R ) (T1 − T4 ) = ( 0.20 mol ) ( 32 ) ( 8.31 J/mol K )( 600 K − 300 K ) = 747.9 K Thus, QH = Q1→ 2 + Q4 →1 = 1439.1 J . The thermal efficiency of the engine is
η=
Wnet 345.6 J = = 0.24 = 24% QH 1439.1 J
19.64. Model: Processes 2 → 1 and 4 → 3 are isobaric. Processes 3 → 2 and 1 → 4 are isochoric. Visualize:
Solve: (a) Except in an adiabatic process, heat must be transferred into the gas to raise its temperature. Thus heat is transferred in during processes 4 → 3 and 3 → 2. This is the reverse of the heat engine in Example 19.2. (b) Heat flows from hot to cold. Since heat energy is transferred into the gas during processes 4 → 3 and 3 → 2, which end with the gas at temperature 2700 K, the reservoir temperature must be T > 2700 K. This is the hot reservoir, so the heat transferred is QH. Similarly, heat energy is transferred out of the gas during processes 2 → 1 and 1 → 4. This requires that the reservoir temperature be T < 300 K. This is the cold reservoir, and the energy transferred during these two processes is QC. (c) The heat energies were calculated in Example 19.2, but now they have the opposite signs. QH = Q43 + Q32 = 7.09 × 105 J + 15.19 × 105 J = 22.28 × 105 J QC = Q21 + Q14 = 21.27 × 105 J + 5.06 × 105 J = 26.33 × 105 J (d) For a counterclockwise cycle in the pV diagram, the work is Win. Its value is the area inside the curve, which is Win = (∆p)(∆V) = (2 × 101,300 Pa)(2 m3) = 4.05 × 105 J. Note that Win = QC – QH, as expected from energy conservation. (e) Since QC > QH, more heat is exhausted to the cold reservoir than is extracted from the hot reservoir. In this device, work is used to transfer energy “downhill,” from hot to cold. The exhaust energy is QC = QH + Win > QH. This is the energy-transfer diagram of Figure 19.19. (f) No. A refrigerator uses work input to transfer heat energy from the cold reservoir to the hot reservoir. This device uses work input to transfer heat energy from the hot reservoir to the cold reservoir.
19.65. Solve: (a) If you wish to build a Carnot engine that is 80% efficient and exhausts heat into a cold reservoir at 0°C, what temperature (in °C) must the hot reservoir be? (b) 0.80 = 1 −
( 0°C + 273) ⇒ 273 = 0.20 ⇒ T = 1092°C H (TH + 273) TH + 273
19.66. Solve: (a) A refrigerator with a coefficient of performance of 4.0 exhausts 100 J of heat in each cycle. What work is required each cycle and how much heat is removed each cycle from the cold reservoir? (b) We have 4.0 = QC Win ⇒ QC = 4Win . This means QH = QC + Win = 4Win + Win = 5Win ⇒ Win = QH = 100 J = 20 J 5 5 Hence, QC = QH − Win = 100 J − 20 J = 80 J.
19.67. Solve: (a) A heat engine operates at 20% efficiency and produces 20 J of work in each cycle. What is the net heat extracted from the hot reservoir and the net heat exhausted in each cycle? (b) We have 0.20 = 1 − QC QH . Using the first law of thermodynamics, Wout = QH − QC = 20 J ⇒ QC = QH − 20 J Substituting into the definition of efficiency,
0.20 = 1 −
QH − 20 J 20 J 20 J 20 J = 1−1+ = ⇒ QH = = 100 J 0.20 QH QH QH
The heat exhausted is QC = QH − 20 J = 100 J − 20 J = 80 J .
19.68.
Solve: (a)
In this heat engine, 400 kJ of work is done each cycle. What is the maximum pressure? (b)
1 ( pmax − 1.0 ×105 Pa )( 2.0 m3 ) = 4.0 × 105 J ⇒ pmax = 5.0 ×105 Pa = 500 kPa 2
19.69. Model: The heat engine follows a closed cycle, starting and ending in the original state. Visualize: The figure indicates the following seven steps. First, the pin is inserted when the heat engine has the initial conditions. Second, heat is turned on and the pressure increases at constant volume from 1 atm to 3 atm. Third, the pin is removed. The flame continues to heat the gas and the volume increases at constant pressure from 50 cm3 to 100 cm3. Fourth, the pin is inserted and some of the weights are removed. Fifth, the container is placed on ice and the gas cools at constant volume to a pressure of 1 atm. Sixth, with the container still on ice, the pin is removed. The gas continues to cool at constant pressure to a volume of 50 cm3. Seventh, with no ice or flame, the pin is inserted back in and the weights returned bringing the engine back to the initial conditions and ready to start over.
Solve:
(a)
(b) The work done per cycle is the area inside the curve: Wout = (∆p)(∆V) = (2 × 101,300 Pa)(50 × 10–6 m3) = 10.13 J (c) Heat energy is input during processes 1 → 2 and 2 → 3, so QH = Q12 + Q23. This is a diatomic gas, with CV = 3 R and CP = 52 R. The number of moles of gas is 2
n=
p1V1 (101,300 Pa)(50 × 10−6 m3 ) = = 0.00208 mol RT1 (8.31 J/mol K) (293 K)
Process 1 → 2 is isochoric, so T2 = (p2/p1)T1 = 3T1 = 879 K. Process 2 → 3 is isobaric, so T3 = (V3/V2)T2 = 2T2 = 1758 K. Thus
Q12 = nCV ΔT = 52 nR(T2 − T1 ) = 52 (0.00208 mol)(8.31 J/mol K)(586 K) = 25.32 J Similarly, Q23 = nCP ΔT = 72 nR (T3 − T2 ) = 72 (0.00208 mol)(8.31 J/mol K)(879 K) = 53.18 J Thus QH = 25.32 J + 53.18 J = 78.50 J and the engine’s efficiency is
η=
Wout 10.13 J = = 0.13 = 13% QH 78.50 J
19.70. Model: System 1 undergoes an isochoric process and system 2 undergoes an isobaric process. Solve: (a) Heat will flow from system 1 to system 2 because system 1 is hotter. Because there is no heat input from (or loss to) the outside world, we have Q1 + Q2 = 0 J. Heat Q1, which is negative, will change the temperature of system 1. Heat Q2 will both change the temperature of system 2 and do work by lifting the piston. But these consequences of heat flow don’t change the fact that Q1 + Q2 = 0 J. System 1 undergoes constant volume cooling from T1i = 600 K to Tf. System 2, whose pressure is controlled by the weight of the piston, undergoes constant pressure heating from T2i = 300 K to Tf. Thus, Q1 + Q2 = 0 J = n1CV (Tf − T1i ) + n2CP (Tf − T2i ) = n1 ( 32 R ) (Tf − T1i ) + n2 ( 52 R ) (Tf − T2i ) Solving this equation for Tf gives
Tf =
3n1T1i + 5n2T2i 3 ( 0.060 mol )( 600 K ) + 5 ( 0.030 mol )( 300 K ) = = 464 K 3n1 + 5n2 3 ( 0.060 mol ) + 5 ( 0.030 mol )
(b) Knowing Tf , we can compute the heat transferred from system 1 to system 2:
Q2 = n2CP (Tf − T2i ) = n2 ( 52 R ) (Tf − T2i ) = 102 J (c) The change of thermal energy in system 2 is
ΔEth = n2CV ΔT = n2 ( 32 R ) (Tf − T2i ) = 53 Q2 = 61.2 J According to the first law of thermodynamics, Q2 = Ws + ΔEth. Thus, the work done by system 2 is Ws = Q − ΔEth = 102.0 J − 61.2 J = 40.8 J. The work is done to lift the weight of the cylinder and the air above it by a height Δy. The weight of the air is wair = pA = pπ r 2 = (101.3 × 10 N/m 2 )π (0.050 m) 2 = 795.6 N. So,
Ws = ( wcyl + wair )Δy ⇒ Δy =
Ws 40.8 J = = 0.050 m ( wcyl + wair ) ( 2.0 kg ) ( 9.8 m/s 2 ) + 795.6 N
(d) The fraction of heat converted to work is
Ws 40.8 J = = 0.40 = 40% Q2 102.0 J
19.71. Model: Process 1 → 2 and process 3 → 4 are adiabatic, and process 2 → 3 and process 4 → 1 are isochoric. Visualize: Please refer to Figure CP19.71. Solve: (a) For adiabatic process 1 → 2, Q12 = 0 J and W12 =
p2V2 − p1V1 nR (T2 − T1 ) = 1−γ 1−γ
For isochoric process 2 → 3, W23 = 0 J and Q23 = nCV (T3 − T2 ) . For adiabatic process 3 → 4, Q34 = 0 J and
W34 =
p4V4 − p3V3 nR (T4 − T3 ) = 1−γ 1−γ
For isochoric process 4 → 1, W41 = 0 J and Q41 = nCV (T1 − T4 ) . The work done per cycle is
Wnet = W12 + W23 + W34 + W41 =
nR (T2 − T1 ) nR (T4 − T3 ) nR +0 J+ +0 J = (T2 − T1 + T4 − T3 ) 1− γ 1−γ 1−γ
(b) The thermal efficiency of the heat engine is
η=
Q nC (T − T ) Wout Q = 1 − C = 1 − 41 = 1 − V 4 1 QH QH Q23 nCV (T3 − T2 )
The last step follows from the fact that T3 > T2 and T4 > T1. We will now simplify this expression further as follows: γ −1
⎛V ⎞ γ −1 pV γ = pVV γ −1 = nRTV γ −1 ⇒ nRTV = nRT2V2γ −1 ⇒ T2 = T1 ⎜ 1 ⎟ 1 1 ⎝ V2 ⎠ Similarly, T3 = T4 r γ −1 . The equation for thermal efficiency now becomes
η =1−
(c)
1 T4 − T1 = 1 − γ −1 γ −1 r − T4 r T1r γ −1
= T1r γ −1
19.72. Model: For the Diesel cycle, process 1 → 2 is an adiabatic compression, process 2 → 3 is an isobaric expansion, process 3 → 4 is an adiabatic expansion, and process 4 → 1 is isochoric. Visualize: Please refer to CP19.72. Solve: (a) It will be useful to do some calculations using the compression ratio, which is r=
Vmax V1 1050 cm3 = = = 21 Vmin V2 50 cm3
The number of moles of gas is
n=
5 −6 3 p1V1 (1.013 × 10 Pa )(1050 × 10 m ) = =0.0430 mol RT1 ( 8.31 J/mol K ) ⎡⎣( 25+273) K ⎤⎦
For an adiabatic process, γ
⎛V ⎞ p1V1 = p2V2 ⇒ p2 = ⎜ 1 ⎟ p1 = r γ p1 = 211.40 × 1 atm = 71.0 atm = 7.19 × 106 Pa ⎝ V2 ⎠ γ
γ
γ −1
⎛V ⎞ γ −1 TV = T2V2γ −1 ⇒ T2 = ⎜ 1 ⎟ 1 1 ⎝ V2 ⎠
T1 = r γ −1T1 = 210.41 × 298 K = 1007 K
Process 2 → 3 is an isobaric heating with Q = 1000 J. Constant pressure heating obeys
Q = nCP ΔT ⇒ ΔT =
Q nCP
The gas has a specific heat ratio γ = 1.40 = 7 / 5, thus CV = 52 R and CP = 72 R. Knowing CP, we can calculate first
ΔT = 800 K and then T3 = T2 + ΔT = 1807 K. Finally, for an isobaric process we have V2 V3 T 1807 K = ⇒ V3 = 3 V2 = ( 50 × 10−6 m3 ) = 89.7 × 10−6 m3 T2 T3 T2 1007 K Process 3 → 4 is an adiabatic expansion to V4 = V1. Thus, γ
1.4
⎛V ⎞ ⎛ 89.7 ×10−6 m 3 ⎞ p3V3γ = p4V4γ ⇒ p4 = ⎜ 3 ⎟ p3 = ⎜ −6 3 ⎟ ⎝ 1050 ×10 m ⎠ ⎝ V4 ⎠ γ −1
γ −1
T3V3
= T4V4
Point 1 2 3 4
γ −1
⎛V ⎞ ⇒ T4 = ⎜ 3 ⎟ ⎝ V4 ⎠ p
( 7.19 ×10
6
⎛ 89.7 ×10−6 m3 ⎞ T3 = ⎜ −6 3 ⎟ ⎝ 1050 ×10 m ⎠
Pa ) = 2.30 ×105 Pa = 2.27 atm 0.4
V 5
1.00 atm = 1.013 × 10 Pa 6
71.0 atm = 7.19 × 10 Pa 6
71.0 atm = 7.19 × 10 Pa 5
2.27 atm = 2.30 × 10 Pa
T −6
3
298 K = 25°C
−6
3
1007 K = 734°C
−6
3
1807 K = 1534°C
−6
3
675 K = 402°C
1050 × 10 m 50.0 × 10 m 89.7 × 10 m
1050 × 10 m
(b) For adiabatic process 1 → 2,
(Ws )12 =
p2V2 − p1V1 = −633 J 1−γ
For isobaric process 2 → 3,
(Ws )23 = p2ΔV = p2 (V3 − V2 ) = 285 J For adiabatic process 3 → 4,
(1807 K ) = 675 K
(Ws )34 =
p4V4 − p3V3 = 1009 J 1−γ
For isochoric process 4 → 1, (Ws )41 = 0 J. Thus,
(Ws )cycle = Wout = (Ws )12 + (Ws )23 + (Ws )34 + (Ws )41 = 661 J (c) The efficiency is
η=
Wout 661 J = = 66.1% ≈ 66% QH 1000 J
(d) The power output of one cylinder is
661 J 2400 cycle 1 min J × × = 26,440 = 26.4 kW cycle min 60 sec s For an 8-cylinder engine the power will be 211 kW or 283 horsepower.
20.1.
Model: The wave is a traveling wave on a stretched string. Solve: The wave speed on a stretched string with linear density μ is vstring = TS / μ . The wave speed if the tension is doubled will be ′ = vstring
2TS
μ
= 2vstring = 2 ( 200 m/s ) = 283 m/s
20.2. Solve:
Model: The wave is a traveling wave on a stretched string. The wave speed on a stretched string with linear density μ is T 75 N ⇒ μ = 3.333 × 10−3 kg/m vstring = S ⇒ 150 m/s =
μ
μ
For a wave speed of 180 m/s, the required tension will be 2 2 TS = μ vstring = ( 3.333 × 10−3 kg/m ) (180 m/s ) = 110 N
20.3.
Model: The wave pulse is a traveling wave on a stretched string. The wave speed on a stretched string with linear density μ is T TS LTS 2.0 m ( 2.0 m )( 20 N ) ⇒ m = 0.025 kg = 25 g = S = = ⇒ = m/L m 50 × 10−3 s m μ
Solve:
vstring
20.4.
Model: This is a wave traveling at constant speed. The pulse moves 1 m to the right every second. Visualize: The snapshot graph shows the wave at all points on the x-axis at t = 0 s. The wave is just reaching x = 5.0. The first part of the wave causes an upward displacement of the medium. The rising portion of the wave is 2 m wide, so it will take 2 s to pass the x = 5.0 m point. The constant part of the wave, whose width is 2 m, will take 2 seconds to pass x = 5.0 m and during this time the displacement of the medium will be a constant (Δy = 1 cm). The trailing edge of the pulse arrives at t = 4 s at x = 5.0 m. The displacement now becomes zero and stays zero for all later times.
20.5.
Model: This is a wave traveling at constant speed. The pulse moves 1 m to the left every second. Visualize: This snapshot graph shows the wave at all points on the x-axis at t = 2 s. You can see that the leading edge of the wave at t = 2 s is precisely at x = 0 m. That is, in the first 2 seconds, the displacement is zero at x = 0 m. The first part of the wave causes a downward displacement of the medium, so immediately after t = 2 s the displacement at x = 0 m will be negative. The negative portion of the wave pulse is 3 m wide and takes 3 s to pass x = 0 m. The positive portion begins to pass through x = 0 m at t = 5 s and until t = 8 s the displacement of the medium is positive. The displacement at x = 0 m returns to zero at t = 8 s and remains zero for all later times.
20.6.
Model: This is a wave traveling at constant speed to the right at 1 m/s. Visualize: This is the history graph of a wave at x = 0 m. The graph shows that the x = 0 m point of the medium first sees the negative portion of the pulse wave at t = 1.0 s. Thus, the snapshot graph of this wave at t = 1.0 s must have the leading negative portion of the wave at x = 0 m.
20.7.
Model: This is a wave traveling at constant speed to the left at 1 m/s. Visualize: This is the history graph of a wave at x = 2 m. Because the wave is moving to the left at 1 m/s, the wave passes the x = 2 m position a distance of 1 m in 1 s. Because the flat part of the history graph takes 2 s to pass the x = 2 m position, its width is 2 m. Similarly, the width of the linearly increasing part of the history graph is 2 m. The center of the flat part of the history graph corresponds to both t = 0 s and x = 2 m.
20.8.
Visualize:
Figure EX20.8 shows a snapshot graph at t = 0 s of a longitudinal wave. This diagram shows a row of particles with an inter-particle separation of 1.0 cm at equilibrium. Because the longitudinal wave has a positive amplitude of 0.5 cm between x = 3 cm and x = 8 cm, the particles at x = 3, 4, 5, 6, 7 and 8 cm are displaced to the right by 0.5 cm.
20.9.
Visualize:
We first draw the particles of the medium in the equilibrium positions, with an inter-particle spacing of 1.0 cm. Just underneath, the positions of the particles as a longitudinal wave is passing through are shown at time t = 0 s. It is clear that relative to the equilibrium the particle positions are displaced negatively on the left side and positively on the right side. For example, the particles at x = 0 cm and x = 1 cm are at equilibrium, the particle at x = 2 cm is displaced left by 0.5 cm, the particle at x = 3 cm is displaced left by 1.0 cm, the particle at x = 4 cm is displaced left by 0.5 cm, and the particle at x = 5 cm is undisplaced. The behavior of particles for x > 5 cm is opposite of that for x < 5 cm.
20.10. Solve: (a) The wave number is k=
2π
λ
=
2π = 3.1 rad/m 2.0 m
(b) The wave speed is
⎛ω v = λ f = λ⎜ ⎝ 2π
⎞ ⎛ 30 rad/s ⎞ ⎟ = (2.0 m) ⎜ ⎟ = 9.5 m/s ⎠ ⎝ 2π ⎠
20.11. Solve: (a) The wavelength is λ=
2π 2π = = 4.2 m k 1.5 rad/m
(b) The frequency is
f =
v
λ
=
200 m/s = 48 Hz 4.19 m
20.12. Model: The wave is a traveling wave.
Solve: (a) A comparison of the wave equation with Equation 20.14 yields: A = 3.5 cm, k = 2.7 rad/m, ω = 124 rad/s, and φ0 = 0 rad. The frequency is
f =
ω 2π
=
124 rad/s = 19.7 Hz ≈ 20 Hz 2π
(b) The wavelength is
λ= (c) The wave speed v = λ f = 46 m/s .
2π 2π = = 2.33 m ≈ 2.3 m k 2.7 rad/m
20.13. Model: The wave is a traveling wave.
Solve: (a) A comparison of the wave equation with Equation 20.14 yields: A = 5.2 cm, k = 5.5 rad/m, ω = 72 rad/s, and φ0 = 0 rad. The frequency is
f =
ω 2π
=
72 rad/s = 11.5 Hz ≈ 11 Hz 2π
(b) The wavelength is
λ= (c) The wave speed v = λ f = 13 m/s.
2π 2π = = 1.14 m ≈ 1.1 m k 5.5 rad/m
20.14.
Solve: The amplitude of the wave is the maximum displacement, which is 6.0 cm. The period of the wave is 0.60 s, so the frequency f = 1 T = 1 0.60 s = 1.67 Hz . The wavelength is
λ=
v 2 m/s = = 1.2 m f 1.667 Hz
20.15.
Solve:
According to Equation 20.28, the phase difference between two points on a wave is
Δφ = φ2 − φ1 = 2π
Δr
λ
=
2π
λ
( r2 − r1 ) ⇒ ( 3π
rad − 0 rad ) =
2π
λ
( 80 cm − 20 cm ) ⇒ λ = 40 cm
20.16.
Solve:
According to Equation 20.28, the phase difference between two points on a wave is
Δφ = φ2 − φ1 =
2π
λ
( r2 − r1 )
If φ1 = π rad at r1 = 4.0 m, we can determine φ2 at any r value at the same instant using this equation. At r2 = 3.5 m,
φ2 = φ1 + At r2 = 4.5 m, φ = 32 π rad.
2π
λ
( r2 − r1 ) = π
rad +
2π π ( 3.5 m − 4.0 m ) = rad 2.0 m 2
20.17.
Solve:
Visualize:
For a sinusoidal wave, the phase difference between two points on the wave is given by Equation 20.28:
Δφ = φ2 − φ1 =
2π
λ
( r2 − r1 ) =
2π
λ
( 40 m − 30 m ) ⇒ λ =
2π (10 m ) Δφ
Δφ = 2π for two points on adjacent wavefronts and Δφ = 4π for two points separated by 2λ. Thus, λ = 10 m
when Δφ = 2π , and λ = 5 m when Δφ = 4π . The crests corresponding to these two wavelengths are shown in the figure. One can see that a crest of the wave passes the 40 m–listener and the 30 m–listener simultaneously. The lowest two possible frequencies will occur for the largest two possible wavelengths, which are 10 m and 5 m. Thus, the lowest frequency is f1 = The next highest frequency is f 2 = 68 Hz.
v
λ
=
340 m/s = 34 Hz 10 m
20.18. Visualize:
Solve:
(a) Because the same wavefront simultaneously reaches listeners at x = −7.0 m and x = + 3.0 m,
Δφ = 0 rad =
2π
λ
( r2 − r1 ) ⇒ r2 = r1
Thus, the source is at x = −2.0 m, so that it is equidistant from the two listeners. (b) The third person is also 5.0 m away from the source. Her y-coordinate is thus y = (5 m) 2 − (2 m) 2 = 4.6 m .
20.19. Solve: Two pulses of sound are detected because one pulse travels through the metal to the microphone while the other travels through the air to the microphone. The time interval for the sound pulse traveling through the air is Δtair =
Δx 4.0 m = = 0.01166 s = 11.66 ms vair 343 m/s
Sound travels faster through solids than gases, so the pulse traveling through the metal will reach the microphone before the pulse traveling through the air. Because the pulses are separated in time by 11.0 ms, the pulse traveling through the metal takes Δtmetal = 0.66 ms to travel the 4.0 m to the microphone. Thus, the speed of sound in the metal is vmetal =
Δx 4.0 m = = 6060 m/s ≈ 6100 m/s Δtmetal 0.00066 s
20.20.
Solve:
(a) In aluminum, the speed of sound is 6420 m/s. The wavelength is thus equal to
λ=
v 6420 m/s = = 3.21 × 10−3 m = 3.21 mm ≈ 3.2 mm f 2.0 × 106 Hz
(b) The speed of an electromagnetic wave is c. The frequency would be
f =
c
λ
=
3.0 × 108 m/s = 9.3 × 1010 Hz 3.21 × 10−3 m
20.21.
Solve:
(a) The frequency is
f =
vair
λ
=
343 m/s = 1715 Hz ≈ 1700 Hz 0.20 m
(b) The frequency is
f =
c
λ
=
3.0 × 108 m/s = 1.5 × 109 Hz = 1.5 GHz 0.20 m
(c) The speed of a sound wave in water is vwater = 1480 m/s. The wavelength of the sound wave would be v 1480 m/s λ = water = = 9.87 × 10−7 m ≈ 990 nm f 1.50 × 109 Hz
20.22. Solve:
Model: Light is an electromagnetic wave that travels with a speed of 3 × 108 m/s. (a) The frequency of the blue light is
f blue =
c
λ
=
3.0 × 108 m/s = 6.67 × 1014 Hz 450 × 10−9 m
(b) The frequency of the red light is
3.0 × 108 m/s = 4.62 × 1014 Hz 650 × 10−9 m (c) Using Equation 20.30 to calculate the index of refraction, f red =
λmaterial =
λvacuum n
⇒n=
λvacuum 650 nm = = 1.44 λmaterial 450 nm
20.23. Solve:
Model: Microwaves are electromagnetic waves that travel with a speed of 3 × 108 m/s. (a) The frequency of the microwave is
f microwaves =
c
λ
=
3.0 × 108 m/s = 1.0 × 1010 Hz = 10 GHz 3.0 × 10−2 m
(b) The refractive index of air is 1.0003, so the speed of microwaves in air is vair = c /1.00 ≈ c. The time for the microwave signal to travel is
t=
50 km 50 × 103 m = = 0.167 ms ≈ 0.17 ms vair ( 3.0 ×108 m 1.00 )
Assess: A small time of 0.17 ms for the microwaves to cover a distance of 50 km shows that the electromagnetic waves travel very fast.
20.24. Solve:
Model: Radio waves are electromagnetic waves that travel with speed c. (a) The wavelength is
λ=
c 3.0 × 108 m/s = = 2.96 m f 101.3 MHz
(b) The speed of sound in air at 20°C is 343 m/s. The frequency is
f =
vsound
λ
=
343 m/s = 116 Hz 2.96 m
20.25. Solve:
Model: Light is an electromagnetic wave. (a) The time light takes is
3.0 mm 3.0 × 10−3 m 3.0 × 10−3 m = = = 1.5 × 10−11 s 8 vglass cn 3.0 × 10 m/s 1.50 ( )
t= (b) The thickness of water is
d = vwatert =
c nwater
t=
3.0 × 108 m/s (1.5 ×10−11 s ) = 3.4 mm 1.33
20.26.
Solve:
(a) The speed of light in a material is given by Equation 20.29:
n=
c c ⇒ vmat = vmat n
The refractive index is
n=
λvac λ 420 nm ⇒ vsolid = c solid = ( 3.0 × 108 m/s ) = 1.88 × 108 m/s λmat λvac 670 nm
(b) The frequency is f =
vsolid
λsolid
=
1.88 × 108 m/s = 4.48 × 1014 Hz 420 nm
20.27. Model: Assume
that
the
glass
has
index
of
refraction
n = 1.5.
This
means
that
vglass = c/n = 2 × 108 m/s. Visualize: We apply v = λ f twice, once in air and then in the glass. The frequency will be the same in both cases. Solve: (a) In the air f air =
vair
λair
=
3.0 ×108 m/s = 8.57 × 108 Hz ≈ 8.6 × 108 Hz 0.35 m
The frequency is the same in both media, so f glass = 8.6 × 108 Hz. (b) Now that we know f glass and vglass , we can find λglass .
λglass = Assess:
vglass f glass
=
2.0 × 108 m/s = 23 cm 8.57 ×108 Hz
We get the same answer from λglass = λair /nglass = 35 cm/1.5 = 23 cm .
20.28.
Solve: The energy delivered to the eardrum in time t is E = Pt, where P is the power of the wave. The intensity of the wave is I = P / a where a is the area of the ear drum. Putting the above information together, we have
E = Pt = ( Ia ) t = Iπ r 2t = ( 2.0 × 10−3 W/m 2 )π ( 3.0 × 10−3 m ) ( 60 s ) = 3.4 × 10−6 J 2
20.29. Solve: The energy delivered to an area a in time t is E = Pt, where the power P is related to the intensity I as I = P / a. Thus, the energy received by your back is E = Pt = Iat = (0.80)(1400 W/m2)(0.30 × 0.50 m2)(3600 s) = 6.0 × 105 J
20.30. Solve: If a source of spherical waves radiates uniformly in all directions, the ratio of the intensities at distances r1 and r2 is 2
I1 r22 I ⎛ 2m ⎞ −3 = 2 ⇒ 50 m = ⎜ ⎟ = 1.6 × 10 I 2 m ⎝ 50 m ⎠ I 2 r1
⇒ I 50 m = I 2 m (1.6 × 10−3 ) = ( 2.0 W/m 2 )(1.6 × 10−3 ) = 3.2 × 10−3 W/m 2 Assess: The power generated by the sound source is P = I2m [4π(2 m)2] = (2.0 W/m2)(50.27) = 101 W. This is a significant amount of power.
20.31.
Solve: (a) The intensity of a uniform spherical source of power Psource a distance r away is I = Psource / 4π r 2 . Thus, the intensity at the position of the microphone is
I 50 m =
35 W 4π ( 50 m )
2
= 1.1 × 10−3 W/m 2
(b) The sound energy impinging on the microphone per second is
P = Ia = (1.1× 10−3 W/m 2 )(1.0 × 10−4 m 2 ) = 1.1× 10−7 W = 1.1× 10−7 J/s ⇒ Energy impinging on the microphone in 1 second = 1.1×10−7 J
20.32. Solve: Because the sun radiates waves uniformly in all directions, the intensity I of the sun’s rays when they impinge upon the earth is I=
Psun P 4 × 1026 W ⇒ I earth = sun2 = = 1400 W/m 2 2 4π r 4π rearth 4π (1.496 × 1011 m )2
With rsun-Venus = 1.082 × 1011 m and rsun-Mars = 2.279 × 1011 m, the intensities of electromagnetic waves at these planets are I venus = 2700 W/m 2 and I Mars = 610 W/m 2 .
20.33. Visualize: Equation 20.35 gives the sound intensity level as ⎛ I ⎞ ⎟ ⎝ I0 ⎠
β = (10 dB)log10 ⎜ where I 0 = 1.0 ×10−12 W/m 2 . Solve: (a)
⎛I ⎞ ⎛ 5.0 ×10−8 W/m 2 ⎞ = 47 dB ⎟ = (10 dB)log10 ⎜ −12 2 ⎟ ⎝ 1.0 ×10 W/m ⎠ ⎝ I0 ⎠
β = (10 dB)log10 ⎜ (b)
⎛I ⎞ ⎛ 5.0 ×10−2 W/m 2 ⎞ = 107 dB ⎟ = (10 dB)log10 ⎜ −12 2 ⎟ ⎝ 1.0 ×10 W/m ⎠ ⎝ I0 ⎠
β = (10 dB)log10 ⎜
Assess: As mentioned in the chapter, each factor of 10 in intensity changes the sound intensity level by 10 dB; between the first and second parts of this problem the intensity changed by a factor of 106 , so we expect the sound intensity level to change by 60 dB.
20.34. Visualize: We can solve Equation 20.35 for the sound intensity, finding I = I 0 ×10 β /10dB . Solve: (a)
I = I 0 ×10 β /10dB = (1.0 ×10−12 W/m 2 ) × 103.6 = 4.0 × 10−9 W/m 2 (b)
I = I 0 ×10 β /10dB = (1.0 ×10−12 W/m 2 ) × 109.6 = 4.0 × 10−3 W/m 2 Assess:
Since the sound intensity levels in the two parts of this problem differ by 60dB we expect the sound
intensities to differ by a factor of 106 .
20.35. Model: Assume the pole is tall enough that we don’t have to worry about the ground absorbing or reflecting sound. Visualize: The area of a sphere of radius R is A = 4π R 2 . Also recall that I = P/A ; we are given P = 5.0 W . We seek R for β = 90dB . Solve:
A=
P P 5.0 W = = = 5000 m 2 I I 0 ×10 β/10dB (1.0 ×10−12 W/m 2 ) ×1090dB/10dB R=
A 5000 m 2 = = 20 m 4π 4π
Assess: This is a reasonable distance from the loudspeaker for a moderately loud sound.
20.36. Model: The frequency of the opera singer’s note is altered by the Doppler effect. Solve:
(a) Using 90 km/h = 25 m/s, the frequency as her convertible approaches the stationary person is f+ =
f0 600 Hz = = 650 Hz 1 − vS v 1 − 25 m/s 343 m/s
(b) The frequency as her convertible recedes from the stationary person is f0 600 Hz f− = = = 560 Hz 1 + vS v 1 + 25 m/s 343 m/s
20.37. Model: Your friend’s frequency is altered by the Doppler effect. The frequency of your friend’s note increases as he races towards you (moving source and a stationary observer). The frequency of your note for your approaching friend is also higher (stationary source and a moving observer). Solve: (a) The frequency of your friend’s note as heard by you is f+ =
f0 400 Hz = = 432 Hz vS 25.0 m/s 1− 1− v 340 m/s
(b) The frequency heard by your friend of your note is
⎛ v ⎞ ⎛ 25.0 m/s ⎞ f + = f 0 ⎜1 + 0 ⎟ = ( 400 Hz ) ⎜1 + ⎟ = 429 Hz v⎠ 340 m/s ⎠ ⎝ ⎝
20.38.
Model: Sound frequency is altered by the Doppler effect. The frequency increases for an observer approaching the source and decreases for an observer receding from a source. Solve: You need to ride your bicycle away from your friend to lower the frequency of the whistle. The minimum speed you need to travel is calculated as follows:
v0 ⎞ ⎛ v ⎞ ⎛ f − = ⎜1 − 0 ⎟ f 0 ⇒ 20 kHz = ⎜ 1 − ⎟ ( 21 kHz ) ⇒ v0 = 16.3 m/s ≈ 16 m/s v⎠ ⎝ ⎝ 343 m/s ⎠ Assess: A speed of 16.3 m/s corresponds to approximately 35 mph. This is a possible but very fast speed on a bicycle.
20.39. Solve:
Model: The mother hawk’s frequency is altered by the Doppler effect. The frequency is f + as the hawk approaches you is
f+ =
Assess:
f0 800 Hz ⇒ 900 Hz = ⇒ vS = 38.1 m/s vS 1 − vS v 1− 343 m/s
The mother hawk’s speed of 38.1 m/s ≈ 80 mph is reasonable.
20.40. Visualize: changing shape. Solve:
The function D(x, t) represents a pulse that travels in the positive x-direction without
(a)
(b) The leading edge of the pulse moves forward 3 m each second. Thus, the wave speed is 3 m/s. (c) x − 3t is a function of the form D(x – vt), so the pulse moves to the right at v = 3 m/s.
20.41. Solve: (a) We see from the history graph that the period T = 0.20 s and the wave speed v = 4.0 m/s. Thus, the wavelength is λ=
v = vT = ( 4.0 m/s )( 0.20 s ) = 0.80 m f
(b) The phase constant φ0 is obtained as follows:
D ( 0 m, 0 s ) = A sin φ0 ⇒ −2 mm = ( 2 mm ) sin φ0 ⇒ sin φ0 = −1 ⇒ φ0 = − 12 π rad (c) The displacement equation for the wave is
2π t π⎞ ⎛ 2π x ⎞ ⎛ 2π x D ( x, t ) = A sin ⎜ − 2π ft + φ0 ⎟ = ( 2.0 mm ) sin ⎜ − − ⎟ = ( 2.0 mm ) sin ( 2.5π x − 10π t − 12 π ) 0.80 m 0.20 s 2⎠ λ ⎝ ⎠ ⎝ where x and t are in m and s, respectively.
20.42. Solve: (a) We can see from the graph that the wavelength is λ = 2.0 m. We are given that the wave’s frequency is f = 5.0 Hz. Thus, the wave speed is v = λf = 10 m/s. (b) The snapshot graph was made at t = 0 s. Reading the graph at x = 0 m, we see that the displacement is D ( x = 0 m, t = 0 s ) = D ( 0 m, 0 s ) = 0.5 mm = 12 A Thus D ( 0 m,0 s ) = 12 A = A sin φ0 ⇒ φ0 = sin −1 ( 12 ) =
π 6
rad or
5π rad 6
Note that the value of D(0 m, 0 s) alone gives two possible values of the phase constant. One of the values will cause the displacement to start at 0.5 mm and increase with distance—as the graph shows—while the other will cause the displacement to start at 0.5 mm but decrease with distance. Which is which? The wave equation for t = 0 s is
⎛ 2π x ⎞ D ( x, t = 0 ) = A sin ⎜ + φ0 ⎟ λ ⎝ ⎠ If x is a point just to the right of the origin and is very small, the angle (2π x / λ + φ0 ) is just slightly bigger than
the angle φ0 . Now sin 31° > sin 30°, but sin151° < sin150°, so the value φ0 = 16 π rad is the phase constant for which the displacement increases as x increases. (c) The equation for a sinusoidal traveling wave can be written as
⎡ ⎛x ⎤ ⎛ 2π x ⎞ ⎞ − 2π ft + φ0 ⎟ = A sin ⎢ 2π ⎜ − ft ⎟ + φ0 ⎥ D ( x, t ) = A sin ⎜ ⎝ λ ⎠ ⎠ ⎣ ⎝λ ⎦ Substituting in the values found above, ⎡ ⎛ x ⎞ π⎤ − ( 5.0 s −1 ) t ⎟ + ⎥ D ( x, t ) = (1.0 mm ) sin ⎢ 2π ⎜ 2.0 m ⎠ 6⎦ ⎣ ⎝
20.43.
Solve:
The time for the sound wave to travel down the tube and back is t = 440 μ s since 1 division
is equal to 100 μs. So, the speed of the sound wave in the liquid is
v=
2 × 25 cm = 1140 m/s ≈ 1100 m/s 440 μ s
20.44. Solve:
Model: The wave is a traveling wave on a stretched string. The wave speed on a string whose radius is R, length is L, and mass density is ρ is vstring = TS / μ with
μ=
m ρV ρπ R 2 L = = = ρπ R 2 L L L
If the string radius doubles, then ′ = vstring
TS
ρπ ( 2R )
2
=
vstring 2
=
280 m/s = 140 m/s 2
20.45. Solve:
Model: The wave pulse is a traveling wave on a stretched string. While the tension TS is the same in both the strings, the wave speeds in the two strings are not. We have
v1 =
TS
μ1
and
v2 =
TS
μ2
⇒ v12 μ1 = v22 μ 2 = TS
Because v1 = L1 / t1 and v2 = L2 / t2 , and because the pulses are to reach the ends of the string simultaneously, the above equation can be simplified to
L12 μ1 L22 μ 2 μ2 4.0 g/m L = 2 ⇒ 1= = = 2 ⇒ L1 = 2 L2 t2 t μ1 L2 2.0 g/m Since L1 + L2 = 4 m, 2 L2 + L2 = 4 m ⇒ L2 = 1.66 m ≈ 1.7 m
and
L1 = 2 (1.66 m ) = 2.34 m ≈ 2.3 m
20.46. Solve: Δt is the time the sound wave takes to travel down to the bottom of the ocean and then up to the ocean surface. The depth of the ocean is 2d = ( vsound in water ) Δt ⇒ d = ( 750 m/s ) Δt Using this relation and the data from Figure P20.46, we can generate the following table for the ocean depth (d ) at various positions (x) of the ship. x (km) d (km) Δt (s) 0 20 40 45 50 60
6 4 4 8 4 2
4.5 3.0 3.0 6.0 3.0 1.5
20.47. Visualize:
Solve: The explosive’s sound travels down the lake and into the granite, and then it is reflected by the oil surface. The echo time is thus equal to
techo = t water down + tgranite down + tgranite up + twater up 0.94 s =
d granite d granite 500 m 500 m + + + ⇒ d granite = 790 m 1480 m/s 6000 m/s 6000 m/s 1480 m/s
20.48. Model: Assume a room temperature of 20°C. Visualize:
Solve:
The distance between the source and the left ear (EL) is
d L = x 2 + ( y + 0.1 m ) = ⎡⎣( 5.0 m ) cos45° ⎤⎦ + ⎡⎣( 5.0 m ) sin45° + 0.1 m ⎤⎦ = 5.0712 m 2
2
2
Similarly d R = 4.9298 m. Thus, d L − d R = Δd = 0.1414 m
For the sound wave with a speed of 343 m/s, the difference in arrival times at your left and right ears is Δt =
Δd 0.1414 m = = 410 μ s 343 m/s 343 m/s
20.49. Solve:
Model: The laser beam is an electromagnetic wave that travels with the speed of light. The speed of light in the liquid is
vliquid =
30 × 10−2 m = 2.174 × 108 m/s 1.38 × 10−9 s
The liquid’s index of refraction is
n=
c vliquid
=
3.0 × 108 = 1.38 2.174 × 108
Thus the wavelength of the laser beam in the liquid is
λliquid =
λvac n
=
633 nm = 459 nm 1.38
20.50. Solve:
Model: The temperature is 20°C for both air and water. For the sound speed of vair = 343 m/s, the wavelength of the sound wave in air is
λair =
343 m/s = 1.340 m 256 Hz
On entering water the frequency does not change, so fwater = fair and fwater /fair = 1. The wave speed in water is vwater = 1480 m/s, so
vwater 1480 m/s = = 4.31 vair 343 m/s Finally, the wavelength in water is v 1480 m/s λ 5.781 m λwater = water = = 5.781 m ⇒ water = = 4.31 f water 256 Hz λair 1.340 m Assess: This last result is expected because v = f λ and the frequency remains unchanged as the wave enters from air into water.
20.51.
Solve:
The difference in the arrival times for the P and S waves is
Δt = tS − tP = Assess:
d d 1 1 ⎛ ⎞ 6 ⇒ 120 s = d ⎜ − − ⎟ ⇒ d = 1.23 × 10 m=1230 km vS vP ⎝ 4500 m/s 8000 m/s ⎠
d is approximately one-fifth of the radius of the earth and is reasonable.
20.52. Solve:
Model: This is a sinusoidal wave. (a) The equation is of the form D( y, t ) = A sin(ky + ω t + φ0 ), so the wave is traveling along the y-axis.
Because it is +ω t rather than −ω t the wave is traveling in the negative y-direction. (b) Sound is a longitudinal wave, meaning that the medium is displaced parallel to the direction of travel. So the air molecules are oscillating back and forth along the y-axis. (c) The wave number is k = 8.96 m -1 , so the wavelength is
λ=
2π 2π = = 0.701 m k 8.96 m −1
The angular frequency is ω = 3140 s −1 , so the wave’s frequency is f =
ω 3140 s −1 = = 500 Hz 2π 2π
Thus, the wave speed v = λ f = (0.70 m)(500 Hz) = 350 m/s. The period T = 1/f = 0.00200 s = 2.00 ms. (d) The interval t = 0 s to t = 4 ms is exactly 2 cycles of the wave. The initial value at y = 1 m is
D ( y = 1 m, t = 0 s ) = ( 0.02 mm ) sin ( 8.96 + 14 π ) = −0.0063 mm
Assess: The wave is a sound wave with speed v = 350 m/s. This is greater than the room-temperature speed of 343 m/s, so the air temperature must be greater than 20°.
20.53. Model: This is a sinusoidal wave. Solve: (a) The displacement of a wave traveling in the positive x-direction with wave speed v must be of the form D(x, t) = D(x− vt). Since the variables x and t in the given wave equation appear together as x + vt, the wave is traveling toward the left, that is, in the −x direction. (b) The speed of the wave is v=
ω k
=
2π 0.20 s = 12 m/s 2π rad 2.4 m
The frequency is f =
ω 2π rad 0.20 s = = 5.0 Hz 2π 2π
The wave number is k=
2π rad = 2.6 rad/m 2.4 m
(c) The displacement is
⎡ ⎛ 0.20 m 0.50 s ⎞ ⎤ D ( 0.20 m, 0.50 s ) = ( 3.0 cm ) sin ⎢ 2π ⎜ + + 1⎟ ⎥ = −1.5 cm ⎣ ⎝ 2.4 m 0.20 s ⎠ ⎦
20.54. Model: This is a sinusoidal wave traveling on a stretched string in the +x direction. Solve: (a) From the displacement equation of the wave, A = 2.0 cm, k = 12.57 rad/m, and ω = 638 rad/s. Using the equation for the wave speed in a stretched string, vstring =
2
2
⎛ω ⎞ ⎛ 638 rad/s ⎞ 2 ⇒ TS = μ vstring = μ ⎜ ⎟ = ( 5.00 × 10−3 kg/m3 ) ⎜ ⎟ = 12.6 N μ ⎝k⎠ ⎝ 12.57 rad/m ⎠
TS
(b) The maximum displacement is the amplitude Dmax ( x, t ) = 2.00 cm . (c) From Equation 20.17, v y max = ω A = ( 638 rad/s ) ( 2.0 × 10−2 m ) = 12.8 m/s
20.55.
Solve:
The wave number and frequency are calculated as follows:
2π rad = 4π rad/m ⇒ ω = vk = ( 4.0 m/s )( 4π rad/m ) = 16π rad/s 0.50 m Thus, the displacement equation for the wave is k=
2π
λ
=
D ( y, t ) = ( 5.0 cm ) sin ⎣⎡( 4π rad/m ) y + (16π rad/s ) t ⎦⎤ Assess:
The positive sign in the sine function’s argument indicates motion along the −y direction.
20.56.
Solve:
The angular frequency and wave number are calculated as follows:
ω = 2π f = 2π ( 200 Hz ) = 400π rad/s ⇒ k =
ω v
=
400π rad/s = π rad/m 400 m/s
The displacement equation for the wave is D ( x, t ) = ( 0.010 mm ) sin ⎣⎡(π rad/m ) x − ( 400π rad/s ) t + 12 π rad ⎦⎤ Assess: Note the negative sign with ω t in the sine function’s argument. This indicates motion along the +x direction.
20.57.
Solve:
A sinusoidal traveling wave is represented as D ( x, t ) = A sin ( kx − ω t ) . Replacing t with t + T
and using the relationship ω = 2π/T between the angular frequency and period, D ( x, t + T ) = A sin ( kx − ω ( t + T ) ) = A sin ( kx − ω t − ωT ) = A sin ( kx − ω t − 2π ) = A ⎡⎣sin ( kx − ω t ) cos ( 2π ) − cos ( kx − ω t ) sin ( 2π ) ⎤⎦ = A sin ( kx − ω t ) = D ( x, t )
20.58. Solve: According to Equation 20.28, the phase difference between two points on a wave is Δφ = 2π ( r2 − r1 ) λ . For the first point and second point, r1 =
(1.00 cm − 0 cm )
r2 =
( −1.00 cm − 0 cm ) + (1.50 cm − 0 cm )
2
+ ( 3.00 cm − 0 cm ) + ( 2.00 cm − 0 cm ) = 3.742 cm 2
2
2
2
+ ( 2.50 cm − 0 cm ) = 3.082 cm 2
The wavelength is
λ= ⇒ Δφ =
v 346 m/s = = 0.02641 m = 2.641 cm f 13,100 Hz
2π ( 3.742 cm − 3.082 cm ) π π 180° = rad = rad × = 90° 2.641 cm 2 2 π rad
20.59. Solve:
Model: We have a sinusoidal traveling wave on a stretched string. (a) The wave speed on a string and the wavelength are calculated as follows:
v=
TS
μ
=
20 N v 100.0 m/s = 100 m/s ⇒ λ = = = 1.0 m 0.002 kg/m f 100 Hz
(b) The amplitude is determined by the oscillator at the end of the string and is A = 1.0 mm. The phase constant can be obtained from Equation 20.15 as follows:
D ( 0 m, 0 s ) = A sin φ0 ⇒ −1.0 mm = (1.0 mm ) sin φ0 ⇒ φ0 = −
π 2
rad
(c) The wave (as distinct from the oscillator) is described by D ( x, t ) = A sin ( kx − ω t + φ0 ) . In this equation the
wave number and angular frequency are k=
2π
λ
=
2π = 2π rad/m 1.0 m
ω = vk = (100.0 m/s )( 2π rad/m ) = 200π rad/s
Thus, the wave’s displacement equation is D ( x, t ) = (1.0 mm ) sin ⎡⎣( 2π rad/m ) x − ( 200π rad/s ) t − 12 π rad ⎤⎦
(d) The displacement is
D ( 0.50 m, 0.015 s ) = (1.0 mm ) sin ⎡⎣( 2π rad/s )( 0.50 m ) − ( 200π rad/s )( 0.015 s ) − 12 π ⎤⎦ = −1.0 mm
20.60. Model: We have a wave traveling to the right on a string. Visualize:
Solve: The snapshot of the wave as it travels to the right for an infinitesimally small time Δt shows that the velocity at point 1 is downward, at point 3 is upward, and at point 2 is zero. Furthermore, the speed at points 1 and 3 is the maximum speed given by Equation 20.17: v1 = v3 = ω A . The frequency of the wave is
ω = 2π f = 2π
v
λ
=
2π ( 45 m/s ) = 300π rad/s ⇒ ω A = ( 300π rad/s ) ( 2.0 × 10−2 m ) = 19 m/s 0.30 m
Thus, v1 = −19 m/s, v2 = 0 m/s, and v3 = +19 m/s.
20.61. Model: This is a wave traveling to the left at a constant speed of 50 cm/s. Solve: The particles at positions between x = 2 cm and x = 7 cm have a speed of 10 cm/s, and the particles between x = 7 cm and x = 9 cm have a speed of −25 cm/s. That is, at the time the snapshot of the velocity is shown, the particles of the medium have upward motion for 2 cm ≤ x ≤ 7 cm, but downward motion for 7 cm ≤ x ≤ 9 cm. The width of the front section of the wave pulse is 7 cm – 2 cm = 5 cm and the width of the rear section is 9 cm – 7 cm = 2 cm. With a wave speed of 50 cm/s, the time taken by the front section to pass through a particular point is 5 cm 50 cm/s = 0.1 s and the time taken by the rear section of the wave to pass through a point is 2 cm 50 cm/s = 0.04 s . Thus the wave causes the upward moving particles to go through a displacement of
A = (10 cm/s )( 0.1 s ) = 1.0 cm . The downward moving particles have a maximum displacement of
( −25 cm/s )( 0.04 s ) = −1.0 cm .
20.62. Solve:
Model: The wave is traveling on a stretched string. The wave speed on the string is
v=
TS
μ
=
50 N = 100 m/s 0.005 kg/m
The speed of the particle on the string, however, is given by Equation 20.17. The maximum speed is calculated as follows:
v y = −ω A cos ( kx − ω t + φ0 ) ⇒ v y
max
= ω A = 2π fA = 2π
⎛ 100 m/s ⎞ A = 2π ⎜ ⎟ ( 0.030 m ) = 9.4 m/s λ ⎝ 2.0 m ⎠
v
20.63. Model: A sinusoidal wave is traveling along a stretched string. Solve: From Equation 20.17 and Equation 20.20, vy max = ωA and ay max = ω2A. These two equations can be combined to give ω=
a y max v y max
=
v 200 m/s 2 ω 2.0 m/s = 100 rad/s ⇒ f = = 15.9 Hz ≈ 16 Hz ⇒ A = y max = = 2.0 cm ω 100 rad/s 2.0 m/s 2π
20.64.
Solve: (a) At a distance r from the bulb, the 5 watts of visible light have spread out to cover the surface of a sphere of radius r. The surface area of a sphere is a = 4πr2. Thus, the intensity at a distance of 2 m is
I=
P P 5.0 W = = = 0.095 W/m 2 a 4π r 2 4π ( 2.0 m )2
Note that the presence of the wall has nothing to do with the intensity. The wall allows you to see the light, but the light wave has the same intensity at all points 2 m from the bulb whether it is striking a surface or moving through empty space. (b) Unlike the light from a light bulb, a laser beam does not spread out. We ignore the small diffraction spread of the beam. The laser beam creates a dot of light on the wall that is 2 mm in diameter. The full 5 watts of light is 2 concentrated in this dot of area a = π r 2 = π ( 0.001 m ) = 3.14 × 10−6 m 2 . The intensity is 5W P = = 1.6 MW/m 2 a 3.14 × 10−6 m 2 Although the power of the light source is the same in both cases, the laser produces light on the wall whose intensity is over 16 million times that of the light bulb. I=
20.65. Model: The radio wave is an electromagnetic wave. Solve: At a distance r, the 25 kW power station spreads out waves to cover the surface of a sphere of radius r. The surface area of a sphere is 4πr2. Thus, the intensity of the radio waves is I=
Psource 25 × 103 W = = 2.0 × 10−5 W/m 2 2 2 4π r 4π (10 × 103 m )
20.66.
Solve:
(a) The peak power of the light pulse is
Ppeak =
ΔE 500 mJ 0.50 J = = = 5.0 × 107 W Δt 10 ns 1.0 × 10−8 s
(b) The average power is Pavg =
Etotal 10 × 500 mJ 5.0 J = = = 5.0 W 1.0 s 1.0 s 1.0 s
The laser delivers pulses of very high power. But the laser spends most of its time “off,” so the average power is very much less than the peak power. (c) The intensity is
I laser =
5.0 × 107 W P 5.0 × 107 W = = = 6.4 × 1017 W/m 2 2 7.85 × 10−11 m 2 a π ( 5.0 μ m )
(d) The ratio is
I laser 6.4 × 1017 W/m 2 = = 5.8 × 1014 I sun 1.1 × 103 W/m 2
20.67. Solve:
Model: We have a traveling wave radiated by the tornado siren. (a) The power of the source is calculated as follows:
I 50 m = 0.10 W/m 2 =
Psource Psource 2 = ⇒ Psource = ( 0.10 W/m 2 ) 4π ( 50 m ) = (1000π ) W 2 2 4π r 4π ( 50 m )
The intensity at 1000 m is I1000 m =
Psource
4π (1000 m )
2
=
(1000π ) W = 250 μ W 2 4π (1000 m )
m
2
(b) The maximum distance is calculated as follows:
I=
Psource (1000π ) W ⇒ r = 16 km ⇒ 1.0 × 10−6 W/m 2 = 4π r 2 4π r 2
20.68. Model: Assume the saw is far enough off the ground that we don’t have to worry about reflected sound. Visualize:
First note that β1 − β 2 = 20dB ⇒ I1/I 2 = 10 ⋅ 10 = 100 (a change of 10 dB corresponds to a change in
intensity by a factor of 10). Then use I1 A1 = P and then P = I 2 A2 ⇒ A2 = P/I 2 , and finally solve for R2 = A2 / 4π . Solve:
Put all of the above together. P
R2 = Assess:
I1 A1
A2 I2 I2 = = = 4π 4π 4π
I1 I2
(4π R12 ) 4π
= R1
The scaling laws help and the answer is reasonable.
I1 = R1 100 = (5.0 m)(10) = 50 m I2
20.69.
Model: Assume the two loudspeakers broadcast the same power and that the platforms are high enough off the ground that we don’t have to worry about reflected sound. Visualize: Call the distance between the loudspeakers d . Call the intensity halfway between the speakers (at d / 2 ) I1 and the sound intensity level there β1 (= 75 dB ); call them I 2 and β 2 at 1/4 the distance from one pole
and 3/4 the distance from the other pole on the line between them. We seek β 2 . We first apply a general approach for different sound intensity levels: ⎡ ⎛I ⎞ ⎛ I ⎞⎤ ⎛ I /I ⎞ ⎛I ⎞ Δβ = β 2 − β1 = (10dB) ⎢ log10 ⎜ 2 ⎟ − log10 ⎜ 1 ⎟ ⎥ = (10 dB)log10 ⎜ 2 0 ⎟ = (10 dB)log10 ⎜ 2 ⎟ I I I I / ⎝ I1 ⎠ ⎝ 0⎠ ⎝ 0 ⎠ ⎦⎥ ⎝ 1 0⎠ ⎣⎢ Solve: Recall that for the general case of spherical symmetry I = P/A , where P is the power emitted by the source and A = 4π R 2 is the area of the sphere. Now we find the ratio of the intensities I 2 /I1 and then plug it in the formula above and add it to 75 dB.
I1 = I2 =
P P 2P + = 4π ( d / 2) 2 4π (d / 2) 2 π d 2
P P 4P 4P (36 + 4) P 40 P 20 + = + = = = I1 4π ( d / 4) 2 4π (3d / 4) 2 π d 2 9π d 2 9π d 2 9π d 2 9 ⎛I ⎞ ⎛ 20 ⎞ Δβ = (10 dB)log10 ⎜ 2 ⎟ = (10 dB)log10 ⎜ ⎟ = 3.48 dB I 9 ⎠ ⎝ ⎝ 1⎠
β 2 = β1 + Δβ = 75 dB + 3.48 dB = 78 dB Assess:
An increase of about 3dB corresponds to a doubling of the intensity. 20/9 is close to double.
20.70.
Model:
As suggested, model the bald head as a hemisphere with radius R = 0.080 m. This means the
surface area of the bald head (hemisphere) is A = 2π R 2 = 0.0402m 2 . Visualize: We are given β = 93 dB and ΔE = 0.10 J . We also know that I = I 0 × 10 β /10 dB and P = IA . Also recall P = ΔE Δt . Solve: Put all of the above together to find Δt .
Δt =
ΔE ΔE ΔE 0.10 J = = = = 1250 s ≈ 21 min P IA ( I 0 × 10 β /10 dB )(2π R 2 ) (10−12 W/m 2 × 109.3 )(0.0402 m 2 )
Assess: 21 min seems like quite a while to deliver 0.10 J of energy, but sounds waves don’t carry a lot of energy unless the intensity is high.
20.71.
Model: The bat’s chirping frequency is altered by the Doppler effect. The frequency is increased as the bat approaches and it decreases as the bat recedes away. Solve: The bat must fly away from you, so that the chirp frequency observed by you is less than 25 kHz. From Equation 20.38,
f− =
Assess:
f0 25,000 Hz ⇒ vS = 85.8 m/s ≈ 86 m/s ⇒ 20,000 Hz = 1 + vS / v ⎛ vS ⎞ 1+ ⎜ ⎟ ⎝ 343 m/s ⎠
This is a rather large speed: 85.8 m/s ≈ 180 mph. This is not possible for a bat.
20.72. Model: The sound generator’s frequency is altered by the Doppler effect. The frequency increases as the generator approaches the student, and it decreases as the generator recedes from the student. Solve: The generator’s speed is ⎛ 100 ⎞ vS = rω = r ( 2π f ) = (1.0 m ) 2π ⎜ rev/s ⎟ = 10.47 m/s ⎝ 60 ⎠ The frequency of the approaching generator is f0 600 Hz f+ = = = 619 Hz ≈ 620 Hz 1 − vS v 1 − 10.47 m/s 343 m/s Doppler effect for the receding generator, on the other hand, is
f0 600 Hz = = 582 Hz ≈ 580 Hz 1 + vS v 1 + 10.47 m/s 343 m/s Thus, the highest and the lowest frequencies heard by the student are 620 Hz and 580 Hz. f− =
20.73. Solve: We will closely follow the details of section 20.7 in the textbook. Figure 20.29 shows that the wave crests are stretched out behind the source. The wavelength detected by Pablo is λ− = 13 d , where d is the distance the wave has moved plus the distance the source has moved at time t = 3T. These distances are Δxwave = vt = 3vT and Δxsource = vSt = 3vST . The wavelength of the wave emitted by a receding source is thus
d Δxwave + Δxsource 3vT + 3vST = = = ( v + vS ) T 3 3 3 The frequency detected in Pablo’s direction is thus v v f0 = = f− = λ− ( v + vs ) T 1 + vS v
λ− =
20.74. Model: We are looking at the Doppler effect for the light of an approaching source. Solve:
(a) The time is
54 × 106 km = 180 s = 3.0 min 3 × 105 km/s (b) Using Equation 20.40, the observed wavelength is t=
λ=
Assess:
1 − vs c 1 − 0.1c c λ0 = ( 540 nm ) = (0.9045)(540 nm) = 488 nm ≈ 490 nm 1 + vs c 1 + 0.1c c
490 nm is slightly blue shifted from green.
20.75. Model: We are looking at the Doppler effect for the light of a receding source. Visualize:
Note that the daredevil’s tail lights are receding away from your rocket’s light detector with a relative speed of 0.2c. Solve: Using Equation 20.40, the observed wavelength is
λ=
1 + vS c 1 + 0.2c c λ0 = ( 650 nm ) = 796 nm ≈ 800 nm 1 − vS c 1 − 0.2c c
This wavelength is in the infrared region.
20.76. Model: The Doppler effect for light of a receding source yields an increased wavelength. Solve: Because the measured wavelengths are 5% longer, that is, λ = 1.05λ0, the distant galaxy is receding away from the earth. Using Equation 20.40,
λ = 1.05λ0 =
1 + vs c 1 + vs c 2 ⇒ vs = 0.049 c = 1.47 × 107 m/s λ0 ⇒ (1.05) = 1 − vs c 1 − vs c
20.77. Model: The Doppler effect for light of an approaching source leads to a decreased wavelength. Solve: The red wavelength (λ0 = 650 nm) is Doppler shifted to green (λ = 540 nm) due to the approaching light
source. In relativity theory, the distinction between the motion of the source and the motion of the observer disappears. What matters is the relative approaching or receding motion between the source and the observer. Thus, we can use Equation 20.40 as follows:
λ = λ0
1 − vs c 1 − vs c ⇒ 540 nm = ( 650 nm ) 1 + vs c 1 + vs c
⇒ vs = 5.5 × 104 km/s = 2.0 × 108 km/h The fine will be ⎛ 1$ ⎞ km/hr − 50 km/hr ) ⎜ ⎟ = $200 million ⎝ 1 km/hr ⎠ The police officer knew his physics.
( 2.0 ×10
8
Assess:
20.78. Model: The wave pulse is a traveling wave on a stretched string. The two masses hanging from the steel wire are in static equilibrium. Visualize:
Solve:
The wave speed along the wire is vwire =
4.0 m = 166.7 m/s 0.024 s
Using Equation 20.2, vwire = 166.7 m/s=
T1
=
T1
μ ( 0.060 kg 8.0 m ) G G Because point 1 is in static equilibrium, with Fnet = 0,
( Fnet ) x = T1 − T2 cos 40° ⇒ T2 =
⇒ T1 = 208.4 N
T1 = 2721 N cos 40°
( Fnet ) y = T2 sin 40° − w = 0 N ⇒ w = mg = T2 sin 40° ⇒ m =
( 272.1 N ) sin 40° = 17.8 kg 9.8 m/s 2
20.79. Solve: The time for the wave to travel from California to the South Pacific is t=
d 8.00 × 106 m = = 5405.4 s v 1480 m/s
A time decrease to 5404.4 s implies the speed has changed to v =
d = 1480.28 m/s. t
Since the 4.0 m/s increase in velocity is due to an increase of 1°C, an increase of 0.28 m/s occurs due to a temperature increase of ⎛ 1°C ⎞ ⎜ ⎟ ( 0.28 m/s ) = 0.07°C ⎝ 4.0 m/s ⎠
Thus, a temperature increase of approximately 0.07°C can be detected by the researchers.
20.80. Solve: The wave speeds along the two metal wires are v1 =
T
μ1
=
2250 N = 500 m/s 0.009 kg/m
T
v2 =
μ2
=
2250 N = 300 m/s 0.025 kg/m
The wavelengths along the two wires are
λ1 =
v1 500 m/s 1 = = m f 1500 Hz 3
λ2 =
v2 300 m/s 1 = = m f 1500 Hz 5
Thus, the number of wavelengths over two sections of the wire are 1.0 m
λ1
=
1.0 m
( 13 m )
=3
1.0 m
λ2
=
1.0 m
( 15 m )
The number of complete cycles of the wave in the 2.00-m-long wire is 8.
=5
20.81. Model: The wave pulse is a traveling wave on a stretched wire. Visualize:
Solve: (a) At a distance y above the lower end of the rope, the point P is in static equilibrium. The upward tension in the rope must balance the weight of the rope that hangs below this point. Thus, at this point
T = w = Mg = ( μ y ) g where μ = m/L is the linear density of the entire rope. Using Equation 20.2, we get
v=
T
μ
=
μ yg = gy μ
(b) The time to travel a distance dy at y, where the wave speed is v = gy , is
dt =
dy dy = v gy
Finding the time for a pulse to travel the length of the rope requires integrating from one end of the rope to the other: T L L dy 1 2 L Δt = ∫ dt = ∫ = L ⇒ Δt = 2 2 y = 0 g gy g g 0 0
(
)
20.82. Visualize:
Solve: as
(a) Using the graph, the refractive index n as a function of distance x can be mathematically expressed n = n1 +
n2 − n1 x L
At position x, the light speed is v = c / n. The time for the light to travel a distance dx at x is
dt =
dx n 1⎛ n −n ⎞ = dx = ⎜ n1 + 2 1 x ⎟ dx v c c⎝ L ⎠
To find the total time for the light to cover a thickness L of a glass we integrate as follows: T
T = ∫ dt = 0
2 n n 1 ⎛ n2 − n1 ⎞ (n − n ) ⎛n −n ⎞L ⎛n +n ⎞ x ⎟ dx = 1 ∫ dx + 2 1 ∫ x dx = 1 L + ⎜ 2 1 ⎟ = ⎜ 1 2 ⎟ L ⎜ n1 + ∫ c 0 cL 0 c c0⎝ L ⎝ cL ⎠ 2 ⎝ 2c ⎠ ⎠ L
L
(b) Substituting the given values into this equation, (1.50 + 1.60 ) × 0.010 m = 5.17 × 10−11 s T= 2 ( 3.0 × 108 m/s )
L
21.1. Model: The principle of superposition comes into play whenever the waves overlap. Visualize:
The graph at t = 1.0 s differs from the graph at t = 0.0 s in that the left wave has moved to the right by 1.0 m and the right wave has moved to the left by 1.0 m. This is because the distance covered by the wave pulse in 1.0 s is 1.0 m. The snapshot graphs at t = 2.0 s, 3.0 s, and 4.0 s are a superposition of the left and the right moving waves. The overlapping parts of the two waves are shown by the dotted lines.
21.2. Model: The principle of superposition comes into play whenever the waves overlap. Visualize:
The snapshot graph at t = 1.0 s differs from the graph t = 0.0 s in that the left wave has moved to the right by 1.0 m and the right wave has moved to the left by 1.0 m. This is because the distance covered by each wave in 1.0 s is 1.0 m. The snapshot graphs at t = 2.0 s, 3.0 s, and 4.0 s are a superposition of the left and the right moving waves. The overlapping parts of the two waves are shown by the dotted lines.
21.3. Model: The principle of superposition comes into play whenever the waves overlap. Visualize:
At t = 4.0 s the shorter pulses overlap and cancel. At t = 6.0 s the longer pulses overlap and cancel.
21.4. Model: The principle of superposition comes into play whenever the waves overlap. Solve: 4 s.
(b)
(a) As graphically illustrated in the figure below, the snapshot graph of Figure EX21.5b was taken at t =
21.5. Model: A wave pulse reflected from the string-wall boundary is inverted and its amplitude is unchanged. Visualize:
The graph at t = 2 s differs from the graph at t = 0 s in that both waves have moved to the right by 2 m. This is because the distance covered by the wave pulse in 2 s is 2 m. The shorter pulse wave encounters the boundary wall at 2.0 s and is inverted on reflection. This reflected pulse wave overlaps with the broader pulse wave, as shown in the snapshot graph at t = 4 s. At t = 6 s, only half of the broad pulse is reflected and hence inverted; the shorter pulse wave continues to move to the left with a speed of 1 m/s. Finally, at t = 8 s both the reflected pulse waves are inverted and they are both moving to the left.
21.6. Model: Reflections at both ends of the string cause the formation of a standing wave. Solve: Figure EX21.6 indicates 5/2 wavelengths on the 2.0-m-long string. Thus, the wavelength of the standing wave is λ = 52 ( 2.0 m ) = 0.80 m . The frequency of the standing wave is f =
v
λ
=
40 m/s = 50 Hz 0.80 m
21.7. Model: Reflections at the string boundaries cause a standing wave on the string. Solve: Figure EX21.7 indicates two full wavelengths on the string. Hence λ = 12 (60 cm) = 30 cm = 0.30 m. Thus
v = λ f = ( 0.30 m )(100 Hz ) = 30 m/s
21.8. Model: Reflections at the string boundaries cause a standing wave on the string. Solve: (a) When the frequency is doubled ( f ′ = 2 f 0 ) , the wavelength is halved ( λ ′ = 12 λ0 ) . This halving of the wavelength will increase the number of antinodes to six. (b) Increasing the tension by a factor of 4 means
v=
T
μ
⇒ v′ =
T′
μ
=
4T
μ
= 2v
For the string to continue to oscillate as a standing wave with three antinodes means λ ′ = λ0 . Hence, v′ = 2v ⇒ f ′λ ′ = 2 f 0λ0 ⇒ f ′λ0 = 2 f 0λ0 ⇒ f ′ = 2 f 0 That is, the new frequency is twice the original frequency.
21.9. Model: A string fixed at both ends supports standing waves.
Solve: (a) We have fa = 24 Hz = mf1 where f1 is the fundamental frequency that corresponds to m = 1. The next successive frequency is fb = 36 Hz = (m + 1) f1. Thus,
24 Hz f b ( m + 1) f1 m + 1 36 Hz = 12 Hz = = = = 1.5 ⇒ m + 1 = 1.5m ⇒ m = 2 ⇒ f1 = fa mf1 m 24 Hz 2 The wave speed is v = λ1 f1 =
2L f1 = ( 2.0 m )(12 Hz ) = 24 m/s 1
(b) The frequency of the third harmonic is 36 Hz. For m = 3, the wavelength is
λm =
2 L 2 (1 m ) 2 = = m 3 3 m
21.10. Model: A string fixed at both ends supports standing waves. Solve:
(a) A standing wave can exist on the string only if its wavelength is
λm =
2L m
m = 1, 2, 3, …
The three longest wavelengths for standing waves will therefore correspond to m = 1, 2, and 3. Thus,
λ1 =
2 ( 2.40 m ) = 4.80 m 1
λ2 =
2 ( 2.40 m ) = 2.40 m 2
λ3 =
2 ( 2.40 m ) = 1.60 m 3
(b) Because the wave speed on the string is unchanged from one m value to the other,
f 2λ2 = f3λ3 ⇒ f 3 =
f 2λ2
λ3
=
( 50 Hz )( 2.40 m ) = 75 Hz 1.60 m
21.11. Model: A string fixed at both ends forms standing waves. Solve:
(a) The wavelength of the third harmonic is calculated as follows:
λm =
2L 2 L 2.42 m ⇒ λ3 = = = 0.807 m ≈ 0.81 m m 3 3
(b) The speed of waves on the string is v = λ3 f3 = (0.807 m)(180 Hz) = 145.3 m/s. The speed is also given by
v = TS / μ , so the tension is TS = μ v 2 =
m 2 0.004 kg 2 v = (145.3 m/s ) = 69.7 N ≈ 70 m L 1.21 m
21.12. Model: For the stretched wire vibrating at its fundamental frequency, the wavelength of the standing wave is λ1 = 2 L. Visualize:
Solve:
The wave speed on the steel wire is vwire = f λ = f ( 2 L ) = ( 80 Hz )( 2 × 0.90 m ) = 144 m/s
and is also equal to
TS μ , where
μ=
mass 5.0 × 10−3 kg = = 5.555 × 10−3 kg/m length 0.90 m
The tension TS in the wire equals the weight of the sculpture or Mg. Thus, 2 2 5.555 × 10−3 kg/m ) (144 m/s ) ( Mg μ vwire vwire = ⇒M = = = 12 kg μ g 9.8 m/s 2
21.13. Model: The laser light forms a standing wave inside the cavity. Solve:
The wavelength of the laser beam is
λm =
2 ( 0.5300 m ) 2L ⇒ λ100,000 = = 10.60 μ m 100,000 m
The frequency is
f100,000 =
c
λ100,000
=
3.000 × 108 m/s = 2.830 × 1013 Hz 10.60 × 10−6 m
21.14. Solve: (a) For the open-open tube, the two open ends exhibit antinodes of a standing wave. The possible wavelengths for this case are 2L λm = m = 1, 2, 3, … m The three longest wavelengths are λ1 =
2 (1.21 m ) = 2.42 m 1
λ2 =
2 (1.21 m ) = 1.21 m 2
λ3 =
2 (1.21 m ) = 0.807 m 3
λ3 =
4 (1.21 m ) = 0.968 m 5
(b) In the case of an open-closed tube,
λm =
4L m
m = 1, 3, 5, …
The three longest wavelengths are
λ1 =
4 (1.21 m ) = 4.84 m 1
λ2 =
4 (1.21 m ) = 1.61 m 3
21.15. Model: We have an open-open tube that forms standing sound waves. Solve: The gas molecules at the ends of the tube exhibit maximum displacement, making antinodes at the ends. There is another antinode in the middle of the tube. Thus, this is the m = 2 mode and the wavelength of the standing wave is equal to the length of the tube, that is, λ = 0.80 m. Since the frequency f = 500 Hz, the speed of = sound in this case is v = fλ (500 Hz)(0.80 m) = 400 m/s. Assess: The experiment yields a reasonable value for the speed of sound.
21.16. Solve: For the open-open tube, the fundamental frequency of the standing wave is f1 = 1500 Hz when the tube is filled with helium gas at 0°C. Using λm = 2 L m , f1 helium =
vhelium
λ1
=
970 m/s 2L
Similarly, when the tube is filled with air, f1 air =
Assess:
vair
λ1
=
331 m/s f 331 m/s ⎛ 331 m/s ⎞ ⇒ 1 air = ⇒ f1 air = ⎜ ⎟ (1500 Hz ) = 512 Hz 2L f1 helium 970 m/s ⎝ 970 m/s ⎠
Note that the length of the tube is one-half the wavelength whether the tube is filled with helium or air.
21.17. Model: An organ pipe has a “sounding” hole where compressed air is blown across the edge of the pipe. This is one end of an open-open tube with the other end at the true “end” of the pipe. Solve: For an open-open tube, the fundamental frequency is f1 = 16.4 Hz. We have
λ1 =
⎞ 1 ⎛ 343 m/s ⎞ 2L λ 1⎛v ⇒ L = 1 = ⎜ sound ⎟ = ⎜ ⎟ = 10.5 m 1 2 2 ⎝ f1 ⎠ 2 ⎝ 16.4 Hz ⎠
Assess: The length of the organ pipe is ≈ 34.5 feet. That is actually somewhat of an overestimate since the antinodes of real tubes are slightly outside the tube. The actual length in a real organ is about 32 feet, and this is the tallest pipe in the so called “32 foot rank” of pipes.
21.18. Model: Reflections at the string boundaries cause a standing wave on a stretched string. Solve: Because the vibrating section of the string is 1.9 m long, the two ends of this vibrating wire are fixed, and the string is vibrating in the fundamental harmonic. The wavelength is
λm =
2L ⇒ λ1 = 2 L = 2 (1.90 m ) = 3.80 m m
The wave speed along the string is v = f1λ1 = (27.5 Hz)(3.80 m) = 104.5 m/s. The tension in the wire can be found as follows: ⎛ mass ⎞ 2 ⎛ 0.400 kg ⎞ T 2 v = S ⇒ TS = μ v 2 = ⎜ ⎟v = ⎜ ⎟ (104.5 m/s ) = 2180 N μ length 2.00 m ⎝ ⎠ ⎝ ⎠
21.19. Model: A string fixed at both ends forms standing waves.
Solve: A simple string sounds the fundamental frequency f1 = v/2L. Initially, when the string is of length LA = 30 cm, the note has the frequency f1A = v/2LA. For a different length, f1B = v/2LB. Taking the ratio of each side of these two equations gives
f1A v / 2 LA LB f = = ⇒ LB = 1A LA f1B v / 2 LB LA f1B We know that the second frequency is desired to be f1B = 523 Hz. The string length must be LB =
440 Hz ( 30 cm ) = 25.2 cm 523 Hz
The question is not how long the string must be, but where must the violinist place his finger. The full string is 30 cm long, so the violinist must place his finger 4.8 cm from the end. Assess: A fingering distance of 4.8 cm from the end is reasonable.
21.20. Model: Interference occurs according to the difference between the phases ( Δφ ) of the two waves. Solve: (a) A separation of 20 cm between the speakers leads to maximum intensity on the x-axis, but a separation of 60 cm leads to zero intensity. That is, the waves are in phase when (Δ x)1 = 20 cm but out of phase
when (Δ x) 2 = 60 cm. Thus,
( Δx)2 − ( Δ x)1 =
λ 2
⇒ λ = 2 ( 60 cm − 20 cm ) = 80 cm
(b) If the distance between the speakers continues to increase, the intensity will again be a maximum when the separation between the speakers that produced a maximum has increased by one wavelength. That is, when the separation between the speakers is 20 cm + 80 cm = 100 cm.
21.21. Model: The interference of two waves depends on the difference between the phases (Δφ ) of the two waves. Solve: (a) Because the speakers are in phase, Δφ0 = 0 rad. Let d represent the path-length difference. Using m = 0 for the smallest d and the condition for destructive interference, we get
Δφ = 2π ⇒ 2π
Δx
λ
Δx
λ
+ Δφ0 = 2 ( m + 12 )π rad
+ Δφ0 = π rad ⇒ 2π
d
λ
+ 0 rad = π rad ⇒ d =
m = 0, 1, 2, 3 …
λ 2
1 ⎛ v ⎞ 1 ⎛ 343 m/s ⎞ = ⎜ ⎟= ⎜ ⎟ = 0.25 m 2 ⎝ f ⎠ 2 ⎝ 686 Hz ⎠
(b) When the speakers are out of phase, Δφ0 = π . Using m = 1 for the smallest d and the condition for constructive interference, we get Δx m = 0, 1, 2, 3, … Δφ = 2π + Δφ0 = 2mπ
λ
λ 1 ⎛ v ⎞ 1 ⎛ 343 m/s ⎞ d ⇒ 2π + π = 2π ⇒ d = = ⎜ ⎟ = ⎜ ⎟ = 0.25 m λ 2 2 ⎝ f ⎠ 2 ⎝ 686 Hz ⎠
21.22. Model: We assume that the speakers are identical and that they are emitting in phase. Solve: Since you don’t hear anything, the separation between the two speakers corresponds to the condition of destructive interference. With Δφ0 = 0 rad, Equation 21.23 becomes
2π
d
λ
= 2 ( m + 12 )π rad ⇒ d = ( m + 12 ) λ ⇒ d =
Since the wavelength is
λ=
v 340 m/s = = 2.0 m f 170 Hz
three possible values for d are 1.0 m, 3.0 m, and 5.0 m.
λ 3λ 5λ 2
,
2
,
2
21.23. Model: Reflection is maximized if the two reflected waves interfere constructively. Solve: The film thickness that causes constructive interference at wavelength λ is given by Equation 21.32: λC =
−9 λ m ( 600 × 10 m ) (1) 2nd ⇒d = C = = 216 nm m 2n ( 2 )(1.39 )
where we have used m = 1 to calculate the thinnest film. Assess: The film thickness is much less than the wavelength of visible light. The above formula is applicable because nair < nfilm < nglass.
21.24. Model: Reflection is maximized if the two reflected waves interfere constructively. Solve: The film thickness that causes constructive interference at wavelength λ is given by Equation 21.32: λC =
−9 λ m ( 500 × 10 m ) (1) 2nd ⇒d = C = = 200 nm m 2n ( 2 )(1.25)
where we have used m = 1 to calculate the thinnest film. Assess: The film thickness is much less than the wavelength of visible light. The above formula is applicable because nair < noil < nwater.
21.25. Solve: (a) The circular wave fronts emitted by the two sources show that the two sources are in phase. This is because the wave fronts of each source have moved the same distance from their sources. (b) Let us label the top source as 1 and the bottom source as 2. Since the sources are in phase, Δφ0 = 0 rad. For
the point P, r1 = 3λ and r2 = 4λ. Thus, Δr = r2 − r1 = 4λ − 3λ = λ. The phase difference is Δφ =
2πΔr
λ
2π ( λ )
=
λ
= 2π
This corresponds to constructive interference. For the point Q, r1 = 72 λ and r2 = 2λ. The phase difference is
Δφ =
2πΔr
λ
=
2π ( 32 λ )
λ
= 3π
This corresponds to destructive interference. For the point R, r1 = 52 λ and r2 = 72 λ . The phase difference is Δφ =
2π ( λ )
λ
= 2π
This corresponds to constructive interference. P Q R
r1 3λ
λ 5 λ 2
7 2
r2 4λ 2λ 7 2
λ
Δr
λ λ
C/D C D
λ
C
3 2
21.26. Solve: (a) The circular wave fronts emitted by the two sources indicate the sources are out of phase. This is because the wave fronts of each source have not moved the same distance from their sources. (b) Let us label the top source as 1 and the bottom source as 2. The phase difference between the sources is Δφ0 = π . For the point P, r1 = 2λ and r2 = 3λ. The phase difference is Δφ =
2πΔr
λ
+ Δφ0 =
2π ( 3λ − 2λ )
λ
+ π = 3π
This corresponds to destructive interference. For the point Q, r1 = 3λ and r2 = 32 λ. The phase difference is
Δφ =
2π ( 32 λ )
λ
+ π = 4π
This corresponds to constructive interference. For the point R, r1 = 52 λ and r2 = 3λ. The phase difference is Δφ =
2π ( 12 λ )
λ
+ π = 2π
This corresponds to constructive interference.
Assess:
P Q
r1 2λ 3λ
r2 3λ
R
5 2
λ
3λ
3 2
λ
Δr
λ λ 1 λ 2 3 2
Note that it is not r1 or r2 that matter, but the difference Δr between them.
C/D D C C
21.27. Model: The two speakers are identical, and so they are emitting circular waves in phase. The overlap of these waves causes interference. Visualize:
Solve:
From the geometry of the figure,
r2 = r12 + ( 2.0 m ) = 2
( 4.0 m )
2
+ ( 2.0 m ) = 4.472 m 2
So, Δr = r2 − r1 = 4.472 m − 4.0 m = 0.472 m. The phase difference between the sources is Δφ0 = 0 rad and the wavelength of the sound waves is v 340 m/s λ= = = 0.1889 m f 1800 Hz
Thus, the phase difference of the waves at the point 4.0 m in front of one source is 2π ( 0.472 m ) + 0 rad = 5π rad = 2.5(2π rad) λ 0.1889 m This is a half-integer multiple of 2π rad, so the interference is perfect destructive. Δφ = 2π
Δr
+ Δφ0 =
21.28. Model: The two radio antennas are emitting out-of-phase, circular waves. The overlap of these waves causes interference. Visualize:
Solve:
From the geometry of the figure, r1 = 800 m and r2 =
( 800 m )
2
+ ( 600 m ) = 1000 m 2
So, Δr = r2 − r1 = 200 m and Δφ0 = π rad. The wavelength of the waves is
λ=
c 3.0 × 108 m/s = = 100 m f 3.0 × 106 Hz
Thus, the phase difference of the waves at the point (300 m, 800 m) is Δφ = 2π
Δr
λ
+ Δφ0 =
2π ( 200 m ) + π rad = 5π rad = 2.5(2π rad) 100 m
This is a half-integer multiple of 2π rad, so the interference is perfect destructive.
21.29. Solve: The beat frequency is f beat = f1 − f 2 ⇒ 3 Hz = f1 − 200 Hz ⇒ f1 = 203 Hz f1 is larger than f2 because the increased tension increases the wave speed and hence the frequency.
21.30. Solve: The flute player’s initial frequency is either 523 Hz + 4 Hz = 527 Hz or 523 Hz − 4 Hz = 519 Hz. Since she matches the tuning fork’s frequency by lengthening her flute, she is increasing the wavelength of the standing wave in the flute. A wavelength increase means a decrease of frequency because v = fλ. Thus, her initial frequency was 527 Hz.
21.31. Model: The superposition of two slightly different frequencies creates beats. Solve: Let λ1 = 780.54510 nm and λ2 > λ1 . This means f 2 < f1 and Δf = f1 − f 2 = ⇒ λ2 =
c
λ1
−
c
λ2
= 98.5 × 106 Hz
1 1 = (1/ λ1 ) − (Δf / c) 1/(780.54510 × 10−9 m) − (98.5 × 106 Hz) /(3.00 × 108 m/s)
= 780.54530 nm Assess:
A small difference in wavelengths, ( λ2 − λ1 ) = 0.00020 nm = 0.20 pm, can yield beats at a relatively
high frequency of 98.5 MHz.
21.32. Model: The principle of superposition applies to overlapping waves. Visualize:
Solve: Because the wave pulses travel along the string at a speed of 100 m/s, they move a distance of d = vt = (100 m/s)(0.05 s) = 5 m in 0.050 s. The front of the wave pulse moving left, which is located at x = 1 m at t = 0.050 s, was thus located at x = 6 m at t = 0 s. This helps us draw the snapshot of the wave pulse moving left at t = 0 s (shown as a dashed line). Subtracting this wave snapshot from the resultant at t = 0 s (shown as a solid line) yields the right-traveling wave’s snapshot at t = 0 s (shown as a dotted line). Finally, the snapshot graph of the wave pulse moving right at t = 0.050 s is the same as at t = 0 s (shown as a dotted line) except that it is shifted to the right by 5 m.
21.33. Model: The wavelength of the standing wave on a string vibrating at its second-harmonic frequency is equal to the string’s length. Visualize:
Solve:
The length of the string L = 2.0 m, so λ = L = 2.0 m. This means the wave number is k=
2π
λ
=
2π = π rad m 2.0 m
According to Equation 21.5, the displacement of a medium when two sinusoidal waves superpose to give a standing wave is D ( x, t ) = A ( x ) cos ω t , where A ( x ) = 2a sin kx = Amax sin kx. The amplitude function gives the
amplitude of oscillation from point to point in the medium. For x = 10 cm, A ( x = 10 cm ) = ( 2.0 cm ) sin ⎡⎣(π rad m )( 0.10 m ) ⎤⎦ = 0.62 cm Similarly,
A( x = 20 cm) = 1.18 cm,
A ( x = 30 cm ) = 1.62 cm,
A ( x = 40 cm ) = 1.90 cm,
A ( x = 50 cm ) = 2.00 cm. Assess:
Consistent with the above figure, the amplitude of oscillation is a maximum at x = 0.50 m.
and
21.34. Model: The wavelength of the standing wave on a string is λm = 2 L m , where m = 1, 2, 3, … We assume that 30 cm is the first place from the left end of the string where A = Amax 2. Visualize:
Solve:
The amplitude of oscillation on the string is A ( x ) = Amax sin kx . Since the string is vibrating in the third
harmonic, the wave number is k=
2π
λ
=
2π
( 2 L 3)
=3
π L
Substituting into the equation for the amplitude, 1 3π π ⎛ 3π ⎞ ⎛ 3π ⎞ 1 Amax = Amax sin ⎜ ( 0.30 m ) ⎟ ⇒ sin ⎜ ( 0.30 m ) ⎟ = ⇒ ( 0.30 m ) = rad ⇒ L = 5.4 m 2 L L 2 L 6 ⎝ ⎠ ⎝ ⎠
21.35. Model: The wavelength of the standing wave on a string vibrating at its fundamental frequency is equal to 2L. Solve: The amplitude of oscillation on the string is A ( x ) = 2a sin kx, where a is the amplitude of the traveling wave and the wave number is
k=
2π
λ
=
2π π = 2L L
Substituting into the above equation,
⎡⎛ π ⎞⎛ L ⎞ ⎤ ⎛ 1 ⎞ A ( x = 14 L ) = 2.0 cm = 2a sin ⎢⎜ ⎟⎜ ⎟ ⎥ ⇒ 1.0 cm = a ⎜ ⎟ ⇒ a = 2 cm = 1.4 cm ⎝ 2⎠ ⎣⎝ L ⎠⎝ 4 ⎠ ⎦
21.36. Solve: You can see in Figure 21.4 that the time between two successive instants when the antinodes are at maximum height is half the period, or 12 T . Thus T = 2(0.25 s) = 0.50 s, and so
f =
v 3.0 m/s 1 1 = = 2.0 Hz ⇒ λ = = = 1.5 m T 0.50 s f 2.0 Hz
21.37. Model: The wave on a stretched string with both ends fixed is a standing wave. For vibration at its fundamental frequency, λ = 2L.
Solve: The wavelength of the wave reaching your ear is 39.1 cm = 0.391 m. So the frequency of the sound wave is
f =
vair
λ
=
344 m/s = 879.8 Hz 0.391 m
This is also the frequency emitted by the wave on the string. Thus,
879.8 Hz =
vstring
λ
=
1 TS
λ
μ
=
1
150 N
λ 0.0006 kg/m
⇒ L = 12 λ = 0.284 m = 28.4 cm
⇒ λ = 0.568 m
21.38. Model: The wave on a stretched string with both ends fixed is a standing wave. Solve: We must distinguish between the sound wave in the air and the wave on the string. The listener hears a sound wave of wavelength λsound = 40 cm = 0.40 m. Thus, the frequency is
f =
vsound
λsound
=
343 m/s = 857.5 Hz 0.40 m
The violin string oscillates at the same frequency, because each oscillation of the string causes one oscillation of the air. But the wavelength of the standing wave on the string is very different because the wave speed on the string is not the same as the wave speed in air. Bowing a string produces sound at the string’s fundamental frequency, so the wavelength of the string is
λstring = λ1 = 2 L = 0.60 m ⇒ vstring = λstring f = ( 0.60 m )( 857.5 Hz ) = 514.5 m/s The tension is the string is found as follows: vstring =
TS
μ
⇒ TS = μ ( vstring ) = ( 0.001 kg/m )( 514.5 m/s ) = 260 N 2
2
21.39. Model: A string fixed at both ends forms standing waves.
Solve: (a) Three antinodes means the string is vibrating as the m = 3 standing wave. The frequency is f3 = 3f1, so the fundamental frequency is f1 = 13 (420 Hz) = 140 Hz. The fifth harmonic will have the frequency f5 = 5f1 = 700 Hz. (b) The wavelength of the fundamental mode is λ1 = 2L = 1.20 m. The wave speed on the string is v = λ1 f1 = (1.20 m) (140 Hz) = 168 m/s. Alternatively, the wavelength of the n = 3 mode is λ3 =
1 3
(2L) = 0.40 m, from
which v = λ3 f3 = (0.40 m)(420 Hz) = 168 m/s. The wave speed on the string is given by
v=
TS
μ
⇒ TS = μ v 2 = ( 0.0020 kg/m )(168 m/s ) = 56 N 2
Assess: You must remember to use the linear density in SI units of kg/m. Also, the speed is the same for all modes, but you must use a matching λ and f to calculate the speed.
21.40. Model: Assume that the extra kilogram doesn’t stretch the wire longer (so L stays the same) nor thinner (so μ stays the same). Also assume that because the wire is thin its own weight is negligible, so Ts is constant throughout the wire and is equal to Mg. Visualize: The wire is fixed at both ends so in the second harmonic L = λ . We are given f 2 = 200 Hz and
f 2′ = 245 Hz and M ′ = M + 1.0 kg. Apply v = λ f and v = Ts /μ . Solve:
Cancel off λ ,
μ , and
g in turn.
T ′/μ f 2′ v′/λ = = s = f 2 v/λ Ts /μ
M ′g Mg
=
( M + 1.0 kg) g Mg
=
M + 1.0 kg M
2
⎛ f 2′ ⎞ M + 1.0 kg ⎜ ⎟ = M ⎝ f2 ⎠
⎛ ⎛ f ′ ⎞2 ⎞ M ⎜ ⎜ 2 ⎟ − 1⎟ = 1.0 kg ⎜ ⎝ f2 ⎠ ⎟ ⎝ ⎠ M=
1.0 kg 2
⎛ f 2′ ⎞ ⎜ ⎟ −1 ⎝ f2 ⎠
=
1.0 kg 2
⎛ 245 Hz ⎞ ⎜ ⎟ −1 ⎝ 200 Hz ⎠
= 2.0 kg
Assess: We did not expect M to be really huge or a) it would have broken the wire, and b) adding one more kilogram wouldn’t have made as big a difference in f 2 as it did.
21.41. Model: The stretched string with both ends fixed forms standing waves. Visualize:
Solve: The astronauts have created a stretched string whose vibrating length is L = 2.0 m. The weight of the hanging mass creates a tension TS = Mg in the string, where M = 1.0 kg. As a consequence, the wave speed on the string is
v=
TS
μ
=
Mg
μ
where μ = (0.0050 kg)/(2.5 m) = 0.0020 kg/m is the linear density. The astronauts then observe standing waves at frequencies of 64 Hz and 80 Hz. The first is not the fundamental frequency of the string because 80 Hz ≠ 2 × 64 Hz. But we can easily show that both are multiples of 16 Hz: 64 Hz = 4 f1 and 80 Hz = 5 f1 . Both frequencies are also multiples of 8 Hz. But 8 Hz cannot be the fundamental frequency because, if it were, there would be a standing wave resonance at 9(8 Hz) = 72 Hz. So the fundamental frequency is f1 = 16 Hz. The fundamental wavelength is λ1 = 2L = 4.0 m. Thus, the wave speed on the string is v = λ1 f1 = 64.0 m/s . Now we can find g on Planet X:
v=
Mg
μ
⇒g=
μ M
v2 =
0.0020 kg/m 2 ( 64 m/s ) = 8.2 m/s 2 1.0 kg
21.42. Model: The stretched bungee cord that forms a standing wave with two antinodes is vibrating at the second harmonic frequency. Visualize:
Solve:
Because the vibrating cord has two antinodes, λ2 = L = 1.80 m. The wave speed on the cord is vcord = f λ = ( 20 Hz )(1.80 m ) = 36 m/s
The tension TS in the cord is equal to kΔL, where k is the bungee’s spring constant and ΔL is the 0.60 m the bungee has been stretched. Thus, T k ΔL vcord = S =
μ
μ
21.43. Solve: (a) Because the frequency of the standing wave on the copper wire is the same as the frequency on the aluminum wire, v v f Cu = f Al ⇒ Cu = Al λCu
λCu
Let nCu be the number of half-wavelength antinodes on the copper wire and nAl be the number of half-wavelength antinodes on the aluminum wire. Thus,
0.44 m ⎛λ ⎞ nCu ⎜ Cu ⎟ = 0.22 m ⇒ λCu = nCu ⎝ 2 ⎠
1.20 m ⎛λ ⎞ nAl ⎜ Al ⎟ = 0.60 m ⇒ λAl = nAl ⎝ 2 ⎠
nCu ⎛ vAl ⎞⎛ 0.44 m ⎞ = ⎜ ⎟ nAl ⎜⎝ vCu ⎟⎠ ⎝ 1.20 m ⎠ We can find vCu and vAl by using the following equations for a stretched wire: ⇒
vCu
=
vAl
( 0.44 m nCu ) (1.20 m vCu =
T
μ Cu
nAl )
⇒
vAl =
T
μ Al
The linear densities are calculated as follows:
μ Cu =
ρ V mCu ⎛ ρ Cu ⎞ 2 = Cu Cu = ⎜ ⎟π r ( 0.22 m ) 0.22 m 0.22 m ⎝ 0.22 m ⎠
= ( 8920 kg/m3 )π ( 5.0 × 10−4 m ) = 7.006 × 10−3 kg/m3 2
μ Al =
ρ V mAl ⎛ ρ Al ⎞ 2 = Al Al = ⎜ ⎟π r ( 0.60 m ) 0.60 m 0.60 m ⎝ 0.60 m ⎠
= ( 2700 kg/m3 )π ( 5.0 × 10−4 m ) = 2.121 × 10−3 kg/m3 2
⇒ vCu =
20 N 20 N = 53.43 m/s vAl = = 97.12 m/s 7.006 × 10−3 kg/m3 2.121 × 10−3 kg/m3
Going back to the nCu nAl equation, we have nCu ⎛ 97.12 m/s ⎞⎛ 0.44 m ⎞ 2 =⎜ ⎟⎜ ⎟ = 0.666 = ⇒ nCu = 2 and nAl = 3 nAl ⎝ 53.43 m/s ⎠⎝ 1.20 m ⎠ 3
Substituting into the expressions for wavelength and frequency,
λCu =
v 0.44 m 0.44 m 53.43 m/s = = 0.22 m ⇒ f Cu = Cu = = 243 Hz ≈ 240 Hz nCu 2 0.22 m λCu
λAl =
v 1.20 m 1.20 m 97.12 m/s = = 0.40 m ⇒ f Al = Al = = 243 Hz ≈ 240 Hz nAl 3 0.40 m λAl
(b) At this frequency of 240 Hz, there are 3 antinodes on the aluminum wire.
21.44. Visualize: Use primed quantities for when the sphere is submerged. We are given f5′ = f3 and M = 1.5 kg . We also know the density of water is ρ = 1000 kg/m3 . In the third mode before the sphere is
submerged L = 32 λ ⇒ λ = 32 L. Likewise, after the sphere is submerged L = 52 λ ′ ⇒ λ ′ = 52 L. The tension in the string before the sphere is submerged is Ts = Mg , but after the sphere is submerged, according to Archimedes’ principle, it is reduced by the weight of the water displaced by the sphere: Ts′ = Mg − ρVg , where V = 34 π R 3 . Solve:
We are looking for R so solve Ts′ = Mg − ρVg for ρVg and later we will isolate R from that.
ρVg = Mg − Ts′ Solve v′ = Ts /μ for Ts′. Also substitute for V.
⎛4⎞ ⎝ ⎠
ρ ⎜ ⎟ π R3 g = Mg − μ v′2 3 Now use v′ = λ ′ f ′. ⎛4⎞ ⎝ ⎠
ρ ⎜ ⎟ π R 3 g = Mg − μ (λ ′ f5′) 2 3 Recall that f 5′ = f3 and λ ′ = 52 L. ⎛ 4⎞ ⎝ ⎠
⎛
2
⎞
2
ρ ⎜ ⎟π R 3 g = Mg − μ ⎜⎜ Lf3 ⎟⎟ ⎜5 ⎟ 3 ⎝
⎠
⎛2 ⎝
v⎞
Substitute f 3 = v/λ . ⎛ 4⎞ ⎝ ⎠
2
ρ ⎜ ⎟π R 3 g = Mg − μ ⎜ L ⎟ 3 5 λ ⎠
Now use λ = 32 L and v = Ts /μ . ⎛2 T /μ ⎞ ⎛4⎞ ρ ⎜ ⎟ π R 3 g = Mg − μ ⎜ L 2 s ⎟ ⎜ ⎟ 3 5 ⎝ ⎠ 3L ⎠ ⎝
2
The 2’s, μ ’s, and L’s cancel. ⎛ 4⎞ ⎝ ⎠
⎛
⎛ 4⎞ ⎝ ⎠
⎛ 3⎞ ⎝ ⎠
⎞
3
2
ρ ⎜ ⎟π R 3 g = Mg − ⎜⎜ Ts ⎟⎟ ⎜5 ⎟ 3 ⎝
⎠
Ts = Mg . 2
ρ ⎜ ⎟π R 3 g = Mg − ⎜ ⎟ Mg 3 5 Cancel g and factor M out on the right side. ⎛
⎛ 4⎞ ⎝ ⎠
2 ⎛ 3⎞ ⎞ ⎝ ⎠ ⎠
⎛ ⎝
9 ⎞ ⎠
⎛ 16 ⎞ ⎝ ⎠
ρ ⎜ ⎟π R 3 = M ⎜1 − ⎜ ⎟ ⎟ = M ⎜ 1 − ⎟ = M ⎜ ⎟ ⎜ 3 5 ⎟ 25 25 ⎝
Now solve for R. ⎛ 3 ⎞⎛ 16 ⎞ M R 3 = ⎜ ⎟⎜ ⎟ ⎝ 4 ⎠⎝ 25 ⎠ πρ
R=
3
12 M 12 (1.5 kg) =3 = 6.1 cm 25 πρ 25 π (1000 kg/m3 )
Assess: The density of the sphere turns out to be about 1.5 × the density of water, which means it sinks and is in a reasonable range for densities.
21.45. Visualize: First compute the length L of the wire from the Pythagorean theorem. L = 2 2 m. Now μ = M/L = 0.075 kg/ 2 2 m = 0.02652 kg/m. Also, in the fundamental mode λ = 2 L; here λ = 4 2 m = 5.657 m. Solve: Apply the principles of statics to the point at the end of the bar. ΣFy = Ts sin 45° − (8 kg)( g ) = 0 N ⇒ Ts =
(8 kg)(9.8 m/s 2 ) = 110.87 N 2/ 2
Use these values for λ , μ , and Ts to find f. f = Assess:
v
λ
=
Ts /μ
λ
=
(110.87 N) / (0.02652 kg/m) = 11 Hz 5.657 m
This seems like a reasonable frequency for a mechanical system like this.
21.46. Model: Assume that while the spring provides the same tension to both strings it also acts as a fixed point for the end of each string so an integral number of half wavelengths fit in each string. Visualize: Use a subscript L for the left string and R for the right string. We are given μ L = 2.0 g/m. From the
assumption above we know (Ts ) L = (Ts ) R = Ts . We are also given f L = f R = f and LL = LR = L. Notice from the diagram that λL = L and λR = 23 L. Solve:
From v = λ f and v = Ts /μ eliminate v and solve for μ : μ = T / (λ f ) 2 .
1 9 μ R (Ts ) R / (λR f R ) 2 Ts / ( 23 Lf ) 2 = = = = μ L (Ts ) L / (λL f L ) 2 Ts / ( Lf ) 2 ( 32 ) 2 4 9 4
9 4
μ R = μ L = (2.0 g/m) = 4.5 g/m Assess:
We expect a slower wave speed in the right string to correspond to a larger mass density.
21.47. Model: The microwave forms a standing wave between the two reflectors. Solve: (a) There are reflectors at both ends, so the electromagnetic standing wave acts just like the standing wave on a string that is tied at both ends. The frequencies of the standing waves are
fm = m
vlight 2L
=m
3.0 × 108 m/s c =m = m (1.5 × 109 Hz ) = 1.5m GHz 2L 2 ( 0.10 m )
where we have noted that electromagnetic waves of all frequencies travel with the speed of light c. The generator can produce standing waves at any frequency between 10 GHz and 20 GHz. These are m 7 8 9 10 11 12 13
fm (GHz) 10.5 12.0 13.5 15.0 16.5 18.0 19.5
(b) There are 7 different standing wave frequencies. Even-numbered values of n create a node at the center, and odd-numbered values of n create an antinode at the center. So the frequencies where the midpoint is an antinode are 10.5, 13.5, 16.5, and 19.5 GHz.
21.48. Model: The fundamental wavelength of an open-open tube is 2L and that of an open-closed tube is 4L. Solve:
We are given that f1 open-closed = f3 open-open = 3f1 open-open
⇒
vair
λ1 open-closed
=3
⇒ Lopen-closed =
vair
λ1 open-open
2 Lopen-open 12
⇒ =
1 3 = 4 Lopen-closed 2 Lopen-open
2 ( 78.0 cm ) = 13.0 cm 12
21.49. Model: A tube forms standing waves. Solve: (a) The fundamental frequency cannot be 390 Hz because 520 Hz and 650 Hz are not integer multiples of it. But we note that the difference between 390 Hz and 520 Hz is 130 Hz as is the difference between 520 Hz and 650 Hz. We see that 390 Hz = 3 × 130 Hz = 3f1, 520 Hz = 4f1, and 650 Hz = 5f1. So we are seeing the third, fourth, and fifth harmonics of a tube whose fundamental frequency is 130 Hz. According to Equation 21.17, this is an open-open tube because fm = mf1 with m = 1, 2, 3, 4, … For an open-closed tube m has only odd values. (b) Knowing f1, we can now find the length of the tube:
L=
v 343 m/s = = 1.32 m 2 f1 2 (130 Hz )
(c) 520 Hz is the fourth harmonic. This is a sound wave, not a wave on a string, so the wave will have four nodes and will have antinodes at the ends, as shown.
(d) With carbon dioxide, the new fundamental frequency is
f1 =
v 280 m/s = = 106 Hz 2 L 2 (1.32 m )
Thus the frequencies of the n = 3, 4, and 5 modes are f3 = 3f1 = 318 Hz, f4 = 4f1 = 424 Hz, and f5 = 5f1 = 530 Hz.
21.50. Model: Particles of the medium at the nodes of a standing wave have zero displacement. Solve: The cork dust settles at the nodes of the sound wave where there is no motion of the air molecules. The separation between the centers of two adjacent piles is 12 λ . Thus, 123 cm λ = ⇒ λ = 82 cm 3 2
Because the piston is driven at a frequency of 400 Hz, the speed of the sound wave in oxygen is
v = f λ = ( 400 Hz )( 0.82 m ) = 328 m/s Assess:
A speed of 328 m/s in oxygen is close to the speed of sound in air, which is 343 m/s at 20°C.
21.51. Model: The nodes of a standing wave are spaced λ/2 apart. Visualize:
Solve: The wavelength of the mth mode of an open-open tube is λm = 2L/m. Or, equivalently, the length of the tube that generates the mth mode is L = m(λ/2). Here λ is the same for all modes because the frequency of the tuning fork is unchanged. Increasing the length of the tube to go from mode m to mode m + 1 requires a length change ∆L = (m + 1)(λ/2) – mλ/2 = λ/2 That is, lengthening the tube by λ/2 adds an additional antinode and creates the next standing wave. This is consistent with the idea that the nodes of a standing wave are spaced λ/2 apart. This tube is first increased ΔL = 56.7 cm − 42.5 cm = 14.2 cm, then by ∆L = 70.9 cm – 56.7 cm = 14.2 cm. Thus λ/2 = 14.2 cm and thus λ = 28.4 cm = 0.284 m. Therefore the frequency of the tuning fork is f =
v
λ
=
343 m/s = 1208 Hz ≈ 1210 Hz 0.284 m
21.52. Model: The open-closed tube forms standing waves. Visualize:
Solve: When the air column length L is the proper length for a 580 Hz standing wave, a standing wave resonance will be created and the sound will be loud. From Equation 21.18, the standing wave frequencies of an open-closed tube are fm = m(v/4L), where v is the speed of sound in air and m is an odd integer: m = 1, 3, 5, … The frequency is fixed at 580 Hz, but as the length L changes, 580 Hz standing waves will occur for different values of m. The length that causes the mth standing wave mode to be at 580 Hz is
L=
m ( 343 m/s ) ( 4 )( 580 Hz )
We can place the values of L, and corresponding values of h = 1 m − L, in a table: m 1 3 5 7
L 0.148 m 0.444 m 0.739 m 1.035 m
h=1m−L 0.852 m = 85.2 cm 0.556 m = 55.6 cm 0.261 m = 26.1 cm h can’t be negative
So water heights of 26 cm, 56 cm, and 85 cm will cause a standing wave resonance at 580 Hz. The figure shows the m = 3 standing wave at h = 56 cm.
21.53. Model: A stretched wire, which is fixed at both ends, forms a standing wave whose fundamental frequency f1 wire is the same as the fundamental frequency f1 open-closed of the open-closed tube. The two frequencies are the same because the oscillations in the wire drive oscillations of the air in the tube. Visualize:
Solve:
The fundamental frequency in the wire is
f1 wire =
vwire 1 TS 1 = = 2 Lwire 2 Lwire μ (1.0 m )
440 N = 469 Hz ( 0.0010 kg/0.050 m )
The fundamental frequency in the open-closed tube is
f1 open-closed = 469 Hz =
vair 340 m/s 340 m/s = ⇒ Ltube = = 0.181 m ≈ 18 cm 4 Ltube 4 Ltube 4 ( 469 Hz )
21.54. Model: A stretched wire, which is fixed at both ends, creates a standing wave whose fundamental frequency is f1 wire . The second vibrational mode of an open-closed tube is f 3 open-closed . These two frequencies are equal because the wire’s vibrations generate the sound wave in the open-closed tube. Visualize:
Solve:
The frequency in the tube is
f 3 open-closed =
3 ( 340 m/s ) 3vair = = 300 Hz 4 Ltube 4 ( 0.85 cm )
⇒ f1 wire = 300 Hz =
1 vwire TS = 2 Lwire 2 Lwire μ
⇒ TS = ( 300 Hz ) ( 2 Lwire ) μ = (300 Hz)2 (2 × 0.25 m)2(0.020 kg/m) = 450 N 2
2
21.55. Model: The standing waves in the tube will have a displacement antinode at the top where the gas molecules are free to move and a node at the water where they are not. Visualize: For the wire, we are given Ts = 400 N. In the fundamental mode with both ends of the wire fixed
Lwire = λwire / 2. Hence, λwire = 2Lwire = 2(50.0 cm) = 1.00 m. We also know μ wire = 0.00100 kg/0.500 m = 0.00200 kg/m. The given information is sufficient to compute the frequency. From v = λ f and v = Ts /μ eliminate v and solve for f .
f =
(Ts ) wire /μ wire
λwire
=
400 N/ (0.00200 kg/m) = 447.2 Hz 1.00 m
Solve: Now we turn our attention to the gas, realizing that the frequency of the wave in the wire will be the same as the frequency of the sound in the gas. There is initially a node of the standing sound wave at the water level in the tube. The water is then lowered until the next standing wave is achieved; this is the next time there is a node at the water level. The distance between adjacent nodes in a standing wave is 12 λ , so
Δh = 12 λgas = 30.5 cm ⇒ λgas = 61.0 cm. vgas = λgas f gas = (0.610 m)(447.2 Hz) = 273 m/s Assess: We were told the gas is more dense than air so it will stay in the tube; for more dense gases we expect a slower sound speed. Our answer bears this out, but is still in the range of the speed of sound for typical gases.
21.56. Model: In a rod in which a longitudinal standing wave can be created, the standing wave is equivalent to a sound standing wave in an open-open tube. Both ends of the rod are antinodes, and the rod is vibrating in the fundamental mode. Solve: Since the rod is in the fundamental mode, λ1 = 2L = 2(2.0 m) = 4.0 m. Using the speed of sound in aluminum, the frequency is v 6420 m/s = 1605 Hz ≈ 1600 Hz f1 = Al = 4.0 m λ1
21.57. Model: Model the tunnel as an open-closed tube. v (m = odd) to find L, but we need to 4L know m first. Since m takes on only odd values for the open-closed tube the next resonance after m is m + 2. We are given f m = 4.5 Hz and f m + 2 = 6.3 Hz. Solve: v f m + 2 (m + 2) 4 L m + 2 = = v fm m ( m) 4L Visualize:
We are given v = 335 m/s. We would like to use f m = m
⎛ f ⎞ m ⎜ m+2 ⎟ = m + 2 ⎝ fm ⎠ ⎛ f ⎞ m ⎜ m + 2 − 1⎟ = 2 ⎝ fm ⎠
m=
2 2 = =5 6.3 Hz fm+2 −1 −1 4.5 Hz fm
Now that we know m we can finish up.
fm = m Assess:
v v 335 m/s ⇒L=m = (5) = 93 m 4L 4 fm 4(4.5 Hz)
93 m seems like a reasonable length for a tunnel.
21.58. Model: A standing wave in an open-closed tube must have a node at the closed end of the tube and an antinode at the open end. Visualize:
Solve: We first draw a series of pictures showing all the possible standing waves. By examination, we see that the first standing wave mode is 14 of a wavelength, so the tube’s length is L = 14 λ . The next mode is 34 of a
wavelength. The tube’s length hasn’t changed, so in this mode L = 34 λ . The next mode is now slightly more than a wavelength. That is, L = 54 λ . The next mode is 74 of a wavelength, so L = 74 λ . We see that there is a pattern. The length of the tube and the possible standing wave wavelengths are related by mλ L= m = 1, 3, 5, 7, … = odd integers 4 Solving for λ, we find that the wavelengths and frequencies of standing waves in an open-closed tube are 4L ⎫ λm = ⎪⎪ m m = 1, 3, 5, 7, … = odd integers v v ⎬ =m ⎪ fm = λm 4 L ⎭⎪
21.59. Model: The amplitude is determined by the interference of the two waves.
Solve: For interference in one dimension, where the speakers are separated by a distance Δx, the amplitude of the net wave is A = 2a cos ( 12 Δφ ) , where a is the amplitude of each wave and Δφ = 2πΔx λ + Δφ0 is the phase
difference between the two waves. The speakers are emitting identical waves so they have identical phase constants and Δφ0 = 0 . Thus,
λ ⎛ πΔx ⎞ −1 ⎛ 1.5 ⎞ A = 1.5a = 2a cos ⎜ ⎟ ⇒ Δx = cos ⎜ ⎟ π ⎝ λ ⎠ ⎝ 2 ⎠ The wavelength of a 1000 Hz tone is λ = vsound f = 0.343 m. Thus the separation must be
Δx =
0.343 m
π
cos −1 ( 0.75 ) = 0.0789 m ≈ 7.9 cm
It is essential to note that the argument of the arccosine is in radians, not in degrees.
21.60. Model: Constructive or destructive interference occurs according to the phases of the two waves. Visualize:
Solve:
(a) To go from destructive to constructive interference requires moving the speaker Δx = 12 λ ,
equivalent to a phase change of π rad. Since Δx = 40 cm, we find λ = 80 cm. (b) Destructive interference at Δx = 10 cm requires 2π
Δx
λ
3π ⎛ 10 cm ⎞ rad + Δφ0 = 2π ⎜ ⎟ rad + Δφ0 = π rad ⇒ Δφ0 = 4 80 cm ⎝ ⎠
(c) When side by side, with Δx = 0, the phase difference is Δφ = Δφ0 = 3π/4 rad. The amplitude of the superposition of the two waves is 3π ⎛ Δφ ⎞ a = 2a cos ⎜ = 0.77a ⎟ = 2a cos 8 ⎝ 2 ⎠
21.61. Model: Interference occurs according to the difference between the phases of the two waves. Visualize:
Solve:
(a) The phase difference between the sound waves from the two speakers is Δφ = 2π
Δx
λ
+ Δφ0
We have a maximum intensity when Δx = 0.50 m and Δx = 0.90 m. This means
2π
( 0.50 m ) + Δφ λ
0
= 2mπ rad
⎛ 0.90 m ⎞ 2π ⎜ ⎟ + Δφ0 = 2 ( m + 1)π rad ⎝ λ ⎠
Taking the difference of the above two equations, vsound 340 m/s ⎛ 0.40 m ⎞ 2π ⎜ = = 850 Hz ⎟ = 2π ⇒ λ = 0.40 m ⇒ f = λ 0.40 m ⎝ λ ⎠
(b) Using again the equations that correspond to constructive interference,
π ⎛ 0.50 m ⎞ 2π ⎜ ⎟ + Δφ0 = 2mπ rad ⇒ Δφ0 = φ20 − φ10 = − rad 2 ⎝ 0.40 m ⎠ We have taken m = 1 in the last equation. This is because we always specify phase constants in the range –π rad to π rad (or 0 rad to 2π rad). m = 1 gives − 12 π rad (or equivalently, m = 2 will give 32 π rad).
21.62. Model: Constructive or destructive interference occurs according to the phases of the two waves. Solve:
The phase difference between the sound waves from the two speakers is
Δφ = 2π
Δx
λ
+ Δφ0
With no delay between the two signals, Δφ0 = 0 rad and
Δφ =
2π ( 2.0 m ) ⎛ 340 Hz ⎞ = 2π ( 2.0 m ) ⎜ ⎟ = 4π rad v f ⎝ 340 m/s ⎠
According to Equation 21.22, this corresponds to constructive interference. A delay of 1.47 ms corresponds to an inherent phase difference of ⎛ 2π ⎞ Δφ 0 = ⎜ ⎟ (1.47 ms ) rad = ⎝ T ⎠
( 2π f )(1.47 ms ) rad = 2π (1.47 ms )( 340 Hz ) rad = π
The phase difference Δφ between the signals is then ⎛ Δx ⎞ Δφ = 2π ⎜ ⎟ + Δφ0 = 4π rad + π rad = 5π rad ⎝ λ ⎠
Thus, the interference along the x-axis will be perfect destructive.
rad
21.63. Model: Reflection is maximized for constructive interference of the two reflected waves, but minimized for destructive interference. Solve: (a) Constructive interference of the reflected waves occurs for wavelengths given by Equation 21.32:
λm =
2nd 2 (1.42 )( 500 nm ) (1420 nm ) = = m m m
Thus, λ1 = 1420 nm, λ2 = 12 (1420 nm ) = 710 nm, λ3 = 473 nm, λ4 = 355 nm, … Only the wavelength of 473
nm is in the visible range. (b) For destructive interference of the reflected waves,
λ=
2 (1.42 )( 500 nm ) 1420 nm 2nd = = m − 12 m − 12 m − 12
Thus, λ1 = 2 × 1420 nm = 2840 nm, λ2 =
2 3
(1420 nm ) = 947 nm, λ3 = 568 nm, λ4 = 406 nm, …
The wavelengths
of 406 nm and 568 nm are in the visible range. (c) Beyond the limits 430 nm and 690 nm the eye’s sensitivity drops to about 1 percent of its maximum value. The reflected light is enhanced in blue (473 nm). The transmitted light at mostly 568 nm will be yellowish green.
21.64. Solve: (a) The intensity of reflected light from the uncoated glass is I 0 = ca 2 , where a is the amplitude of the reflected light. We will assume that the amplitude of the reflected light from both the bottom and the top of the coated film is a. The interference of the two reflected waves determines the amplitude of the resultant wave which is given by A = 2a cos ( Δφ 2 )
where
Δφ = 2π
Δx
λcoat
+ Δφ0
With Δφ0 = 0 rad, Δx = 2d , and λcoat = λair n , we have Δφ =
⇒ A = 2a cos
2π ( 2d ) 4π dn 4π ( 92 nm )(1.39 ) 1607 nm + 0 rad = = = λair n λ λ λ
803.5 nm
λ
Iλ ⎛ 803.5 nm ⎞ 2 ⎛ 803.5 nm ⎞ ⇒ I λ = cA2 = c 4a 2 cos 2 ⎜ ⎟ ⇒ = 4cos ⎜ ⎟ λ λ I0 ⎝ ⎠ ⎝ ⎠
(b) The values of ( I λ I 0 ) at λ = 400, 450, 500, 550, 600, 650, and 700 nm are 0.719, 0.182, 0.005, 0.048, 0.211,
0.431, and 0.674, respectively. (c)
21.65. Model: Reflection is minimized when the two reflected waves interfere destructively. Solve:
Equation 21.2 gives the condition for perfect destructive interference between the two waves:
Δφ = 2π
Δx
λ
+ Δφ0 = 2 ( m + 12 )π rad
The wavelength of the sound is
λ=
v 343 m/s = = 0.2858 m f 1200 Hz
Let d be the separation between the mesh and the wall. Substituting Δφ0 = 0 rad, Δx = 2d , m = 0, and the above value for the wavelength, 2π ( 2d ) 0.2858 m + 0 rad = π rad ⇒ d = = 0.0715 m = 7.15 cm 4 0.2858 m
21.66. Model: A light wave that reflects from a boundary at which the index of refraction increases has a phase shift of π rad. Solve: (a) Because nfilm > nair, the wave reflected from the outer surface of the film (called 1) is inverted due to the phase shift of π rad. The second reflected wave does not go through any phase shift of π rad because the index of refraction decreases at the boundary where this wave is reflected, which is on the inside of the soap film. We can write for the phases
φ1 = kx1 + φ10 + π rad
φ2 = kx2 + φ20 + 0 rad
⇒ Δφ = φ2 − φ1 = k ( x2 − x1 ) + (φ20 − φ10 ) − π rad = k Δx + Δφ0 − π rad = k Δx − π rad Δφ0 = 0 rad because the sources are identical. For constructive interference, ⎛ 2π ⎞ Δφ = 2mπ rad ⇒ k Δx − π rad = 2mπ rad ⇒ ⎜ ⎟ ( 2d ) = ( 2m + 1) π rad ⎝ λfilm ⎠ ⇒ λfilm =
λC n
=
2d 2nd 2.66d m = 0, 1, 2, 3, … ⇒ λC = = 1 1 m+ 2 m + 2 m + 12
(b) For m = 0 the wavelength for constructive interference is
( 2.66 )( 390 nm ) = 2075 nm ( 12 ) λC = 415 nm ( ~violet ) . Red and violet together give a purplish color.
λC = For m = 1 and 2, λC = 692 nm ( ~red ) and
21.67. Model: The two radio antennas are sources of in-phase, circular waves. The overlap of these waves causes interference. Visualize:
Solve: Maxima occur along lines such that the path difference to the two antennas is Δr = mλ . The 750 MHz = 7.50 × 108 Hz wave has a wavelength λ = c f = 0.40 m. Thus, the antenna spacing d = 2.0 m is exactly 5λ. The maximum possible intensity is on the line connecting the antennas, where Δr = d = 5λ . So this is a line of maximum intensity. Similarly, the line that bisects the two antennas is the Δr = 0 line of maximum intensity. In between, in each of the four quadrants, are four lines of maximum intensity with Δr = λ, 2λ, 3λ, and 4λ. Although we have drawn a fairly accurate picture, you do not need to know precisely where these lines are located to know that you have to cross them if you walk all the way around the antennas. Thus, you will cross 20 lines where Δr = mλ and will detect 20 maxima.
21.68. Model: The changing sound intensity is due to the interference of two overlapped sound waves. Solve: Minimum intensity implies destructive interference. Destructive interference occurs where the path length difference for the two waves is Δr = ( m + 12 ) λ . We have assumed Δφ0 = 0 rad for two speakers playing
“exactly the same” tone. The wavelength of the sound is λ = vsound f = ( 343 m/s ) 686 Hz = 0.500 m . Consider a
point that is a distance x in front of the top speaker. Let r1 be the distance from the top speaker to the point and r2 the distance from the bottom speaker to the point. We have r2 = x 2 + ( 3 m )
r1 = x
2
Destructive interference occurs at distances x such that
Δr = x 2 + 9 m 2 − x = ( m + 12 ) λ To solve for x, isolate the square root on one side of the equation and then square: x 2 + 9 m = ⎡⎣ x + ( m + 12 ) λ ⎤⎦ = x 2 + 2 ( m + 12 ) λ x + ( m + 12 ) λ 2 ⇒ x = 2
2
9 m − ( m + 12 ) λ 2 2
2 ( m + 12 ) λ
Evaluating x for different values of m: m 0 1 2 3
x (m) 17.88 5.62 2.98 1.79
Because you start at x = 2.5 m and walk away from the speakers, you will only hear minima for values x > 2.5 m. Thus, to correct significant figures, minima will occur at distances of 3.0 m, 5.6 m, and 18 m.
21.69. Model: The changing sound intensity is due to the interference of two overlapped sound waves. Visualize:
The listener moving relative to the speakers changes the phase difference between the waves.
Solve: (a) Initially when you are at P, equidistant from the speakers, you hear a sound of maximum intensity. This implies that the two speakers are in phase (Δφ0 = 0). However, on moving to Q you hear a minimum of sound intensity implying that the path length difference from the two speakers to Q is λ /2. Thus, 1 2
λ = Δr =
( r1 )
2
+ ( 5.0 m ) − r1 = 2
(12.0 m )
⇒ λ = 2.0 m ⇒ f =
v
λ
=
2
+ ( 5.0 m ) − 12.0 m = 1.0 m 2
340 m/s = 170 Hz 2.0 m
(b) At Q, the condition for perfect destructive interference is
Δφ =
2π ( Δr )
λ
+ 0 rad = 2 ( m − 12 )π rad ⇒
⇒ f = ( m − 12 )
2πΔr = 2 ( m − 12 )π rad v f
v ⎛ 340 m/s ⎞ = ( m − 12 ) ⎜ ⎟ Δr ⎝ 1.0 m ⎠
For m = 1, 2, and 3, f1 = 170 Hz, f 2 = 510 Hz, and f 3 = 850 Hz.
21.70. Model: The amplitude is determined by the interference of the two waves. Visualize:
Solve:
The amplitude of the sound wave is A = 2a cos ( 12 Δφ ) . With Δφ0 = 0 rad, the phase difference between
the waves is Δφ = φ2 − φ1 = 2π
Δr
λ
= 2π
Δr ⎛ πΔr ⎞ ⇒ A = 2a cos ⎜ ⎟ 2.0 m ⎝ 2.0 m ⎠
At the coordinates (0.0 m, 0.0 m), Δr = 0 m, so A = 2a. At the coordinates (0.0 m, 0.5 m),
Δr =
( 3.0 m )
2
+ ( 2.5 m ) − 2
( 3.0 m )
2
+ (1.5 m ) = 0.551 m ⇒ A = 2a cos 2
( 0.551 m )π 2.0 m
= 1.30a
At the coordinates (0.0 m, 1.0 m), Δr =
( 3.0 m )
2
+ ( 3.0 m ) − 2
( 3.0 m )
2
⎛ (1.08 m )π ⎞ 2 + (1.0 m ) = 1.08 m ⇒ A = 2a cos ⎜ ⎟ = 0.25a ⎝ 2.0 m ⎠
At the coordinates (0.0 m, 1.5 m), Δr = 1.568 m and A = 1.56a. At the coordinates (0.0 m, 2.0 m), Δr = 2.0 m and A = 2a.
21.71. Model: The two radio transmitters are sources of out-of-phase, circular waves. The overlap of these waves causes interference. Visualize:
Solve:
The phase difference of the waves at point P is given by
Δφ = 2π Δr =
( 3000 m )
2
+ ( 85 m ) − 2
Δr
λ
+ Δφ0
( 3000 m )
2
+ ( 35 m ) = 0.99976 m 2
The intensity at P is a maximum. Using m = 1 for the first maximum, and Δφ0 = π rad since the transmitters are out of phase, the condition for constructive interference is Δφ = 2mπ = 2π . Thus, 2π rad = 2π
Δr
λ
+ π rad ⇒ λ = 2Δr = 2 ( 0.99976 m ) ⇒ f =
c
λ
=
3 × 108 m/s = 150 MHz 2 ( 0.99976 m )
21.72. Model: The two radio antennas are sources of in-phase waves. The overlap of these waves causes interference. Visualize:
Solve:
(a) The phase difference of the two waves at point P is given by
Δφ = 2π
Δr
λ
+ Δφ0
Δr =
(800 m )
2
+ ( 650 m ) − 2
(800 m )
2
+ ( 550 m ) = 59.96 m 2
The wavelength of the radio wave is
λ=
c 3.0 × 108 m/s = = 100 m f 3.0 × 106 m
Since the sources are identical, Δφ0 = 0 rad. The phase difference at P due to the two waves is
⎛ 59.96 m ⎞ Δφ = 2π ⎜ ⎟ + 0 rad = 1.2π rad ⎝ 100 m ⎠ (b) Since Δφ = 1.20π = 0.6 ( 2π ) , which is neither m2π nor ( m + 12 ) 2π , the interference at P is somewhere in
between maximum constructive and perfect destructive. (c) At a point 10 m further north we have Δr =
( 800 m )
2
+ ( 660 m ) − 2
( 800 m )
2
+ ( 560 m ) = 60.58 m
⎛ 60.58 m ⎞ ⇒ Δφ = 2π ⎜ ⎟ + 0 rad = 1.21π rad = ⎝ 100 m ⎠
2
( 0.605) 2π
Because the phase difference is increasing as you move north, you are moving from a destructive interference condition Δφ = ( m + 12 ) 2π with m = 0 toward a constructive interference condition Δφ = m ( 2π ) with m = 1. The signal strength will therefore increase.
21.73. Model: The amplitude is determined by the interference of the two waves.
Solve: (a) We have three identical loudspeakers as sources. Δr between speakers 1 and 2 is 1.0 m and λ = 2.0 m. Thus Δr = 12 λ , which gives perfect destructive interference for in-phase sources. That is, the interference of the waves from loudspeakers 1 and 2 is perfect destructive, leaving only the contribution due to speaker 3. Thus the amplitude is a. (b) If loudspeaker 2 is moved away by one-half of a wavelength or 1.0 m, then all three waves will reach you in phase. The amplitude of the superposed waves will therefore be maximum and equal to A = 3a. (c) The maximum intensity is I max = CA2 = 9Ca 2 . The ratio of the intensity to the intensity of a single speaker is
I max I single speaker
=
9Ca 2 =9 Ca 2
21.74. Model: The superposition of two slightly different frequencies gives rise to beats. Solve:
The third harmonic of note A and the second harmonic of note E are
f3A = 3 f1A = 3( 440 Hz ) = 1320 Hz
f 2E = 2 f1E = 2 ( 659 Hz ) = 1318 Hz
⇒ f 3A − f 2E = 1320 Hz − 1318 Hz = 2 Hz (b) The beat frequency between the first harmonics is f1E − f1A = 659 Hz − 440 Hz = 219 Hz The beat frequency between the second harmonics is f2E − f2A = 1318 Hz − 880 Hz = 438 Hz
The beat frequency between f3A and f2E is 2 Hz. It therefore emerges that the tuner looks for a beat frequency of 2 Hz. (c) If the beat frequency is 4 Hz, then the second harmonic frequency of the E string is
f 2E = 1320 Hz − 4 Hz = 1316 Hz ⇒ f1E =
1 2
(1316 Hz )
= 658 Hz
Note that the second harmonic frequency of the E string could also be f 2E = 1320 Hz + 4 Hz = 1324 Hz ⇒ f1E = 662 Hz
This higher frequency can be ruled out because the tuner started with low tension in the E string and we know that T ⇒ f ∝ T vstring = λ f =
μ
21.75. Model: The superposition of two slightly different frequencies creates beats. Solve:
(a) The wavelength of the sound initially created by the flutist is
λ=
342 m/s = 0.77727 m 440 Hz
When the speed of sound inside her flute has increased due to the warming up of the air, the new frequency of the A note is f′=
346 m/s = 445 Hz 0.77727 m
Thus the flutist will hear beats at the following frequency: f ′ − f = 445 Hz − 440 Hz = 5 beats/s
Note that the wavelength of the A note is determined by the length of the flute rather than the temperature of air or the increased sound speed. (b) The initial length of the flute is L = 12 λ = 12 ( 0.77727 m ) = 0.3886 m. The new length to eliminate beats needs
to be L′ =
λ′ 2
1 ⎛ v′ ⎞ 1 ⎛ 346 m/s ⎞ = ⎜ ⎟= ⎜ ⎟ = 0.3932 m 2 ⎝ f ⎠ 2 ⎝ 440 Hz ⎠
Thus, she will have to extend the “tuning joint” of her flute by 0.3932 m − 0.3886 m = 0.0046 m = 4.6 mm
21.76. Solve: (a) Yvette’s speed is the width of the room divided by time. This means vY =
n ( 12 λ ) t
⇒
n 2vY = t λ
Note that λ is the distance between two consecutive antinodes, and n is the number of such half wavelengths that fill the entire width of the room. (b) Yvette observes a higher frequency f+ of the source she is moving toward and a lower frequency f− of the source she is receding from. If v is the speed of sound and f is the sound wave’s frequency, we have 1 2
⎛ v ⎞ f + = f ⎜1 + Y ⎟ v ⎠ ⎝
⎛ v ⎞ f − = f ⎜1 − Y ⎟ v ⎠ ⎝
The expression for the beat frequency is
⎛ v ⎞ f + − f − = f ⎜1 + Y ⎟ − v ⎠ ⎝
v vv 2v ⎛ v ⎞ f ⎜1 − Y ⎟ = 2 f Y = 2 Y = Y v v v λ λ ⎝ ⎠
(c) The answers to part (a) and (b) are the same. Even though you and Yvette have different perspectives, you should agree as to how many modulations per second she hears.
21.77. Model: The frequency of the loudspeaker’s sound in the back of the pick-up truck is Doppler shifted. As the truck moves away from you, its frequency is decreased. Solve: Because you hear 8 beats per second as the truck drives away from you, the frequency of the sound from the speaker in the pick-up truck is f − = 400 Hz − 8 Hz = 392 Hz. This frequency is
f− =
f0 vS 400 Hz ⇒1+ = = 1.020408 ⇒ vS = 7.0 m/s 343 m/s 392 Hz 1 + vS v
That is, the velocity of the source vS and hence the pick-up truck is 7.0 m/s.
21.78. Model: A stretched string under tension supports standing waves. Solve:
(a) The wave speed on a stretched string is
vstring =
T
μ
= fλ ⇒ f =
1 T
λ μ
The wavelength λ cannot change if the length of the string does not change. So, df 1 1 = dT λ μ
1 2
(T )
−1/ 2
=
1 2λ
1
μT
=
Δf ΔT 1 ⎛1 T ⎞ 1 = f ⇒ ⎜⎜ ⎟⎟ = f 2T 2T ⎝ λ μ ⎠ 2T
(b) Since there are 5 beats per second, Δf = 5 Hz =
f ΔT ΔT 10 Hz 10 Hz ⇒ = = = 0.020 = 2.0% 2 T T f 500 Hz
That is, an increase of 2.0% in the tension of one of the strings will cause 5 beats per second.
21.79. Model: The microphone will detect a loud sound only if there is a standing wave resonance in the tube. The sound frequency does not change, but changing the length of the tube can create a standing wave. Solve: The standing wave condition is f = 280 Hz = m
v 2L
m = 1, 2, 3,…
where L is the total length of the tube. When the slide is extended a distance s, the tube has two straight sides, each of length s + 80 cm, plus a semicircular bend of length 12 ( 2π r ) . The radius is r = 12 (10 cm ) = 5.0 cm. The
tube’s total length is
L = 2 ( s + 80 cm ) + 12 ( 2π × 5.0 cm ) = 175.7 cm + 2s = 1.757 m + 2s A standing wave resonance will be created if ⎡
⎤ v 343 m/s =m = 0.6125m ⎥ 2 ( 280 Hz ) ⎢⎣ 2 f ⎥⎦
[ L = 1.757 + 2s ] = ⎢ m
⇒ s = 0.3063m − 0.8785 meters We can tabulate the different extensions s that correspond to standing wave modes m = 1, m = 2, m = 3, and so on. m 1 2 3 4 5 6
s −0.572 m −0.266 m 0.040 m = 4.0 cm 0.347 m = 34.7 cm 0.653 m = 65.3 cm 1.959 m
Physically, the extension must be greater than 0 cm and less than 80 cm. Thus, the three slide extensions that create a standing wave resonance at 280 Hz are 4.0 cm, 35 cm, and 65 cm to two significant figures.
21.80. Model: The stretched wire is vibrating at its second harmonic frequency. Visualize: Let l be the full length of the wire, and L be the vibrating length of the wire. That is, L = ( 12 ) l .
Solve:
The wave speed on a stretched wire is
Ts
vwire =
μ
= fλ
The frequency f = 100 Hz and the wavelength λ = 12 l because it is a second harmonic wave. The tension Ts =
(1.25 kg)g because the hanging mass is in static equilibrium and μ = 1.00 × 10−3 kg/m. Substituting in these values, −3 2 kg/m ) l 2 ⎛ l ⎞ (1.00 × 10 g = 100 Hz ⇒ = 100 Hz = ( 2.00 m −1s −2 ) l 2 ( ) ( ) ⎜ ⎟ −3 2 (1.25 kg ) ⎝ 2⎠ (1.00 ×10 kg/m )
(1.25 kg ) g
To find l we can use the equation for the time period of a simple pendulum: l T2 ( 314 s/100 ) g = 0.250 s 2 g ⇒l = 2 g = ( ) 4π 4π 2 g 2
T = 2π
Substituting this expression for l into the equation for g, we get g = ( 2.000 m −1s −2 )( 0.025 s 2 ) g 2 ⇒ ( 0.125 m −1s 2 ) g 2 − g = 0 2
⇒ ⎡⎣( 0.125 m −1s 2 ) g − 1⎤⎦ g = 0 ⇒ g = 8.00 m/s 2 Assess:
A value of 8.0 m/s2 is reasonable for the information given in the problem.
21.81. Model: The steel wire is under tension and it vibrates with three antinodes. Solve: When the spring is stretched 8.0 cm, the standing wave on the wire has three antinodes. This means λ3 = 32 L and the tension TS in the wire is TS = k (0.080 m), where k is the spring constant. For this tension,
vwire =
TS
μ
⇒ f λ3 =
TS
μ
⇒ f =
3 k ( 0.08 m ) μ 2L
We will let the stretching of the spring be Δx when the standing wave on the wire displays two antinodes. This means λ2 = L and TS′ = kx. For the tension TS′, v′wire =
TS′
μ
⇒ f λ2 =
TS′
μ
⇒ f =
1 k Δx L μ
The frequency f is the same in the above two situations because the wire is driven by the same oscillating magnetic field. Now, equating the two frequency equations, 1 k Δx 3 k ( 0.080 m ) ⇒ Δx = 0.18 m = 18 cm = μ L μ 2L
21.82. Model: The frequency is Doppler shifted to higher values for a detector moving toward the source. The frequency is also shifted to higher values for a source moving toward the detector. Visualize:
Solve: (a) We will derive the formula in two steps. First, the object acts like a moving detector and “observes” a frequency that is given by f 0′ = f 0 (1 + v0 v ) . Second, as this moving object reflects (or acts as a “source” of −1 ultrasound waves), the frequency fecho as observed by the original source So is f echo = f 0′ (1 − v0 v ) . Combining
these two equations gives f echo =
f (1 + v0 v ) v + v0 f 0′ = 0 = f0 v − v0 1 − v0 v 1 − v0 v
(b) If v0 << v, then
⎛ v ⎞⎛ v ⎞ f echo = f 0 ⎜1 + 0 ⎟⎜1 − 0 ⎟ v ⎠⎝ v⎠ ⎝
−1
⎛ v ⎞⎛ v ⎞ ⎛ 2v ⎞ = f 0 ⎜ 1 + 0 ⎟⎜ 1 + 0 + … ⎟ = f 0 ⎜1 + 0 + … ⎟ v ⎠⎝ v v ⎝ ⎠ ⎝ ⎠
⇒ f beat = f echo − f 0 ≈
2v0 f0 v
(c) Using part (b) for the beat frequency, ⎛ 2v0 ⎞ 6 65 Hz = ⎜ ⎟ ( 2.40 × 10 Hz ) ⇒ v0 = 2.09 cm/s ⎝ 1540 m/s ⎠
(d) Assuming the heart rate is 90 beats per minute the angular frequency is
ω = 2π f = 2π (1.5 beats/s ) = 9.425 rad/s Using v0 = vmax = ω A , A=
T=
v0
ω
=
2.09 cm/s = 2.2 mm 9.425 rad/s
(1.50 + 1.60 )
2 ( 3.0 × 108 m/s )
× 0.010 m = 5.17 × 10−11 s
21.83. Solve: (a) The wavelengths of the standing wave modes are λm = ⇒ λ1 =
2 (10.0 m ) = 20.0 m 1
2L m = 1, 2, 3, …. m
λ2 =
2 (10.0 m ) = 10.0 m 2
λ3 =
2 (10.0 m ) = 6.67 m 3
The depth of the pool is 5.0 m. Clearly the standing waves with λ2 and λ3 are “deep water waves” because the 20 m depth is larger than one-quarter of the wavelength. The wave with λ1 barely qualifies to be a deep water standing wave.
(b) The wave speed for the first standing wave mode is
v1 =
g λ1 = 2π
( 9.8 m/s ) ( 20.0 m ) = 5.59 m/s 2
2π
Likewise, v2 = 3.95 m/s and v3 = 3.22 m/s . (c) We have v=
g λm g mg = f m λm ⇒ f m = = 2π 2πλm 4π L
Please note that m is the mode and not the mass. (d) The period of oscillation for the first standing wave mode is calculated as follows f1 = Likewise, T2 = 2.53 s and T3 = 2.07 s.
(1) ( 9.8 m/s 2 ) = 0.279 Hz 4π (10.0 m )
⇒ T1 = 3.58 s
21.84. Model: The overlap of the waves causes interference. Solve:
(a) The waves traveling to the left are
⎡ ⎛ x t ⎞⎤ D1 = a sin ⎢ 2π ⎜ − − ⎟ ⎥ ⎣ ⎝ λ T ⎠⎦
⎡ ⎛ ( x − L) t ⎞ ⎤ D2 = a sin ⎢ 2π ⎜ − − ⎟ + φ20 ⎥ λ T⎠ ⎣⎢ ⎝ ⎦⎥
The phase difference between the waves on the left side of the antenna is thus
⎛ ( x − L) t ⎞ L ⎛ x t⎞ ΔφL = φ2 − φ1 = 2π ⎜ − − ⎟ + φ20 − 2π ⎜ − − ⎟ = 2π + φ20 λ λ T⎠ ⎝ λ T⎠ ⎝ On the right side of the antennas, where x1 = x2 + L, the two waves are ⎡ ⎛ x t ⎞⎤ D1 = a sin ⎢ 2π ⎜ − ⎟ ⎥ ⎣ ⎝ λ T ⎠⎦
⎡ ⎛ ( x − L) t ⎞ ⎤ − ⎟ + φ20 ⎥ D2 = a sin ⎢ 2π ⎜ T⎠ ⎣⎢ ⎝ λ ⎦⎥
Thus, the phase difference between the waves on the right is
⎛ ( x − L) t ⎞ 2π L ⎛x t⎞ + φ20 ΔφR = φ2 − φ1 = 2π ⎜ − ⎟ + φ20 − 2π ⎜ − ⎟ = − λ T T λ λ ⎝ ⎠ ⎝ ⎠ We want to have destructive interference on the country side or on the left and constructive interference on the right. This requires ΔφL = φ20 +
2π L
λ
= 2π ( m + 12 ) ΔφR = φ20 −
2π L
λ
= 2π n
These are two simultaneous equations, and we can satisfy them both if L and φ20 are properly chosen. Subtracting the second equation from the first to eliminate φ20 , 4π L
λ
= 2π ( m + 12 − n ) ⇒ L = ( m + 12 − n )
λ 2
The smallest value of L that works is for n = m, in which case L = 14 λ . (b) From the ΔφR equation,
⎡ 2π ( 14 λ ) π 2π L π⎤ = φ20 − = φ20 − ⎥ = 2π n ⇒ φ20 = + 2π n rad ⎢φ20 − 2 λ λ 2 ⎣ ⎦ Adding integer multiples of 2π to the phase constant doesn’t really change the wave, so the physically significant phase constant is for n = 0, namely φ20 = 12 π rad.
(c) We have φ20 = 12 π rad = 14 ( 2π ) . If the wave from antenna 2 was delayed by one full period T, it would shift the wave by
one full cycle. We would describe this by a phase constant of 2π rad. So a phase constant of
1 4
( 2π )
rad can be achieved by
delaying the wave by Δt = T . 1 4
(d) A wave with frequency f = 1000 kHz = 1.00 × 106 Hz has a period T = 1.00 × 10−6 s = 1.00 μ s and wavelength
λ = c / f = 300 m. So this broadcast scheme will work if the antennas are spaced L = 75 m apart and if the broadcast from antenna 2 is delayed by Δt = 0.25 μ s = 250 ns.
21-1
22.1. Model: Two closely spaced slits produce a double-slit interference pattern.
Visualize: The interference pattern looks like the photograph of Figure 22.3(b). It is symmetrical with the m = 2 fringes on both sides of and equally distant from the central maximum. Solve: The bright fringes occur at angles θm such that
d sin θ m = mλ ⇒ sin θ 2 =
2 ( 500 ×10−9 m )
( 50 ×10
−6
m)
m = 0, 1, 2, 3, …
= 0.02 ⇒ θ 2 = 0.020 rad = 0.020 rad ×
180°
π rad
= 1.15°
22.2. Model: Two closely spaced slits produce a double-slit interference pattern.
Visualize: The interference pattern looks like the photograph of Figure 22.3(b). It is symmetrical, with the m = 2 fringes on both sides of and equally distant from the central maximum. Solve: The two paths from the two slits to the m = 2 bright fringe differ by Δr = r2 − r1 , where
Δr = mλ = 2λ = 2 ( 500 nm ) = 1000 nm Thus, the position of the m = 2 bright fringe is 1000 nm farther away from the more distant slit than from the nearer slit.
22.3. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). Solve: The bright fringes are located at positions given by Equation 22.4, d sin θ m = mλ . For the m = 3 bright
orange fringe, the interference condition is d sinθ 3 = 3 ( 600 ×10−9 m ) . For the m = 4 bright fringe the condition
is d sinθ 4 = 4λ . Because the position of the fringes is the same, d sin θ 3 = d sin θ 4 = 4λ = 3 ( 600 × 10−9 m ) ⇒ λ =
3 4
( 600 × 10
−9
m ) = 450 nm
22.4. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). Solve: The formula for fringe spacing is L λL L Δy = ⇒ 1.8 × 10−3 m = ( 600 × 10−9 m ) ⇒ = 3000 d d d The wavelength is now changed to 400 nm, and L d , being a part of the experimental setup, stays the same. Applying the above equation once again, λL Δy = = ( 400 ×10−9 m ) ( 3000 ) = 1.2 mm d
22.5. Visualize: The fringe spacing for a double slit pattern is Δy = λ = 600 nm. We also see from the figure that Δy = 13 cm. Solve:
d
Solve the equation for d. d=
Assess:
λL
λL Δy
=
0.36 mm is a typical slit spacing.
(600 × 10−9 m)(2.0 m) = 0.36 mm 1 × 10 −2 m 3
. We are given L = 2.0 m and
22.6. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). Solve: The fringe spacing is λL λ L (589 × 10−9 m)(150 × 10−2 m) Δy = ⇒d = = = 0.22 mm d Δy 4.0 × 10−3 m
22.7. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). Solve: The dark fringes are located at positions given by Equation 22.9: λL m = 0, 1, 2, 3, … ym′ = ( m + 12 ) d
⇒ y5′ − y1′ = ( 5 + 12 )
λL d
− (1 + 12 )
λL d
⇒ 6.0 × 10−3 m =
4λ (60 × 10−2 m) ⇒ λ = 500 nm 0.20 × 10−3 m
22.8. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). Solve: In a span of 12 fringes, there are 11 gaps between them. The formula for the fringe spacing is Δy =
Assess:
⎛ 52 × 10−3 m ⎞ (633 × 10 −9 m)(3.0 m) ⇒ d = 0.40 mm ⇒ ⎜ ⎟= d 11 d ⎝ ⎠
λL
This is a reasonable distance between the slits, ensuring d L = 1.34 ×10−4 << 1 .
22.9. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram in Figure 22.8. Solve: The bright constructive-interference fringes are given by Equation 22.15:
d sin θ m = mλ ⇒ sin θ1 =
Likewise, sin θ 2 = 0.110 and θ 2 = 6.3°.
m = 0, 1, 2, … −9
(1)(550 × 10 m) = 0.055 ⇒ θ1 = 3.2° (1.0 × 10−2 m) /1000
22.10. Model: A diffraction grating produces a series of constructive-interference fringes at values of θ m determined by Equation 22.15. Solve: We have d sin θ m = mλ
m = 0, 1, 2, 3, … ⇒ d sin 20.0° = 1λ and d sin θ 2 = 2λ
Dividing these two equations,
sin θ 2 = 2sin 20.0° = 0.6840 ⇒ θ 2 = 43.2°
22.11. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram in Figure 22.8. Solve: The bright constructive-interference fringes are given by Equation 22.15:
d sin θ m = mλ ⇒ d =
mλ (2)(600 × 10−9 m) = = 1.89 × 10−6 m sin θ m sin(39.5°)
The number of lines in per millimeter is (1 × 10−3 m) (1.89 × 10−6 m) = 530.
22.12. Model: A diffraction grating produces an interference pattern. Visualize:
Solve:
The interference pattern looks like the diagram in Figure 22.8.
The bright fringes are given by Equation 22.15:
d sin θ m = mλ
m = 0, 1, 2, 3, … ⇒ d sinθ1 = (1) λ ⇒ d = λ sin θ1
The angle θ1 can be obtained from geometry as follows: tanθ1 =
( 0.32 m ) 2.0 m
2
= 0.080 ⇒ θ1 = tan–1 (0.080) 4.57°
Using sin θ1 = sin 4.57° = 0.07968 , d=
633 ×10−9 m = 7.9 μ m 0.07968
22.13. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram of Figure 22.8. Solve: The bright interference fringes are given by
d sin θ m = mλ The slit spacing is d = 1 mm 500 = 2.00 ×10
−6
m and m = 1. For the red and blue light,
⎛ 656 ×10−9 m ⎞ ⎟ = 19.15° −6 ⎝ 2.00 ×10 m ⎠
θ1 red = sin −1 ⎜
m = 0, 1, 2, 3, …
⎛ 486 ×10−9 m ⎞ ⎟ = 14.06° −6 ⎝ 2.00 ×10 m ⎠
θ1 blue = sin −1 ⎜
The distance between the fringes, then, is Δy = y1 red − y1 blue where y1 red = (1.5 m ) tan19.15° = 0.521 m y1 blue = (1.5 m ) tan14.06° = 0.376 m
So, Δy = 0.145 m = 14.5 cm.
22.14. Model: Assume the screen is centered behind the slit. We actually want to solve for m, but given the other data, it is unlikely that we will get an integer from the equations for the edge of the screen, so we will have to truncate our answer to get the largest order fringe on the screen. Visualize: Refer to Figure 22.7. Use Equation 22.15: d sin θ m = mλ , and Equation 22.16: ym = L tanθ m . We 1 mm. As mentioned above, we are not guaranteed that a bright are given λ = 510 nm, L = 2.0 m, and d = 500
fringe will occur exactly at the edge of the screen, but we will kind of assume that one does and set ym = 1.0 m; if we do not get an integer for m then the fringe was not quite at the edge of the screen and we will truncate our answer to get an integer m. Solve: Solve Equation 22.16 for θ m and insert it in Equation 22.15.
θ m = tan −1
ym L
Solve Equation 22.15 for m. m=
d
λ
sin θ m =
1 mm 1. 0 m ⎞ y ⎞ ⎛ ⎛ sin ⎜ tan −1 m ⎟ = 500 sin ⎜ tan −1 ⎟ = 1.8 2.0 m ⎠ λ ⎝ L ⎠ 510 nm ⎝
d
Indeed, we did not get an integer, so truncate 1.8 to get m = 1. This means we will see three fringes, one for m = 0, and one on each side for m = ±1. Assess: Our answer fits with the statement in the text: “Practical gratings, with very small values for d , display only a few orders.”
22.15. Model: A narrow single slit produces a single-slit diffraction pattern. Visualize: The intensity pattern for single-slit diffraction will look like Figure 22.14. Solve: The minima occur at positions yp = p So Δy = y2 − y1 =
Lλ a
−9 2λ L 1λ L λ L λ L ( 633 ×10 m ) (1.5 m ) − = ⇒ a= = = 2.0 ×10−4 m = 0.20 mm a a a Δy 0.00475 m
22.16. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The intensity pattern for single-slit diffraction will look like Figure 22.14. Solve: The width of the central maximum for a slit of width a = 200λ is
w=
2λ L 2λ ( 2.0 m ) = = 20 mm 200λ a
22.17. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The intensity pattern for single-slit diffraction will look like Figure 22.14. Solve: Angle θ = 0.70° = 0.0122 rad is a small angle (<< 1 rad). Thus we use Equation 22.20 to find the wavelength of light. The angles of the minima of intensity are
θp = p ⇒λ =
aθ p p
=
λ
p = 1, 2, 3, κ a ( 0.10 ×10−3 m ) ( 0.0122 rad ) 2
= 610 nm
22.18. Visualize: Use Equation 22.22: w = the figure that w = 1.0 cm. Solve: Solve the equation for a. a= Assess:
2λ L . We are given λ = 600 nm and L = 2.0 m. We see from a
2λ L 2(600 × 10−9 m)(2.0 m) = = 0.24 mm w 0.010 m
0.24 mm is in the range of typical slit widths.
22.19. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The intensity pattern for single-slit diffraction will look like Figure 22.14. Solve: The width of the central maximum for a slit of width a = 200λ is
w=
−9 2λ L 2 ( 500 ×10 m ) ( 2.0 m ) = = 0.0040 m = 4.0 mm 0.0005 m a
22.20. Model: The spacing between the two buildings is like a single slit and will cause the radio waves to be diffracted. Solve: Radio waves are electromagnetic waves that travel with the speed of light. The wavelength of the 800 MHz waves is
λ=
3 ×108 m/s = 0.375 m 800 ×106 Hz
To investigate the diffraction of these waves through the spacing between the two buildings, we can use the general condition for complete destructive interference: a sin θ p = pλ (p = 1, 2, 3, …) where a is the spacing
between the buildings. Because the width of the central maximum is defined as the distance between the two p = 1 minima on either side of the central maximum, we will use p = 1 and obtain the angular width Δθ = 2θ1 from
⎛λ⎞ ⎛ 0.375 m ⎞ a sin θ1 = λ ⇒ θ1 = sin −1 ⎜ ⎟ = sin −1 ⎜ ⎟ = 1.43° a ⎝ ⎠ ⎝ 15 m ⎠ Thus, the angular width of the wave after it emerges from between the buildings is Δθ = 2 (1.43° ) = 2.86° ≈ 2.9°.
22.21. Model: The crack in the cave is like a single slit that causes the ultrasonic sound beam to diffract. Visualize:
Solve:
The wavelength of the ultrasound wave is
λ=
340 m/s = 0.0113 m 30 kHz
Using the condition for complete destructive interference with p = 1,
⎛ 0.0113 m ⎞ a sin θ1 = λ ⇒ θ1 = sin −1 ⎜ ⎟ = 2.165° ⎝ 0.30 m ⎠ From the geometry of the diagram, the width of the sound beam is
w = 2 y1 = 2 (100 m × tanθ1 ) = 200 m × tan 2.165° = 7.6 m Assess: The small-angle approximation is almost always valid for the diffraction of light, but may not be valid for the diffraction of sound waves, which have a much larger wavelength.
22.22. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Solve: The width of the central maximum for a circular aperture of diameter D is
w=
2.44λ L (2.44)(500 × 10−9 m)(2.0 m) = = 4.9 mm D 0.50 × 10−3 m
22.23. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Visualize: The intensity pattern will look like Figure 22.15. Solve: According to Equation 22.23, the angle that locates the first minimum in intensity is
θ1 =
1.22λ 1.22(2.5 × 10−6 m) = = 0.01525 rad = 0.874° D 0.20 × 10−3 m
These should be rounded to 0.015 rad = 0.87°.
22.24. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Visualize: The intensity pattern will look like Figure 22.15. Solve: From Equation 22.24, the diameter of the circular aperture is
D=
2.44λ L 2.44(633 × 10−9 m)(4.0 m) = = 0.25 mm w 2.5 × 10−2 m
22.25. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Visualize: The intensity pattern will look like Figure 22.15. Solve: From Equation 22.24, L=
Dw (0.12 × 10−3 m)(1.0 × 10−2 m) = = 78 cm 2.44λ 2.44(633 × 10−9 m)
22.26. Model: An interferometer produces a new maximum each time L2 increases by 12 λ causing the pathlength difference Δr to increase by λ. Visualize: Please refer to the interferometer in Figure 22.20. Solve: From Equation 22.33, the wavelength is 2ΔL2 2(100 × 10−6 m) λ= = = 4.0 × 10−7 m = 400 nm Δm 500
22.27. Model: An interferometer produces a new maximum each time L2 increases by 12 λ causing the pathlength difference Δr to increase by λ. Visualize: Please refer to the interferometer in Figure 22.20. Solve: From Equation 22.33, the number of fringe shifts is
Δm =
2ΔL2
λ
=
2(1.00 × 10−2 m) = 30,467 656.45 × 10−9 m
22.28. Model: An interferometer produces a new maximum each time L2 increases by 12 λ causing the pathlength difference Δr to increase by λ. Visualize: Please refer to the interferometer in Figure 22.20. Solve: From Equation 22.33, the distance the mirror moves is
ΔL2 =
Δmλ (33,198)(602.446 × 10−9 m) = = 0.0100000 m = 1.00000 cm 2 2
Assess: Because the wavelength is known to 6 significant figures and the fringes are counted exactly, we can determine ΔL to 6 significant figures.
22.29. Model: An interferometer produces a new maximum each time L2 increases by 12 λ causing the pathlength difference Δr to increase by λ. Visualize: Please refer to the interferometer in Figure 22.20. Solve: For sodium light of the longer wavelength (λ1) and of the shorter wavelength (λ2), ΔL = m
λ1 2
ΔL = ( m + 1)
λ2 2
We want the same path difference 2(L2 − L1) to correspond to one extra wavelength for the sodium light of shorter wavelength (λ2) . Thus, we combine the two equations to obtain: m
λ1 2
= ( m + 1)
λ2 2
⇒ m ( λ1 − λ2 ) = λ2 ⇒ m =
λ2
λ1 − λ2
=
589.0 nm = 981.67 ≅ 982 589.6 nm − 589.0 nm
Thus, the distance by which M2 is to be moved is
ΔL = m
λ1 2
⎛ 589.6 nm ⎞ = 982 ⎜ ⎟ = 0.2895 mm 2 ⎝ ⎠
22.30.
Model: Two closely spaced slits produce a double-slit interference pattern with the intensity graph looking like Figure 22.3(b). The intensity pattern due to a single slit diffraction looks like Figure 22.14. Both the spectra consist of a central maximum flanked by a series of secondary maxima and dark fringes. Solve: (a) The light intensity shown in Figure P22.30 corresponds to a double-slit aperture. This is because the fringes are equally spaced and the decrease in intensity with increasing fringe order occurs slowly. (b) From Figure P22.30, the fringe spacing is Δy = 1.0 cm = 1.0 × 10−2 m. Therefore,
Δy =
λL
d λ L (6.00 ×10−9 m)(2.5 m) ⇒d = = = 0.15 mm 0.010 m Δy
22.31. Model: Two closely spaced slits produce a double-slit interference pattern with the intensity graph looking like Figure 22.3(b). The intensity pattern due to a single slit diffraction looks like Figure 22.14. Both the spectra consist of a central maximum flanked by a series of secondary maxima and dark fringes. Solve: (a) The light intensity shown in Figure P22.31 corresponds to a single slit aperture. This is because the central maximum is twice the width and much brighter than the secondary maximum. (b) From Figure P22.31, the separation between the central maximum and the first minimum is y1 = 1.0 cm = 1.0 × 10−2 m. Therefore, using the small-angle approximation, Equation 22.21 gives the condition for the dark minimum: yp =
−9 pLλ Lλ ( 2.5 m ) ( 600 × 10 m ) ⇒a= = = 0.15 mm −2 1.0 ×10 m d y1
22.32. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference fringes are equally spaced on both sides of the central maximum. The interference pattern looks like Figure 22.3(b). Solve: In the small-angle approximation
Δθ = θ m +1 − θ m = (m + 1) Since d = 200λ , we have Δθ =
λ d
=
λ d
−m
λ d
1 rad = 0.286° 200
=
λ d
22.33. Model: Two closely spaced slits produce a double-slit interference pattern. Solve:
The light intensity of a double-slit interference pattern at a position y on the screen is
⎛πd I double = 4 I1 cos 2 ⎜ ⎝ λL
⎛ y ⎞ ⎞ y ⎟ = 4 I1 cos 2 ⎜ π ⎟ ⎠ ⎝ Δy ⎠
where Δy = λ L / d = 4.0 mm is the fringe spacing. Using this value for λ L d , we can find the position on the interference pattern where Idouble = I1 as follows:
π π π 4.0 ×10−3 m ⎛ ⎞ ⎛ ⎞ −1 ⎛ 1 ⎞ y cos y = = 1.3 mm 4 I1 cos 2 ⎜ y I ⇒ = = rad ⇒ = ⎜ ⎟ ⎜ ⎟ ⎟ 1 −3 −3 3 ⎝ 4.0 ×10 m ⎠ ⎝ 4.0 ×10 m ⎠ ⎝ 2⎠ 3
22.34. Solve: According to Equation 22.7, the fringe spacing between the m fringe and the m + 1 fringe is Δy = λ L d . Δy can be obtained from Figure P22.34. The separation between the m = 2 fringes is 2.0 cm implying that the separation between the two consecutive fringes is
Δy = 0.50 × 10−2 m = Assess:
λL d
⇒ L=
d Δy
λ
=
( 0.20 ×10
−3
1 4
(2.0 cm) = 0.50 cm. Thus,
m )( 0.50 ×10−2 m )
600 ×10−9 m
A distance of 167 cm from the slits to the screen is reasonable.
= 167 cm
22.35. Solve: According to Equation 22.7, the fringe spacing between the m fringe and the m + 1 fringe is Δy = λ L d . Δy can be obtained from Figure P22.34. Because the separation between the m = 2 fringes is 2.0 cm, two consecutive fringes are Δy = 14 ( 2.0 cm ) = 0.50 cm apart. Thus, Δy = 0.50 × 10−2 m =
λL d
⇒ λ=
−3 −2 d Δy ( 0.20 ×10 m )( 0.50 ×10 m ) = = 500 nm L 2.0 m
22.36. Solve: The intensity of light of a double-slit interference pattern at a position y on the screen is ⎛πd I double = 4 I1 cos 2 ⎜ ⎝ λL
⎞ y⎟ ⎠
where I1 is the intensity of the light from each slit alone. At the center of the screen, that is, at y = 0 m, I1 = 14 I double . From Figure P22.34, Idouble at the central maximum is 12 mW/m2. So, the intensity due to a single slit is I1 = 3 mW/m 2 .
22.37. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram in Figure 22.8. Solve: 500 lines per mm on the diffraction grating gives a spacing between the two lines of d = 1 mm 500 =
(1×10−3 m) 500 = 2.0 ×10−6 m . The wavelength diffracted at angle θ m = 30° in order m is
λ=
−6 d sinθ m ( 2.0 ×10 m ) sin 30° 1000 nm = = m m m
We’re told it is visible light that is diffracted at 30°, and the wavelength range for visible light is 400–700 nm. Only m = 2 gives a visible light wavelength, so λ = 500 nm.
22.38. Model: Assume the screen is centered behind the slit. Visualize: Refer to Figure 22.7. Think carefully about the situation. The longest wavelength that will show three fringes on each side occurs when y3 = 0.50 m; but the shortest wavelength that will show three fringes on each side
occurs when we almost let the fourth fringe on the screen, i.e., when y4 = 0.50 m. Use Equation 22.15: d sin θ m = mλ , and Equation 22.16: ym = L tan θ m . We are given L = 1.0 m, and d = Solve:
1 200
mm.
Solve Equation 22.16 for θ m and insert it in Equation 22.15.
θ m = tan −1
ym L
Solve Equation 22.15 for λ.
λ=
d d y ⎞ ⎛ sin θ m = sin ⎜ tan −1 m ⎟ m m ⎝ L ⎠
λmax occurs when y3 = 0.50 m; and λmin occurs when y4 = 0.50 m.
Assess:
λmin =
d d y ⎞ ⎛ sin θ 4 = sin ⎜ tan −1 4 ⎟ = m m ⎝ L⎠
1 200
mm ⎛ −1 0.50 m ⎞ sin ⎜ tan ⎟ = 560 nm 4 1.0 m ⎠ ⎝
λmax =
d d y ⎞ ⎛ sin θ 3 = sin ⎜ tan −1 3 ⎟ = m m ⎝ L⎠
1 200
mm ⎛ −1 0.50 m ⎞ sin ⎜ tan ⎟ = 750 nm 3 1.0 m ⎠ ⎝
These are reasonable wavelengths, either in or close to the visible range.
22.39. Model: Each wavelength of light is diffracted at a different angle by a diffraction grating. Solve: Light with a wavelength of 501.5 nm creates a first-order fringe at y = 21.90 cm. This light is diffracted at angle ⎛ 21.90 cm ⎞ ⎟ = 23.65° ⎝ 50.00 cm ⎠
θ1 = tan −1 ⎜
We can then use the diffraction equation dsinθm = mλ, with m = 1, to find the slit spacing: d=
λ sin θ1
=
501.5 nm = 1250 nm sin(23.65°)
The unknown wavelength creates a first order fringe at y = 31.60 cm, or at angle
⎛ 31.60 cm ⎞ ⎟ = 32.29° ⎝ 50.00 cm ⎠
θ1 = tan −1 ⎜
With the split spacing now known, we find that the wavelength is
λ = d sinθ1 = (1250 nm)sin(32.29°) = 667.8 nm Assess: The distances to the fringes and the first wavelength were given to 4 significant figures. Consequently, we can determine the unknown wavelength to 4 significant figures.
22.40. Model: A diffraction grating produces an interference pattern like the diagram of Figure 22.8. Visualize:
Solve: (a) A key statement is that the lines are seen on the screen. This means that the light is visible light, in the range 400 nm–700 nm. We can determine where the entire visible spectrum falls on the screen for different values of m. We do this by finding the angles θm at which 400 nm light and 700 nm light are diffracted. We then use ym = L tan θ m to find their positions on the screen which is at a distance L = 75 cm. The slit spacing is
d = 1 mm 1200 = 8.333 × 10−7 m. For m = 1,
λ = 400 nm: θ1 = sin −1 ( 400 nm/d ) = 28.7° ⇒ y1 = 75 cm tan 28.7° = 41 cm λ = 700 nm: θ1 = sin −1 ( 700 nm/d ) = 57.1° ⇒ y1 = 75 cm tan 57.1° = 116 cm For m = 2,
λ = 400 nm: θ1 = sin −1 ( 2 ⋅ 400 nm/d ) = 73.8° ⇒ y2 = 75 cm tan 73.8° = 257 cm For the 700 nm wavelength at m = 2, θ 2 = sin −1 ( 2 ⋅ 700 nm/d ) = sin −1 (1.68 ) is not defined, so y2 → ∞ . We see that visible light diffracted at m = 1 will fall in the range 41 cm ≤ y ≤ 116 and that visible light diffracted at m = 2 will fall in the range y ≥ 257 cm. These ranges do not overlap, so we can conclude with certainty that the observed diffraction lines are all m = 1. (b) To determine the wavelengths, we first find the diffraction angle from the observed position by using
⎛ y⎞ ⎝ ⎠
⎛ ⎝
y
⎞ ⎠
θ = tan −1 ⎜ ⎟ = tan −1 ⎜ ⎟ L 75 cm This angle is then used in the diffraction grating equation for the wavelength with m = 1, λ = d sin θ1 . Line
θ
λ
56.2 cm 65.9 cm 93.5 cm
36.85° 41.30° 51.27°
500 nm 550 nm 650 nm
22.41.
Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram of Figure 22.8. Solve: (a) A grating diffracts light at angles sin θ m = mλ d . The distance between adjacent slits is
d = 1 mm 600 = 1.667 ×10−6 m = 1667 nm. The angle of the m = 1 fringe is
⎛ 500 nm ⎞ ⎟ = 17.46° ⎝ 1667 nm ⎠
θ1 = sin −1 ⎜
The distance from the central maximum to the m = 1 bright fringe on a screen at distance L is Error! Objects cannot be created from editing field codes.
(Note that the small angle approximation is not valid for the maxima of diffraction gratings, which almost always have angles > 10°. ) There are two m = 1 bright fringes, one on either side of the central maximum. The distance between them is Δy = 2 y1 = 1.258 m ≈ 1.3 m. (b) The maximum number of fringes is determined by the maximum value of m for which sinθm does not exceed 1 because there are no physical angles for which sinθ > 1. In this case,
sinθ m =
mλ m ( 500 nm ) = d 1667 nm
We can see by inspection that m = 1, m = 2, and m = 3 are acceptable, but m = 4 would require a physically impossible sinθ 4 > 1. Thus, there are three bright fringes on either side of the central maximum plus the central maximum itself for a total of seven bright fringes.
22.42. Model: A diffraction grating produces a series of constructive-interference fringes at values of θ m that are determined by Equation 22.15. Solve: For the m = 3 maximum of the red light and the m = 5 maximum of the unknown wavelength, Equation 22.15 gives
d sinθ 3 = 3 ( 660 ×10−9 m ) d sin θ 5 = 5λunknown The m = 5 fringe and the m = 3 fringe have the same angular positions. This means θ5 = θ3. Dividing the two equations, 5λunknown = 3 ( 660 ×10−9 m ) ⇒ λunknown = 396 nm
22.43. Model: A diffraction grating produces an interference pattern that is determined by both the slit spacing and the wavelength used. The visible spectrum spans the wavelengths 400 nm to 700 nm. Solve: According to Equation 22.16, the distance ym from the center to the mth maximum is ym = L tan θ m . The
angle of diffraction is determined by the constructive-interference condition d sin θ m = mλ , where m = 0, 1, 2, 3, … The width of the rainbow for a given fringe order is thus w = yred − yviolet. The slit spacing is d=
1 mm 1.0 ×10−3 m = = 1.6667 ×10−6 m 600 600
For the red wavelength and for the m = 1 order, d sinθ1 = (1) λ ⇒ θ1 = sin −1
λ d
= sin −1
700 ×10−9 m = 24.83° 1.6667 ×10−6 m
From the equation for the distance of the fringe,
yred = L tanθ1 = ( 2.0 m ) tan ( 24.83° ) = 92.56 cm Likewise for the violet wavelength,
⎛ 400 ×10−9 m ⎞ ⎟ = 13.88° ⇒ yviolet = ( 2.0 m ) tan (13.88° ) = 49.42 cm −6 ⎝ 1.6667 ×10 m ⎠
θ1 = sin −1 ⎜
The width of the rainbow is thus 92.56 cm − 49.42 cm = 43.14 cm ≈ 43 cm .
22.44. Model: A diffraction grating produces an interference pattern that is determined by both the slit spacing and the wavelength used. Solve: (a) If blue light (the shortest wavelengths) is diffracted at angle θ, then red light (the longest wavelengths) is diffracted at angle θ + 30°. In the first order, the equations for the blue and red wavelengths are
sin θ =
λb d
d sin (θ + 30° ) = λr
Combining the two equations we get for the red wavelength, ⎛ λb2 ⎞ ⎛ λb ⎞ ⎟ 0.8660 + d ( 0.50 ) ⎜1 − 2 ⎟ ⎝d ⎠ ⎝ d ⎠
λr = d ( sin θ cos30° + cosθ sin 30° ) = d ( 0.8660sinθ + 0.50cosθ ) = d ⎜ ⇒ ( 0.50 ) d 1 −
λb2 d
2
= λr − 0.8660λb ⇒ ( 0.50 ) ( d 2 − λb2 ) = ( λr − 0.8660λb ) 2
2
2
⎛ λ − 0.8660λb ⎞ 2 ⇒ d= ⎜ r ⎟ + λb 0.50 ⎝ ⎠ Using λb = 400 × 10−9 m and λr = 700 ×10−9 m , we get d = 8.125 × 10−7 m. This value of d corresponds to 1 mm 1.0 ×10−3 m = = 1230 lines/mm d 8.125 ×10−7 m (b) Using the value of d from part (a) and λ = 589 ×10−9 m , we can calculate the angle of diffraction as follows: d sin θ1 = (1) λ ⇒ ( 8.125 ×10−7 m ) sin θ1 = 589 ×10−9 m ⇒ θ1 = 46.5°
22.45. Model: A diffraction grating produces an interference pattern that is determined by both the slit spacing and the wavelength used. Solve: An 800 line/mm diffraction grating has a slit spacing d = (1.0 × 10−3 m) 800 = 1.25 × 10−6 m. Referring to Figure P22.45, the angle of diffraction is given by
tan θ1 =
y1 0.436 m = = 0.436 ⇒ θ1 = 23.557° ⇒ sin θ1 = 0.400 L 1.0 m
Using the constructive-interference condition d sin θ m = mλ ,
λ=
d sin θ1 = (1.25 × 10−6 m ) ( 0.400 ) = 500 nm 1
We can obtain the same value of λ by using the second-order interference fringe. We first obtain θ2: tan θ 2 =
y2 0.436 m + 0.897 m = = 1.333 ⇒ θ 2 = 53.12° ⇒ sinθ 2 = 0.800 L 1.0 m
Using the constructive-interference condition,
λ=
−6 d sin θ 2 (1.25 ×10 m ) ( 0.800 ) = = 500 nm 2 2
Assess: Calculations with the first-order and second-order fringes of the interference pattern give the same value for the wavelength.
22.46. Model: A diffraction grating produces an interference pattern that is determined by both the slit spacing and the wavelength used. Solve: From Figure P22.45, tan θ1 =
0.436 m = 0.436 ⇒ θ1 = 23.557° ⇒ sin θ1 = 0.400 1.0 m
Using the constructive-interference condition d sin θ m = mλ ,
d sin 23.557° = (1) ( 600 ×10−9 m ) ⇒ d =
600 ×10−9 m = 1.50 × 10−6 m sin ( 23.557° )
Thus, the number of lines per millimeter is 1.0 ×10−3 m = 670 lines/mm 1.50 ×10−6 m Assess: The same answer is obtained if we perform calculations using information about the second-order bright constructive-interference fringe.
22.47. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The diffraction-intensity pattern from a single slit will look like Figure 22.14. Solve: The dark fringes are located at
yp =
pλ L p = 0, 1, 2, 3, … a
The locations of the first and third dark fringes are y1 =
λL a
y3 =
3λ L a
Subtracting the two equations,
( y3 − y1 ) =
2 ( 589 ×10−9 m ) ( 0.75 m ) 2λ L 2λ L ⇒ a= = = 0.12 mm a y3 − y1 7.5 ×10−3 m
22.48. Visualize: The relationship between the diffraction grating spacing d, the angle at which a particular order of constructive interference occurs θ m , the wavelength of the light, and the order of the constructive interference m is d sin θ m = mλ . Also note N = 1/d . Solve: The first order diffraction angle for green light is
θ1 = sin −1 (λ/d ) = sin −1 (5.5 × 10−7 m/ 2.0 × 10−6 m) = sin −1 (0.275) = 0.278 rad = 16° Assess: This is a reasonable angle for a first order maximum.
22.49. Visualize: We are given L = 1.50 m and m = 2. We also know that for λ = 610 nm, y2 = 1.611 m, and we seek λ for y2 = 1.606 m. We will need to solve for d and use it to find the new λ. Use Equation 22.15: d sin θ m = mλ , and Equation 22.16: ym = L tanθ m . Solve:
d=
mλ = sin θ m
mλ (2)(610 nm) = = 1.667 mm ⎛ −1 ym ⎞ ⎛ −1 1.611 m ⎞ sin ⎜ tan sin tan ⎟ ⎜ ⎟ L ⎠ 1.50 m ⎠ ⎝ ⎝
Now that we know d we can use it to find the new wavelength.
λ= Assess:
d d y ⎞ 1.667 mm ⎛ −1 1.606 m ⎞ ⎛ sin θ m = sin ⎜ tan −1 m ⎟ = sin ⎜ tan ⎟ = 609 nm 2 1.50 m ⎠ m m ⎝ L ⎠ ⎝
The two bright peaks are caused by specific wavelengths of light that differ by only 1 nm.
22.50. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The diffraction-intensity pattern from a single slit will look like Figure 22.14. Solve: These are not small angles, so we can’t use equations based on the small-angle approximation. As given by Equation 22.19, the dark fringes in the pattern are located at a sin θ p = pλ , where p = 1, 2, 3, … For the first minimum of the pattern, p = 1. Thus, a
λ
=
p 1 = sin θ p sin θ1
For the three given angles the slit width to wavelength ratios are
1 ⎛a⎞ = 2, (a): ⎜ ⎟ = ⎝ λ ⎠30° sin 30°
1 ⎛a⎞ (b): ⎜ ⎟ = = 1.15, ⎝ λ ⎠60° sin 60°
1 ⎛a⎞ =1 (c): ⎜ ⎟ = ⎝ λ ⎠90° sin 90°
Assess: It is clear that the smaller the a/λ ratio, the wider the diffraction pattern. This is a conclusion that is contrary to what one might expect.
22.51. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The diffraction-intensity pattern from a single slit will look like Figure 22.14. Solve: 45 ° is not a small angle, so we can’t use equations based on the small-angle approximation. As given by Equation 22.19, the dark fringes in the pattern are located at a sin θ p = pλ , where p = 1, 2, 3, … For the first minimum to be at 45°, p = 1 and the width of the slit is
a=
pλ 633 ×10−9 m = = 895 nm sin θ p sin 45°
22.52. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The diffraction-intensity pattern from a single slit will look like Figure 22.14. Solve: As given by Equation 22.19, the dark fringes in the pattern are located at a sinθ p = pλ , where p = 1, 2, 3, … For the diffraction pattern to have no minima, the first minimum must be located at least at θ1 = 90°. From the constructive-interference condition a sin θ p = pλ , we have
a=
pλ λ ⇒a= = λ = 633 nm sin θ p sin 90°
22.53. Model: A narrow slit produces a single-slit diffraction pattern. Visualize:
The dark fringes in this diffraction pattern are given by Equation 22.21:
yp =
pλ L a
p = 1, 2, 3, …
We note that the first minimum in the figure is 0.50 cm away from the central maximum. We are given a = 0.02 nm and L = 1.5 m. Solve: Solve the above equation for λ .
λ=
y pa pL
Assess: 670 nm is in the visible range.
=
(0.50 × 10−2 m)(0.20 × 10−3 m) = 670 nm (1)(1.5 m)
22.54. Model: A narrow slit produces a single-slit diffraction pattern. Solve:
The dark fringes in this diffraction pattern are given by Equation 22.21:
pλ L p = 1, 2, 3, … a We note from Figure P22.53 that the first minimum is 0.50 cm away from the central maximum. Thus, yp =
L= Assess:
ay p pλ
( 0.15 ×10 m )( 0.50 ×10 (1) ( 600 ×10 m ) −3
=
This is a typical slit to screen separation.
−9
−2
m)
= 1.3 m
22.55. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Solve: Within the small angle approximation, which is almost always valid for the diffraction of light, the width of the central maximum is λL w = 2 y1 = 2.44 D
From Figure P22.53, w = 1.0 cm, so
D= Assess:
2.44λ L 2.44(500 × 10−9 m)(1.0 m) = = 0.12 mm w (1.0 × 10−2 m)
This is a typical size for an aperture to show diffraction.
22.56. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Solve: Within the small-angle approximation, the width of the central maximum is w = 2.44
λL D
Because w = D, we have
D = 2.44
λL D
⇒ D = 2.44λ L ⇒ D = (2.44)(633 × 10−9 m)(0.50 m) = 0.88 mm
22.57. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Solve: (a) Because the visible spectrum spans wavelengths from 400 nm to 700 nm, we take the average wavelength of sunlight to be 550 nm. (b) Within the small-angle approximation, the width of the central maximum is
w = 2.44
λL D
⇒ (1 × 10−2 m) = (2.44)
(550 × 10−9 m)(3 m) ⇒ D = 4.03 × 10−4 m = 0.40 mm D
22.58. Model: The antenna is a circular aperture through which the microwaves diffract. Solve: (a) Within the small-angle approximation, the width of the central maximum of the diffraction pattern is w = 2.44λL/D. The wavelength of the radiation is
λ=
c 3 × 108 m/s 2.44(0.025 m)(30 × 103 m) = = 0.025 m ⇒ w = = 920 m 9 f 12 × 10 Hz 2.0 m
That is, the diameter of the beam has increased from 2.0 m to 915 m, a factor of 458. (b) The average microwave intensity is
I=
100 ×103 W P = = 0.15 W/m 2 area π ⎡ 12 ( 915 m ) ⎤ 2 ⎣ ⎦
22.59. Model: The laser beam is diffracted through a circular aperture. Visualize:
Solve: (a) No. The laser light emerges through a circular aperture at the end of the laser. This aperture causes diffraction, hence the laser beam must gradually spread out. The diffraction angle is small enough that the laser beam appears to be parallel over short distances. But if you observe the laser beam at a large distance it is easy to see that the diameter of the beam is slowly increasing. (b) The position of the first minimum in the diffraction pattern is more or less the “edge” of the laser beam. For diffraction through a circular aperture, the first minimum is at an angle
θ1 =
−9 1.22λ 1.22 ( 633 × 10 m ) = = 5.15 × 10−4 rad = 0.029° D 0.0015 m
(c) The diameter of the laser beam is the width of the diffraction pattern:
w=
−9 2.44λ L 2.44 ( 633 × 10 m ) ( 3 m ) = = 0.00309 m = 0.31 cm D 0.0015 m
(d) At L = 1 km = 1000 m, the diameter is
w=
−9 2.44λ L 2.44 ( 633 × 10 m ) (1000 m ) = = 1.03 m ≈ 1.0 m D 0.0015 m
22.60. Model: The laser light is diffracted by the circular opening of the laser from which the beam emerges. Solve:
The diameter of the laser beam is the width of the central maximum. We have
w=
−9 8 2.44λ L 2.44 ( 532 × 10 m )( 3.84 × 10 m ) 2.44λ L = = 0.50 m ⇒ D= D w 1000 m
In other words, the laser beam must emerge from a laser of diameter 50 cm.
22.61. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize:
Solve:
(a) The m = 1 bright fringes are separated from the m = 0 central maximum by
Δy =
λL d
=
( 600 × 10
−9
m ) (1.0 m )
0.0002 m
= 0.0030 m = 3.0 mm
(b) The light’s frequency is f = c/λ = 5.00 × 1014 Hz. Thus, the period is T = 1/f = 2.00 × 10−15 s. A delay of 5.0 × 10−16 s = 0.50 × 10−15 s is 14 T . (c) The wave passing through the glass is delayed by
1 4
of a cycle. Consequently, the two waves are not in phase as
they emerge from the slits. The slits are the sources of the waves, so there is now a phase difference Δφ0 between the two sources. A delay of a full cycle (Δt = T) would have no effect at all on the interference because it corresponds to a phase difference Δφ0 = 2π rad. Thus a delay of 14 of a cycle introduces a phase difference Δφ0 = 1 4
( 2π ) = 12 π
rad.
(d) The text’s analysis of the double-slit interference experiment assumed that the waves emerging from the two slits were in phase, with Δφ0 = 0 rad. Thus, there is a central maximum at a point on the screen exactly halfway
between the two slits, where Δφ = 0 rad and Δr = 0 m. Now that there is a phase difference between the sources, the central maximum—still defined as the point of constructive interference where Δφ = 0 rad—will shift to one side. The wave leaving the slit with the glass was delayed by
1 4
of a period. If it travels a shorter distance to the
screen, taking of a period less than the wave coming from the other slit, it will make up for the previous delay and the two waves will arrive in phase for constructive interference. Thus, the central maximum will shift toward the slit with the glass. How far? A phase difference Δφ0 = 2π would shift the fringe pattern by Δy = 3.0 mm, 1 4
making the central maximum fall exactly where the m = 1 bright fringe had been previously. This is the point where Δr = (1)λ, exactly compensating for a phase shift of 2π at the slits. Thus, a phase shift of Δφ0 = 12 π = 14 ( 2π ) will shift the fringe pattern by 14 ( 3 mm ) = 0.75 mm . The net effect of placing the glass in the
slit is that the central maximum (and the entire fringe pattern) will shift 0.75 mm toward the slit with the glass.
22.62. Model: A diffraction grating produces an interference pattern, which looks like the diagram of Figure 22.8. Solve: (a) Nothing has changed while the aquarium is empty. The order of a bright (constructive interference) fringe is related to the diffraction angle θm by d sin θ m = mλ , where m = 0, 1, 2, 3, … The space between the slits is d=
1.0 mm = 1.6667 × 10−6 m 600
For m = 1, sin θ1 =
λ
⎛ 633 × 10−9 m ⎞ ⇒ θ1 = sin −1 ⎜ ⎟ = 22.3° −6 d ⎝ 1.6667 ×10 m ⎠
(b) The path-difference between the waves that leads to constructive interference is an integral multiple of the wavelength in the medium in which the waves are traveling, that is, water. Thus, 633 nm 633 nm λ 4.759 ×10−7 m λ= = = 4.759 ×10−7 m ⇒ sinθ1 = = = 0.2855 ⇒ θ1 = 16.6° nwater 1.33 d 1.6667 ×10−6 m
22.63. Model: An interferometer produces a new maximum each time L2 increases by 12 λ causing the pathlength difference Δr to increase by λ. Visualize: Please refer to the interferometer in Figure 22.20. Solve: The path-length difference between the two waves is Δr = 2L2 − 2L1. The condition for constructive interference is Δr = mλ, hence constructive interference occurs when
2 ( L2 − L1 ) = mλ ⇒ L2 − L1 = 12 mλ2 = 1200 ( 12 λ ) = 600λ where λ = 632.8 nm is the wavelength of the helium-neon laser. When the mirror M2 is moved back and a hydrogen discharge lamp is used, 1200 fringes shift again. Thus, L2′ − L1 = 1200 ( 12 λ ′ ) = 600λ ′ where λ ′ = 656.5 nm. Subtracting the two equations,
( L2 − L1 ) − ( L2′ − L1 ) = 600 ( λ − λ ′ ) = 600 ( 632.8 ×10−9 m − 656.5 ×10−9 m ) ⇒ L2′ = L2 + 14.2 ×10−6 m That is, M2 is now 14.2 μm closer to the beam splitter.
22.64.
Model: The gas increases the number of wavelengths in one arm of the interferometer. Each additional wavelength causes one bright-dark-bright fringe shift. Solve: From Equation 22.36, the number of fringe shifts is
Δm = m2 − m1 = ( n − 1)
2d
λvac
= (1.00028 − 1)
( 2 ) ( 2.00 ×10−2 m ) 600 ×10−9 m
= 19
22.65. Model: The water increases the number of wavelengths in one arm of the interferometer. Solve: (a) The incident light has λvac = 500 nm and f = c / λvac = 6.00 × 1014 Hz. Water doesn’t affect the frequency, which is still f w = f = 6.00 ×1014 Hz . The wavelength changes to λw = λvac / nw = 376 nm. (b) Light travels in a layer of thickness L twice—once to the right and then, after reflecting, once to the left—for a total distance 2L. The number of wavelengths in distance 2L is N = 2L/λ. If the 1 mm layer is a vacuum, the number of wavelengths is N vac =
2L
λvac
=
2 ( 0.001 m ) = 4000 500 ×10−9 m
If the vacuum is replaced by 1 mm of water, the number of wavelengths is Nw =
2L
λw
=
2 ( 0.001 m ) = 5319 376 × 10−9 m
So with the water added, the light travels ≈ 1320 extra wavelengths. (c) Each additional wavelength of travel shifts the pattern by 1 fringe. So the addition of the water shifts the interference pattern by ≈ 1320 fringes.
22.66. Model: The piece of glass increases the number of wavelengths in one arm of the interferometer. Each additional wavelength causes one bright-dark-bright fringe shift. Solve: We can rearrange Equation 22.36 to find that the index of refraction of glass is
n = 1+
λvac Δm 2d
( 500 ×10 nm ) ( 200 ) = 1.50 = 1+ 2 ( 0.10 ×10 m ) −9
−3
22.67. Model: The arms of the interferometer are of equal length, so without the crystal the output would be bright. Visualize: We need to consider how many more wavelengths fit in the electro-optic crystal than would have occupied that space (6.70 μ m ) without the crystal; if it is an integer then the interferometer will produce a bright output; if it is a half-integer then the interferometer will produce a dark output. But the wavelength we need to consider is the wavelength inside the crystal, not the wavelength in air.
λn =
λ n
We are told the initial n with no applied voltage is 1.522, and the wavelength in air is λ = 1.000 μ m. Solve: The number of wavelengths that would have been in that space without the crystal is
6.70 μ m = 6.70 1.000 μ m (a) With the crystal in place (and n = 1.522) the number of wavelengths in the crystal is
6.70 μ m = 10.20 1.000 μ m/1.522 10.20 – 6.70 = 3.50 which shows there are a half-integer number more wavelengths with the crystal in place than if it weren't there. Consequently the output is dark with the crystal in place but no applied voltage. (b) Since the output was dark in the previous part, we want it to be bright in the new case with the voltage on. That means we want to have just one half more extra wavelengths in the crystal (than if it weren't there) than we did in the previous part. That is, we want 4.00 extra wavelengths in the crystal instead of 3.5, so we want 6.70 + 4.00 = 10.70 wavelengths in the crystal.
6.70 μ m = 10.70 1.000 μ m/n
⇒
n=
10.70(1.000 μ m) = 1.597 6.70 μ m
Assess: It seems reasonable to be able to change the index of refraction of a crystal from 1.522 to 1.597 by applying a voltage.
22.68. Model: A diffraction grating produces an interference pattern like the one shown in Figure 22.8. We also assume that the small-angle approximation is valid for this grating. Solve: (a) The general condition for constructive-interference fringes is
d sin θ m = mλ
m = 0, 1, 2, 3, …
When this happens, we say that the light is diffracted at an angle θm. Since it is usually easier to measure distances rather than angles, we will consider the distance ym from the center to the mth maximum. This distance is ym = L tan θ m . In the small-angle approximation, sinθ m ≈ tanθ m , so we can write the condition for constructive interference as d
ym mλ L = mλ ⇒ ym = L d
The fringe separation is ym +1 − ym = Δy =
λL d
(b) Now the laser light falls on a film that has a series of “slits” (i.e., bright and dark stripes), with spacing
d′ =
λL d
Applying once again the condition for constructive interference: d ′ sin θ m = mλ ⇒ d ′
mλ L mλ L y′m = mλ ⇒ y′m = = = md d′ λL d L
The fringe separation is ym′ +1 − ym′ = Δy′ = d . That is, using the film as a diffraction grating produces a diffraction pattern whose fringe spacing is d, the spacing of the original slits.
22.69. Model: Two closely spaced slits produce a double-slit interference pattern. The interference pattern is symmetrical on both sides of the central maximum. Visualize:
In figure (a), the interference of laser light from the two slits forms a constructive-interference central (m = 0) fringe at P. The paths 1P and 2P are equal. When a glass piece of thickness t is inserted in the path 1P, the interference between the two waves yields the 10th dark fringe at P. Note that the glass piece is not present in figure (a). Solve: The number of wavelengths in the air-segment of thickness t is t m1 =
λ
The number of wavelengths in the glass piece of thickness t is t t nt m2 = = = λglass λ n λ The path length has thus increased by Δm wavelengths, where
Δm = m2 − m1 = ( n − 1)
t
λ
From the 10th dark fringe to the 1st dark fringe is 9 fringes and from the 1st dark fringe to the 0th bright fringe is one-half of a fringe. Hence,
1 19 19 t 19 λ 19 ( 633 ×10 m ) = ⇒ = ( n − 1) ⇒ t = = = 12.0 μm 2 2 2 2 ( n − 1) 2 1.5 − 1.0 λ −9
Δm = 9 +
22.70. Visualize: To I double
is I1 use Equation 22.14: 1 λL ⎛πd ⎞ to get the fraction desired. y ⎟ . Then divide by the distance to the first minimum y0 = = 4 I1 cos 2 ⎜ 2 d ⎝ λL ⎠
Solve:
find
the
y
location
where
the
intensity
First set I double = I1 .
⎛πd ⎞ I double = 4 I1 cos 2 ⎜ y⎟ ⎝ λL ⎠ ⎛πd I1 = 4 I1 cos 2 ⎜ ⎝ λL
⎞ y⎟ ⎠
1 ⎛πd = cos 2 ⎜ 4 ⎝ λL
⎞ y⎟ ⎠
1 ⎛πd = cos ⎜ 2 ⎝ λL
⎞ y⎟ ⎠
y=
λ L −1 ⎛ 1 ⎞ cos ⎜ ⎟ πd ⎝ 2⎠
Now set up the ratio that will give the desired fraction.
λ L −1 ⎛ 1 ⎞ cos ⎜ ⎟ y πd ⎝ 2 ⎠ = 2 cos −1 ⎛ 1 ⎞ = 2 π = 2 = ⎜ ⎟ 1 λL π y0′ ⎝ 2⎠ π 3 3 2 d Assess: The fraction must be less than 1, and 23 seems reasonable.
22.71. Model: A diffraction grating produces a series of constructive-interference fringes at values of θm that are determined by Equation 22.15. Solve: (a) The condition for bright fringes is d sin θ = mλ . If λ changes by a very small amount Δλ, such that θ changes by Δθ, then we can approximate Δλ/Δθ as the derivative dλ/dθ: 2
λ=
2
d d d Δλ d λ d ⎛ mλ ⎞ ⎛d⎞ 2 ≈ = cosθ = sin θ ⇒ 1 − sin 2 θ = 1− ⎜ ⎟ = ⎜ ⎟ −λ m m m Δθ dθ m ⎝ d ⎠ ⎝m⎠ Δλ ⇒ Δθ = 2 ⎛d⎞ 2 ⎜ ⎟ −λ ⎝m⎠
(b) We can now obtain the first-order and second-order angular separations for the wavelengths λ = 589.0 nm and λ + Δλ = 589.6 nm. The slit spacing is
d=
1.0 ×10−3 m = 1.6667 × 10−6 m 600
The first-order (m = 1) angular separation is Δθ =
0.6 ×10−9 m
2.7778 ×10−12 m 2 − 0.3476 ×10−12 m 2 = 3.85 ×10−4 rad = 0.022°
=
The second order (m = 2) angular separation is 0.6 × 10−9 m Δθ = = 1.02 × 10 −3 rad = 0.058° −12 2 −12 2 0.6945 × 10 m − 0.3476 × 10 m
0.6 ×10−9 m 1.5589 ×10−6 m
22.72. Model: The intensity in a double-slit interference pattern is determined by diffraction effects from the slits. Solve: (a) For the two wavelengths λ and λ + Δλ passing simultaneously through the grating, their first-order peaks are at y1 =
λL d
y1′ =
(λ + Δλ )L d
Subtracting the two equations gives an expression for the separation of the peaks: Δy = y1′ − y1 =
Δλ L d
(b) For a double-slit, the intensity pattern is
⎛πd ⎞ I double = 4 I1 cos 2 ⎜ y⎟ ⎝ λL ⎠ The intensity oscillates between zero and 4I1, so the maximum intensity is 4I1. The width is measured at the point where the intensity is half of its maximum value. For the intensity to be 12 I max = 2 I1 for the m = 1 peak:
π λL ⎛πd ⎞ ⎛πd ⎞ 1 πd 2 I1 = 4 I1 cos 2 ⎜ yhalf ⎟ ⇒ cos 2 ⎜ yhalf ⎟ = ⇒ yhalf = ⇒ yhalf = λ λ λ L L L 2 4 4d ⎝ ⎠ ⎝ ⎠ The width of the fringe is twice yhalf. This means w = 2 yhalf =
λL 2d
But the location of the m = 1 peak is y1 = λ L d , so we get w = 12 y1 . (c) We can extend the result obtained in (b) for two slits to w = y1 N . The condition for barely resolving two
diffraction fringes or peaks is w = Δymin. From part (a) we have an expression for the separation of the first-order peaks and from part (b) we have an expression for the width. Thus, combining these two pieces of information,
y1 Δλ L y d ⎛ λL ⎞ d λ = Δymin = min ⇒ Δλmin = 1 = ⎜ = ⎟ N d LN ⎝ d ⎠ NL N (d) Using the result of part (c), N=
λ 656.27 ×10−9 m = = 3646 lines Δλmin 0.18 ×10−9 m
22.73. Model: A diffraction grating produces an interference pattern like the one shown in Figure 22.8, when the incident light is normal to the grating. The equation for constructive interference will change when the light is incident at a nonzero angle.
Solve:
(a) The path difference between waves 1 and 2 on the incident side is Δr1 = d sin φ . The path
difference between the waves 1 and 2 on the diffracted side, however, is Δr2 = d sin θ . The total path difference between waves 1 and 2 is thus Δr1 + Δr2 = d ( sinθ + sin φ ) . Because the path difference for
constructive interference must be equal to mλ, d ( sin θ + sin φ ) = mλ
m = 0, ±1, ±2, …
(b) For φ = 30° the angles of diffraction are sin θ1 =
sin θ −1 =
(1) ( 500 ×10−9 m )
(1.0 ×10
−3
m ) 600
( −1) ( 500 ×10−9 m )
(1.0 ×10
−3
m ) 600
− sin 30° = −0.20 ⇒ θ1 = −11.5°
− sin 30° = −0.80 ⇒ θ −1 = −53.1°
22.74. Solve:
(a)
We have two incoming and two diffracted light rays at angles φ and θ and two wave fronts perpendicular to the rays. We can see from the figure that the wave 1 travels an extra distance Δr = d sin φ to reach the reflection spot. Wave 2 travels an extra distance Δr = d sin θ from the reflection spot to the outgoing wave front. The path difference between the two waves is
Δr = Δr1 − Δr2 = d ( sin θ − sin φ ) (b) The condition for diffraction, with all the waves in phase, is still Δr = mλ. Using the results from part (a), the diffraction condition is d sinθ m = mλ + d sin φ m = … −2, −1, 0, 1, 2, …
Negative values of m will give a different diffraction angle than the corresponding positive values. (c) The “zeroth order” diffraction from the reflection grating is m = 0. From the diffraction condition of part (b), this implies d sin θ 0 = d sin φ and hence θ 0 = φ . That is, the zeroth order diffraction obeys the law of reflection—the angle of reflection equals the angle of incidence. 1 (d) A 700 lines per millimeter grating has spacing d = 700 mm = 1.429 × 10−6 m = 1429 nm. The diffraction angles are given by
⎛ m ( 500 nm ) ⎞ ⎛ mλ ⎞ + sin φ ⎟ = sin −1 ⎜ + sin 40° ⎟ 1429 nm ⎝ d ⎠ ⎝ ⎠
θ m = sin −1 ⎜
(e)
M
θm
≤−5 −4 −3 −2 −1 0 1 ≥2
not defined −49.2° −24.0° −3.3° 17.0° 40.0° 83.1° not defined
22.75. Model: Diffraction patterns from two objects can just barely be resolved if the central maximum of one image falls on the first dark fringe of the other image. Solve: (a) Using Equation 22.24 with the width equal to the aperture diameter,
w=D=
2.44λ L ⇒ D = 2.44λ L = D
( 2.44 ) ( 550 ×10−7
m ) ( 0.20 m ) = 0.52 mm
(b) We can now use Equation 22.23 to find the angle between two distant sources that can be resolved. The angle is
α = 1.22
λ D
=
1.22 ( 550 ×10−9 m ) 0.52 ×10−3 m
= 1.29 ×10 −3 rad = 0.074°
(c) The distance that can be resolved is
(1000 m )α = (1000 m ) (1.29 × 10−3 rad ) = 1.3 m
22-1
23.1. Model: Light rays travel in straight lines. Solve:
(a) The time is t=
Δx 1.0 m = = 3.3 × 10−9 s = 3.3 ns c 3 ×108 m/s
(b) The refractive indices for water, glass, and cubic zirconia are 1.33, 1.50, and 1.96, respectively. In a time of 3.33 ns, light will travel the following distance in water:
⎛ c ⎞ ⎛ 3 × 108 m/s ⎞ −9 Δxwater = vwatert = ⎜ ⎟t = ⎜ ⎟ ( 3.33 × 10 s ) = 0.75 m 1.33 n ⎠ ⎝ water ⎠ ⎝ Likewise, the distances traveled in the glass and cubic zirconia are Δxglass = 0.667 m and Δxcubic zirconia = 0.458 m. Assess: The higher the refractive index of a medium, the slower the speed of light and hence smaller the distance it travels in that medium in a given time.
23.2. Model: Light rays travel in straight lines. Solve: Let tglass, toil, and tplastic be the times light takes to pass through the layers of glass, oil, and plastic. The time for glass is
tglass =
−2 Δx nglass (1.0 ×10 m ) (1.50 ) Δx Δx = = = = 0.050 ns vglass c nglass c 3.0 × 108 m/s
Likewise, toil = 0.243 ns and tplastic = 0.106 ns. Thus, ttotal = tglass + toil + tplastic = 0.050 ns + 0.243 ns + 0.106 ns = 0.399 ns = 0.40 ns. Assess: The small time is due to the high value for the speed of light.
23.3. Model: Light rays travel in straight lines. The light source is a point source. Visualize:
Solve:
Let w be the width of the aperture. Then from the geometry of the figure,
w 12.0 cm = ⇒ w = 8.0 cm 2.0 m 2.0 m + 1.0 m
23.4. Model: Light rays travel in straight lines. Also, the red and green light bulbs are point sources. Visualize:
Solve:
The width of the aperture is w = 1 m. From the geometry of the figure for red light,
w2 x′ = ⇒ x′ = 2w = 2 (1.0 m ) = 2.0 m 1m 3m + 1m
The red light illuminates the wall from x = 0.50 m to x = 4.50 m. For the green light, w4 x1 = ⇒ x1 = 1.0 m 1m 3m + 1m
3w 4 x2 = ⇒ x2 = 3.0 m 1m 3m + 1m
Because the back wall exists only for 2.75 m to the left of the green light source, the green light has a range from x = 0 m to x = 3.75 m.
23.5.
Visualize:
Note the similar triangles in this figure.
Solve:
15 cm d = 5 cm 180 cm d=
15 cm (180 cm) = 540 cm = 5.4 m 5 cm
Assess: This is a typical distance for photographs of people.
23.6. Model: Use the ray model of light. Visualize:
According to the law of reflection, θ r = θ i . Solve: From the geometry of the diagram,
θ i + φ = 90° Using the law of reflection, we get
θ r + ( 60° − φ ) = 90°
90° − φ = 90° − ( 60° − φ ) ⇒ φ = 30°
Assess: The above result leads to a general result for plane mirrors: If a plane mirror rotates by an angle φ relative to the horizontal, the reflected ray makes an angle of 2φ with the horizontal.
23.7. Model: Light rays travel in straight lines and follow the law of reflection. Visualize:
Solve:
We are asked to obtain the distance h = x1 + 5.0 cm. From the geometry of the diagram,
tan θ i =
x1 10 cm
tan θ r =
x2 15 cm
x1 + x2 = 10 cm
Because θ r = θ i , we have x1 x2 10 cm − x1 = = ⇒ (15 cm ) x1 = 100 cm 2 − 10 x1 ⇒ x1 = 4.0 cm 10 cm 15 cm 15 cm Thus, the ray strikes a distance 9.0 cm below the top edge of the mirror.
23.8. Model: Think of the view in the figure as a horizontal view of a vertical wall and the laser beam and hexagonal mirror in a vertical plane for ease of labeling. The laser beam will strike the highest spot on the wall when a new corner rotates into the laser beam and the angle the laser makes with the normal is greatest. We will compute how high on the wall this highest spot is from the center spot behind the laser; then we will multiply by two because symmetry says the reflected beam will hit the lowest spot just as the face rotates out of the laser beam (and the beam makes the largest angle with the normal in the downward direction), and then a new corner enters the beam with the reflection at the top again. Visualize: From the small right triangle inside the hexagon we deduce d = 0.20 m / sin 60°. Therefore, the distance from the wall to the corner of the hexagon just as it enters the laser beam is 2.0 m − d . This becomes the base of a large right triangle whose side on the wall is x and whose angle opposite x is 60°.
Solve:
Solve the large right triangle for x. tan 60° =
x 2.0 m − d
⎛ 0.20 m ⎞ x = (tan 60°)(2.0 m − d ) = (tan 60°) ⎜ 2.0 m − ⎟ = 3.06 m sin 60° ⎠ ⎝ Now, because of symmetry, double x to get the total length of the streak of laser light: 2 x = 6.1 m. Assess: The 50 cm distance from the laser to the center of the hexagon is irrelevant.
23.9. Model: Light rays travel in straight lines and follow the law of reflection. Visualize:
To determine the angle φ , we must know the point P on the mirror where the ray is incident. P is a distance x2 from the far wall and a horizontal distance x1 from the laser source. The ray from the source must strike P so that the angle of incidence θi is equal to the angle of reflection θr. Solve: From the geometry of the diagram, 1.5 m 3 m tan φ = = x1 + x2 = 5 m x2 x1 ⇒
⇒ tan φ =
1.5 m 3m 10 m = ⇒ (1.5 m ) x1 = 15 m 2 − ( 3 m ) x1 ⇒ x1 = 5 m − x1 x1 3
3m 9 = = 0.90 ⇒ φ = 42° x1 10
23.10. Model: Use the ray model of light. Visualize:
The arrow is at a distance s from the mirror, so its image is at a distance s′ behind the mirror. When you are at x = 0 m, a ray from the arrow’s head, after reflection from the mirror, is able to enter your eye. Similarly, a ray from the arrow’s tail, after normal incidence, is reflected into the eye. That is, the eye is able to see the arrow’s head and tail. While walking toward the right, a ray from the arrow’s head will reflect from the mirror’s right edge and enter your eye at P. A ray starting from the arrow’s tail will also enter your eye when you are at P. That is, while at P you will be able to see the entire image of the arrow. However, the light from the arrow’s head can never reach beyond point P. Solve: Point P is a distance x from the origin. From the geometry of the diagram, tan φ =
1 m 1 m x−2 m x−2 m 1 x−2m = = = ⇒ = ⇒ x = 3.5 m s 2 m s +1 m 3m 2 3m
Thus, the range of x over which you can see the entire arrow in the mirror is 0 m to 3.5 m.
23.11. Model: Use the ray model of light and the law of reflection. Visualize:
We only need one ray of light that leaves your toes and reflects in your eye. Solve: From the geometry of the diagram, the distance from your eye to the toes’ image is
2d = (400 cm) 2 + (165 cm) 2 = 433 cm Assess:
The light appears to come from your toes’ image.
23.12. Model: Use the ray model of light and Snell’s law. Visualize:
Solve:
According to Snell’s law for the air-water and water-glass boundaries,
nair sinθ air = nwater sin θ water
nwater sin θ water = nglass sin θ glass
From these two equations, we have nair sinθ air = nglass sin θ glass ⇒ sin θ glass =
nair ⎛ 1.0 ⎞ −1 ⎛ sin 60° ⎞ sin θ air = ⎜ ⎟ sin 60° ⇒ θ glass = sin ⎜ ⎟ = 35° nglass ⎝ 1.50 ⎠ ⎝ 1.5 ⎠
23.13. Visualize: Use Snell’s law n1 sinθ1 = n2 sinθ 2 . We are given θ 2 = θ CZ = 25°. We look up in Table 23.1 n1 = noil = 1.46 and n2 = nCZ = 1.96. Solve:
Solve the equation for θ1.
⎛ n2 ⎞ ⎛ 1.96 ⎞ sinθ 2 ⎟ = sin −1 ⎜ sin 25° ⎟ = 35° n 1.46 ⎝ ⎠ ⎝ 1 ⎠
θ1 = sin −1 ⎜
Assess: Since the ray goes into a material with higher index of refraction we know it bends toward the normal, so we expect θ1 > θ 2 ; this is the case.
23.14. Model: Use the ray model of light. The sun is a point source of light. Visualize:
A ray that arrives at the diver 50° above horizontal refracted into the water at θwater = 40°. Solve: Using Snell’s law at the water-air boundary
nair sinθ air = nwater sin θ water ⇒ sin θ air =
nwater ⎛ 1.33 ⎞ sin θ water = ⎜ ⎟ sin 40° nair ⎝ 1.0 ⎠
⇒ θair = 58.7° Thus the height above the horizon is θ = 90° − θair = 31.3° ≈ 31°. Because the sun is far away from the fisherman (and the diver), the fisherman will see the sun at the same angle of 31° above the horizon.
23.15. Model: Represent the laser beam with a single ray and use the ray model of light. Solve:
Using Snell’s law at the air-water boundary,
nair sinθ air = nliquid sinθ liquid ⇒ nliquid = nair Assess:
As expected, nliquid is larger than nair.
sin θ air ⎛ sin 37° ⎞ = 1.0 ⎜ ⎟ = 1.37 sin θ liquid ⎝ sin 26° ⎠
23.16. Model: Use the ray model of light. For an angle of incidence greater than the critical angle, the ray of light undergoes total internal reflection. Visualize:
Solve:
The critical angle of incidence is given by Equation 23.9:
⎛ ncladding ⎞ −1 ⎛ 1.48 ⎞ ⎟ = sin ⎜ ⎟ = 67.7° ⎝ 1.60 ⎠ ⎝ ncore ⎠
θ c = sin −1 ⎜
Thus, the maximum angle a light ray can make with the wall of the core to remain inside the fiber is 90° − 67.7° = 23.3°. Assess: We can have total internal reflection because ncore > ncladding.
23.17. Model: Use the ray model of light. For an angle of incidence greater than the critical angle, the ray of light undergoes total internal reflection. Visualize:
Solve:
The critical angle of incidence is given by Equation 23.9:
⎛ noil ⎞ ⎛ 1.46 ⎞ = sin −1 ⎜ ⎟ = 76.7° ⎜ n ⎟⎟ ⎝ 1.50 ⎠ ⎝ glass ⎠
θ c = sin −1 ⎜ Assess:
The critical angle exists because noil < nglass.
23.18. Model: Represent the can as a point source and use the ray model of light. Visualize:
Paraxial rays from the can refract into the water and enter into the fish’s eye. Solve: The object distance from the edge of the aquarium is s. From the water side, the can appears to be at an image distance s′ = 30 cm. Using Equation 23.13,
s′ =
30 cm n2 n ⎛ 1.33 ⎞ = 23 cm s = water s = ⎜ ⎟s ⇒ s = 1.33 n1 nair ⎝ 1.0 ⎠
23.19. Model: Represent the beetle as a point source and use the ray model of light. Visualize:
Paraxial rays from the beetle refract into the air and then enter into the observer’s eye. The rays in the air when extended into the plastic appear to be coming from the beetle at a shallower location, a distance s′ from the plastic-air boundary. Solve: The actual object distance is s and the image distance is s′ = 2.0 cm. Using Equation 23.13,
s′ = Assess:
n2 n 1.0 s = air s ⇒ 2.0 cm = s ⇒ s = 3.2 cm n1 nplastic 1.59
The beetle is much deeper in the plastic than it appears to be.
23.20. Model: Represent the diver’s head and toes as point sources. Use the ray model of light. Visualize:
Paraxial rays from the head and the toes of the diver refract into the air and then enter into your eyes. When these refracted rays are extended into the water, the head and the toes appear elevated toward you. Solve: Using Equation 23.13, n n n sT′ = 2 sT = air sT sH′ = air sH n1 nwater nwater Subtracting the two equations, her apparent height is
s′H − sT′ =
nair 1.0 ( sH − sT ) = (150 cm ) = 113 cm nwater 1.33
23.21. Model: Represent the aquarium’s wall as a point source, and use the ray model of light. Visualize:
Paraxial rays from the outer edge (O) are refracted into the water and then enter into the fish’s eye. When extended into the wall, these rays will appear to be coming from O′ rather from O. The point on the inside edge (I) of the wall will not change its apparent location. Solve: We are given that sO − sI = 4.00 mm and sO′ − sI′ = 3.50 mm. Using Equation 23.13,
sO′ = ⇒ sO′ − s′I =
nwater sO nwall
sI′ =
nwater sI nwall
nwater 1.33 4.00 mm ⎞ ( sO − sI ) ⇒ 3.50 mm = ( 4.00 mm ) ⇒ nwall = (1.33) ⎛⎜ ⎟ = 1.52 nwall nwall ⎝ 3.50 mm ⎠
23.22. Model: Use the ray model of light. Visualize:
Solve:
Using Snell’s law,
⎛ sin 30° ⎞ nair sin 30° = nred sin θ red ⇒ θ red = sin −1 ⎜ ⎟ = 19.2° ⎝ 1.52 ⎠ ⎛ sin 30° ⎞ nair sin 30° = nviolet sin θ violet ⇒ θ violet = sin −1 ⎜ ⎟ = 18.8° ⎝ 1.55 ⎠ Thus the angular spread is Δθ = θ red − θ violet = 19.2° − 18.8° = 0.4°
23.23. Model: Use the ray model of light and the phenomenon of dispersion. Visualize:
Solve: (a) From the graph in Figure 23.29, we estimate the index of refraction for the red light (656 nm) to be nred = 1.572 and for the blue light (456 nm) to be nblue = 1.587. (b) The angle of incidence onto the rear of the prism is 35°. Using these values for the refractive index and Snell’s law,
⎛ 1.572sin 35° ⎞ nred sin 35° = nair sin θ red ⇒ θ red = sin −1 ⎜ ⎟ = 64.4° 1.0 ⎝ ⎠
⎛ 1.587sin 35° ⎞ nblue sin 35° = nair sin θ blue ⇒ θ blue = sin −1 ⎜ ⎟ = 65.5° 1.0 ⎝ ⎠ ⇒ Δθ = θ blue − θ red = 1.1°
23.24. Model: Use the ray model of light and the phenomenon of dispersion. Visualize:
Solve:
Using Snell’s law for the red light,
nair sinθ air = nred sin θ red ⇒ 1.0sin θ air = 1.45sin 26.3° ⇒ θ air = sin −1 (1.45sin 26.3° ) = 40.0° Now using Snell’s law for the violet light, nair sin θ air = nviolet sinθ violet ⇒ 1.0sin 40.0° = nviolet sin 25.7° ⇒ nviolet = 1.48 Assess:
As expected, nviolet is slightly larger than nred.
23.25. Model: The intensity of scattered light is inversely proportional to the fourth power of the wavelength. −4 Solve: We want to find the wavelength of infrared light such that I IR = 0.01I 500 . Because I 500 ∝ ( 500 nm ) and I IR ∝ λ −4 , we have 4
I 500 ⎛ λ ⎞ =⎜ ⎟ = 100 ⇒ λ = 1580 nm I IR ⎝ 500 nm ⎠
23.26. Model: Use ray tracing to locate the image. Solve:
The figure shows the ray-tracing diagram using the steps of Tactics Box 23.2. You can see from the diagram that the image is in the plane where the three special rays converge. The image is inverted and is located at s′ = 20.0 cm to the right of the converging lens.
23.27. Model: Use ray tracing to locate the image. Solve:
The figure shows the ray-tracing diagram using the steps of Tactics Box 23.2. You can see from the diagram that the image is in the plane where the three special rays converge. The image is located at s′ = 15 cm to the right of the converging lens, and is inverted.
23.28. Model: Use ray tracing to locate the image. Solve:
The figure shows the ray-tracing diagram using the steps of Tactics Box 23.2. You can see that the rays after refraction do not converge at a point on the refraction side of the lens. On the other hand, the three special rays, when extrapolated backward toward the incidence side of the lens, meet at P′, which is 15 cm from the lens. That is, s′ = −15 cm. The image is upright.
23.29. Model: Use ray tracing to locate the image. Solve:
The figure shows the ray-tracing diagram using the steps of Tactics Box 23.3. The three rays after refraction do not converge at a point, but they appear to come from P′. P′ is 6 cm from the diverging lens, so s′ = −6 cm. The image is upright.
23.30. Model: Assume the biconvex lens is a thin lens.
Solve: If the object is on the left, then the first surface has R1 = +40 cm (convex toward the object) and the second surface has R2 = −40 cm (concave toward the object). The index of refraction of glass is n = 1.50, so the lensmaker’s equation is
⎛1 1 ⎞ 1 1 ⎞ ⎛ 1 = ( n − 1) ⎜ − ⎟ = (1.50 − 1) ⎜ − ⎟ ⇒ f = 40 cm f R R 40 cm 40 cm ⎠ − ⎝ 2 ⎠ ⎝ 1
23.31. Model: Assume the planoconvex lens is a thin lens.
Solve: If the object is on the left, then the first surface has R1 = ∞ and the second surface has R2 = −40 cm (concave toward the object). The index of refraction of polystyrene plastic is 1.59, so the lensmaker’s equation is
⎛1 1 ⎞ 1 1 ⎞ 1 0.59 ⎛1 ⇒ f = 68 cm = ( n − 1) ⎜ − ⎟ = (1.59 − 1) ⎜ − ⎟⇒ = f 40 cm f R R 40 cm ∞ − ⎝ ⎠ 2 ⎠ ⎝ 1
23.32. Model: Assume the biconcave lens is a thin lens.
Solve: If the object is on the left, then the first surface has R1 = −40 cm (concave toward the object) and the second surface has R2 = +40 cm (convex toward the object). The index of refraction of glass is 1.50, so the lensmaker’s equation is
⎛1 1 ⎞ 1 1 ⎞ 1 ⎞ ⎛ 1 ⎛ = ( n − 1) ⎜ − ⎟ = (1.50 − 1) ⎜ − ⎟ = ( 0.50 ) ⎜ − ⎟ ⇒ f = −40 cm f R R 40 cm 40 cm 20 cm ⎠ − + ⎝ ⎠ ⎝ 2 ⎠ ⎝ 1
23.33. Model: Assume the meniscus lens is a thin lens.
Solve: If the object is on the left, then the first surface has R1 = 30 cm (convex toward the object) and the second surface has R2 = 40 cm (convex toward the object). The index of refraction of polystyrene plastic is 1.59, so the lensmaker’s equation is
⎛1 1 ⎞ 1 1 ⎞ ⎛ 1 = ( n − 1) ⎜ − ⎟ = (1.59 − 1) ⎜ − ⎟ ⇒ f = 203 cm ≈ 200 cm f R R 30 cm 40 cm ⎠ ⎝ 2 ⎠ ⎝ 1
23.34. Model: The water is a spherical refracting surface. Consider the paraxial rays that refract from the air into the water. Solve: If the cat’s face is 20 cm from the edge of the bowl, then s = +20 cm. The spherical fish bowl surface has R = +25 cm, because it is the convex surface that is toward the object. Also n1 = 1 (air) and n2 = 1.33 (water). Using Equation 23.21,
n1 n2 n2 − n1 1 1.33 1.33 − 1 0.33 ⇒ + = = = 0.0132 cm −1 + = 20 cm s′ 25 cm 25 cm s s′ R
⇒
1.33 = ( 0.0132 − 0.050 ) cm −1 ⇒ s′ = −36 cm s′
This is a virtual image located 36 cm outside the fishbowl. The fish, inside the bowl, sees the virtual image. That is, the fish sees the cat’s face 36 cm from the bowl.
23.35. Model: Model the bubble as a point source and consider the paraxial rays that refract from the plastic into the air. The edge of the plastic is a spherical refracting surface. Visualize:
Solve: The bubble is at P, a distance of 2.0 cm from the surface. So, s = 2.0 cm. A ray from P after refracting from the plastic-air boundary bends away from the normal axis and enters the eye. This ray appears to come from P′, so the image of P is at P′ and it is a virtual image. Because P faces the concave side of the refracting surface, R = −4.0 cm. Furthermore, n1 = 1.59 and n2 = 1.0. Using Equation 23.21, n1 n2 n2 − n1 1.59 1.0 1.0 − 1.59 0.59 + = ⇒ + = =+ = 0.1475 cm −1 −4.0 cm s s′ R 2.0 cm s′ 4.0 cm
⇒
1 = 0.1475 cm −1 − 0.795 cm −1 ⇒ s′ = −1.54 cm s′
That is, the bubble appears 1.54 cm ≈ 1.5 cm beneath the surface.
23.36. Model: Assume the lens is thin. Visualize:
1 1 1 fs + = ⇒ s′ = s s′ f s− f Solve:
s′ =
fs (20 cm)(60 cm) = = 30 cm s− f 60 cm − 20 cm
The magnification is m = − s′ s = − 30 cm 60 cm = −1 2. This means the image is inverted and has a height of 0.50 cm. Assess: Ray tracing will confirm these results.
23.37. Solve: The image is at 40 cm as seen in the figure. It is inverted.
Assess:
When the object is outside the focal length we get an inverted image.
23.38. Solve: The image is at −30 cm as seen in the figure. It is upright.
Assess:
When the object is within the focal length we get a magnified upright image.
23.39. Solve: The image is at −12 cm as seen in the figure. It is upright.
Assess:
We expected an upright virtual image from the convex mirror.
23.40. Model: The speed of light in a material is determined by the refractive index as v = c n . Solve: To acquire data from memory, a total time of only 2.0 ns is allowed. This time includes 0.5 ns that the memory unit takes to process a request. Thus, the travel time for an infrared light pulse from the central processing unit to the memory unit and back is 1.5 ns. Let d be the distance between the central processing unit and the memory unit. The refractive index of silicon for infrared light is nSi = 3.5. Then,
1.5 ns =
(1.5 ns ) c = (1.5 ×10−9 s )( 3.0 ×108 m/s ) ⇒ d = 6.4 cm 2d 2d 2dnSi = = ⇒d = vSi c nSi c 2nSi 2 ( 3.5 )
23.41.
Model: Treat the red ball as a point source and use the ray model of light. Solve: (a) Using the law of reflection, we can obtain 3 images of the red ball. (b) The images of the ball are located at B, C, and D. Relative to the intersection point of the two mirrors, the coordinates of B, C, and D are: B(+1 m, −2 m), C(−1 m, +2 m), and D(+1 m, +2 m). (c)
23.42. Model: Treat the laser beam as a ray and use the ray model of light. Visualize:
Solve:
From the geometry of the mirrors and the rays, β = 50°, α =30°, and φ = 20°.
23.43. Model: For a mirror, the image distance behind the mirror equals the object’s distance in front of the mirror. Visualize:
Solve: Your face is 2.0 from the mirror into which you are looking. The image of your face (image 1) is 2.0 m behind the mirror, or 4.0 m away. Behind you, the image of the back of your head (image 2) is 3.0 m behind the mirror on the other wall. You can’t see this image because you’re looking to the right. However, the reflected rays that appear to come from image 2 (a virtual image) act just like the rays from an object—that is, just as the rays would if the back of your head were really at the position of image 2. These rays reflect from the mirror 2.0 m in front of you into which you’re staring and form an image (image 3) 8.0 m behind the mirror. This is the image of the back of your head that you see in the mirror in front of you. Since you’re 2.0 m from the mirror, the image of the back of your head is 10 m away.
23.44. Model: Treat the laser beam as a ray and use the ray model of light. Visualize:
As the cylinder rotates by an angle θ, the path of the reflected laser beam changes by an angle 2θ relative to the direction of incidence. Solve: Because the angle 2θ is very small,
tan 2θ ≅ 2θ =
1 180 2.0 ×10−3 m rad = degrees = 0.011° ⇒θ= 5.0 m 5000 π ( 5000 )
23.45. Model: Use the ray model of light. For an angle of incidence greater than the critical angle, the ray of light undergoes total internal reflection. Visualize:
For angles θwater that are less than the critical angle, light will be refracted into the air. Solve: Snell’s law at the water-air boundary is nair sinθ air = nwater sinθ water . Because the maximum angle of θair is 90°, we have 1 ⎞ (1.0 ) sin 90° = 1.33sinθ water ⇒ θ water = sin −1 ⎛⎜ ⎟ = 48.75° 1.33 ⎝ ⎠
Applying Snell’s law again to the glass-water boundary, ⎛n ⎞ ⎛ 1.33 ( sin 48.75° ) ⎞ nglass sinθ glass = nwater sinθ water ⇒ θ glass = sin −1 ⎜ water sin θ water ⎟ = sin −1 ⎜ ⎟ = 42° ⎜ nglass ⎟ 1.50 ⎝ ⎠ ⎝ ⎠
Thus 42° is the maximum angle of incidence onto the glass for which the ray emerges into the air.
23.46. Model: Use the ray model of light. Visualize:
Solve: When the plastic is in place, the microscope focuses on the virtual image of the dot. From the figure, we note that s = 1.0 cm and s′ = 1.0 cm − 0.4 cm = 0.6 cm. The rays are paraxial, and the object and image distances are measured relative to the plastic-air boundary. Using Equation 23.13,
s′ =
nair 1.0 1.0 cm = 1.67 s ⇒ 0.6 cm = (1.0 cm ) ⇒ nplastic = nplastic nplastic 0.6 cm
23.47. Model: Use the ray model of light and the law of refraction. Solve: Snell’s law at the air-glass boundary is nair sinθ air = nglass sin θ glass . We require θ glass = 12 θ air , so nair sin ( 2θ glass ) = nglass sin θ glass ⇒ nair ( 2sin θ glass cosθ glass ) = nglass sin θ glass
⎛n ⎞ ⎛ 1.50 ⎞ ⇒ θ glass = cos −1 ⎜ glass ⎟ = cos −1 ⎜ ⎟ = 41.4° ⇒ θ air = 82.8° ⎝ 2 × 1.0 ⎠ ⎝ 2nair ⎠
23.48. Model: Use the ray model of light and the law of refraction. Visualize:
Solve: (a) The ray of light strikes the meter stick at Pempty , which is a distance L from the zero mark of the meter
stick. So, tan 60° =
L ⇒ L = ( 50 cm ) tan 60° = 86.6 cm 50 cm
(b) The ray of light refracts at Phalf and strikes the meter stick a distance x1 + x2 from the zero of the meter stick.
We can find x1 from the triangle PfullPhalf O′: tan 60° =
x1 ⇒ x1 = ( 25 cm ) tan 60° = 43.30 cm 25 cm
We also have x2 = ( 25 cm ) tan φhalf . Using Snell’s law,
⎛ sin 60° ⎞ nair sin 60° = nwater sin φhalf ⇒ φhalf = sin −1 ⎜ ⎟ = 40.63° ⎝ 1.33 ⎠
⇒ x2 = ( 25 cm ) tan 40.63° = 21.45 cm ⇒ x1 + x2 = 43.30 cm + 21.45 cm = 64.8 cm (c) The ray of light experiences refraction at Pfull and the angle of refraction is the same as in part (b). We get
tan φfull =
x3 ⇒ x3 = ( 50 cm ) tan 40.63° = 42.9 cm 50 cm
23.49.
Model: Use the ray model of light. Light undergoes total internal reflection if it is incident on a boundary at an angle greater than the critical angle. Visualize:
Solve: (a) To reach your eye, a light ray must refract through the top surface of the water and into the air. You can see in the figure that rays coming from the bottom of the tank are incident on the top surface at fairly small angles, but rays from the marks near the top of the tank are incident at very large angles—greater than the critical angle. These rays undergo total internal reflection in the water and do not exit into the air where they can be seen. Thus you can see the marks from the bottom of the tank upward. (b) The highest point you can see is the one from which the ray reaches the top surface at the critical angle θc. For a water-air boundary, the critical angle is θc = sin–1(1/1.33) = 48.75°. You can see from the figure that the depth of this point is such that L L 65.0 cm = tanθ c ⇒ d = = = 57.0 cm d tanθ c tan(48.75°)
Since the marks are every 10 cm, the high mark you can see is the one at 60 cm.
23.50. Model: Use the ray model of light and the law of refraction. Assume the sun is a point source of light. Visualize:
When the bottom of the pool becomes completely shaded, a ray of light that is incident at the top edge of the swimming pool does not reach the bottom of the pool after refraction. Solve: The depth of the swimming pool is d = 4.0 m tan θ water . We will find the angle by using Snell’s law. We have 4.0 m ⎛ sin 70° ⎞ nwater sin θ water = nair sin 70° ⇒ θ water = sin −1 ⎜ = 4.0 m ⎟ = 44.95° ⇒ d = tan 44.95° ⎝ 1.33 ⎠
23.51. Model: Use the ray model of light and the law of refraction. Assume that the laser beam is a ray of light. Visualize:
The laser beam enters the water 2.0 m from the edge, undergoes refraction, and illuminates the goggles. The ray of light from the goggles then retraces its path and enters your eyes. Solve: From the geometry of the diagram, tan φ =
1.0 m ⇒ φ = tan −1 ( 0.50 ) = 26.57° ⇒ θ air = 90° − 26.57° = 63.43° 2.0 m
Snell’s law at the air-water boundary is nair sinθ air = nwater sinθ water . Using the above result,
(1.0 ) sin 63.43° = 1.33sinθ water
⎛ sin 63.43° ⎞ ⇒ θ water = sin −1 ⎜ ⎟ = 42.26° ⎝ 1.33 ⎠
Taking advantage of the geometry in the diagram again, x = tan θ water ⇒ x = ( 3.0 m ) tan 42.26° = 2.73 m 3.0 m
The distance of the goggles from the edge of the pool is 2.73 m + 2.0 m = 4.73 m ≈ 4.7 m.
23.52. Model: Use the ray model of light and the law of refraction. Assume that the laser beam is a ray of light. Visualize:
Solve:
(a) From the geometry of the diagram at side A,
tan φ =
10 cm ⎛ 10 ⎞ ⇒ φ = tan −1 ⎜ ⎟ ⇒ φ = 33.69° 15 cm ⎝ 15 ⎠
This means the angle of incidence at side A is θair = 90° − 33.69° = 56.31°. Using Snell’s law at side A, ⎛ 1.0sin 56.31° ⎞ nair sin θ air = nwater sinθ water A ⇒ θ water A = sin −1 ⎜ ⎟ = 38.73° 1.330 ⎝ ⎠ This ray of light now strikes side B. The angle of incidence at this water-air boundary is θ water B = 90° − θ water A = 51.3°. The critical angle for the water-air boundary is
⎛ nair ⎞ −1 ⎛ 1.0 ⎞ ⎟ = sin ⎜ ⎟ = 48.8° ⎝ 1.33 ⎠ ⎝ nwater ⎠
θ c = sin −1 ⎜
Because the angle θwater B is larger than θc, the ray will experience total internal reflection. (b) We will now repeat the above calculation with x = 25 cm. From the geometry of the diagram at side A, φ = 21.80° and θ air = 68.20°. Using Snell’s law at the air-water boundary, θ water A = 44.28° and θ water B = 45.72°. Because θ water B < θ c , the ray will be refracted into the air. The angle of refraction is calculated as follows: ⎛ 1.33sin 45.72° ⎞ nair sin θ air B = nwater sinθ water B ⇒ θ air B = sin −1 ⎜ ⎟ = 72.2° 1 ⎝ ⎠ (c) Using the critical angle for the water-air boundary found in part (a), θwater A = 90° − 48.75° = 41.25°. According to Snell’s law, ⎛ 1.33sin 41.25° ⎞ nair sin θ air = nwater sinθ water A ⇒ θ air = sin −1 ⎜ ⎟ = 61.27° 1.0 ⎝ ⎠
⇒ φ = 90° − 61.27° = 28.73° The minimum value of x for which the laser beam passes through side B and emerges into the air is calculated as follows: tan φ =
10 cm 10 cm ⇒ x= = 18.2 cm ≈ 18 cm tan 28.73° x
23.53. Model: Use the ray model of light. Assume the bonfire is a point source right at the corner of the lake. Visualize:
Solve: (a) Light rays from the fire enter the lake right at the edge. Even though the rays in air are incident on the surface at a range of angles from ≈0° up to 90°, the larger index of refraction of water causes the rays to travel downward in the water with angles ≤θc, the critical angle. Some of these rays can reach a fish that is deep in the lake, but a shallow fish out from shore is in the “exclusion zone” that is not reached by any rays from the fire. Thus a fish needs to be deep to see the light from the fire. (b) The shallowest fish that can see the fire is one that receives light rays refracting into the water at the critical angle θc. These are rays that were incident on the water’s surface at ≈90°. The critical angle for a water-air boundary is
⎛ 1.00 ⎞ ⎟ = 48.75° ⎝ 1.33 ⎠
θ c = sin −1 ⎜ The fish is 20 m from shore, so its depth is d=
20 m = 17.5 m ≈ 18 m tan(48.75°)
That is, a fish 20 m from shore must be at least 18 m deep to see the fire.
23.54. Model: Use the ray model of light. Assume that the target is a point source of light. Visualize:
Solve:
From the geometry of the figure with θair = 60°,
tan θ air =
x1 ⇒ x1 = ( 2.0 m )( tan 60° ) = 3.464 m 2.0 m
Let us find the horizontal distance x2 by applying Snell’s law to the air-water boundary. We have
⎛ sin 60° ⎞ nwater sinθ water = nair sin θ air ⇒ θ water = sin −1 ⎜ ⎟ = 40.63° ⎝ 1.33 ⎠ Using the geometry of the diagram, x2 = tanθ water ⇒ x2 = (1.0 m ) tan 40.63° = 0.858 m 1.0 m
To determine θtarget, we note that tanθ target =
3.0 m 3.0 m = = 0.6941 ⇒ θtarget = 35° x1 + x2 3.464 m + 0.858 m
23.55. Model: Use the ray model of light and the phenomena of refraction and dispersion. Visualize:
The refractive index of violet light is greater than the refractive index of red light. The violet wavelength thus gets refracted more than the red wavelength. Solve: Using Snell’s law for the red light at the air-glass boundary,
⎛ n sinθ air ⎞ −1 ⎛ 1.0sin 30° ⎞ nair sinθ air = nred sin θ red ⇒ θ red = sin −1 ⎜ air ⎟ = sin ⎜ ⎟ = 19.30° n ⎝ 1.513 ⎠ red ⎝ ⎠ From the geometry of the diagram, d red = tan θ red 10.0 cm
d violet = tanθ violet 10.0 cm
⇒ d red = (10.0 cm ) tan (19.30° ) = 3.502 cm ⇒ d violet = 3.502 cm − 0.1 cm = 3.402 cm
⎛ d ⎞ ⎛ 3.402 cm ⎞ ⇒ θ violet = tan −1 ⎜ violet ⎟ = tan −1 ⎜ ⎟ = 18.79° ⎝ 10.0 cm ⎠ ⎝ 10.0 cm ⎠ That is, white light is incident on a piece of glass at 30°, and it gets dispersed. The violet light makes an angle of 18.79° with the vertical. Using Snell’s law, nviolet sinθ violet = nair sin θ air ⇒ nviolet =
(1.0 ) sin 30° = 1.552 sin18.79°
23.56. Model: Use the ray model of light and the phenomena of refraction and dispersion. Visualize:
Solve: Since violet light is perpendicular to the second surface, it must reflect at θviolet = 30° at the first surface. Using Snell’s law at the air-glass boundary where the ray is incident,
nair sin θ air = nviolet sinθ violet ⇒ nviolet =
nair sin θ air (1.0 ) sin 50° = = 1.5321 sin θ violet sin 30°
Since nviolet = 1.02 nred, nred = 1.5021. Using Snell’s law for the red light at the first surface
⎛ 1.0sin 50° ⎞ nred sin θ red = nair sin θ air ⇒ θ red = sin −1 ⎜ ⎟ = 30.664° ⎝ 1.5021 ⎠ The angle of incidence on the rear face of the prism is thus θr glass = 30.664° − 30° = 0.664°. Using Snell’s law once again for the rear face and for the red wavelength, ⎛ n sin θ r glass ⎞ −1 ⎛ 1.5021sin 0.664° ⎞ nred sin θ r glass = nair sinθ r air ⇒ θ r air = sin −1 ⎜ red ⎟ = sin ⎜ ⎟ = 0.997° n 1.0 ⎝ ⎠ air ⎝ ⎠
Because θv air = 0° and θr air = 0.997°, φ = θ r air − θ v air = 0.997° ≅ 1.00°.
23.57. Model: Use the ray model of light and the phenomenon of refraction. Visualize:
Solve:
(a) The critical angle θc for the glass-air boundary is
⎛ 1.0 ⎞ nglass sin θ c = nair sin 90° ⇒ θ c = sin −1 ⎜ ⎟ = 41.81° ⎝ 1.50 ⎠ For the triangle ABC,
θ glass 1 + 120° + ( 90° − θ c ) = 180° ⇒ θ glass 1 = 180° − 120° − ( 90° − 41.81° ) = 11.81° Having determined θglass 1, we can now find θair 1 by using Snell’s law: ⎛ 1.50 × sin11.81° ⎞ nair sin θ air 1 = nglass sinθ glass 1 ⇒ θ air 1 = sin −1 ⎜ ⎟ = 17.88° 1.0 ⎝ ⎠ Thus, the smallest angle θ1 for which a laser beam will undergo TIR on the hypotenuse of this glass prism is 17.9°. (b) After reflecting from the hypotenuse (face 3) the ray of light strikes the base (face 2) and refracts into the air. From the triangle BDE,
( 90° − θ
glass 2
) + 60° + ( 90° − θ ) = 180° ⇒ θ c
glass 2
= 90° + 60° + 90° − 41.81° − 180° = 18.19°
Snell’s law at the glass-air boundary of face 2 is
⎛ n sinθ glass 2 ⎞ −1 ⎛ 1.50sin18.19° ⎞ nglass sinθ glass 2 = nair sin θ air 2 ⇒ θ air 2 = sin −1 ⎜ glass ⎟ = sin ⎜ ⎟ = 27.9° 1.0 n ⎝ ⎠ air ⎝ ⎠ Thus the ray exits 27.9° left of the normal.
23.58. Model: Use the ray model of light. Solve:
(a) Using Snell’s law at the air-glass boundary, with φ being the angle of refraction inside the prism,
nair sin β = n sin φ ⇒ sin β = n sin φ Considering the triangle made by the apex angle and the refracted ray,
( 90° − φ ) + ( 90° − φ ) + α = 180° ⇒ φ = 12 α Thus sin β = n sin ( 12 α ) ⇒ β = sin −1 ( n sin( 12 )α )
(b) Using the above expression, we obtain
n=
sin β sin 52.2° = = 1.58 sin( 12 α ) sin 30°
23.59. Model: The bubble is a point source of light. The surface is a spherical refracting surface.
Solve: The bubble is in zircon, so n1 = 1.96 and n2 = 1.00. The surface is concave (object facing into a “cave”) as seen from the bubble (the object), so R = −3.0 cm. Equation 23.21 is n1 n2 n2 − n1 1.96 1.00 1.00 − 1.96 + = ⇒ + = ⇒ s′ = −3.0 cm s s′ R 3.0 cm s′ −3.0 cm
Thus, seen from outside, the bubble appears to be 3.0 cm beneath the surface. That is, a bubble at the center actually appears to be at the center.
23.60. Model: Use the ray model of light. The surface is a spherically refracting surface. Visualize:
Solve: Because the rays are parallel, s = ∞. The rays come to focus on the rear surface of the sphere, so s′ = 2R, where R is the radius of curvature of the sphere. Using Equation 23.21,
n1 n2 n2 − n1 1 n n −1 + = ⇒ + = ⇒ n = 2.00 s s′ R ∞ 2R R
23.61. Model: Assume that the converging lens is a thin lens. Use ray tracing to locate the image. Solve:
(a)
The figure shows the ray-tracing diagram using the steps of Tactics Box 23.2. The three rays after refraction converge to give an image at s′ = 40 cm. The height of the image is h′ = 2 cm. (b) Using the thin-lens formula,
1 1 1 1 1 1 1 1 + = ⇒ + = ⇒ = ⇒ s′ = 40 cm s s′ f 40 cm s′ 20 cm s′ 40 cm The image height is obtained from M =−
s′ 40 cm =− = −1 s 40 cm
The image is inverted and as tall as the object, that is, h′ = 2.0 cm. The values for h′ and s′ obtained in parts (a) and (b) agree.
23.62. Model: Use ray tracing to locate the image. Assume that the converging lens is a thin lens. Solve:
(a)
The figure shows the ray-tracing diagram using the steps of Tactics Box 23.2. The three special rays that experience refraction do not converge at a point. Instead they appear to come from a point that is 15 cm on the same side as the object itself. Thus s′ = −15 cm. The image is upright and has a height of h′ = 1.5 cm. (b) Using the thin-lens formula,
1 1 1 1 1 1 1 1 =− ⇒ s′ = −15 cm + = ⇒ + = ⇒ ′ ′ ′ s s f 10 cm s 30 cm s 15 m The image height is obtained from M =−
s′ −15 cm =− = +1.5 s 10 cm
The image is upright and 1.5 times the object, that is, 1.5 cm high. These values agree with those obtained in part (a).
23.63. Model: Use ray tracing to locate the image. Assume that the converging lens is a thin lens. Solve:
(a)
The figure shows the ray-tracing diagram using the steps of Tactics Box 23.2. The three special rays after refracting do not converge. Instead the rays appear to come from a point that is 60 cm on the same side of the lens as the object, so s′ = −60 cm. The image is upright and has a height of 8.0 cm. (b) Using the thin-lens formula,
1 1 1 1 1 1 1 1 + = ⇒ =− ⇒ s′ = −60 cm + = ⇒ s s′ f 15 cm s′ 20 cm s′ 60 cm The image height is obtained from M =−
s′ −60 cm =− = +4 s 15 cm
Thus, the image is 4 times larger than the object or h′ = Mh = 4h = 4 ( 2.0 cm ) = 8.0 cm. The image is upright. These values agree with those obtained in part (a).
23.64. Model: Use ray tracing to locate the image. Assume the converging lens is a thin lens. Solve:
(a)
The figure shows the ray-tracing diagram using the steps of Tactics Box 23.2. After refraction, the three special rays converge and give an image 50 cm away from the converging lens. Thus, s′ = +50 cm. The image is inverted and its height is 0.65 cm. (b) Using the thin-lens formula,
1 1 1 1 1 1 1 1 ⇒ = + = ⇒ + = ⇒ s′ = 50 cm ′ ′ ′ s s f 75 cm s 30 cm s 50 cm The image height is obtained from M =−
s′ 50 cm 2 =− =− 75 cm 3 s
The image height is h′ = Mh = ( −2 / 3)(1 cm ) = −0.67 cm. Because of the negative sign, the image is inverted.
These results agree with those obtained in part (a).
23.65. Model: Use ray tracing to locate the image. Assume the diverging lens is a thin lens. Solve:
(a)
The figure shows the ray-tracing diagram using the steps of Tactics Box 23.3. After refraction, the three special rays do not converge. The rays, on the other hand, appear to meet at a point that is 8.5 cm on the same side of the lens as the object. So s′ = −8.5 cm. The image is upright and has a height of 1.1 cm. (b) Using the thin-lens formula, 1 7 60 1 1 1 1 1 1 ⇒ =− ⇒ s′ = − cm = −8.6 cm + = ⇒ + = 15 cm s′ −20 cm s s′ f s′ 60 cm 7
The image height is obtained from
( −60/7 cm ) = + 4 = 0.57 s′ =− 15 cm 7 s Thus, the image is 0.57 times larger than the object, or h′ = Mh = (0.57)(2.0 cm) = 1.14 cm. The image is upright because M is positive. These values agree, within measurement accuracy, with those obtained in part (a). M =−
23.66. Model: Use ray tracing to locate the image. Assume the diverging lens is a thin lens. Solve:
(a)
The figure shows the ray-tracing diagram using the steps of Tactics Box 23.3. After refraction from the diverging lens, the three special rays do not converge. However, the rays appear to meet at a point that is 20 cm on the same side as the object. So s′ = −20 cm. The image is upright and has a height of 0.3 cm. (b) Using the thin-lens formula,
1 1 1 1 1 1 = − = − =− ⇒ s′ = −20 cm ′ s f s −30 cm 60 cm 20 cm The image height is obtained from M =−
s′ −20 cm 1 =− = = 0.33 s 60 cm 3
Thus, h′ = Mh = ( 0.33)(1.0 cm ) = 0.33 cm, and the image is upright because M is positive. These values for s′ and h′ agree with those obtained in part (a).
23.67. Visualize: Refer to Figure 23.57. 1 1 1 fs + = ⇒ s′ = s s′ f s− f We are given f = 60 cm, s = 20 cm, and h = 1.0 cm. Solve: fs (60 cm)(20 cm) s′ = = = −30 cm s− f 20 cm − 60 cm The negative sign means the image is behind the mirror; it is a virtual image. The magnification is m = − s′ s = 30 cm 20 cm = 1.5. This means the image is upright and has a height of h′ = mh = (1.5)(1.0 cm) = 1.5 cm. Assess: Ray tracing will confirm these results.
23.68. Visualize: Refer to Figure 23.55. 1 1 1 fs + = ⇒ s′ = s s′ f s− f We are given f = −60 cm, s = 20 cm, and h = 1.0 cm. Solve:
s′ =
fs (−60 cm)(20 cm) = = −15 cm s− f 20 cm + 60 cm
The negative sign means the image is behind the mirror; it is a virtual image. The magnification is m = − s′ s = 15 cm 20 cm = 0.75. This means the image is upright and has a height of h′ = mh = (0.75)(1.0 cm) = 0.75 cm. Assess: Ray tracing will confirm these results.
23.69. Model: Assume the lens is a thin lens and the thin-lens formula applies. Solve: Because we want to form an image of the spider on the wall, the image is real and we need a converging lens. That is, both s′ and s are positive. This also implies that the spider’s image is inverted, so M = − s′ s = − 12 . Using the thin-lens formula with s′ = 12 s,
1 1 1 1 1 1 3 1 s + = ⇒ +1 = ⇒ = ⇒ f = s f 3 s s′ f s 2s f We also know that the spider is 2.0 m from the wall, so s + s′ = 2.0 m = s +
1 2
s ⇒ s = 13 ( 4.0 m ) = 133.3 cm
Thus, f = 13 s = 44 cm and s′ = 2.0 m − 1.33 m = 0.67 m = 67 cm. We need a 44 cm focal length lens placed 67 cm from the wall.
23.70. Model: Assume the lens to be a thin lens. Solve: Because we want to form an image of the candle on the wall, we need a converging lens. We have s + s′ = 200 cm. Using the thin-lens formula,
1 1 1 1 1 1 + = ⇒ + = ⇒ s 2 − ( 200 cm ) s + 6400 cm 2 = 0 ′ s s f s 200 cm − s 32 cm The two solutions to this equation are s = 160 cm and 40 cm. When s = 160 cm, then s′ = 200 cm − 160 cm = 40 cm. The magnification is s′ 40 cm M =− =− = −0.25 s 160 cm
so the image is inverted and its height is (2.0 cm)(0.25) = 0.50 cm. When s = 40 cm, then s′ = 200 cm − 40 cm = 160 cm. The magnification is s′ 160 cm M =− =− = −4 s 40 cm so the image is again inverted and its height is (2.0 cm)(4) = 8.0 cm.
23.71. Model: The eye is a converging lens and assume it is a thin lens. Solve: (a) The diameter of an adult eyeball is typically 4.0 cm. (b) The near point distance is approximately 10 inches ≈ 25 cm. (c) Using the thin-lens formula,
1 1 1 1 1 1 1 29 + = ⇒ = ⇒ f ≈ 3.4 cm + = ⇒ 25 cm 4.0 cm f f 100 cm s s′ f
23.72. Model: Assume the projector lens is a thin lens. Solve:
(a) The absolute value of the magnification of the lens is
M =
h′ 98 cm = = 49 h 2 cm
Because the projector forms a real image of a real object, the image will be inverted. Thus, M = −49 = −
s′ ⇒ s′ = 49 s s
We also have s + s′ = 300 cm ⇒ s + 49s = 300 cm ⇒ s = 6.0 cm ⇒ s′ = 294 cm
Using these values of s and s′, we can find the focal length of the lens:
1 1 1 1 1 = + = + ⇒ f = 5.9 cm ′ f s s 6.0 cm 294 cm (b) From part (a) the lens should be 6.0 cm from the slide.
23.73. Visualize: The lens must be a converging lens for this scenario to happen, so we expect f to be positive. In the first case the upright image is virtual ( s′2 < 0) and the object must be closer to the lens than the focal point. The lens is then moved backward past the focal point and the image becomes real ( s′2 > 0).
1 1 1 ss′ + = ⇒ f = s s′ f s + s′ We are given s1 = 10cm and m1 = 2. Solve: Since the first image is virtual, s′ < 0. We are told ′ m1 = 2 = − s1 s1 ⇒ s1′ = −20 cm. We can now find the focal length of the lens. f =
the
first
magnification
is
s1s1′ (10 cm)(−20 cm) = = 20 cm s1 + s1′ 10 cm − 20 cm
After the lens is moved, m2 = −2 = − s2′ s2 . Start with the thin lens equation again.
1 1 1 + = s2 s′2 f Replace s′2 with − m2 s2 . 1 1 1 + = s2 − m2 s2 f Now solve for s2 . −m2 s2 + s2 1 = s2 (−m2 s2 ) f −m2 s2 + s2 1 = f − m2 s22 Cancel one s2 . m2 − 1 1 = m2 s2 f Multiply both sides by fs2 . ⎛ m −1⎞ ⎛ −2 − 1 ⎞ s2 = f ⎜ 2 ⎟ = (20 cm) ⎜ ⎟ = 30 cm ⎝ −2 ⎠ ⎝ m2 ⎠ The distance the lens moved is s2 − s1 = 30 cm − 10 cm = 20 cm. Assess: We knew s2 needed to be bigger than f; it is, and is in a reasonable range. The final answer for the distance the lens moved also seems reasonable.
23.74. Model: Assume the symmetric converging lens is a thin lens. Solve: Because the lens forms a real image on the screen of a real object, the image is inverted. Thus, M = −2 = − s′ s . Also, s + s′ = 60 cm ⇒ s +2s = 60 cm ⇒ s = 20 cm ⇒ s′ = 40 cm
We can use the thin-lens formula to determine the radius of curvature of the symmetric converging lens ( R1 = R2 ) as follows:
⎛1 1 ⎞ 1 1 1 + = = ( n − 1) ⎜ − ⎟ s s′ f ⎝ R1 R2 ⎠ Using R1 = + R (convex toward the object), R2 = − R (concave toward the object), and n = 1.59,
⎛ 1 1 1 1 + = (1.59 − 1) ⎜⎜ − R − R 20 cm 40 cm ⎝
⎞ 3 1.18 ⇒ R = 15.7 cm ≈ 16 cm = ⎟⎟ ⇒ 40 cm R ⎠
23.75. Visualize: We are given f = R 2 = 40 cm 2 = 20 cm. We are also given m = − s′ s = 3. Solve:
Solve the thin lens equation for s′.
1 1 1 + = s s′ f s′ =
fs s− f
Plug this into the magnification equation, s = − s′ m . s=−
1 fs m s− f
Cancel an s from the numerator of each side, and multiply both sides by s − f . s− f =− s= f −
f m
f 1⎞ ⎛ ⎛ 1⎞ = f ⎜1 − ⎟ = ( 20 cm ) ⎜ 1 − ⎟ = 13.3 cm m ⎝ m⎠ ⎝ 3⎠
Assess: This answer is within the focal length of the concave mirror as we expect for an upright, magnified, virtual image.
23.76. Visualize: We are given h = 2.0 cm and h′ = 1.0 cm so we know m = 0.5. Solve:
Solve m = − s′ s for s . s=
Plug this result into the thin lens equation.
− s′ m
1 1 1 ss′ + = ⇒ f = s s′ f s + s′
⎛ − s′ ⎞ ′ ⎛ − s′ ⎞ ′ ⎜ ⎟( s ) ⎜ ⎟(s ) ss′ m ⎠ m ⎠ f = =⎝ =⎝ − s′ 1⎞ ⎛ s + s′ + s′ s′ ⎜ 1 − ⎟ m ⎝ m⎠ Cancel one s′ and distribute the m in the denominator. − s′ − s′ 150 cm = = = −300 cm f = 1 ⎞ m − 1 0.5 − 1 ⎛ m ⎜1 − ⎟ ⎝ m⎠ Assess: The negative value for f tells us this is a convex mirror.
23.77. Visualize: We are given f = 25 cm. From the thin mirror equation we know that s′ =
fs . The five s− f
points under consideration are s = 50 cm, 75 cm, 100 cm, 125 cm, 150 cm. fs Solve: (a) Use s′ = for each of the five points. s− f s (cm) s′ (cm) h (cm) h′ (cm) 50 50 10 10 75 37.5 10 5.0 100 33.3 10 3.3 125 31.3 10 2.5 150 30 10 2.0 (b)
(c) The image is curved and not parallel to the axis. Assess: For further reading, see “Longitudinal Magnification Drawing Mistake” by Héctor Rabal, Nelly Cam, and Marcelo Trivi in The Physics Teacher, vol. 42, January 2004, pp. 31–33, but know that Equation 4 there is missing a couple of minus signs.
23.78. Visualize: First concentrate on the optic axis and the ray parallel to it. Geometry says if parallel lines are both cut by a diagonal (in this case the line through the center of curvature and normal to the mirror at the point of incidence) the interior angles are equal; so φ = θ i . The law of reflection says that θ i = θ r , so we conclude φ = θ r . Now concentrate on the triangle whose sides are R, a, and b. Because two of the angles are equal then it is isosceles; therefore b = a. Apply the law of cosines to this triangle. Solve:
b 2 = a 2 + R 2 − 2aR cos φ Because a = b, they drop out. R 2 = 2aR cos φ R = 2a cos φ
We want to know how big a is in terms of R, so solve for a. a=
R 2cos φ
If φ 1 then cos φ ≈ 1, so in the limit of small φ , a = R 2, and then since f = R − a it must also be that f = Assess:
R 2
Many textbooks forget to stress that f = R 2 only in the limit of small φ , i.e., for paraxial rays.
23.79. Model: Use the ray model of light and assume the lens is a thin lens. Visualize: Please refer to Figure 23.47. Solve: Let n1 be the refractive index of the fluid and n2 the refractive index of the lens. The lens consists of two spherical surfaces having radii of curvature R1 and R2 and the lens thickness t → 0. For the refraction from the surface with radius R1, we use Equation 23.21:
n1 n2 n2 − n1 + = s1 s1′ R1 For the refraction from surface with radius R2,
n2 n1 n1 − n2 + = R2 − s1′ s′2 A negative sign is used with s1′ because the image formed by the first surface of the lens is a virtual image. This virtual image is the object for the second surface. Adding the two equations,
⎛1 1 ⎞ n1 n1 1 1 1 (n − n ) ⎛ 1 1 ⎞ + = n2 − n1 ⎜ − ⎟ ⇒ + = = 2 1 ⎜ − ⎟ ′ s1 s2 s1 s′2 f n1 ⎝ R1 R2 ⎠ ⎝ R1 R2 ⎠ (b) In air, R1 = +40 cm (convex toward the object), R2 = −40 cm (concave toward the object), n1 = 1.0, and n2 = 1.50. So,
1 ⎛ 1.50 − 1.0 ⎞⎛ 1 1 ⎞ =⎜ − ⎟⎜ ⎟ ⇒ f = 40 cm f ⎝ 1.0 ⎠⎝ 40 cm −40 cm ⎠ In water, n1 = 1.33 and n2 = 1.50. So, 1 ⎛ 1.50 − 1.33 ⎞⎛ 1 1 ⎞ =⎜ − ⎟⎜ ⎟ ⇒ f = 156 cm f ⎝ 1.33 ⎠⎝ 40 cm −40 cm ⎠
23.80. Model: Use the ray model of light. Solve:
(a) The time (t) is the time to travel from A to the interface (t1) and from the interface to B (t2). That is,
t = t1 + t2 =
d1 d 2 d d nd n d n n + = 1 + 2 = 1 1 + 2 2 = 1 x2 + a2 + 2 v1 v2 c n1 c n2 c c c c
( w − x)
2
+ b2
(b) Because t depends on x and there is only one value of x for which the light travels from A to B in the least possible amount of time, we have
n2 ( w − x ) dt n1 x =0= − 2 2 2 dx c x +a c ( w − x ) + b2 The solution (hard to do!) would give xmin. (c) From the geometry of the figure, x x2 + a2
=
x = sin θ1 d1
w− x
( w − x)
2
+b
2
=
w− x = sin θ 2 d2
Thus, the condition of part (b) becomes n1 n sin θ1 − 2 sinθ 2 = 0 ⇒ n1 sin θ1 = n2 sinθ 2 c c
23.81. Model: Assume the ray model of light. The ball is not a thin lens. However, the image due to refraction from the first surface is the object for the second surface. Visualize:
Solve:
(a) For refraction from the first surface, R = +5 cm (convex toward the object). Thus,
n1 n2 n2 − n1 1.0 1.50 0.50 1.50 1 + = ⇒ + = ⇒ =− ⇒ s1′ = −22.5 cm s1 s1′ R 6 cm s1′ 5 cm s1′ 15 cm The image is virtual (to the left of the surface) and upright. For refraction from the second surface, s2 = 22.5 cm + 10 cm = 32.5 cm and R = −5.0 cm (concave toward the object). Thus,
1.50 1.0 1 − 1.50 1 1 1 1.50 + = = = − ⇒ s2′ = 18.6 cm ⇒ 32.5 cm s′2 −5.0 cm 10 cm s2′ 10 cm 32.5 cm The image is 18.6 cm from the right edge of the ball and thus 23.6 cm from the center. (b) The ray diagram showing the formation of the image is shown above. (c) Using the thin-lens equation, 1 1 1 1 1 1 + = ⇒ + = ⇒ f = 7.5 cm 6 cm + 5 cm 18.6 cm + 5 cm f s s′ f
23.82. Model: Use the ray model of light. Visualize:
The angle of refraction is θ 2 + δθ for those wavelengths that have a refractive index of n + δn. Solve: (a) Applying Snell’s law to the diagrams,
(1) sinθ1 = n sinθ 2
(1) sinθ1 = ( n + δ n ) sin (θ 2 + δθ )
Equating the right hand sides of the above two equations and using the formula for the sine of a sum, n sin θ 2 = ( n + δ n )( sin θ 2 cos δθ + cosθ 2 sin δθ ) = ( n + δ n )( sin θ 2 + cosθ 2δθ ) where we have assumed that δθ θ . Multiplying the expressions, n sinθ 2 = n sinθ 2 + n cosθ 2δθ + δ n sinθ 2 + δ nδθ cosθ 2 We can ignore the last term on the right-hand side because it is the product of two small terms. The equation becomes
⎛δn⎞ n cosθ 2δθ = −δ n sin θ 2 ⇒ δθ = − tan θ 2 ⎜ ⎟ ⎝ n ⎠ Note that δθ has to be in radians. We can obtain the same result in the following way as well. From Snell’s law, sin θ 2 =
sin θ1 n
Differentiating relative to n
δθ n sin θ 2 − sinθ 2 ⎛ sinθ 2 ⎞ ⎛ 1⎞ = ( sinθ1 ) ⎜ − 2 ⎟ = − = ⎟ = cosθ 2 δ n δ n n n2 n ⎝ ⎠ ⎝ ⎠ ⎛δn⎞ ⇒ δθ = − tanθ 2 ⎜ ⎟ ⎝ n ⎠
δ⎜
(b) We have θ1 = 30° and nred = 1.552. Because the red wavelength is larger than the violet wavelength, nred < nviolet. Also, if the refraction angle for the red light is θ2, the refraction angle for the violet is less than θ2. Thus, δθ = −0.28°. From the formula obtained in part (a),
n ⎛δn⎞ δθ ⎟ ⇒ δn = − θ2 tan n ⎝ ⎠
δθ = − tanθ 2 ⎜ To determine tanθ2, we note that
⎛ sin 30° ⎞ nred sinθ 2 = nair sin 30° ⇒ θ 2 = sin −1 ⎜ ⎟ = 18.794° ⇒ tan θ 2 = 0.3403 ⎝ 1.552 ⎠ Thus, the expression for the change in the index of refraction is
δn = −
1.552 π rad ⎞ ( −0.28° ) ⎛⎜ ⎟ = 0.0223 ⇒ nviolet = nred + δn = 1.552 + 0.022 = 1.574 0.3403 ⎝ 180° ⎠
23.83. Visualize: The lateral magnification is m = − s′ s . Solve: We need to solve the thin-lens equation for s′.
1 1 1 + = s s′ f 1 1 1 = − s′ f s
s′ =
1 fs = 1 1 s− f − f s
Now insert this expression for s′ into the expression for m. fs s′ f s− f m=− =− =− s s s− f Now define the longitudinal magnification as the rate of change of s′ with respect to s. M=
ds′ d d ⎛ fs ⎞ = ( s′ ) = ⎜ ⎟ ds ds ds ⎝ s − f ⎠
Use the quotient rule of differentiation. M=
d ⎛ fs ⎞ ( s − f ) f − fs −f2 = ⎜ ⎟= 2 2 ds ⎝ s − f ⎠ (s − f ) (s − f )
This last result is equal to − m2 , so M = − m 2 . Assess: For further reading, see “Longitudinal Magnification Drawing Mistake” by Héctor Rabal, Nelly Cam, and Marcelo Trivi in The Physics Teacher, vol. 42, January 2004, pp. 31–33, but know that Equation 4 there is missing a couple of minus signs.
23.84. Model: Assume the lens is thin so we can solve the thin-lens equation for s′ = Visualize:
fs . s− f
We are given f = 0.150 m and s = 10 m. We are also given ds dt = 5.0 m s. Refer to Problem
23.83 to learn about longitudinal magnification: M = ds′ ds = −m 2 . Solve:
The speed of the image is ds′ dt . Use the chain rule. Then apply m = − s′ s . ⎛ fs ⎜ s− f ′ ds′ ds′ ds ds d s s d s ⎛ ⎞ = =M = −m2 = − ⎜ ⎟ = −⎜ dt ds dt dt dt ⎝ s ⎝ s ⎠ dt 2
2
⎛ f ⎞ ds ds ′ = −⎜ ⎟ dt s =10 m ⎝ s − f ⎠ dt
Assess:
2
⎞ 2 ⎟ ds ⎛ f ⎞ ds ⎟ = −⎜ ⎟ ⎠ dt ⎝ s − f ⎠ dt
2
s =10 m
0.15 m ⎛ ⎞ =⎜ ⎟ ( 5.0 m s ) = −1.16 mm s ⎝ 10 m − 0.15 m ⎠
The negative answer indicates the image is moving toward the lens.
24.1. Model: Each lens is a thin lens. The image of the first lens is the object for the second lens. Visualize:
The figure shows the two lenses and a ray-tracing diagram. The ray-tracing shows that the lens combination will produce a real, inverted image behind the second lens. Solve: (a) From the ray-tracing diagram, we find that the image is ≈ 50 cm from the second lens and the height of the final image is 4.5 cm. (b) s1 = 15 cm is the object distance of the first lens. Its image, which is a virtual image, is found from the thinlens equation:
1 1 1 1 1 5 = − = − =− ⇒ s1′ = −24 cm s1′ f1 s1 40 cm 15 cm 120 cm The magnification of the first lens is m1 = −
s1′ ( −24 cm ) = 1.6 =− s1 15 cm
The image of the first lens is now the object for the second lens. The object distance is s2 = 24 cm + 10 cm = 34 cm. A second application of the thin-lens equation yields: 1 1 1 1 1 680 cm = − = − ⇒ s2′ = = 48.6 cm s2′ f 2 s2 20 cm 34 cm 14 The magnification of the second lens is m2 = −
s′2 48.6 cm =− = −1.429 s2 34 cm
The combined magnification is m = m1m2 = (1.6 )( −1.429 ) = −2.286 . The height of the final image is (2.286)(2.0 cm) =
4.57 cm. These calculated values are in agreement with those found in part (a).
24.2. Model: Each lens is a thin lens. The image of the first lens is the object for the second lens. Visualize:
The figure shows the two lenses and a ray-tracing diagram. The ray tracing shows that the lens combination will produce a virtual, inverted image in front of the second lens. Solve: (a) From the ray-tracing diagram, we find that the image is 20 cm in front of the second lens and the height of the final image is 2.0 cm. (b) s1 = 60 cm is the object distance of the first lens. Its image, which is a real image, is found from the thin-lens equation:
1 1 1 1 1 1 = − = − = ⇒ s1′ = 120 cm s1′ f s s1 40 cm 60 cm 120 cm The magnification of the first lens is
m1 = −
s1′ 120 cm =− = −2 s1 60 cm
The image of the first lens is now the object for the second lens. The object distance is s2 = 160 cm −120 cm = 40 cm. A second application of the thin-lens equation yields: 1 1 1 −1 1 =− + = + ⇒ s′2 = −20 cm s2′ s2 f 2 +40 cm −40 cm The magnification of the second lens is m2 = −
s′2 −20 cm =− = +0.5 s2 40 cm
The overall magnification is m = m1m2 = ( −2 )( 0.5 ) = −1.0 . The height of the final image is (+1.0)(2.0 cm) = 2.0
cm. The image is inverted because m has a negative sign. These calculated values are in agreement with those found in part (a).
24.3. Model: Each lens is a thin lens. The image of the first lens is the object for the second lens. Visualize:
The figure shows the two lenses and a ray-tracing diagram. The ray tracing shows that the lens combination will produce a real, upright image behind the second lens. Solve: (a) From the ray-tracing diagram, we find that the image is 10 cm behind the second lens and the height of the final image is 2 cm. (b) s1 = 20 cm is the object distance of the first lens. Its image, which is real and inverted, is found from the thin lens equation:
1 1 1 fs (10 cm)(20 cm) + = ⇒ s1′ = 1 1 = = 20 cm s1 s1′ f1 s1 − f1 20 cm − 10 cm The magnification of the first lens is m1 = −
s1′ 20 cm =− = −1 s1 20 cm
The image of the first lens is now the object for the second lens. The object distance is s2 = 30 cm − 20 cm = 10 cm. A second application of the thin-lens equation yields 1 1 1 fs (5 cm)(10 cm) + = ⇒ s2′ = 2 2 = = 10 cm s2 s2′ f 2 s2 − f 2 10 cm − 5 cm The magnification of the second lens is
m2 = −
s′2 10 cm =− = −1 s2 10 cm
The combined magnification is m = m1m2 = ( −1)(−1) = 1. The height of the final image is (1)(2.0 cm) = 2.0 cm. These calculated values are in agreement with those found in part (a). Assess: The thin-lens equation agrees with the ray tracing.
24.4. Model: Each lens is a thin lens. The image of the first lens is the object for the second lens. Visualize:
The figure shows the two lenses and a ray-tracing diagram. The ray tracing shows that the lens combination will produce a virtual, inverted image at the first lens. Solve: (a) From the ray-tracing diagram, we find that the image is 30 cm in front of the second lens and the height of the final image is 6 cm. (b) s1 = 20 cm is the object distance of the first lens. Its image, which is real and inverted, is found from the thin lens equation:
1 1 1 fs (10 cm)(20 cm) + = ⇒ s1′ = 1 1 = = 20 cm s1 s1′ f1 s1 − f1 20 cm − 10 cm The magnification of the first lens is m1 = −
s1′ 20 cm =− = −1 20 cm s1
The image of the first lens is now the object for the second lens. The object distance is s2 = 30 cm − 20 cm = 10 cm. A second application of the thin lens equation yields 1 1 1 fs (15 cm)(10 cm) + = ⇒ s′2 = 2 2 = = −30 cm s2 s2′ f 2 s2 − f 2 10 cm − 15 cm The magnification of the second lens is m2 = −
s′2 ( −30 cm) =− =3 10 cm s2
The combined magnification is m = m1m2 = ( −1)(3) = −3. The height of the final image is (3)(2.0 cm) = 6.0 cm. The image is inverted because m has a negative sign. These calculated values are in agreement with those found in part (a). Assess: The thin lens equation agrees with the ray tracing.
24.5. Model: Each lens is a thin lens. The image of the first lens is the object for the second lens. Visualize:
The figure shows the two lenses and a ray-tracing diagram. The ray tracing shows that the lens combination will produce a virtual, inverted image in front of the second lens. Solve: (a) From the ray-tracing diagram, we find that the image is 3.3 cm behind the second lens and the height of the final image is ≈ 0.7cm. (b) s1 = 20 cm is the object distance of the first lens. Its image, which is real and inverted, is found from the thin-lens equation:
1 1 1 fs (10 cm)(20 cm) + = ⇒ s1′ = 1 1 = = 20 cm s1 s1′ f1 s1 − f1 20 cm − 10 cm The magnification of the first lens is m1 = −
s1′ 20 cm =− = −1 s1 20 cm
The image of the first lens is now the object for the second lens. The object distance is s2 = 30 cm − 20 cm = 10 cm. A second application of the thin-lens equation yields 1 1 1 fs ( −5 cm)(10 cm) + = ⇒ s2′ = 2 2 = = −3.33 cm s2 s2′ f 2 s2 − f 2 10 cm + 5 cm The magnification of the second lens is m2 = −
s′2 (−3.3 cm) =− = 0.33 s2 10 cm
The combined magnification is m = m1m2 = (−1)(0.33) = −0.33. The height of the final image is (0.33)(2.0 cm) = 0.66 cm. The image is inverted because m has a negative sign. These calculated values are in agreement with those found in part (a). Assess: The thin-lens equation agrees with the ray tracing.
24.6. Model:
s W f so we can use Equation 24.1: m = − f / s.
Solve:
h′ = mh = −
f 15 mm h=− (2.0 m) = −3.0 mm s 10 m
The height of the image on the detector is 3.0 mm. Assess: This seems reasonable given typical focal lengths and detector sizes.
24.7. Visualize: Equation 24.2 gives f-number = f / D. Solve:
f-number = Assess:
f 35 mm = = 5.0 D 7.0 mm
This is in the range of f-numbers for typical camera lenses.
24.8. Visualize: Solve Equation 24.2 for D. Solve:
D=
f 12 mm = = 3.0 mm f-number 4.0
Assess: This is in the same ballpark as the example after Equation 24.2.
24.9. Visualize: First we compute the f-number of the first lens and then the diameter of the second. Solve:
f-number =
f 12 mm = = 3.0 D 4.0 mm
Now for the new lens.
D= Assess:
f 18 mm = = 6.0 mm f-number 3.0
Given the same f-number, the longer focal length lens has a larger diameter.
24.10. Visualize: We want the same exposure in both cases. The exposure depends on I Δtshutter . We'll also use Equation 24.3. Solve:
exposure = I Δt ∝
1 Δt ( f -number) 2
1 1 Δt = Δt ′ 2 ( f -number) ( f -number)′2 Δt ′ =
( f-number)′2 (4.0) 2 ⎛ 1 ⎞ 1 1 Δt = s⎟ = s≈ s ⎜ 2 ( f-number) (5.6) 2 ⎝ 125 ⎠ 245 250
Assess: An alternate approach without a lot of calculation is that since we changed the lens (opened) by one f stop that doubles the intensity so we need half the time interval to achieve the same exposure.
24.11. Visualize: We want the same exposure in both cases. The exposure depends on I Δtshutter . We'll also use Equation 24.3. The lens is the same lens in both cases, so f = f ′. Solve: exposure = I Δt ∝
D2 Δt f2
D2 D′2 Δ = Δt ′ t f2 f ′2
Solve for D′; then simplify. 2
⎛ f ′ ⎞ ⎛ Δt ⎞ Δt 1 125s = (3.0 mm) = (3.0 mm) 4 = 6.0 mm D′ = D 2 ⎜ ⎟ ⎜ ⎟=D ′ ′ Δt 1 500s ⎝ f ⎠ ⎝ Δt ⎠ Assess: Since we decreased the shutter speed by a factor of 4 we need to increase the aperture area by a factor of 4, and this means increase the diameter by a factor of 2.
24.12. Model: Ignore the small space between the lens and the eye.
Visualize: Refer to Example 24.4, but we want to solve for s′, the near point. Solve: (a) The power of the lens is positive which means the focal length is positive, so Ramon wears converging lenses. This is the remedy for hyperopia. (b) We want to know where the image should be for an object s = 25 cm given 1 f = 2.0 m −1. f =
s′ =
1 = 0.50 m P
fs (0.50 m)(0.25 m) = = −0.50 m s− f 0.25 m + 0.50 m
So the near point is 50 cm. Assess: The negative sign on s′ is expected because we need the image to be virtual.
24.13. Model: Ignore the small space between the lens and the eye.
Visualize: Refer to Example 24.5, but we want to solve for s′, the far point. Solve: (a) The power of the lens is negative which means the focal length is negative, so Ellen wears diverging lenses. This is the remedy for myopia. (b) We want to know where the image should be for an object s = ∞ m given 1 f = −1.0 m −1. f =
1 = −1.0 m P
1 1 1 + = s s′ f When s = ∞ m, 1 1 1 + = ⇒ s′ = f = −1.0 m ∞ m s′ f So the far point is 100 cm. Assess: The negative sign on s′ is expected because we need the image to be virtual.
24.14. Model: With normal vision the farthest distance at which a relaxed eye can focus (the far point) is infinity. Rays coming from infinity are nearly parallel, so the focal length of the lens would be 24 mm, the same as the length of the eye. However, the far point is less than infinity for many people, and most sources quote the focal length of the eye as 17 mm–22 mm. However, for simplicity of calculation, assume the vision is normal and the focal length of the lens/cornea combination is 24 mm Visualize: Equation 24.2 gives f-number = f D . Solve: (a) For the fully dilated pupil (dark-adapted eye):
f-number =
f 24 mm = = 3.0 D 8.0 mm
(b) For the fully contracted pupil (eye in bright light):
f 24 mm = = 16 D 1.5 mm These answers correspond to the values given in the text. f-number =
Assess:
24.15. Model: The angle subtended by the image is 8 × the angle subtended by the object. Visualize: The angle subtended by the object is h s . Solve:
h s
θ = (8×) = (8×) Assess: The binoculars do indeed help.
14 cm = 0.0622 rad = 3.6° 1800 cm
24.16. Visualize: Equation 24.10 relates the variables in question: M =−
L 25 cm fobj feye
We are given M = 500×, L = 20 cm, and feye = 5.0 cm Solve:
Solve for f obj.
fobj = − Assess:
L 25 cm 20 cm 25 cm =− = 0.20 cm = 2.0 mm −500 5.0 cm M feye
This is in the same ball park as the case in the book.
24.17. Visualize: Equation 24.10 relates the variables in question: M =−
L 25 cm fobj feye
We are given M = 100 ×, L = 160 mm, and fobj = 8.0 cm. Solve:
Solve for f obj. feye = −
Assess:
L 25 cm 160 mm 25 cm =− = 5.0 cm f obj M 8.0 mm ( −100)
This is the same feye as in the previous exercise.
24.18. Model: Assume the thin-lens equation is valid. For part (b) refer to Equation 24.11 and the definition of α . Visualize:
We are given fobj = 9.0 mm and m=−
Solve:
s′ = −40 ⇒ s′ = 40 s s
(a) Use the thin-lens equation.
1 1 1 + = s s′ fobj
1 1 1 + = s 40s fobj 41 1 = 40s fobj s=
41 41 f obj = (9.0 mm) = 9.2 mm 40 40
(b) We are given n = 1.00. Refer to Figure 24.14a to determine α .
α = tan −1 (3.0 mm / 9.0 mm) = tan -1 (1/ 3) ⎛ ⎛ 1 ⎞⎞ NA = n sin α = (1.00) sin ⎜ tan −1 ⎜ ⎟ ⎟ = 0.32 ⎝ 3 ⎠⎠ ⎝ Assess: We used s = f in this calculation as suggested in the text. If we had used s = 9.2 mm (from part (a)) we would get NA = 0.31 , only a little different. These values of NA are typical for a simple microscope.
24.19. Visualize: We are given NA = 0.90, L = 160 mm, mobj = −20, and the book says n = 1.46. Solve:
Start with Equation 24.11. NA = n sin α ⎛ NA ⎞ ⎟ ⎝ n ⎠ ⎛ ⎛ NA ⎞ ⎞ tan α = tan ⎜ sin −1 ⎜ ⎟⎟ ⎝ n ⎠⎠ ⎝
α = sin −1 ⎜
⎛1⎞ But from Figure 24.14a we also have tan α = ⎜ ⎟ D / f obj , where D is the diameter of the lens. So combine those ⎝ 2⎠ two expressions for tan α and solve for D. ⎛ ⎛ NA ⎞ ⎞ D = 2 f obj tan ⎜ sin −1 ⎜ ⎟⎟ ⎝ n ⎠⎠ ⎝ We need the side calculation using Equation 24.9: mobj = −
L L ⇒ f obj = − f obj mobj
Insert this back in the equation for D. ⎛ L ⎞ ⎛ −1 ⎛ NA ⎞ ⎞ D = 2⎜ − tan sin ⎜ ⎟⎟ ⎜ mobj ⎟⎟ ⎜⎝ ⎝ n ⎠⎠ ⎝ ⎠ ⎛ 160 mm ⎞ ⎛ −1 ⎛ 0.90 ⎞ ⎞ D = 2⎜ − ⎟ tan ⎜ sin ⎜ ⎟ ⎟ = 13 mm −20 ⎠ ⎝ ⎝ ⎝ 1.46 ⎠ ⎠ Assess:
The answer seems to be in a reasonable range for objective lens diameter.
24.20. Visualize: Figure 24.15 shows from similar triangles that for the eyepiece lens to collect all the light Dobj f obj
=
Deye f eye
We also see from Equation 24.12 that M = − f obj / f eye . We are given M = −20 and Dobj = 12 cm. Solve:
Deye = Dobj
f eye f obj
=
Dobj −M
=
12cm = 0.60 cm = 6.0 mm 20
Assess: The answer is almost as wide as a dark-adapted eye.
24.21. Model: Assume the eyepiece is a simple magnifier with M eye = 25 cm / f eye . Visualize:
f eye = 25 cm /10 = 2.5 cm.
Solve: (a) The magnification of a telescope is
M =−
f obj f eye
=
100 cm = 40 2.5 cm
(b)
f-number =
f 1.00 m = = 5.0 D 0.20 m
Assess: These results are in reasonable ranges for magnification and f-number.
24.22. Model: Diffraction prevents focusing light to an arbitrarily small point. Model the lens of diameter D as an aperture in front of an ideal lens with an 8.0 cm focal length. Solve: Assuming that the incoming laser beam is parallel, the focal length of the lens should be 8.0 cm. From Equation 24.13, the minimum spot size in the focal plane of this lens is w=
2.44 ( 633 ×10−9 m )( 8.0 ×10−2 m ) 2.44λ f ⇒ 10 × 10−6 m = ⇒ D = 0.012 m = 1.2 cm D D
24.23. Model: Two objects are marginally resolvable if the angular separation between the objects, as seen from the lens, is α = 1.22λ / D. Solve:
Let Δy be the separation between the two light bulbs, and let L be their distance from a telescope. Thus,
α=
−2 Δy D (1.0 m ) ( 4.0 ×10 m ) Δy 1.22λ = = 55 km ⇒ L= = 1.22λ L D 1.22 ( 600 ×10−9 m )
24.24. Visualize: Equation 24.15 gives the smallest resolvable distance: d min = 0.61λ / NA. We are given λ = 500 nm and NA = 1.0. d min =
Assess:
0.61λ (0.61)(500 nm) = = 305 nm ≈ 310 nm NA 1.0
The smallest object one can see is on the order of the wavelength.
24.25. Visualize: Equation 24.15 gives the smallest resolvable distance: d min = 0.61λ / NA. We are given λ = 600 nm and d min = 0.75μ m = 750 nm. Solve:
NA = Assess:
0.61λ (0.61)(600 nm) = = 0.49 d min 750 nm
This is in the normal range for NA.
24.26. Visualize: We are given h1 = 1.0 cm, s1 = 4.0 cm, f1 = 5.0 cm, and f 2 = −8.0 cm. Solve:
First compute the image from the first lens. s1′ =
f1s1 ( 5.0 cm )( 4.0 cm ) = −20 cm = s1 − f1 4.0 cm − 5.0 cm
h1′ = −h1
s1′ −20 cm = −(1.0 cm) = 5.0 cm s1 4.0 cm
This is a virtual, upright image 20 cm to the left of the first lens. The second lens is 12 cm to the right of the first one, so s2 = 32 cm. s′2 =
( −8.0 cm )( 32 cm ) = −6.4 cm f 2 s2 = s2 − f 2 32 cm − ( −8.0 cm) h2 = h1′ = 5.0 cm
h2′ = − h2
s′2 −6.4 cm = − ( 5.0 cm ) = 1.0 cm s2 32 cm
This is a virtual, upright image 6.4 cm to the left of the second lens (5.6 cm to the right of the first lens). The image is 1.0 cm tall (the same size as the object). Assess: Ray tracing confirms these results.
24.27. Model: The parallel rays can be considered to come from an object infinitely far away: s1 = ∞ . The lens is a diverging lens. Visualize: If s1 = ∞ the thin lens equation tells us that s1′ = f1′; we are given that f1 = −10 cm. We are also given for the mirror f 2 = 10 cm. Solve: Since s1′ = −10 cm the image is virtual 10 cm to the left of the lens. The image from the lens becomes the
object for the mirror ⇒ s2 = 30 cm; this is three times the mirror's focal length, or s2 = 3 f 2 . s2′ =
f ( 3 f2 ) 3 f 2 s2 = 2 = f 2 = 15 cm s2 − f 2 3 f 2 − f 2 2
Therefore the initial parallel rays are brought to a focus 15 cm to the left of the mirror, or 5 cm to the right of the lens. Assess: The answer is reasonable and can be verified by ray tracing.
24.28. Visualize: The object is within the focal length of the converging lens, so we expect the image to be upright, virtual, and to the left of the lens. The image of the lens becomes the object for the mirror, and we expect the second image to be upright and virtual behind (to the right of) the mirror. Solve: s1′ =
f1s1 (10 cm )( 5 cm ) = −10 cm = s1 − f1 5 cm − 10 cm
The image of the lens becomes the object for the mirror ⇒ s2 = 15 cm, and we expect the second image to be upright and virtual behind (to the right of) the mirror.
s1′ =
h′ = h
( −30 cm )(15 cm ) = −10 cm f 2 s2 = s2 − f 2 15 cm − ( −30 cm )
s1′ s′2 ⎛ −10 cm ⎞⎛ −10 cm ⎞ = (1.0 cm ) ⎜ ⎟⎜ ⎟ = 1.3 cm s1 s2 ⎝ 5.0 cm ⎠⎝ 15 cm ⎠
The final image is 10 cm to the right of the mirror, or 15 cm to the right of the lens. It is upright with a height of 1.3 cm. Assess: Ray tracing will verify the answer.
24.29. Solve: (a) The image location from the first lens is s1′ =
f1s1 ( −2.5 cm)(2.5 cm) = = −1.25 cm s1 − f1 2.5 cm − ( −2.5 cm)
So the image from the first lens is 1.25 cm to the left of the first lens, upright and virtual. Now, s2 = d + 1.25 cm. We are told the final image is at infinity: s′2 = ∞ ⇒ s2 = f 2 ⇒ f 2 = d + 1.25 cm d = f 2 − 1.25cm = 3.75 cm (b)
(c)
h′ = −h
s1′ = 0.50 cm s1
The angular size is
θ = tan
h′ h′ 0.50 cm ≈ = = 0.10 rad f 2 f 2 5.0 cm
(d) If the object were held at the eye’s near point, it would subtend:
θ NP =
h 1.0 cm = = 0.040 rad 25 cm 25 cm
The angular magnification is M= Assess:
θ 0.10 rad = = 2.5 θ NP 0.040 rad
The numerical answers seem to agree with the drawing.
24.30. Visualize: See Figure 24.15. Parallel rays coming into the first lens will focus at the focal point of the first lens. If that position is also the focal point of the second lens then the rays will also leave the second lens parallel.
Solve: (a) This is similar to a telescope.
d = f1 + f 2 (b) Looking at the similar triangles in the diagram shows that
w1 w2 = f1 f2 w2 =
f2 w1 f1
Assess: Figure P24.30 says f 2 > f1 and our answer then shows that w2 > w1 which is the goal of a beam expander.
24.31. Visualize: Hard thought shows that if the left focal points for both lenses coincide then the parallel rays before and after the beam splitter are reproduced. The first lens diverges the rays as if they had come from the focal point of the converging lens.
Solve: (a)
d = f 2 − | f1 | But since we are given f1 < 0, this is equivalent to d = f 2 + f1 (b) Looking at the similar triangles in the diagram shows that
w1 w2 = | f1 | f 2 w2 =
f2 w1 | f1 |
Assess: Figure P24.31 says f 2 > | f1 | and our answer then shows that w2 > w1 which is the goal of a beam expander.
24.32. Visualize: We simply need to work backwards. We are given f1 = 7.0 cm and f 2 = 15 cm. We are also given s2′ = −10 cm. We use this to find s2 . Solve: (a)
s2 =
f 2 s′2 (15 cm)(−10 cm) = = 6.0 cm s2′ − f 2 −10 cm − 15 cm
So the final image is 6.0 cm to the left of the second lens, or 14 cm to the right of the first lens. That is, the object for the second lens is the image from the first lens, so s1′ = 20 cm = −6.0 cm = 14 cm. s1 =
f1s1′ (7.0 cm)(14 cm) = = 14 cm s1′ − f1 14 cm − 7.0 cm
Thus, L = 14 cm. (b) To find the height and orientation we need to look at the magnification. ⎛ s′ ⎞⎛ s′ ⎞ ⎛ 14 cm ⎞⎛ −10 cm ⎞ m = m1m2 = ⎜ − 1 ⎟⎜ − 2 ⎟ = ⎜ − ⎟⎜ − ⎟ = −1.7 ⎝ s1 ⎠⎝ s2 ⎠ ⎝ 14 cm ⎠⎝ 6.0 cm ⎠ h′ = hm = (1.0 cm)( −1.7) = −1.7 cm The negative sign indicates that the image is inverted. Assess: Ray tracing would verify the answers.
24.33. Model: The plane faces have an infinite radius of curvature. We are assuming the lens(es) are thin. Visualize: Use Equation 23.27, the lens makers’ equation: ⎛1 1 ⎞ 1 = (n − 1) ⎜ − ⎟ f ⎝ R1 R2 ⎠ Solve:
(a) For a symmetric convex lens call R = | R1 | = | R2 | where the sign convention says R1 > 0 and R2 < 0.
⎛1 1 ⎞ 1 ⎛1 1⎞ ⎛2⎞ = ( n − 1) ⎜ − ⎟ = (n − 1) ⎜ + ⎟ = (n − 1) ⎜ ⎟ f R R R R ⎝ ⎠ ⎝ R⎠ 2 ⎠ ⎝ 1 Invert both sides.
f =
1 R ( n − 1) 2
Now for the plano-convex halves: ⎛ 1 1⎞ 1 ⎛1 ⎞ ⎛1⎞ = ( n − 1) ⎜ − ⎟ = ( n − 1) ⎜ + 0 ⎟ = (n − 1) ⎜ ⎟ f1 R ∞ R ⎝ ⎠ ⎝R⎠ ⎝ 1 ⎠ Invert both sides.
f1 =
1 R ( n − 1)
⎛1 1 ⎞ 1 1⎞ ⎛ ⎛1⎞ = ( n − 1) ⎜ − ⎟ = (n − 1) ⎜ 0 + ⎟ = (n − 1) ⎜ ⎟ ∞ f2 R R ⎝ ⎠ ⎝ R⎠ 2 ⎠ ⎝ Invert both sides. f2 =
1 R (n − 1)
From the three results we see that f1 = f 2 = 2 f . (b) Use the thin-lens equation. As a single lens with s = 12 f : s′ =
fs f (1 f ) =1 2 =−f s− f 2 f − f
As a two-lens system with f1 = f 2 = 2 f and s1 = 12 f = 14 f1 : s1′ =
f1s1 (2 f )( 12 f ) f2 2 = 1 = 3 =− f s1 − f1 −2 f 3 2 f −2f
Since this is to the left of the lens s2 = 23 f s′2 =
(2 f )( 23 f ) 43 f 2 f 2 s2 = 2 = 4 =−f s2 − f 2 −3 f 3 f −2f
We get the same result treating it as one symmetric convex lens or as two plano-convex halves (with zero separation). Assess: We can think of each plano-convex half as providing half the refraction. The answers are consistent.
24.34. Model: Use the ray model of light. Assume both the lenses are thin lenses. Visualize:
Solve:
Begin by finding the image of the diverging lens,
1 1 1 1 1 1 + = + = ⇒ ⇒ s1′ = −10 cm 20 cm s1′ −20 cm s1 s1′ f1 This image is the object for the second lens. Its distance from the screen is s2 + s′2 = 110 cm − 10 cm = 100 cm. The overall magnification is M = m1m2 = −
h′ = −2 h
The magnification of the diverging lens is m1 = −
s1′ ( −10 cm ) = 1 =− s1 20 cm 2
Thus the magnification of the converging lens needs to be
m2 = −
s′2 = −4 ⇒ s′2 = 4s2 s2
Substituting this result into s2 + s2′ = 100 cm, we have s2 + 4s2 = 100 cm, which means s2 = 20 cm and s′2 = 80 cm. We can find the focal length by using the thin-lens equation for the converging lens:
1 1 1 1 1 1 + = ⇒ = + ⇒ f2 = 16 cm s2 s′2 f 2 f 2 20 cm 80 cm Hence, the second lens is a converging lens of focal length 16 cm. It must be placed 10 cm in front of the diverging lens, toward the screen, or 80 cm from the screen.
24.35. Model: Yang has myopia. Normal vision will allow Yang to focus on a very distant object. In measuring distances, we'll ignore the small space between the lens and her eye. Solve: Because Yang can see objects at 150 cm with a fully relaxed eye, we want a lens that creates a virtual image at s′ = −150 cm (negative because it's a virtual image) of an object at s = ∞ cm. From the thin-lens equation,
1 1 1 1 1 = + = + = −0.67 D f s s′ ∞ m −1.5 m So Yang gets a prescription for a −0.67 D lens which has f = −150 cm. Since Yang can accommodate to see things as close as 20 cm we need to create a virtual image at 20 cm of objects that are at s = new near point. That is, we want to solve the thin-lens equation for s when s′ = −20 cm and f = −150 cm. s= Assess:
fs′ ( −150 cm )( −20 cm ) = 23 cm = s′ − f −20 cm − ( −150 cm )
Diverging lenses are always used to correct myopia.
24.36. Visualize: Use Equation 23.21: n1 n2 n2 − n1 + = s s′ R where n1 = 1.00 for air and n2 = 1.34 for aqueous humor. If we think of incoming parallel rays coming to a focus in the humor then we have s = ∞ and s′ = f . Solve:
1.0 n2 n2 − n1 + = f R ∞ Solve for R. R= f
n2 − n1 1.34 − 1.00 = ( 3.0 cm ) = 0.76 cm n2 1.34
Assess: If you think about the dimensions of an eye, this answer seems physically possible.
24.37. Visualize: Use Equation 23.27, the lens makers' equation: ⎛1 1 ⎞ 1 = ( n − 1) ⎜ − ⎟ f ⎝ R1 R2 ⎠ For a symmetric lens R1 = R2 and f =
R and R = 2 ( n − 1) f 2 ( n − 1)
Also needed will be the magnification of a telescope: M = − f obj / f eye ⇒ f eye = − f obj / M (but we will drop the negative sign). We are given Robj = 100 cm and M = 20. Solve:
Reye = 2 ( n − 1) f eye Assess:
Robj 2 ( n − 1) Robj 100 cm = 2(n − 1) = 2 ( n − 1) = = = 5.0 cm 20 M M M f obj
We expect a short focal length and small radius of curvature for telescope eyepieces.
24.38. Model: Assume that each lens is a simple magnifier with M = 25 cm / f . Visualize:
M obj =
25 cm 25 cm ⇒ f obj = f obj M obj
M eye =
25 cm 25 cm ⇒ f eye = f eye M eye
Solve: (a) The magnification of a telescope is
M =−
f obj f eye
25 cm M obj M =− = − eye 25 cm M obj M eye
The way to maximize the magnitude of this is to have M eye > M obj .
M =−
5.0 = −2.5 2.0
The magnification is usually given without the negative sign, so it is 2.5 × . (b) To achieve this we used the 2.0× lens as the objective, which coincides with the text which says the objective should have a long focal length and the eyepiece a short focal length. (c)
L = f obj + f eye = Assess:
25 cm 25 cm 25 cm 25 cm + = + = 17.5 cm M obj M eye 2.0 5.0
This is not a very powerful telescope.
24.39. Model: To make a telescope you need an objective with a long focal length and an eyepiece with a short focal length. Visualize:
f =
1 P
Solve: (a) The lens with the smaller refractive power has the longer focal length and should be used as the object— that’s the lens with P = +3.0 D. The +4.5 D lens should be used as the eyepiece. (b)
M =−
f obj f eye
=−
Peye Pobj
=
4.5 D = −1.5 3.0 D
(c) In a telescope the lenses should be a distance apart equal to the sum of their focal lengths.
d = f obj + f eye =
1 1 1 1 + = + = 0.56m Pobj Peye 3.0 D 4.5 D
Assess: These numbers are reasonable, although a +4.5 D lens is fairly strong. You really could make yourself a telescope by holding the lenses a half-meter apart and get a little (1.5×) magnification. This cannot be done with the glasses of nearsighted people since they wear diverging lenses.
24.40. Model: Assume thin lenses and treat each as a simple magnifier with M = 25cm/f . Visualize:
Equation 24.10 gives the magnification of a microscope.
M = mobjM eye = −
L 25cm f obj f eye
Solve: (a) The more powerful lens (4×) with the shorter focal length should be used as the objective. (b) Solve the equation above for L (drop the negative sign). Mf f (12)( 254cm )( 252cm ) L = obj eye = = 37.5 cm 25 cm 25 cm Assess: This is a long microscope tube.
24.41. Visualize: We’ll use the thin-lens equation and also Equation 24.10: M =− Also recall that P = 1/f . Solve: The power of each lens is 1 1 P1 = = = 50 D f1 0.020 m
L 25 cm f obj f eye
P2 =
1 2 = = 100 D f1 0.010 m
Since we want to use the more powerful lens as the objective, the lens labeled P2 will be the objective. This means the focal lengths of the objective and eyepiece are f obj = 1.0 cm and f eye = 2.0 cm. (a) We want the eyepiece to L = 16 cm from the objective, so s′ = 16 cm − f eye = 16 cm − 2.0 cm = 14 cm. The
object distance for the objective is
s= (b) M = mobjM eye = Assess: answer:
fs′ (1.0 cm)(14 cm) = = 1.1 cm ′ s−f 14 cm − 1.0 cm
− s′ 25 cm (14 cm) 25 cm =− = −160 s f eye (1.08 cm) (2.0 cm)
As expected, s is just beyond the focal point. We can use approximation in Equation 24.9 to get a similar M=
L 25 cm (16 cm) 25 cm =− = −200 f obj f eye (1.0 cm) (2.0 cm)
but the approximation isn’t very good for this microscope.
24.42. Model: While s ≈ f obj we will not assume they are equal. Visualize:
Equation 24.9 says mobj ≈ − L / f obj. We are given L = 180 mm and mobj = −40, where the negative
sign means the image is inverted. Solve: Solve for f obj.
f obj = −
L 180 mm = = 4.5 mm mobj 40
From Equation 24.8, M eye = (25 cm) / f eye ⇒ f eye = 25 cm / 20 = 1.25 cm. For relaxed eye viewing the image of the objective must be 1.25 cm = 12.5 mm from the eyepiece, so s′ = 180 mm − 12.5 mm = 167.5 mm. Thus the sample distance is −1
1 ⎛ 1 ⎞ − s=⎜ ⎟ = 4.6 mm ⎝ 4.5 mm 167.5 mm ⎠ Assess:
You need a short focal length to achieve 800× magnification. We can also verify that s ≈ f obj .
24.43. Model: The width of the central maximum that accounts for a significant amount of diffracted light intensity is inversely proportional to the size of the aperture. The lens is an aperture that focuses light. Solve: To focus a laser beam, which consists of parallel rays from s = ∞, the focal length needs to match the distance to the target: f = L = 5.0 cm. The minimum spot size to which a lens can focus is
w=
2.44 (1.06 ×10−6 m )( 5.0 ×10−2 m ) 2.44λ f ⇒ 5.0 ×10−6 m = ⇒ D = 2.6 cm. D D
24.44. Model: Two objects are marginally resolved if the angular separation between the objects, as seen from your eye lens, is α = 1.22λ D , but the λ we want to use is the λ in the eye: λ = λair / n. Let Δx be the separation between the two headlights of the oncoming car and let L be the distance of these lights from your eyes. For small angles, Δx = α L. We are given D = 7 × 10−3 m, Δx = 1.2 m, and λ = λair / n = 600 nm /1.33 = 450 nm. Solve: Let Δy be the separation between the two headlights of the incoming car and let L be the distance of these lights from your eyes. Then,
α=
(1.20 m ) ( 7.0 × 10−3 m ) Δx 1.20 m 1.22λ 1.22 ( 450 nm ) ⇒ = = = L = = 15 km L L D (1.22 ) ( 450 × 10−9 m ) ( 7.0 ×10−3 m )
Assess: The two headlights are not resolvable if L > 15 km, marginally resolvable at 15 km, and resolvable at L < 15 km.
24.45. Visualize: The angle subtended at the eye due to a circle of diameter d at a near point of 25 cm is α = d / 25 cm. The angle to the first dark minimum in a circular diffraction pattern is θ1 = 1.22λ / D, where λ = λair / n is the wavelength of the light in the eye and D is the pupil diameter. To just barely see the circle as a circle the condition α = θ1 must be met. We are given D = 2.0 ×10−3 m and λ = λair / n = 600 nm /1.33 = 450 mm. Solve:
Equating α and θ1 we have d / 25 cm = 1.22λ / D, which may be solved for the diameter of the circle.
(25 cm)(1.22λ) (25 cm)(1.22)(450 ×10−9 m) = = 6.9 ×10−5 m = 0.069 mm D 2.0 ×10−3 m Assess: The above number is about the size of a needle point. d=
24.46. Visualize: For telescopes the angular resolution is 1.22λ θ= D
And for small angles, s = θ r. We want to know s. We are given λ = 650 nm, D = 2.4 m, and r = 33,000 ly = 2.8 × 1017 km. Solve: Combine the two equations. (a)
s =θr =
1.22λ (1.22)(650 × 10−9 m) r= 2.8 × 1017 km = 9.4 × 1010 km D 2.4 m
(b) The distance from the sun to Jupiter is 7.8 ×1011 m. So we divide the answer from part (a) by this number.
9.4 × 1010 km = 120 7.8 × 1011 m Assess: The HST is good but not good enough to resolve two objects as close together as the sun and Jupiter from a distance of 30,000 ly.
24.47. Model: α = 1.22λ / D.
Two objects are marginally resolved if the angular separation between the objects is
Visualize:
Solve:
(a) The angular separation between the sun and Jupiter is 780 × 109 m 780 × 109 m α= = = 1.92 × 10 −5 rad 4.3 light years 4.3 × ( 3.0 × 108 ) × ( 365 × 24 × 3600 ) m
−9 1.22λ 1.22 ( 600 ×10 m ) = ⇒ D = 0.038 m = 3.8 cm D D (b) The sun is vastly brighter than Jupiter, which is much smaller and seen only dimly by reflected light. In theory it may be possible to resolve Jupiter and the sun, but in practice the extremely bright light from the sun will overwhelm the very dim light from Jupiter.
α=
24.48. Visualize: We’ll start with Equation 24.15 and substitute in Equation 24.11 and an expression for α from Figure 24.14a. We are given f obj = 1.6 mm, d min = 400 nm, n = 1.0, and λ = 550 nm. Solve:
Solve for D.
d min =
0.61λ 0.61λ = = NA n sin α
0.61λ = ⎛ −1 D / 2 ⎞ n sin ⎜ tan ⎟ ⎜ f obj ⎟⎠ ⎝
⎛ D / 2 ⎞ 0.61λ sin ⎜ tan −1 ⎟= ⎜ f obj ⎟⎠ n d min ⎝
tan −1
D/2 0.61λ = sin −1 f obj n d min
⎛ D/2 0.61λ ⎞ = tan ⎜ sin −1 ⎟ f obj n d min ⎠ ⎝
⎛ ⎛ −1 0.61(550 nm) ⎞ 0.61λ ⎞ D = 2 f obj tan ⎜ sin −1 ⎟ = 2(1.6 mm) tan ⎜ sin ⎟ = 4.9 mm n d (1.0)(400 nm) ⎠ ⎝ ⎝ min ⎠ The diameter of the lens must exceed 4.9 mm. Assess: A half centimeter is in the ballpark for microscope lens diameters.
24.49. Model: For a diffraction-limited lens, the minimum focal length is the same size as its diameter. The smallest spot diameter over which you can focus light is wmin ≈ 2.5λ. Solve:
(a) The smallest spot size is wmin ≈ 2.5λ = 2.5 ( 800 × 10−9 m ) = 2 μ m.
(b) The total usable area of the optical disk is 2 2 π ⎡⎢( 5.5 × 10−2 m ) − ( 2 × 10−2 m ) ⎤⎥ = 0.00825 m 2 ⎣ ⎦ The area of each pit is the area of one bit of information and is ⎡⎣(1.25 )( 2 μ m ) ⎤⎦ = [ 2.5 μ m ] . The area of 1 byte is 8 times this quantity and the area of 1 megabyte (MB) of information is 106 times more. This means the number of megabytes (MB) of data that can be stored on the disk is 0.00825 m 2 = 165 MB 2 8 × ( 2.5 × 10−6 m ) × 106 MB−1 2
Assess:
A memory storage capacity of 165 MB is reasonable.
2
24.50. Visualize: Physically, the light rays can either go directly through the lens or they can reflect from the mirror and then go through the lens. We can consider the image from the lens alone and then consider the image from mirror becoming the object for the lens. Solve: (a) First case: the lens gets the subscript 1’s and the mirror the 2’s. The location of the image from the lens is fs (10 cm)(5 cm) s1′ = 1 1 = = −10 cm s1 − f1 5 cm − 10 cm The image is right at the mirror plane and a calculation for a mirror shows that when s2 = 0 then s2′ = 0, too. So the final image is at the mirror, 10 cm to the left of the lens. s′ −10 cm m=− 1 =− = 2.0 s1 5 cm
so h′ = hm = (1.0 cm)(2.0) = 2.0 cm. Second case: the mirror gets the subscript 1’s and the lens the 2’s. The location of the image from the mirror is fs (10 cm)(5 cm) s1′ = 1 1 = = −10 cm s1 − f1 5 cm − 10 cm or 10 cm behind (to the left of) the mirror. This image now becomes the object for the lens and s2 = 20 cm.
s′2 =
f 2 s2 (10 cm)(20 cm) = = 20 cm s2 − f 2 20 cm − 10 cm
So the image is 20 cm to the right of the lens.
⎛ s′ ⎞⎛ s′ ⎞ ⎛ −10 cm ⎞⎛ 20 cm ⎞ m = ⎜ − 1 ⎟⎜ − 2 ⎟ = ⎜ − ⎟⎜ − ⎟ = −2.0 ⎝ s1 ⎠⎝ s2 ⎠ ⎝ 5 cm ⎠⎝ 20 cm ⎠ so h′ = hm = (1.0 cm)(−2.0) = −2.0 cm, where the negative sign indicates the image is inverted. In summary, both images are 2.0 cm tall; one is upright 10 cm left of the lens, the other is inverted 20 cm to the right of the lens. (b)
Assess: The ray tracing verifies the calculations.
24.51. Model: In the small angle approximation the angle subtended by Mars without the telescope is θ obj = D / d where D is the diameter and d is the distance from the earth. Visualize:
We are given f eye = 2.5 cm.
Solve:
M=
θ eye 0.50° 0.50° ⎛ π rad ⎞ = = = 141 θ obj D / d 6800 ×103 m /1.1×1011 m ⎜⎝ 180° ⎟⎠
Ignoring the negative sign, M=
f obj f eye
f obj = Mf eye = 141(2.5 cm) = 353 cm The length of the telescope is L = f obj + feye = 352.5 cm + 2.5 cm = 355 cm = 3.55 m ≈ 3.5 m Assess:
This is longer than most amateur telescopes.
24.52. Visualize: The plane left face will not refract any of the rays (which are parallel to each other and perpendicular to the face), so nothing happens until the rays hit the first curved surface between lens 1 and lens 2. We’ll need to twice (once for each curved surface) apply Equation 23.21: n1 n2 n2 − n1 + = s s′ R
Solve: (a) For the first curved surface we say s = ∞ because the incoming rays are parallel.
n1 n2 n2 − n1 n2 + = ⇒ s′ = R ∞ s′ R n2 − n1 For the second curved surface (where the rays exit into the air) n2 is on the left and nair = 1.0 is on the right (so those will take the place of n1 and n2, respectively). Since the distance between the lenses is zero, s′ from the previous result will be plugged in for s for the second case. The final thing to note is that the magnitude of the new s′ = f because the doublet brings parallel rays to a focus at f, but s′ is negative due to the sign convention in Table 23.3.
n2 n n −n + air = air 2 n2 f R − R n2 − n1 Now solve for f.
n2 − n1 1 1 − n2 + = −f R R 1 1 − n2 − n2 + n1 = −f R
f =
R 2n2 − n1 − 1
(b)
f blue =
R 2 ( n2 )blue − ( n1 )blue − 1
f red =
R 2 ( n2 )red − ( n1 )red − 1
In the condition we desire f blue = f red , so the two denominators must be equal.
2 ( n2 )blue − ( n1 )blue − 1 = 2 ( n2 )red − ( n1 )red − 1 2 ⎡⎣( n2 )blue − ( n1 )red ⎤⎦ = ( n1 )blue − ( n1 )red 1 Δn2 = Δn1 2
(c) Simply find the Δn for each type of glass and hope one is twice the other.
Δncrown = 1.525 − 1.517 = 0.008 Δnflint = 1.632 − 1.616 = 0.016 1 Since Δncrown = Δnflint then crown glass must be the second material, or the converging lens, while flint glass 2 must be the first material, or the diverging lens. (d) Solve the original focal length expression for R. R = f ( 2n2 − n1 − 1)
Since f blue = f red , it doesn’t matter which color we choose for the n’s (as long as we are consistent). Say we pick blue, so n1 = 1.632 and n2 = 1.525. We are given f = 10.0 cm. R = (10.0 cm ) ⎡⎣ 2 (1.525) − 1.632 − 1⎤⎦ = 4.18 cm Assess: The answers to the various parts fit together and the final result is reasonable.
24.53. Visualize: The effective focal length is defined as the distance from the midpoint between
the two lenses to the point that initially parallel rays come to a focus. Because d < f the image from the first lens is to the right of the second lens, so we will use a negative object distance when we analyze the second lens. Solve: (a) If the original object distance is very large ( s1 ≈ ∞) then s1′ = f . This image is to the right of the right lens by
an amount f − d , but since we are treating this as a negative object distance we will put in s2 = −( f − d ) = d − f . The thin-lens equation for the second lens (the diverging one, with a negative focal length) becomes:
1 1 1 + = d − f s′2 − f Solve for s′2 .
s2′ =
− f (d − f ) f − fd = 2 (d − f ) − (− f ) d
This is the distance of the focus to the right of the second lens, however, we want the distance from the midpoint between the lenses, so we add 12 d to the answer. 1 f 2 − fd 1 f − fd 12 d 2 f 2 − fd + 12 d 2 f eff = s′2 + d = + d= 2 + = 2 d 2 d d d (b) We'll plug d = 12 f and d = 14 f in turn into the previous result. Then Equation 24.1 shows that if we take the ratios of
the resulting f eff ’s we'll have the zoom. zoom =
( f eff ) d =1 4 f ( f eff ) d =1 2 f
( = (
f 2 − f ( 14 f ) + 12 ( 14 f ) 2 1 4
f
f 2 − f ( 12 f ) + 12 ( 12 f ) 2 1 2
f
) = ⎛⎜ ) ⎝
1 2 1 4
⎞ f 2 − 14 f 2 + 321 f 2 ⎟ 2 1 2 1 2 ⎠ f −2 f +8 f
Cancel f 2 from each term. 25 5 5 ⎛ 4 ⎞ 1 − 14 + 321 = (2) 325 − (2) = = 2.5 ⎜ ⎟ 1 1 4 2 ⎝ 2 ⎠ 1− 2 + 8 8
So the lens is a 2.5× zoom lens. Assess: This is a reasonable amount of zoom. The magnification spans a factor of 2.5.
24-1
25.1. Model: Balmer’s formula predicts a series of spectral lines in the hydrogen spectrum. Solve:
Substituting into the formula for the Balmer series,
λ=
91.18 nm 91.18 nm ⇒λ = = 410.3 nm 1 1 ⎛ 1 1⎞ − − ⎜ 2 2 ⎟ 22 62 ⎝2 n ⎠
where n = 3, 4, 5, 6, … and where we have used n = 6. Likewise for n = 8 and n = 10, λ = 389.0 nm and λ = 379.9 nm.
25.2. Model: Balmer’s formula predicts a series of spectral lines in the hydrogen spectrum. Solve:
Balmer’s formula is
λ=
91.18 nm ⎛ 1 1⎞ ⎜ 2 − 2⎟ ⎝2 n ⎠
n = 3, 4, 5, 6, …
As n → ∞, 1 n 2 → 0. Thus, λn→∞ = 4 ( 91.18 nm ) = 364.7 nm.
25.3. Model: Balmer’s formula predicts a series of spectral lines in the hydrogen spectrum. Solve:
Using Balmer’s formula,
λ = 389.0 nm =
91.18 nm 1 1 ⇒ − 2 = 0.2344 ⇒ n = 8 4 n ⎛ 1 1⎞ ⎜ 2− 2⎟ ⎝2 n ⎠
25.4. Model: The angles of incidence for which diffraction from parallel planes occurs satisfy the Bragg condition. Solve: The Bragg condition is 2d cosθ m = mλ , where m = 1, 2, 3, … For first and second order diffraction,
2d cosθ1 = (1) λ
2d cosθ 2 = ( 2 ) λ
Dividing these two equations,
cosθ 2 = 2 ⇒ θ 2 = cos −1 ( 2cosθ1 ) = cos −1 ( 2cos68° ) = 41° cosθ1
25.5. Model: The angles of incidence for which diffraction from parallel planes occurs satisfy the Bragg condition. Solve: The Bragg condition is 2d cosθ m = mλ. For m = 1 and for two different wavelengths,
2d cosθ1 = (1) λ1
2d cosθ1′ = (1) λ1′
Dividing these two equations,
cosθ1′ λ1′ cosθ1′ 0.15 nm = ⇒ = ⇒ θ1′ = cos −1 ( 0.4408 ) = 64° cosθ1 λ1 cos54° 0.20 nm
25.6. Model: The angles corresponding to the various orders of diffraction satisfy the Bragg condition. Solve:
The Bragg condition for m = 1 and m = 2 gives
2d cosθ1 = (1) λ
2d cosθ 2 = ( 2 ) λ
Dividing these two equations, cosθ1 =
cosθ 2 cos 45° ⎛ cos 45° ⎞ = ⇒ θ1 = cos −1 ⎜ ⎟ = 69.3° 2 2 ⎝ 2 ⎠
25.7. Model: The angles corresponding to the various diffraction orders satisfy the Bragg condition. Solve: The Bragg condition is 2d cosθ m = mλ , where m = 1, 2, 3, … The maximum possible value of m is the number of possible diffraction orders. The maximum value of cosθm is 1. Thus, 2d = mλ ⇒ m = We can observe up to the fourth diffraction order.
2d
λ
=
2 ( 0.180 nm ) = 4.2 ( 0.085 nm )
25.8. Model: Use the photon model of light. Solve:
The energy of the photon is
Ephoton = hf = h Assess:
⎛ 3.0 ×108 m/s ⎞ −19 = ( 6.63 × 10−34 Js ) ⎜ ⎟ = 3.98 × 10 J −9 λ ⎝ 500 ×10 m ⎠ c
The energy of a single photon in the visible light region is extremely small.
25.9. Model: Use the photon model of light. Solve:
The energy of the single photon is −34 8 ⎛ c ⎞ ( 6.63 ×10 Js )( 3.0 ×10 m/s ) = 1.99 ×10−19 J Ephoton = hf = h ⎜ ⎟ = 1.0 ×10−6 m ⎝λ⎠
⇒ Emol = N A Ephoton = ( 6.023 ×1023 )(1.99 ×10−19 J ) = 1.2 ×105 J Assess: energy.
Although the energy of a single photon is very small, a mole of photons has a significant amount of
25.10. Model: Use the photon model of light. Solve: The energy of a photon with wavelength λ1 is E1 = hf1 = hc λ1 . Similarly, E2 = hc λ2 . Since E2 is equal to 2E1,
hc
λ2
=2
hc
λ1
⇒ λ2 =
λ1 2
=
600 nm = 300 nm 2
Assess: A photon with λ = 300 nm has twice the energy of a photon with λ = 600 nm. This is an expected result, because energy is inversely proportional to the wavelength.
25.11. Model: Use the photon model of light. Solve:
The energy of the x-ray photon is
⎛ 3.0 × 108 m/s ⎞ ⎛c⎞ −16 E = hf = h ⎜ ⎟ = ( 6.63 × 10−34 Js ) ⎜ ⎟ = 2.0 × 10 J −9 ⎝λ⎠ ⎝ 1.0 ×10 m ⎠ Assess: This is a very small amount of energy, but it is larger than the energy of a photon in the visible wavelength region.
25.12. Solve: Your mass is, say, m ≈ 70 kg and your velocity is 1 m/s. Thus, your momentum is p = mv ≈ (70 kg)(1 m/s) = 70 kg m/s. Your de Broglie wavelength is
λ=
h 6.63 ×10−34 Js = ≈ 9 × 10−36 m p 70 kg m/s
25.13. Solve: (a) The baseball’s momentum is p = mv = (0.200 kg)(30 m/s) = 6.0 kg m/s. The baseball’s de Broglie wavelength is λ=
h 6.63 ×10−34 Js = = 1.1×10−34 m p 6.0 kg m/s
(b) Using λ = h p = h mv , we have v=
h 6.63 ×10−34 Js = = 1.7 × 10−23 m/s mλ ( 0.200 kg ) ( 0.20 ×10−9 m )
25.14. Visualize: We'll employ Equations 25.8 (λ = h / p) and 25.9 ( E = p 2 /2m) to express the wavelength in terms of kinetic energy. Solve: First solve Equation 25.9 for p : p = 2mE .
λ=
h h 6.63 × 10−34 J ⋅ s = = = 1.0 nm p 2mE 2 ( 9.11× 10−31 kg )( 2.4 × 10−19 J )
Assess: The energy given is about 1.5 eV, which is a reasonable amount of energy. The resulting wavelength is a few to a few dozen times the size of an atom.
25.15. Solve: (a) For an electron, the momentum p = mv = ( 9.11×10−31 kg ) v . The de Broglie wavelength is λ=
h = 0.20 ×10−9 m ⇒ 0.20 ×10−9 m = p
6.63 ×10−34 Js ⇒ v = 3.6 × 106 m/s ( 9.11×10−31 kg ) v
(b) For a proton, p = mv = (1.67 × 10−9 kg ) v. The de Broglie wavelength is
λ=
h 6.63 ×10−34 Js = 0.20 ×10−9 m ⇒ 0.20 × 10−9 m = ⇒ v = 2.0 × 103 m/s p (1.67 ×10−27 kg ) v
25.16. Model: The momentum of a wave-like particle has discrete values given by pn = n ( h 2 L ) where n = 1, 2, 3, …. Solve: Because we want the smallest box and the momentum of the electron can not exceed a given value, n must be minimum. Thus, p1 = mv =
6.63 ×10−34 Js h h ⇒L= = = 0.036 mm 2L 2mv 2 ( 9.11× 10−31 kg ) (10 m/s )
25.17. Model: A confined particle can have only discrete values of energy. Solve:
From Equation 24.14, the energy of a confined electron is
En =
h2 2 n 8mL2
n = 1, 2, 3, 4, …
The minimum energy is E1 =
h2 h 6.63 ×10−34 Js ⇒L= = = 2.0 × 10−10 m = 0.20 nm 2 −31 −18 8mL 8mE1 8 ( 9.11×10 kg )(1.5 ×10 J )
25.18. Model: Model the 5.0-fm-diameter nucleus as a box of length L = 5.0 fm. Solve:
The proton’s energy is restricted to the discrete values
( 6.63 ×10−34 Js ) n2 h2 2 = = (1.316 × 10−12 J ) n 2 n 2 −27 −15 8mL2 8 (1.67 ×10 kg )( 5.0 ×10 m ) 2
En =
where n = 1, 2, 3, … For n = 1, E1 = 1.3 ×10−12 J , for n = 2, E2 = (1.316 × 10−12 J ) 4 = 5.3 ×10−12 J , and for n = 3,
E3 = 9 E1 = 1.2 ×10−11 J .
25.19.
Model: The generalized formula of Balmer predicts a series of spectral lines in the hydrogen spectrum. Solve: (a) The generalized formula of Balmer
λ=
91.18 m 1⎞ ⎛ 1 ⎜ 2− 2⎟ ⎝m n ⎠
with m = 1 and n > 1 accounts for a series of spectral lines. This series is called the Lyman series and the first two members are
λ1 =
91.18 m = 121.6 nm 1⎞ ⎛ 1 − ⎜ 2 ⎟ ⎝ 2 ⎠
λ2 =
91.18 nm = 102.6 nm 1⎞ ⎛ 1 − ⎜ 2 ⎟ ⎝ 3 ⎠
For n = 4 and n = 5, λ3 = 97.3 nm and λ4 = 95.0 nm . (b) The Lyman series converges when n → ∞ . This means 1 n 2 → 0 and λ → 91.18 nm. (c) For a diffraction grating, the condition for bright (constructive interference) fringes is d sin θ p = pλ , where p =
1, 2, 3, … For first-order diffraction, this equation simplifies to d sin θ = λ . For the first and second members of the Lyman series, the above condition is d sin θ1 = λ1 = 121.6 nm and d sin θ 2 = λ2 = 102.6 nm. Dividing these two equations yields ⎛ 102.6 nm ⎞ sin θ 2 = ⎜ ⎟ sin θ1 = ( 0.84375 ) sinθ1 ⎝ 121.6 nm ⎠ The distance from the center to the first maximum is y = L tan θ . Thus, tan θ1 =
y1 0.376 m = ⇒ θ1 = 14.072° ⇒ sin θ 2 = ( 0.84375 ) sin (14.072° ) ⇒ θ 2 = 11.84° L 1.5 m
Applying the position formula once again, y2 = L tan θ 2 = (1.5 m ) tan (11.84° ) = 0.314 m = 31.4 cm
25.20.
Model: The generalized formula of Balmer predicts a series of spectral lines in the hydrogen spectrum. Solve: (a) The generalized formula of Balmer
λ=
91.18 m 1⎞ ⎛ 1 ⎜ 2− 2⎟ ⎝m n ⎠
with m = 3, and n > 3 accounts for a series of spectral lines. This series is called the Paschen series and the wavelengths are
λ=
91.18 nm 820.62 n 2 = n2 − 9 ⎛1 1⎞ − ⎜ 2 2 ⎟ ⎝3 n ⎠
The first four members are λ1 = 1876 nm, λ2 = 1282 nm, λ3 = 1094 nm, and λ4 = 1005 nm (b) The Paschen series converges when n → ∞. This means
1 91.18 nm →0⇒λ → = 820.6 nm 2 2 n ( 13 ) (c) For a diffraction grating, the condition for bright (constructive interference) fringes is d sin θ p = pλ , where p =
1, 2, 3, … For first-order diffraction, this equation simplifies to d sin θ = λ . For the first and second members of the Paschen series, the condition is d sin θ1 = λ1 and d sinθ 2 = λ2 . Dividing these two equations yields
⎛λ ⎞ ⎛ 1282 nm ⎞ sin θ 2 = sin θ1 ⎜ 2 ⎟ = sin θ1 ⎜ ⎟ = ( 0.6834 ) sinθ1 λ ⎝ 1876 nm ⎠ ⎝ 1⎠ The distance from the center to the first maximum is y = L tan θ . Thus, tan θ1 =
y1 0.607 m = = 0.4047 ⇒ θ1 = 22.03° ⇒ sin θ 2 = ( 0.6834 ) sin 22.03° ⇒ θ 2 = 14.85° L 1.5 m
Applying the position formula once again, y2 = L tan θ 2 = (1.5 m ) tan14.85° = 0.398 m = 39.8 cm
25.21. Model: Use the photon model of light. Solve:
(a) The wavelength is calculated as follows:
( 6.63 ×10−34 Js )( 3.0 ×108 m/s ) = 2.0 ×10−12 m ⎛c⎞ Egamma = hf = h ⎜ ⎟ ⇒ λ = 1.0 ×10−13 J ⎝λ⎠ (b) The energy of a visible-light photon of wavelength 500 nm is −34 8 ⎛ c ⎞ ( 6.63 ×10 Js )( 3.0 ×10 m/s ) Evisible = h ⎜ ⎟ = = 3.978 × 10−19 J 500 ×10−9 m ⎝λ⎠
The number of photons n such that Egamma = nEvisible is n=
Egamma Evisible
=
1.0 ×10−13 J = 2.5 ×105 3.978 ×10−19 J
25.22. Model: Use the photon model. Solve:
The energy of a 1000 kHz photon is
Ephoton = hf = ( 6.63 ×10−34 Js )(1000 ×103 Hz ) = 6.63 × 10−28 J The energy transmitted each second is 20 × 103 J. The number of photons transmitted each second is 20 × 103 J/6.63 × 10 −28 J = 3.0 × 1031.
25.23. Model: Use the photon model for the laser light. Solve:
(a) The energy is
⎛ 3 × 108 m/s ⎞ ⎛c⎞ −19 Ephoton = hf = h ⎜ ⎟ = ( 6.63 × 10−34 Js ) ⎜ ⎟ = 3.1× 10 J −9 ⎝λ⎠ ⎝ 633 ×10 m ⎠ (b) The energy emitted each second is 1.0 ×10 −3 J. The number of photons emitted each second is 1.0 ×10 −3 J/3.14 ×10 −19 J = 3.2 × 1015.
25.24. Model: Use the photon model for the incandescent light. Solve: The photons travel in all directions. At a distance of r from the light bulb, the photons spread over a sphere of surface area 4πr2. The number of photons per second per unit area at the location of your retina is
3 ×1018 s −1
4π (10 ×10 m ) 3
2
= 2.387 ×109 s −1 m −2
The number of photons that enter your pupil per second is 2.387 ×109 s −1m −2 × π ( 3.5 ×10−3 m ) = 9.2 ×104 s −1 2
25.25. Model: Use the photon model of light and the Bragg condition for diffraction. Solve: The Bragg condition for the reflection of x-rays from a crystal is 2d cosθ m = mλ. To determine the angles of incidence θm, we need to first calculate λ. The wavelength is related to the photon’s energy as E = hc λ . Thus, λ=
−34 8 hc ( 6.63 ×10 Js )( 3.0 ×10 m/s ) = = 1.326 × 10−10 m E 1.50 ×10−15 J
From the Bragg condition, ⎡ 1.326 ×10−10 m ) m ⎤ ⎛ mλ ⎞ −1 ( ⎢ ⎥ = cos −1 ( 0.3157m ) ⇒ θ1 = cos −1 ( 0.3157 ) = 71.6° = cos ⎟ −9 ⎝ 2d ⎠ ⎢⎣ 2 ( 0.21×10 m ) ⎥⎦
θ m = cos −1 ⎜
Likewise, θ 2 = cos −1 ( 0.3157 × 2 ) = 50.8° and θ 3 = 18.7°. Note that θ 4 = cos −1 ( 0.3157 × 4 ) is not allowed because the cosθ cannot be larger than 1. Thus, the x-rays will be diffracted at angles of incidence equal to 18.7°, 50.8°, and 71.6°.
25.26. Model: The angles for which diffraction from parallel planes occurs satisfy the Bragg condition. Solve: We cannot assume that these are the first and second order diffractions. The Bragg condition is 2d cosθ m = mλ. We have 2d cos 45.6° = mλ
2d cos 21.0° = ( m + 1) λ
Notice that θm decreases as m increases, so 21.6° corresponds to the larger value of m. Dividing these two equations, cos 45.6° m = = 0.7494 ⇒ m = 3 cos 21.0° m + 1 Thus these are the third and fourth order diffractions. Substituting into the Bragg condition, d=
3 × 0.0700 × 10−9 m = 1.50 × 10−10 m = 0.150 nm 2cos 45.6°
25.27. Model: The x-ray diffraction angles satisfy the Bragg condition. Solve: (a) The Bragg condition ( 2d cosθ m = mλ ) for normal incidence, θm = 0°, simplifies to 2d = mλ. For a thin film of a material on a substrate where nair < nmaterial < nsubstrate, constructive interference between the two reflected waves occurs when 2d = mλ, where λ is the wavelength inside the material. (b) From a thin film with a period of 1.2 nm, that is, with d = 1.2 nm, the two longest x-ray wavelengths that will reflect at normal incidence are
λ1 =
2d 1
This means that λ1 = 2 (1.2 nm ) = 2.4 nm and λ2 = 1.2 nm.
λ2 =
2d 2
25.28. Solve: A small fraction of the light wave of an appropriate wavelength is reflected from each little “bump” in the refractive index. These little bumps act like the atomic planes in a crystal. The light will be strongly reflected (and hence blocked in transmission) if it satisfies the Bragg condition at normal incidence (θ = 0). 2d = mλglass =
−6 2dnglass 2 ( 0.45 ×10 m ) (1.50 ) mλ ⇒λ = = = 1.35 μm nglass m 1
25.29. Model: Particles have a de Broglie wavelength given by λ = h p . The wave nature of the particles causes an interference pattern in a double-slit apparatus. Solve: (a) Since the speed of the neutron and electron are the same, the neutron’s momentum is
pn = mn vn =
mn m m mevn = n meve = n pe me me me
where mn and me are the neutron’s and electron’s masses. The de Broglie wavelength for the neutron is
λn =
h h pe m = = λe e pn pe pn mn
From Section 22.2 on double-slit interference, the fringe spacing is Δy = λ L / d . Thus, the fringe spacing for the electron and neutron are related by
Δy n =
⎛ 9.11× 10−31 kg ⎞ λn m −3 −7 Δye = e Δye = ⎜ ⎟ (1.5 × 10 m ) = 8.18 × 10 m = 0.818 μ m −27 λe mn ⎝ 1.67 ×10 kg ⎠
(b) If the fringe spacing has to be the same for the neutrons and the electrons,
Δye = Δyn ⇒ λe = λn ⇒
⎛ 9.11×10−31 kg ⎞ h h m 3 = ⇒ vn = ve e = ( 2.0 × 106 m/s ) ⎜ ⎟ = 1.1× 10 m/s −27 meve mn vn mn ⎝ 1.67 ×10 kg ⎠
25.30. Model: Electrons have a de Broglie wavelength given by λ = h p . The wave nature of the electrons causes a diffraction pattern. Solve: The width of the central maximum of a single-slit diffraction pattern is given by Equation 22.22: w=
2 (1.0 m ) ( 6.63 × 10−34 Js ) 2λ L 2 Lh 2 Lh = = = = 9.7 × 10−4 m = 0.97 mm −6 a ap amv (1.0 ×10 m )( 9.11×10−31 kg )(1.5 ×106 m/s )
25.31. Model: Neutrons have a de Broglie wavelength given by λ = h p . The wave nature of the neutrons causes a double-slit interference pattern. Solve: Measurements show that the spacing between the m = 1 and m = –1 peaks is 1.4 times as long as the length of the reference bar, which gives the real fringe separation Δy = 70 μm. Similarly, the spacing between the m = 2 and m = –2 is 2.8 times as long as the length of the reference bar and yields Δy = 70 μm. The fringe separation in a double-slit experiment is Δy = λ L d . Hence,
λ=
( 6.63 ×10−34 Js ) ( 3.0 m ) Δy d Δy d h h hL ⇒ = = ⇒v= = = 170 m/s −6 Δy md ( 70 ×10 m )(1.67 × 10−27 kg )( 0.10 ×10−3 m ) L p mv L
25.32. Model: Electrons have a de Broglie wavelength given by λ = h p . Visualize: Please refer to Figure 25.11. Notice that a scattering angle φ = 60° corresponds to an angle of incidence θ = 30°. Solve: Equation 25.6 describes the Davisson-Germer experiment: D sin ( 2θ m ) = mλ . Assuming m = 1, this equation simplifies to D sin 2θ = λ . Using λ = h mv , we have D=
h 6.63 ×10−34 Js = = 1.95 × 10−10 m = 0.195 nm −31 mv sin 2θ ( 9.11×10 kg )( 4.30 × 106 m/s ) sin ( 60° )
25.33. Model: A confined particle can have only discrete values of energy. Solve:
(a) Equation 25.14 simplifies to
( 6.63 ×10−34 Js ) h2 2 En = n = = (1.231× 10−19 J ) n 2 2 2 8mL 8 ( 9.11× 10−31 kg )( 0.70 × 10−9 m ) 2
Thus, E1 = (1.231×10−19 J )(12 ) = 1.2 × 10−19 J, E2 = (1.231×10−19 J )( 22 ) = 4.9 ×10−19 J, and E3 = 1.1×10−18 J. (b) The energy is E2 − E1 = 4.9 × 10−19 J − 1.2 ×10−19 J = 3.7 ×10−19 J. (c) Because energy is conserved, the photon will carry an energy of E2 − E1 = 3.69 ×10−19 J . That is, E2 − E1 = Ephoton = hf =
hc
λ
⇒λ =
( 6.63 ×10−34 Js )( 3.0 ×108 m/s ) = 540 nm hc = 3.69 ×10−19 J E2 − E1
25.34. Model: A particle confined in a one-dimensional box has discrete energy levels. Solve:
(a) Equation 24.14 for the n = 1 state is
( 6.63 ×10−34 Js ) = 5.5 ×10−64 J h2 En = = 8mL2 8 (10 ×10−3 kg ) ( 0.10 m )2 2
The minimum energy of the Ping-Pong ball is E1 = 5.5 × 10−64 J. (b) The speed is calculated as follows: E1 = 5.50 ×10−64 J =
1 2
mv 2 =
1 2
(10 ×10
−3
kg ) v 2 ⇒ v =
2 ( 5.50 ×10−64 J ) 10 ×10−3 kg
= 3.3 × 10−31 m/s
25.35. Model: A particle confined in a one-dimensional box has discrete energy levels. Solve:
Using Equation 24.14 for n = 1 and 2,
( 6.63 ×10−34 Js ) ( 3) = 1.809 ×10−37 Jm2 h2 22 − 12 ) ⇒ 1.0 ×10−19 J = E2 − E1 = 2 ( L2 8mL 8 ( 9.11×10−31 kg ) L2 2
⇒L=
1.809 × 10−37 Jm 2 = 1.3 ×10−9 m = 1.3 nm 1.0 ×10−19 J
25.36. Visualize: From the figure we see that the wavelength is 2.0 nm. We'll employ Equations 25.8 (λ = h / p ) and 25.9 ( E = p 2 /2m) to express the kinetic energy in terms of wavelength. Solve:
2 ( 6.63 ×10−34 J ⋅ s 2.0 nm ) = 6.0 ×10−20 J p2 ( h λ ) = = 2m 2m 2 ( 9.11×10−31 Kg ) 2
E= Assess:
This energy is a little less than one eV, which is reasonable.
25.37. Visualize: The strategy is to take ratios to find n and then plug it back in to find L. Solve:
h2 (n + 1) 2 En +1 8mL2 ( n + 1) 2 6.4 × 10−13 J = = = 2 h ( n) 2 3.6 × 10−13 J En 2 ( ) n 2 8mL Cancel ×10−13 J and take square roots. n +1 6.4 4 = = n 3.6 3
⇒
n=3
So E3 = 3.6 ×10−13 J. Now solve for L.
L= Assess:
( 6.63 ×10−34 J ⋅ s ) ( 3) h2n2 hn = = = 29 fm 8mEn 8mEn 8 (1.67 × 10−27 kg )( 3.6 × 10−13 J )
This is not an atomic-sized box, but a nuclear-sized box; that’s OK for neutrons.
25.38. Model: The allowed energies of a particle of mass m in a two-dimensional square box of side L are Enm = Solve:
h2 ( n2 + m2 ) 8mL2
(a) The minimum energy for a particle is for n = m = 1:
Emin = E11 = (b) The five lowest allowed energies are Emin ,
5 2
5Emin (for n = 1, m = 3 and n = 3, m = 1), and
13 2
h2 h2 12 + 12 ) = 2 ( 8mL 4mL2
Emin (for n = 1, m = 2 and n = 2, m = 1), 4Emin (for n = 2, m = 2), Emin (for n = 2, m = 3 and n = 3, m = 2).
25.39. Model: A particle confined in a one-dimensional box of length L has the discrete energy levels given by Equation 24.14. Solve: (a) Since the energy is entirely kinetic energy,
En =
h2 2 p 2 1 2 h n = = mvn ⇒ vn = n 8mL2 2m 2 2mL
n = 1, 2, 3, …
(b) The first allowed velocity is
v1 =
6.63 ×10−34 Js = 1.82 ×106 m/s 2 ( 9.11×10−31 kg )( 0.20 ×10−9 m )
For n = 2 and n = 3, v2 = 3.64 × 106 m/s and v3 = 5.46 × 106 m/s.
25.40. Model: Sets of parallel planes in a crystal diffract x-rays. Visualize: Please refer to Figure CP25.40. Solve: The Bragg diffraction condition is 2d cosθ m = mλ , where d is the interplanar separation. Because smaller m values correspond to higher angles of incidence, the diffraction angle of 71.3° in the x-ray intensity plot must correspond to m = 1. This means 2d cos71.3° = 1( 0.10 × 10−9 m ) ⇒ d =
0.10 ×10−9 m = 1.56 ×10−10 m 2 ( cos71.3° )
The cosines of the three angles in the x-ray intensity plot are cos71.3° = 0.321, cos50.1° = 0.642, and cos15.8° = 0.962. These are in the ratio 1 : 2 : 3, which tells us that these are the m = 1, 2, and 3 diffraction peaks from a single set of planes with d = 0.156 nm. We can see from the figure that the atomic spacing D of this crystal is related to the interplanar separation d by D=
d 0.156 nm = = 0.18 nm sin 60° sin60°
25.41.
Model: Sets of parallel planes in a crystal diffract x-rays. Visualize: Please refer to Figure 25.7. Solve: (a) The Bragg diffraction condition is 2d cosθ m = mλ. The plane spacing is dA = 0.20 nm and the x-ray wavelength is λ = 0.12 nm. Thus
cosθ m =
−9 mλ m ( 0.12 ×10 m ) = = ( 0.3) m ⇒ θ A1 = cos −1 ( 0.3) = 72.5° 2d A 2 ( 0.20 ×10−9 m )
Likewise for m = 2 and m = 3, θ A2 = cos −1 ( 0.6 ) = 53.1° and θ A3 = cos −1 ( 0.9 ) = 25.8°. These three angles for the x-ray diffraction peaks match the peaks shown in Figure 25.7c. (b) The new interplaner spacing is d B = d A 2 = 0.141 nm (see Figure 25.7b). The Bragg condition for the tilted atomic planes becomes
cosθ m =
mλ = 0.4243m 2d B
For m = 1, θ B1 = cos −1 ( 0.4243) = 64.9°. For m = 2, θ B2 = cos −1 ( 0.8486 ) = 31.9°. (c) The crystal is already tipped by 45° to get the tilted planes (see Figure 25.7b). So, for m = 1, θ1 = 64.9° − 45° = 19.9°. θ = 64.9° + 45° = 109.9° also, but we can’t see beyond 90°. For m = 2,
θ 2 = 31.9° + 45° = 76.9°. These two angles match the angles in the diffraction peaks of the tilted planes.
25.42. Model: This is an integrated problem that uses concepts from Chapter 22. There are two L’s in the problem: L in Chapter 22 refers to the screen distance from the slits, and the L we want here is the length of the box. The wavelength of the neutron determined by the two-slit pattern is the same as the wavelength in the confined box. Visualize: The figure shows Lbox = 2λ .
We also need Equation 22.6: ym =
mλ Lscreen . Also from the figure we see that y2 = 0.20 ×10−3 m. We are given d
Lscreen = 2.0 m and d = 15 ×10−6 m. Solve: Solve Equation 22.6 for λ.
λ= Lbox = 2λ = 2 Assess:
dym mLscreen
(15 × 10−6 m )( 0.20 ×10−3 m ) = 1.5 nm dym =2 mLscreen ( 2 )( 2.0 m )
The two pieces of this problem fit together and make sense together.
Model: Electrons have a de Broglie wavelength given by λ = h p . Trapped electrons in the confinement layer behave like a de Broglie wave in a closed-closed tube or like a string fixed at both ends. Solve: (a) The four longest standing-wave wavelengths in the layer are λ = 2L , L, 32 L , and 12 L . This follows
25.43.
from the general relation for closed-closed tubes: λ = 2 L / n. Thus, λ = 10.0 nm, 5.00 nm, 3.33 nm, and 2.50 nm. (b) We have p = mv =
h
λ
⇒v=
h 6.63 × 10−34 Js 0.7278 ×10−3 m 2 /s = = λ mλ ( 9.11×10−31 kg ) λ
Using the above four longest values of λ we get the four smallest values of v. Thus, v1 =
0.7278 ×10−3 m 2 /s = 7.28 × 104 m/s 10.0 ×10−9 m
v2 = 1.46 ×105 m/s , v3 = 2.18 ×105 m/s , and v4 = 2.91×105 m/s .
25.44. Model: As light is diffracted by matter, matter can also be diffracted by light. Solve:
The de Broglie wavelength of the sodium atoms is
λ=
h h 6.63 ×10−34 Js = = = 3.45 × 10−10 m p mv ( 3.84 × 10−26 kg ) ( 50 m/s )
The slit spacing of the “diffraction grating” is d = 12 λlaser = 12 600 nm = 300 nm. Using the diffraction grating equation with m = 1, we have
d sin θ = (1) λ ⇒ sinθ =
λ d
= 1.151×10−3 ≈ 1.2 ×10−3
In the small-angle approximation, sin θ ≅ tanθ = y L . We get
y = L sinθ = (1.0 m ) (1.151× 10−3 ) = 1.2 mm
25-1
26.1. Model: Use the charge model. Solve: (a) In the process of charging by rubbing, electrons are removed from one material and transferred to the other because they are relatively free to move. Protons, on the other hand, are tightly bound in nuclei. So, electrons have been removed from the glass rod to make it positively charged. (b) Because each electron has a charge of 1.60 × 10−19 C, the number of electrons removed is
8.0 ×10−9 C = 5.0 ×1010 1.60 × 10−19 C
26.2. Model: Use the charge model. Solve: (a) In the process of charging by rubbing, electrons are removed from one material and transferred into the other because they are relatively free to move. Protons, on the other hand, are tightly bound in the nuclei. So, electrons have been added to the plastic rod to make it negatively charged. (b) Because each electron has a charge of 1.60 × 10−19 C, the number of electrons added is
12 ×10−9 C = 7.5 ×1010 1.60 ×10−19 C
26.3. Model: Use the charge model and the model of a conductor as material through which electrons move.
Solve: (a) The charge of a plastic rod decreases from −15 nC to −10 nC. That is, −5 nC charge has been removed from the plastic. Because it is the negatively charged electrons that are transferred, −5 nC has been added to the metal sphere. (b) Because each electron has a charge of 1.60 ×10−19 C and a charge of 5 nC was transferred, the number of electrons transferred from the plastic rod to the metal sphere is
5 ×10−9 C = 3.1×1010 1.60 ×10−19 C
26.4. Model: Use the charge model and the model of a conductor as a material through which electrons move. Solve: (a) The charge of the glass rod decreases from +12 nC to +8.0 nC. Because it is the electrons that are transferred, −4.0 nC of electrons has been added to the glass rod. Thus, electrons are removed from the metal sphere and added to the glass rod. (b) Because each electron has a charge of 1.60 ×10−19 C and a charge of 4.0 nC was transferred, number of electrons transferred from the metal sphere to the glass rod is 4.0 ×10−9 C = 2.5 ×1010 1.60 × 10−19 C
26.5. Model: Use the charge model. Solve: Each oxygen molecule has 16 protons (8 per atom), and there are 6.02 ×1023 oxygen molecules in 1.0 mole of oxygen. Because one proton has a charge of +1.60 × 10−19 C, the amount of positive charge in 1.0 mole of oxygen is
6.022 ×1023 ×16 ×1.6 × 10−19 C = 1.54 ×106 C
26.6. Model: Use the charge model.
Since the density of water is 1 g/cm3, the mass of 1 L of water is 1000 mL ⎞ ⎛ 1 cm3 ⎞ ⎛ 1 g ⎞ (1.0 L ) ⎜⎛ ⎟ ⎜ 3 ⎟ = 1.0 kg. Each water molecule (H20) has 10 protons (8 in the oxygen atom and ⎟⎜ L ⎝ ⎠ ⎝ mL ⎠ ⎝ cm ⎠
Solve:
one per hydrogen atom), and thus 10 electrons. The number of moles is
1.0 ×103 g = 100 mole. There are 6.02 ×1023 10 g/mole
water molecules in 1.0 mole of water. Because one electron has a charge magnitude of 1.60 × 10−19 C, the amount of negative charge in 100 mole of water is 100 × 6.022 ×1023 ×10 × 1.6 × 10−19 C = 9.6 × 107 C
26.7. Model: move. Visualize:
Use the charge model and the model of a conductor as a material through which electrons
The charge carriers in a metal electroscope are the negative electrons. As the positive rod is brought near, electrons are attracted toward it and move to the top of the electroscope. The electroscope leaves now have a net positive charge, due to the missing electrons, and thus repel each other. At this point, the electroscope as a whole is still neutral (no net charge) but has been polarized. On contact, some of the electrons move to the positive rod to neutralize some (but not necessarily all) of the rod’s positive charge. After contact, the electroscope does have a net positive charge. When the rod is removed, the net positive charge on the electroscope quickly spreads to cover the entire electroscope. The net positive charge on the leaves causes them to continue to repel.
26.8. Model: Use the charge model. Solve: (a) No, we cannot conclude that the wall is charged. Attractive electric forces occur between (i) two opposite charges, or (ii) a charge and a neutral object that is polarized by the charge. Rubbing the balloon does charge the balloon. Since the balloon is rubber, its charge is negative. As the balloon is brought near the wall, the wall becomes polarized. The positive side of the wall is closer to the balloon than the negative side, so there is a net attractive electric force between the wall and the balloon. This causes the balloon to stick to the wall, with a normal force balancing the attractive electric force and an upward frictional force balancing the very small gravitational force on the balloon. (b)
26.9. Model: Use the charge model and the model of a conductor as a material through which electrons move. Solve:
The first step shows two neutral metal spheres touching each other. In the second step, the negative rod repels the negative charges which will retreat as far as possible from the top of the left sphere. Note that the two spheres are touching and the net charge on these two spheres is still zero. While the rod is there on top of the left sphere, the right sphere is moved away from the left sphere. Because the right sphere has an excess negative charge by the same amount as the left sphere has an excess positive charge, the separated right sphere is negatively charged as shown in the third step. As the two spheres are moved apart further and the negatively charged rod is moved away from the spheres, the charges on the two spheres redistribute uniformly over the entire sphere surface. Thus, we have oppositely charged the two spheres.
26.10. Model: Use the charge model and the model of a conductor as a material through which electrons move. Solve:
Charging two neutral spheres with opposite but equal charges can be done through the following four steps. (i) Touch the two neutral metal spheres together. (ii) Bring a charged rod (say, positive) close (but not touching) to one of the spheres (say, the left sphere). Note that the two spheres are still touching and the net charge on them is zero. The right sphere has an excess positive charge of exactly the same amount as the left sphere’s negative charge. (iii) Separate the spheres while the charged rod remains close to the left sphere. The charge separation remains on the spheres. (iv) Take the charged rod away from the two spheres. The separated charges redistribute uniformly over the metal sphere surfaces.
26.11. Model: Use the charge model and the model of a conductor as a material through which electrons move. Solve:
Charging two neutral spheres with like charges of exactly equal magnitude can be achieved through the following six steps. (i) Bring a charged rod (say, negative) near a neutral metal sphere. (ii) Touch the neutral sphere with the negatively charged rod, so that the rod-sphere system has a net negative charge. (iii) Move the rod away from the sphere. The sphere is now negatively charged. (iv) Bring this negatively charged sphere close to the second neutral sphere. (v) Touch these two spheres. The excess negative charge is distributed evenly over the two spheres. (vi) Separate the spheres. The excess charge will have the same sign as the charge on the charging rod and will be evenly distributed between the two spheres.
26.12. Model: Use the charge model. Solve:
(a)
(b) A charged object causes the charges inside a molecule to move apart while still being a part of the molecule. This is because electrons, that is, the negative charges in an insulator such as a piece of paper cannot move freely. Thus, the negative part of each molecule is slightly closer to a positively charged object and the positive part of each molecule is slightly farther from the positively charged object. This charge polarization leads to a slightly greater positive-to-negative attractive force than the positive-to-positive repulsive force. Because there are many molecules in a piece of paper, the total force of attraction is enough to cause a piece of paper to be attracted to the object. The test will also work for a negatively charged object because then the charges inside the molecules of paper will separate in the other direction.
26.13. Model: Model the charged masses as point charges. Visualize:
G G Solve: (a) The charge q1 exerts a force F1 on 2 on q2 to the right, and the charge q2 exerts a force F2 on 1 on q1 to the left. Using Coulomb’s law, F1 on 2 = F2 on 1 =
9 2 2 −6 −6 K q1 q2 ( 9.0 ×10 N m /C )(10 ×10 C )(10 ×10 C ) = = 0.90 N 2 r122 (1.0 m )
(b) Newton’s second law on either q1 or q2 is
F1 on 2 = m1a1 ⇒ a1 = Assess:
0.90 N = 0.90 m/s 2 1.0 kg
Even a micro-Couomb is a lot of charge. That is why F1 on 2 ( or F2 on 1 ) is a measurable force.
26.14. Model: Model the plastic spheres as point charges. Visualize:
G (a) The charge q1 = −50.0 nC exerts a force F1 on 2 on q2 = −50.0 nC to the right, and the charge q2 exerts G a force F2 on 1 on q1 to the left. Using Coulomb’s law, Solve:
F1 on 2 = F2 on 1 =
9 2 2 −9 −9 K q1 q2 ( 9.0 ×10 N m /C )( 50.0 × 10 C )( 50.0 ×10 C ) = = 0.056 N 2 r122 ( 2.0 ×10−2 m )
(b) The ratio is
0.056 N F1 on 2 = = 2.9 mg ( 2.0 ×10−3 kg )( 9.8 m/s2 )
26.15. Model: Model the glass bead and the ball bearing as point charges. Visualize:
G The ball bearing experiences a downward electric force F1 on 2 . By Newton’s third law, F2 on 1 = F1 on 2 . Solve: Using Coulomb’s law, F1 on 2 = K
q1 q2 ⇒ 0.018 N = r122
( 9.0 ×10
9
N m 2 /C 2 )( 20 ×10−9 C ) q2
(1.0 ×10−2 m )
2
⇒ q2 = 1.0 ×10−8 C
Because the force F1 on 2 is attractive and q1 is a positive charge, the charge q2 is a negative charge. Thus, q2 = −1.0 ×10−8 C = −10 nC.
26.16. Model: Charges A, B, and C are point charges.
G Please refer to Figure EX26.16. Charge A experiences an electric force FB on A due to charge B and G G G an electric force FC on A due to charge C. The force FB on A is directed to the right and the force FC on A is directed to the left. Solve: Coulomb’s law yields: Visualize:
FB on A =
FC on A = The net force on A is
( 9.0 ×10
9
N m 2 /C2 )(1.0 ×10−9 C )(1.0 × 10−9 C )
(1.0 ×10 m ) N m /C )(1.0 ×10 C )( 4.0 ×10 ( 2.0 ×10 m ) 2
−2
( 9.0 ×10
9
2
−9
2
−2
2
−9
C)
= 9.0 ×10−5 N
= 9.0 ×10−5 N
G G G Fon A = FB on A + FC on A = ( 9.0 × 10−5 N ) iˆ + ( 9.0 ×10−5 N ) −iˆ = 0 N
( )
26.17. Model: Charges A, B, and C are point charges. Visualize: Please refer to Figure EX26.17. Solve: The force on A from charge B is directed upward since the two negative charges repel.
FB on A =
1.0 ×10−9 C )( 2.0 ×10−9 C ) q A qB 9 2 2 ( = 9.0 × 10 Nm /C = 4.5 ×10−5 N ( ) 2 −2 4πε 0 r 2 ( 2.0 ×10 m ) 1
G So FB on A = ( 4.5 × 10−5 N, up ) . The force on A from charge C is directed downwards since the two opposite charges attract.
FC on A = ( 9.0 ×109 Nm 2 /C 2 )
(1.0 ×10 C )( 2.0 ×10 (3.0 ×10 m ) −9
−2
2
−9
C)
= 2.0 ×10−5 N
G So FC on A = (2.0 ×10−5 N, down). The net electric force on charge A is G G G FA = FB on A + FC on A = ( 4.5 ×10−5 N − 2.0 × 10−5 N, up )
= ( 2.5 ×10−5 N, up )
26.18. Model: Objects A and B are point charges. Visualize:
Because there are only two charges A and B, the force on charge A is due to charge B only, and the force on B is due to charge A only. Solve: Coulomb’s law gives the magnitude of the forces between the charge. Thus,
FA on B = FB on A =
( 9.0 ×10
9
N m 2 /C 2 )( 20.0 ×10−9 C )(10.0 ×10−9 C )
( 2.0 ×10−2 m )
2
= 4.5 ×10−3 N
Because the charge on object A is positive and on object B is negative, FB on A is upward and FA on B is downward. Thus, G G FB on A = +4.5 ×10−3 ˆj N FA on B = −4.5 × 10−3 ˆj N
26.19. Model: Assume the glass bead, the proton, and the electron are point charges. Visualize:
Solve:
Coulomb’s law gives
Fbead on electron = Fbead on proton =
( 9 ×10
9
N m 2 /C2 )( 20 × 10−9 C )(1.60 ×10−19 C )
(1.0 ×10
−2
m)
2
(a) Newton’s second law is F = ma, so
aproton = In vector form
Fbead on proton mproton
=
2.88 × 10−13 N = 1.72 ×1014 m/s 2 1.67 ×10−27 kg
G aproton = (1.72 × 1014 m/s 2, away from bead )
(b) Similarly, aelectron =
Fbead on electron 2.88 ×10−13 N = = 3.2 × 1017 m/s 2 melectron 9.11×10−31 kg
G Thus aelectron = ( 3.2 ×1017 m/s 2 , toward bead ) .
= 2.88 ×10−13 N
26.20. Model: Protons and electrons produce electric fields. Solve:
(a) The electric field of the proton is
⎡ ⎤ G +1.60 ×10−19 C ⎥ 1 q E= rˆ = ( 9.0 ×109 N m 2 /C 2 ) ⎢ rˆ = (1.44 × 10−3 N/C, away from proton ) 2 2 ⎢ (1.0 ×10−3 m ) ⎥ 4πε 0 r ⎣ ⎦ (b) The electric field of the electron is G E = (1.44 ×10−3 N/C ) ( − rˆ ) = (1.44 ×10−3 N/C, toward electron )
26.21. Model: Model the proton and theGelectron as point charges.G Solve:
G (a) The force that an electric field E exerts on a charge q is F = qE. A proton has q = e. Thus, G Fproton = e 200iˆ + 400ˆj N/C = 3.20iˆ + 6.40ˆj × 10−17 N
(
where we used e = 1.60 ×10
−19
)
(
C.
)
G G (b) The charge on an electron is q = −e. Thus, Felectron = − Fproton = −3.20iˆ − 6.40 ˆj × 10−17 N
(
)
(c) From Newton’s second law, aproton =
Fproton mproton
=
Fx2 + Fy2 mproton
=
7.15 × 10−17 N = 4.28 ×1010 m/s 2 1.67 ×10−27 kg
(d) The electron experiences a force of the same magnitude but it has a different mass. Thus,
aelectron =
7.15 ×10−17 N = 7.85 ×1013 m/s 2 9.11×10−31 kg
The forces may be the same, but the electron has a much larger acceleration due to its much smaller mass. Assess: The two forces in parts (a) and (b) are equal in magnitude but opposite in direction.
26.22. Model: The electric field is due to a charge and extends to all points in space. Solve:
The magnitude of the electric field at a distance r from a charge q is
E=
q q ⇒ 1.0 N/C = ( 9.0 ×109 N m 2 /C 2 ) ⇒ q = 1.11×10−10 C = 0.111 nC 2 2 4πε 0 r (1.0 m ) 1
26.23. Model: The electric field is that of a positive charge on the glass bead. The charge is assumed to be a point charge. Solve: The electric field is
G 8.0 × 10−9 C E = ( 9.0 ×109 N m 2 /C 2 ) rˆ = 1.80 ×105 rˆ N/C 2 −2 ( 2.0 ×10 m ) where rˆ is the unit vector from the charge to the point at which we calculated the field. That is, the direction of the electric field is away from the bead.
26.24. Model: The electric field is that of the charge on the object. Assume the charge on the object is a point charge. Solve: The electric field at a distance r from a point charge q is
G 1 q E= rˆ 4πε 0 r 2 G Because the electric field points toward the object, E = (180,000 N/C )( − rˆ ) . Thus, − (180,000 N/C ) = ( 9.0 ×109 N m 2 /C2 )
q
( 2.0 ×10−2 m )
2
⇒ q = −8.0 × 10−9 C = −8.0 nC
26.25. Model: A field is the agent that exerts an electric force on a charge. Visualize:
Solve:
G Newton’s second law on the plastic ball is Σ Fnet
( )
the electric force,
Fon q = FG ⇒ q E = mg ⇒ E =
y
G G = Fon q − FG . To balance the gravitational force with
−3 mg (1.0 × 10 kg ) ( 9.8 N/kg ) = = 3.3 × 106 N/C q 3.0 ×10−9 C
Because Fon q must be upward and the charge is negative, the electric field at the location of the plastic ball must G be pointing downward. Thus E = ( 3.3 ×106 N/C, downward ). G G G G G G Assess: F = qE means the sign of the charge q determines the direction of F or E. For positive q, E and F G G are pointing in the same direction. But E and F point in opposite directions when q is negative.
26.26. Model: A field is the agent that exerts an electric force on a charge. Visualize:
Solve:
G G (a) To balance the gravitational force on a proton Σ ( Fnet ) y = Fon p − FG
( )
Fon p = ( FG )p ⇒ q E = mg ⇒ E =
p
= 0 N. This means
−27 mg (1.67 ×10 kg ) ( 9.8 N/kg ) = = 1.02 × 10−7 N/C q 1.60 ×10−19 C
Because Fon p must be upward and the proton charge is positive, the electric field at the location of the proton G must also be pointing upward. Thus E = (1.02 × 10−7 N/C, downward ) . (b) In the case of the electron,
E=
−31 mg ( 9.11×10 kg ) ( 9.8 N/kg ) = = 5.58 ×10−11 N/C q 1.60 ×10−19 C
Because Fon e must be upward and the electron has a negative charge, the electric field at the location of the G electron must be pointing downward. Thus E = ( 5.58 × 10−11 N/C, downward ) .
26.27. Model: The electric field is that of a positive point charge located at the origin. Visualize:
The positions (5.0 cm, 0.0 cm), (−5.0 cm, 5.0 cm), and (−5.0 cm, −5.0 cm) are denoted by A, B, and C, respectively. Solve: (a) The electric field for a positive charge is G ⎛ 1 q ⎞ E =⎜ , away from q ⎟ 2 ⎝ 4πε 0 r ⎠
Using 1 4πε 0 = 9.0 ×109 N m 2 /C2 and q = 12 ×10−9 C, G ⎛ 108 N m 2 /C ⎞ E =⎜ , away from q ⎟ 2 r ⎝ ⎠ The electric fields at points A, B, and C are G 108 N m 2 /C ˆ EA = i = 4.3 ×104 iˆ N/C 2 −2 (5.0 ×10 m ) G EB =
EC =
( −5.0 ×10
( −5.0 ×10
108 N m 2 /C −2
m ) + ( 5.0 ×10 2
−2
108 N m 2 /C −2
m ) + ( −5.0 ×10 2
−2
⎡ 1 4 ˆ 4 ˆ ˆ ˆ ⎤ ⎢ 2 −i + j ⎥ = −1.53 ×10 i + 1.53 × 10 j N/C ⎦ m) ⎣ 2
(
) (
)
⎡ 1 4 ˆ 4 ˆ ˆ ˆ ⎤ ⎢ 2 −i − j ⎥ = −1.53 ×10 i − 1.53 ×10 j N/C ⎦ m) ⎣ 2
(
) (
(b) The three vectors are shown in the diagram. G G G Assess: The vectors EA , EB , and EC are pointing away from the positive charge.
)
26.28. Model: The electric field is that of a negative charge located at the origin. Visualize:
The positions (5.0 cm, 0.0 cm), (−5.0 cm, −5.0 cm), and (−5.0 cm, 5.0 cm) are denoted by A, B, and C, respectively. Solve: (a) The electric field for a positive charge is
G ⎛ 1 q ⎞ E =⎜ , away from q ⎟ 2 ⎝ 4πε 0 r ⎠
Using 1 4πε 0 = 9.0 ×109 N m 2 /C2 and q = −12 ×10−9 C, G ⎛ 108 N m 2 /C ⎞ E =⎜ , toward q ⎟ 2 r ⎝ ⎠ The electric fields at points A, B, and C are G 108 N m 2 /C EA = − ˆj = −4.3 ×104 ˆj N/C 2 ( 5.0 ×10−2 m )
( )
G EB =
( −5.0 ×10
G EC =
108 N m 2 /C −2
( −5.0 ×10
m ) + ( −5.0 ×10 2
−2
108 N m 2 /C −2
m ) + ( 5.0 ×10 2
−2
⎡ 1 ˆ ˆ ⎤ 4 ˆ 4 ˆ ⎢ 2 i + j ⎥ = 1.53 × 10 i + 1.53 × 10 j N/C ⎣ ⎦ m) 2
(
) (
)
⎡ 1 ˆ ˆ ⎤ i − j ⎥ = 1.53 × 104 iˆ − 1.53 × 104 ˆj N/C ⎢ ⎦ m) ⎣ 2 2
(
) (
(b) The three vectors are shown in the diagram. G G G Assess: Note that the vectors EA , EB , and EC are pointing toward the negative charge.
)
26.29. Model: Use the charge model and the model of a conductor as a material through which electrons move. Solve: Because the metal spheres are identical, the total charge is split equally between the two spheres. That is, qA = qB = 5.0 × 1011 electrons. Thus, the charge on metal spheres A and B is
( 5.0 ×10 )( −1.60 ×10 11
−19
C ) = −80 nC.
26.30. Model: Use the charge model and the model of a conductor as a material through which electrons move. Solve: Plastic is an insulator and does not transfer charge from one sphere to the other. The charge of metal sphere A is (1.0 ×1012 )( −1.60 ×10−19 C ) = −160 nC and the charge of metal sphere B is 0 C.
26.31. Model: Use the charge model. Solve:
The number of moles in the penny is M 3.1 g n= = = 0.04882 mol A 63.5 g/mol
The number of copper atoms in the penny is N = nN A = ( 0.04882 mol ) ( 6.02 ×1023 mol−1 ) = 2.939 ×1022 Since each copper atom has 29 electrons and 29 protons, the total positive charge in the copper penny is
( 29 × 2.939 ×10 )(1.60 ×10 22
−19
C ) = 1.36 ×105 C
Similarly, the total negative charge is −1.36 × 105 C. Assess: Total positive and negative charges are equal in magnitude.
26.32. Model: The beads are point charges. Visualize:
Solve: The beads are oppositely charged so are attracted to one another. The force on each is the same by Newton’s third law, and is
F=K
4.0 ×10−9 C )( 8.0 ×10−9 C ) q1 q2 −9 2 2 ( = 9.0 × 10 Nm /C = 7.2 ×10−4 N ( ) 2 −2 r2 2.0 × 10 m ( )
The beads accelerate at different rates because their masses are different. By Newton’s second law, the acceleration of the bead on the left is aleft =
−4 F ( 7.2 ×10 N ) = = 0.36 m/s 2 to the right. m ( 2.0 ×10−3 kg )
For the bead on the right, the acceleration is aright =
( 7.2 ×10 ( 4.0 ×10
−4
−3
N)
kg )
= 0.180 m/s 2 to the left.
26.33. Model: The protons are point charges. Solve:
(a) The electric force between the protons is
FE = K
9 2 2 −19 −19 q1 q2 ( 9.0 ×10 N m /C )(1.60 × 10 C )(1.60 ×10 C ) = = 58 N 2 r2 ( 2.0 ×10−15 m )
(b) The gravitational force between the protons is
FG =
G m1m2 ( 6.67 ×10 = r2
−11
N m 2 /kg 2 )(1.67 ×10−27 kg )(1.67 ×10−27 kg )
( 2.0 ×10
−15
m)
2
(c) The ratio of the electric force to the gravitational force is
FE 58 N = = 1.23 × 1036 FG 4.7 ×10−35 N
= 4.7 ×10−35 N
26.34. Model: The 125Xe nucleus and the proton are point charges. That is, all the charge on the Xe nucleus is assumed to be at its center. Visualize:
Solve:
(a) The magnitude of the force between the nucleus and the proton is given by Coulomb’s law:
Fnucleus on proton =
K qnucleus qproton r
2
=
( 9.0 ×10
9
N m 2 /C2 )( 54 ×1.60 ×10−19 C )(1.60 ×10−19 C )
( 5.0 ×10
−15
m)
2
(b) Applying Newton’s second law to the proton,
Fon proton = mproton aproton ⇒ aproton =
5.0 ×102 N = 3.0 × 1029 m/s 2 1.67 × 10−27 kg
= 5.0 ×102 N
26.35. Model: The two charged spheres are point charges. Solve: The electric force on one charged sphere due to the other charged sphere is equal to the sphere’s mass times its acceleration. Because the spheres are identical and equally charged, m1 = m2 = m and q1 = q2 = q. We have
F2 on 1 = F1 on 2 =
Kq1q2 Kq 2 = 2 = ma r2 r
−3 −2 2 mar 2 (1.0 × 10 kg )(150 m/s )( 2.0 × 10 m ) ⇒q = = = 6.7 × 10−15 C 2 K 9.0 ×109 N m 2 /C 2 ⇒ q = 8.2 × 10−8 C = 82 nC 2
2
26.36. Model: Objects A and B are point charges. Visualize:
Solve: (a) It is given that FA on B = 0.45 N. By Newton’s third law, FB on A = FA on B = 0.45 N. (b) Coulomb’s law is
FB on A = FA on B = 0.45 N = ⇒ qB =
( 0.45 N ) r 2 2K
=
KqA qB K ( 2qB )( qB ) = r2 r2
( 0.45 N ) (10 ×10−2 m )
2
2 ( 9.0 ×109 N m 2 /C 2 )
= 5.0 × 10−7 C ⇒ qA = 2qB = 1.0 × 10−6 C
(c) Newton’s second law is FB on A = mAaA. Hence,
aA =
FB on A FA on B 0.45 N = = = 4.5 m/s 2 mA mB 0.100 kg
26.37. Model: The charges are point charges. Visualize: Please refer to Figure P26.37. G G Solve: The electric force on charge q1 is the vector sum of the forces F2 on 1 and F3 on 1 , where q1 is the 1.0 nC charge, q2 is the 2.0 nC charge, and q3 is the other 2.0 nC charge. We have
G ⎛ K q1 q2 ⎞ , away from q2 ⎟ F2 on 1 = ⎜ 2 r ⎝ ⎠ ⎛ ( 9.0 ×109 N m 2 /C 2 )(1.0 ×10−9 C )( 2.0 ×10−9 C ) ⎞ ⎟ , away from q =⎜ 2 2 −2 ⎜ ⎟ (1.0 ×10 m ) ⎝ ⎠ = (1.8 ×10−4 N, away from q2 ) = (1.8 ×10−4 N ) cos60°iˆ + sin 60° ˆj
(
)
G ⎛ K q1 q3 ⎞ F3 on 1 = ⎜ , away from q3 ⎟ = (1.8 × 10−4 N, away from q3 ) = (1.8 × 10−4 N ) − cos60°iˆ + sin 60 ˆj r2 ⎝ ⎠ G G G ⇒ Fon 1 = F2 on 1 + F3 on 1 = 2 (1.8 ×10−4 N ) sin 60° ˆj = 3.1×10−4 ˆj N
(
So the force on the 1.0 nC charge is 3.1×10−4 N directed upward.
)
26.38. Model: The charges are point charges. Visualize: Please refer to Figure P26.38. G G Solve: The electric force on charge q1 is the vector sum of the forces F2 on 1 and F3 on 1 , where q1 is the 1.0 nC charge, q2 is the 2.0 nC charge, and q3 is the −2.0 nC charges. We have
G ⎛ K q1 q2 ⎞ , away from q2 ⎟ F2 on 1 = ⎜ 2 r ⎝ ⎠ 9 2 2 ⎛ ( 9.0 ×10 N m /C )(1.0 × 10−19 C )( 2.0 × 10−19 C ) ⎞ =⎜ , away from q2 ⎟ 2 ⎜ ⎟ (1.0 ×10−2 m ) ⎝ ⎠
(
= (1.80 × 10−4 N, away from q2 ) = (1.80 × 10−4 N ) cos60°iˆ + sin 60° ˆj
)
G ⎛ K q1 q2 ⎞ F3 on 1 = ⎜ , toward q3 ⎟ = (1.80 ×10−4 N, toward q3 ) = (1.80 × 10−4 N ) cos60°iˆ − sin 60° ˆj . 2 ⎝ r ⎠ G G G ⇒ Fon 1 = F2 on 1 + F3 on 1 = (1.80 ×10−4 N ) ( 2cos60° ) iˆ = 1.80 ×10−4 iˆ N
(
So, the force on the 1.0 nC charge is 1.80 × 10–4 N and it is directed to the right.
)
26.39. Model: The charges are point charges. Visualize:
G G The electric force on charge q1 is the vector sum of the forces F2 on 1 and F3 on 1. We have
Solve:
G ⎛ K q1 q2 ⎞ F2 on 1 = ⎜ , away from q2 ⎟ 2 r ⎝ ⎠ ⎛ ( 9.0 ×109 N m 2 /C 2 )(10 × 10−9 C )( 5.0 ×10−9 C ) ⎞ ⎟ , away from =⎜ q 2 2 −2 ⎜ ⎟ 1.0 10 m × ( ) ⎝ ⎠ = ( 4.5 ×10−3 N, away from q2 ) = −4.5 × 10−3 ˆj N
G ⎛ K q1 q2 ⎞ , toward q3 ⎟ F3 on 1 = ⎜ 2 r ⎝ ⎠ 9 2 2 ⎛ ( 9.0 ×10 N m /C )(10 ×10−9 C )(15 ×10−9 C ) ⎞ , toward q3 ⎟ =⎜ 2 2 ⎜ ⎟ ( 3.0 ×10−2 m ) + (1.0 ×10−2 m ) ⎝ ⎠
(
= (1.35 × 10−3 N, toward q3 ) = (1.35 ×10−3 N ) − cosθ iˆ + sin θ ˆj
From the geometry of the figure,
⎛ 1.0 cm ⎞ ⎟ = 18.4° ⎝ 3.0 cm ⎠
θ = tan −1 ⎜
This means cos θ = 0.949 and sin θ = 0.316. Therefore, G F3 on 1 = −1.28 ×10−3 iˆ + 0.43 ×10−3 ˆj N
(
)
G G G ⇒ Fon 1 = F2 on 1 + F3 on 1 = −1.28 × 10−3 iˆ − 4.07 × 10−3 ˆj N
(
The magnitude and direction of the resultant force vector are
( −1.28 ×10
Fon 1 =
N ) + ( −4.07 × 10−3 N ) = 4.3 × 10−3 N 2
2
4.07 ×10−3 N = 3.180 ⇒ φ = tan −1 ( 3.180 ) = 72.5° below the − x axis, 1.28 ×10−3 N points 252.5° counterclockwise from the +x-axis.
tanφ = or Fon 1
−3
)
)
26.40. Model: The charges are point charges. Visualize:
Solve:
G G The electric force on charge q1 is the vector sum of the forces F2 on 1 and F3 on 1. We have G ⎛ K q1 q2 ⎞ F2 on 1 = ⎜ , toward q2 ⎟ 2 ⎝ r ⎠ 9 2 2 ⎛ ( 9.0 × 10 N m /C )(10 × 10−9 C )(10 × 10−9 C ) ⎞ , toward q2 ⎟ =⎜ 2 ⎜ ⎟ (1.0 ×10−2 m ) ⎝ ⎠ −3 −3 ˆ = ( 0.90 × 10 N, toward q2 ) = 0.90 × 10 j N G ⎛ K q1 q3 ⎞ , toward q3 ⎟ F3 on 1 = ⎜ 2 ⎝ r ⎠ ⎛ ( 9.0 × 109 N m 2 /C 2 )(10 × 10−9 C ) ( 8.0 × 10−9 C ) ⎞ , toward q3 ⎟ =⎜ 2 ⎜ ⎟ ( 3.0 ×10−2 m ) ⎝ ⎠ −4 −3 ˆ = ( 8.0 × 10 N, toward q3 ) = −0.80 × 10 i N
G G G ⇒ Fon 1 = F2 on 1 + F3 on 1 = −0.80 × 10−3 iˆ + 0.90 × 10−3 ˆj N
(
)
The magnitude and direction of the resultant force vector are Fon 1 = tan φ =
( −0.80 ×10
−3
N ) + ( 0.90 × 10−3 N ) = 1.20 × 10−3 N 2
2
0.90 × 10−3 N ⇒ φ = tan −1 (1.13) = 48° above the − x -axis, 0.80 × 10−3 N
or Fon 1 points 132° counterclockwise from the +x-axis.
26.41. Model: The charges are point charges. Visualize:
Solve:
G G The electric force on charge q1 is the vector sum of the forces F2 on 1 and F3 on 1. We have G ⎛ K q1 q2 ⎞ , away from q2 ⎟ F2 on 1 = ⎜ 2 r ⎝ ⎠ ⎛ ( 9.0 ×109 N m 2 /C2 )( 5 ×10−9 C )(10 ×10−9 C ) ⎞ ⎟ q , away from =⎜ 2 2 2 −2 −2 ⎜ ⎟ 4.0 10 m 3.0 10 m × + × ( ) ( ) ⎝ ⎠ = (1.8 ×10−4 N, away from q2 ) = (1.8 × 10−4 N ) − cosθ iˆ − sin θ ˆj
(
)
From the geometry of the figure, tan θ =
G 4.0 cm ⇒ θ = 53.13° ⇒ F2 on 1 = (1.8 ×10−4 N ) −0.6iˆ − 0.8 ˆj 3.0 cm
(
)
⎛ ( 9.0 ×109 N m 2 /C 2 )( 5 ×10−9 C )( 5 × 10−9 C ) ⎞ G −4 ⎟ F3 on 1 = ⎜ , toward q N, toward q3 ) = 2.5 × 10−4 iˆ N 3 = ( 2.5 × 10 2 −2 ⎜ ⎟ ( 3.0 ×10 m ) ⎝ ⎠ G G G ⇒ Fon 1 = F2 on 1 + F3 on 1 = 1.42 × 10−4 iˆ − 1.44 × 10−4 ˆj N
(
)
The magnitude and direction of the resultant force vector are Fon 1 =
(1.42 ×10
−4
N ) + ( −1.44 ×10−4 N ) = 2.0 × 10−4 N 2
2
⎛ 1.44 ×10−4 N ⎞ ⎟ = 45° clockwise from the +x-axis. −4 ⎝ 1.42 ×10 N ⎠
φ = tan −1 ⎜
26.42. Model: The charges are point charges. Visualize:
Solve:
G G The electric force on q1 is the vector sum of the forces F2 on 1 and F3 on 1. We have Error! Objects cannot be created from editing field codes.
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From the geometry of the figure, 4.0 cm ⇒ cosθ = 0.6 and sinθ = 0.8 3.0 cm G G G G ⇒ F3 on 1 = −1.08 × 10−4 iˆ − 1.44 ×10−4 ˆj N ⇒ Fon 1 = F2 on 1 + F3 on 1 = −1.08 × 10−4 iˆ + 1.36 × 10−4 ˆj N tan θ =
(
)
(
The magnitude and direction of the resultant force vector are Fon 1 = tan φ =
(1.08 ×10
−4
N ) + ( −1.36 × 10−4 N ) = 1.74 × 10−4 N 2
2
1.36 ×10−4 N ⇒ φ = 52° clockwise from the −x-axis, or 128° CCW from +x-axis. 1.08 ×10−4 N
)
26.43. Model: The charges are point charges. Visualize: Please refer to Figure P26.43. Solve: Placing the 1.0 nC charge at the origin and calling it q1, the q2 charge is in the first quadrant, the q3 charge is in the fourth quadrant, the q4 charge is in the third quadrant, and the q5 charge is in the second quadrant. G G G G The electric force on q1 is the vector sum of the forces F2 on 1 , F3 on 1 , F4 on 1 , and F5 on 1. The magnitude of these four forces is the same because all four charges are equal and equidistant from q1. So,
( 9.0 ×10 N m /C )( 2.0 ×10 C )(1.0 ×10 ( 0.50 ×10 m ) + ( 0.50 ×10 m ) 9
F2 on 1 = F3 on 1 = F4 on 1 = F5 on 1 =
2
−9
2
−2
2
−2
2
−9
C)
= 3.6 ×10−4 N
G Thus, F on 1 = (3.6 × 10−4 N, toward q2) + (3.6 × 10−4 N, toward q3) + (3.6 × 10−4 N, toward q4) + (3.6 × 10−4 N, toward q5). In component form, G K Fon 1 = Fon 1 ⎡ cos 45°iˆ + sin 45° ˆj + cos 45°iˆ − sin 45° ˆj + − cos 45°iˆ − sin 45° ˆj + − cos 45°iˆ + sin 45° ˆj ⎤ = 0 N ⎣ ⎦
(
) (
) (
) (
)
26.44. Model: The charges are point charges. Visualize: Please refer to Figure P26.44. Solve: Placing the 1.0 nC charge at the origin and calling it q1, the q2 charge is in the first quadrant, the q3 charge is in the fourth quadrant, the q4 charge is in the third quadrant, and the q5 charge is in the second quadrant. The electric force on q1 is the vector sum of the electric forces from the other four charges q2, q3, q4, and q5. The magnitude of these four forces is the same because all four charges are equal in magnitude and are equidistant from q1. So,
( 9.0 ×10 N m /C )( 2.0 ×10 C )(1.0 ×10 ( 0.50 ×10 m ) + ( 0.50 ×10 m ) 9
F2 on 1 = F3 on 1 = F4 on 1 = F5 on 1 =
2
−9
2
−2
2
−2
−9
C)
2
= 3.6 ×10−4 N
G Thus, Fon 1 = (3.6 × 10−4 N, away from q2) + (3.6 × 10−4 N, away from q3) + (3.6 × 10−4 N, toward q4) + (3.6 ×
10−4 N, toward q5). In component form, K Fon 1 = Fon 1 ⎡ − cos 45°iˆ − sin 45° ˆj + − cos 45°iˆ + sin 45° ˆj + − cos 45°iˆ − sin 45° ˆj + − cos 45°iˆ + sin 45° ˆj ⎤ ⎣ ⎦ 3 −4 − = ( 3.6 ×10 N ) −4cos 45°iˆ = −1.02 × 10 iˆ N
(
(
) ( )
) (
) (
)
26.45. Model: The charges are point charges. Visualize: Please refer to Figure P26.45. Solve: Placing the 1.0 nC charge at the origin and calling it q1, the −6.0 nC is q3, the q2 charge is in the first quadrant, and the q4 charge is in the second quadrant. The net electric force on q1 is the vector sum of the electric forces from the other three charges q2, q3, and q4. We have G ⎛ K q1 q2 ⎞ , away from q2 ⎟ F2 on 1 = ⎜ 2 ⎝ r ⎠
⎛ ( 9.0 ×109 N m 2 /C 2 )(1.0 ×10−9 C )( 2.0 ×10−9 C ) ⎞ =⎜ , away from q2 ⎟ 2 ⎜ ⎟ ( 5.0 ×10−2 m ) ⎝ ⎠ −5 −5 ˆ = ( 0.72 ×10 N, away from q2 ) = ( 0.72 ×10 N ) − cos 45°i − sin 45° ˆj
(
)
G ⎛ K q1 q3 ⎞ F3 on 1 = ⎜ , toward q3 ⎟ 2 ⎝ r ⎠ ⎛ ( 9.0 ×109 N m 2 /C2 )(1.0 ×10−9 C )( 6.0 ×10−9 C ) ⎞ ⎟ =⎜ , toward q 3 2 −2 ⎜ ⎟ 5.0 10 m × ( ) ⎝ ⎠ −5 −5 ˆ = ( 2.16 ×10 N, away from q3 ) = 2.16 ×10 j N G ⎛ K q1 q4 ⎞ F4 on 1 = ⎜ , away from q4 ⎟ = ( 0.72 × 10−5 N ) cos 45°iˆ − sin 45° ˆj 2 r ⎝ ⎠ G G G G ⇒ Fon 1 = F2 on 1 + F3 on 1 + F4 on 1 = ⎡⎣( 2.16 ×10 −5 N ) − ( 0.72 ×10 −5 N ) ( 2sin 45° ) ⎤⎦ ˆj = 1.14 × 10−5 ˆj N
(
)
26.46. Model: The charges are point charges. Visualize: Please refer to Figure P26.46. Solve: Placing the 1.0 nC charge at the origin and calling it q1, the −6.0 nC is q3, the q2 charge is in the first quadrant, and the q4 charge is in the second quadrant. The net electric force on q1 is the vector sum of the electric forces from the other three charges q2, q3, and q4. We have
G ⎛ K q1 q2 ⎞ F2 on 1 = ⎜ , toward q2 ⎟ 2 ⎝ r ⎠ ⎛ ( 9.0 ×109 N m 2 /C 2 )(1.0 ×10−9 C )( 2.0 ×10−9 C ) ⎞ =⎜ , toward q2 ⎟ 2 ⎜ ⎟ ( 5.0 ×10−2 m ) ⎝ ⎠ −5 −5 ˆ = ( 0.72 × 10 N, toward q2 ) = ( 0.72 × 10 N ) cos 45°i + sin 45° ˆj
(
)
G ⎛ K q1 q3 ⎞ F3 on 1 = ⎜ , toward q3 ⎟ = ( 2.16 ×10−5 N, toward q3 ) = 2.16 × 10−5 ˆj N 2 r ⎝ ⎠ G ⎛ K q1 q4 ⎞ F4 on 1 = ⎜ , away from q4 ⎟ = ( 0.72 × 10−5 N ) cos 45°iˆ − sin 45° ˆj 2 ⎝ r ⎠ G G G G ⇒ Fon 1 = F2 on 1 + F3 on 1 + F4 on 1 = ( 0.72 ×10−5 N ) ( 2cos 45° ) iˆ + ( 2.16 ×10−5 N ) j
(
(
)
= 1.02 × 10−5 iˆ + 2.2 × 10−5 ˆj N
)
26.47. Model: The charged particles are point charges. Visualize:
Solve:
G G (a) The mathematical problem is to find the position for which the forces F1 on p and F2 on p are equal in
magnitude and opposite in direction. If the proton is at position x, it is a distance x from q1 and d − x from q2, where d = 1.0 cm. The magnitudes of the forces are
F1 on p =
K q1 qp r1p2
=
K q1 qp x2
F2 on p =
K q2 qp
(d − x)
2
Equating the two forces and using d = 1.0 cm, K q1 qp x2
=
K q2 qp
(d − x)
2
⇒
2.0 nC 4.0 nC ⇒ 1 + x2 − 2x = 2x2 ⇒ x2 + 2x − 1 = 0 = 2 x2 (d − x)
The solutions to the equation are x = +0.414 cm and −2.41 cm. Both are points where the magnitudes of the two forces are equal, but +0.414 cm is a point where the magnitudes are equal but the directions are the same. The solution we want is that the proton should be placed at x = 2.4 cm. (b) Yes, the net force on the electron located at x = −2.4 cm will be zero. This is because the solution in part (a) does not depend specifically on the type of the charge that experiences zero force from the other two charges. G Furthermore, if Fon p is zero for a proton, the electric field at that point must be zero. Thus, there will be no force on any charged particle at that point.
26.48. Model:
The charged particles are point charges.
Visualize:
Solve: The two 2.0 nC charges exert an upward force on the 1.0 nC charge. Since the net force on the 1 nC charge is zero, the unknown charge must exert a downward force of equal magnitude. This implies that q is a positive charge. The force of charges 2 on charge 1 is
G ⎛ K q1 q2 ⎞ F2 on 1 = ⎜ , away from q2 ⎟ 2 ⎝ r12 ⎠ 9 2 2 (9.0 ×10 N m / C )(1.0 × 10−9 C)(2.0 ×10−9 C) = (cosθ iˆ + sinθ ˆj ) (0.020 m) 2 + (0.030 m) 2 From the figure, θ = tan–1(2/3) = 33.69°. Thus G F2 on 1 = (1.152 ×10−5 iˆ + 0.768 ×10−5 ˆj ) N G G G From symmetry, F3 on 1 is the same except the x-component is reversed. When we add F2 on 1 and F3 on 1 , the xcomponents cancel and the y-components add to give JG JG F 2 on 1 + F 3 on 1 = 1.536 × 10−5 ˆj N G Fq on 1 must have the same magnitude, pointing in the − ˆj direction, so G K q q1 (1.536 ×10−5 N)(0.020 m) 2 ⇒q= = 0.68 nC Fq on 1 = 1.536 ×10−5 N = 2 (9.0 ×109 N m 2 / C 2 )(1.0 × 10−9 C) r
A positive charge q = 0.68 nC will cause the net force on the 1.0 nC charge to be zero.
26.49. Model: The charged particles are point charges. Visualize: Please refer to Figure P26.49. Solve: The charge q2 is in static equilibrium, so the net electric field at the location of q2 is zero. We have
G G G 3.0 ×10−9 C ) q1 1 ˆ + 1 ( Enet = Eq1 + E−3 nC = i ± −iˆ = 0 N/C 4πε 0 ( 0.20 m )2 4πε 0 ( 0.10 m )2
( )
( )
We have used the ± sign to indicate that a positive charge on q1 leads to an electric field along +iˆ and a negative charge on q1 leads to an electric field along −iˆ. Because the above equation can only be satisfied if we use +iˆ, we infer that the charge q1 is a positive charge. Thus, 3.0 ×10−9 C ˆ q1 +iˆ − i = 0 N/C ⇒ q1 = 12.0 nC 2 2 ( 0.20 m ) ( 0.10 m )
( )
()
26.50. Model: The charged particles are point charges. Visualize:
Solve:
The force on q is the vector sum of the force from −Q and +Q. We have
G ⎛ −Q + q ⎞ KQq , toward − Q ⎟ = 2 F−Q on + q = ⎜ K 2 − cosθ iˆ − sinθ ˆj 2 a + y a + y2 ⎝ ⎠
(
)
G ⎛ K +Q + q ⎞ KQq , away from + Q ⎟ = 2 F+ Q on + q = ⎜ 2 − cosθ iˆ + sin θ ˆj 2 a + y a + y2 ⎝ ⎠
(
)
G KQq ⇒ Fnet = 2 ( −2cosθ ) iˆ + 0 ˆj N a + y2 From the figure we see that cosθ = a
a 2 + y 2 . Thus
( Fnet ) x = Assess:
−2 KQqa
(a
2
+ y2 )
3/ 2
Note that (Fnet)x = Fnet because the y-components of the two forces cancel each other out.
26.51. Model: The charged particles are point charges. Visualize:
Solve:
(a) The force on q is the vector sum of the force from −Q and +Q. We have
⎛ K +Q + q ⎞ G KQq F+ Q on + q = ⎜ , away from + Q ⎟ = −iˆ 2 ⎜ (a − x) ⎟ ( a − x )2 ⎝ ⎠
( )
⎛ K −Q + q ⎞ G KQq F− Q on + q = ⎜ , toward − Q ⎟ = −iˆ ⎜ ( a + x )2 ⎟ ( a + x )2 ⎝ ⎠
( )
⎡ 1 2 KQq ( a 2 + x 2 ) 1 ⎤ ⇒ ( Fnet ) x = − KQq ⎢ + = − ⎥ 2 2 2 ⎢⎣ ( a − x ) ( a + x ) ⎥⎦ ( a2 − x2 ) To arrive at the final expression we used ( a − x ) ( a + x ) = ⎡⎣( a − x )( a + x ) ⎤⎦ = ( a 2 − x 2 ) . (b) There are two cases when x > a. For x > a, 2
2
2
2
⎛ K +Q + q ⎞ G KQq +iˆ , away from + Q ⎟ = F+ Q on + q = ⎜ 2 ⎜ ( x − a) ⎟ ( x − a )2 ⎝ ⎠
( )
⎛ K −Q + q ⎞ G KQq F− Q on + q = ⎜ −iˆ , toward − Q ⎟ = ⎜ ( x + a )2 ⎟ ( x + a )2 ⎝ ⎠
( )
⎡ 1 1 ⎤ −4 KQqax ⇒ ( Fnet ) x = KQq ⎢ − = 2 2⎥ 2 ⎢⎣ ( x − a ) ( x + a ) ⎥⎦ ( x 2 − a 2 ) For x < − a (that is, for negative values of x), G KQq F+ Q on + q = −iˆ 2 x ( − a)
G KQq F− Q on + q = +iˆ 2 a ( + x)
( )
⇒ ( Fnet ) x = −
( )
4 KQqax
(x
2
− a2 )
2
That is, the net force is to the right when x > a and to the right when x < − a. We can combine these two cases into a single equation for x > a :
( Fnet ) x = Here, the force is always to the right when x > a.
4 KQqa x
(x
2
− a2 )
2
26.52. Model: The charges are point charges. Solve:
We will denote the charges −Q, 4Q and −Q by 1, 2, and 3, respectively.
G ⎛ K −Q q ⎞ KQq F1 on q = ⎜ , toward − Q ⎟ = 2 −iˆ 2 L L ⎝ ⎠
( )
⎛ ⎞ G K 4Q q 2 KQq ⎟ = 4 KQq +cos 45°iˆ + sin 45° ˆj ⇒ F = 2 F1 on q F2 on q = ⎜⎜ , away from 4 Q 2 on q = 2 2 ⎟ L2 2 L ⎜ 2L ⎟ ⎝ ⎠
(
(
)
)
G ⎛ K −Q q ⎞ KQq , toward − Q ⎟ = 2 − ˆj F3 on q = ⎜ 2 L L ⎝ ⎠
( )
The net electric force on the charge +q is the vector sum of the electric forces from the other three charges. The net force is G ˆj ⎞ KQq 2 KQq ⎛ iˆ KQq KQq KQq Fnet = 2 −iˆ + + ⎜+ ⎟ + 2 − ˆj = − 2 iˆ 1 − 2 − 2 ˆj 1 − 2 L L2 ⎝ L L L 2 2⎠
( )
(
)
2
( )
(
)
2
(
)
KQq ⎡ KQq ⎤ ⎡ KQq ⎤ ⇒ Fnet = ⎢ 2 1 − 2 ⎥ + ⎢ 2 1 − 2 ⎥ = 2 − 2 L2 ⎣ L ⎦ ⎣ L ⎦
(
)
(
)
26.53. Model: The charges are point charges. Visualize:
We must first identify the region of space where the third charge q3 is located. You can see from the figure that the forces can’t possibly add to zero if q3 is above or below the axis or outside the charges. However, at some point on the x-axis between the two charges the forces from the two charges will be oppositely directed. G G Solve: The mathematical problem is to find the position for which the forces F1 on 3 and F2 on 3 are equal in magnitude. If q3 is the distance x from q1, it is the distance L − x from q2. The magnitudes of the forces are F1 on 3 =
K q1 q3 Kq q3 = r132 x2
F2 on 3 =
K q2 q3 K ( 4q ) q3 = 2 r232 ( L − x)
Equating the two forces, Kq q3 K ( 4q ) q3 L 2 = ⇒ ( L − x ) = 4 x 2 ⇒ x = and − L 2 2 x 3 ( L − x) The solution x = −L is not allowed as you can see from the figure. To find the magnitude of the charge q3, we apply the equilibrium condition to charge q1: F2 on 1 = F3 on 1 ⇒
K q2 q1 K q3 q1 4 = ⇒ 4q = 9 q3 ⇒ q3 = q 2 1 L2 9 ( 3 L)
We are now able to check the static equilibrium condition for the charge 4q (or q2): F1 on 2 = F3 on 2 ⇒ K
4 q1 q2 K q3 q2 q q q = ⇒ 2= 9 2= 2 2 2 2 L L L L − x L ( ) (3 )
The sign of the third charge q3 must be negative. A positive sign on q3 will not have a net force of zero either on the charge q or the charge 4q. In summary, a charge of − 94 q placed x = 13 L from the charge q will cause the 3charge system to be in static equilibrium.
26.54. Model: Use the charge model and assume the copper spheres are point objects with point charges. Solve:
(a) The mass of the copper sphere is 3⎤ ⎛ 4π 3 ⎞ ⎡ 4π M = ρV = ⎜ r ⎟ = ( 8920 kg/m3 ) ⎢ (1.0 ×10−3 m ) ⎥ = 3.736 ×10−5 kg = 0.03736 g ⎝ 3 ⎠ ⎣ 3 ⎦
The number of moles in the sphere is n=
M 0.03742 g = = 5.884 ×10−4 mol A 63.5 g/mol
The number of copper atoms in the sphere is
N = nN A = ( 5.884 × 10−4 mol )( 6.02 ×1023 mol−1 ) = 5.542 ×1020 The number of electrons in the copper sphere is thus 29 × 3.542 × 1020 = 1.027 × 1022. The total positive or
negative charge in the sphere is (1.027 × 1022 )(1.60 × 10−19 C ) = 1643 C. Hence, the spheres will have a net charge of 10−9 × 1643 C = 1.643 × 10−6 C. The force between two such spheres is F=
Kq1q2
(1.0 ×10
−2
m)
2
(9.0 ×10 =
9
N m 2 /C2 )(1.643 × 10−6 C )
(1.0 ×10
−2
m)
2
2
= 2.4 × 102 N
(b) This is a force that is easily detectable. Since we don’t observe such forces, any difference between the proton charge and the electron charge must be smaller than 1 part in 109.
26.55. Model: The electron and the proton are point charges. Solve: The electric Coulomb force between the electron and the proton provides the centripetal acceleration for the electron’s circular motion. Thus, K ( e )( e ) mv 2 = = mrω 2 r2 r
⇒ f =
Ke 2 = mr 3
( 9.0 ×10 N m /C )(1.60 ×10 C ) ( 9.11×10 kg )( 5.3 ×10 ) 9
2
−31
2
−19
−11 3
2
= 4.12 ×1016 rad/s ×
1 rev = 6.6 × 1015 rev/s 2π rad
26.56. Model: Model the metal plate and yourself as point charges. Solve: At the beginning, both you and the metal plates are neutral. As the electrons are pumped from the metal plate into you, the plate becomes as much positively charged as you become negatively charged. When enough charge difference builds up between you and the metal plate, the gravitational force on you will be counterbalanced by the upward electrical force. You will slowly begin to hang suspended in the air when
Fplate on you = FG ⇒ mg ( 2.0 m ) = K 2
⇒q =
K q −q
( 2.0 m )
2
= mg
( 60 kg )( 9.8 N/kg )( 2.0 m ) 9.0 ×10 N m /C 9
2
2
2
Dividing this charge by the charge on an electron yields 3.2 ×1015 electrons.
= 5.11× 10−4 C
26.57. Model: The charged plastic beads are point charges and the spring is an ideal spring that obeys Hooke’s law. Solve: Let q be the charge on each plastic bead. The repulsive force between the beads pushes the beads apart. The spring is stretched until the restoring spring force on either bead is equal to the repulsive Coulomb force. That is,
Kq 2 k Δx r 2 = k Δx ⇒ q = 2 r K The spring constant k is obtained by noting that the weight of a 1.0 g mass stretches the spring 1.0 cm. Thus
mg = k (1.0 × 10−2 m) ⇒ k =
⇒q=
(1.0 ×10
−3
kg ) ( 9.8 N/kg )
1.0 ×10−2 m
= 0.98 N/m
( 0.98 N/m ) ( 4.5 ×10−2 m − 4.0 ×10−2 m )( 4.5 ×10−2 m ) 9.0 ×109 N m 2 /C2
2
= 33 nC
26.58. Solve: (a) Kinetic energy is K = 12 mv 2 , so the velocity squared is v2 = 2K/m. From kinematics, a particle moving through distance Δx with acceleration a, starting from rest, finishes with v2 = 2aΔx. To gain K = 2 × 10−18 J of kinetic energy in Δx = 2.0 μm requires an acceleration
a=
2K / m 2.0 × 10−18 J v2 K = = = = 1.10 × 1018 m/s 2 2Δx 2 Δx mΔx ( 9.11×10−31 kg )( 2.0 ×10−6 m )
(b) The force that produces this acceleration is
F = ma = ( 9.11×10−31 kg )(1.10 ×1018 m/s 2 ) = 1.00 × 10−12 N (c) The electric field is
E=
F 1.00 ×10−12 N = = 6.3 × 106 N/C e 1.6 ×10−19 C
(d) The force on an electron due to charge q is F = K q e r 2. To have a breakdown, the force on the electron
must be at least 1.00 × 10−12 N. The minimum charge that could cause a breakdown will be the charge that causes exactly a force of 1.00 × 10−12 N:
G K qe ( 0.010 m ) (1.00 ×10−12 N ) r 2F F = 2 = 1.00 ×10−12 N ⇒ q = = = 6.9 × 10−8 C = 68 nC r Ke ( 9.0 ×109 N m 2 / C 2 )(1.6 ×10−19 C ) 2
26.59. Model: The charged spheres are point charges. Visualize:
Each sphere is in static equilibrium and the string makes an angle θ with the vertical. The three forces acting on each sphere are the electric force, the gravitational force on the sphere, and the tension force. G G G G G Solve: In static equilibrium, Newton’s first law is Fnet = T + FG + Fe = 0. In component form,
( Fnet ) x = Tx + ( FG ) x + ( Fe ) x = 0 N ⇒ −T sin θ + 0 N +
⇒ T sin θ =
( Fnet ) = Ty + ( FG ) y + ( Fe ) y = 0 N
Kq 2 =0 N d2
Kq 2 Kq 2 = 2 d2 ( 2 L sinθ )
T cosθ − mg + 0 N = 0 N T cosθ = + mg
Dividing the two equations,
( 9.0 ×109 N m2/C2 )(100 ×10−9 C ) = 4.59 ×10−4 Kq 2 sin θ tan θ = 2 = 4 L mg 4 (1.0 m )2 ( 5.0 ×10−3 kg ) ( 9.8 N/kg ) 2
2
For small-angles, tan θ ≈ sin θ . With this approximation we obtain sinθ = 0.07714 rad and θ = 4.4°.
26.60. Model: The charged spheres are point charges. Visualize:
Each sphere is in static equilibrium when the string makes an angle of 20° with the vertical. The three forces acting on each sphere are the electric force, the gravitational force on the sphere, and the tension force. G G G G G Solve: In the static equilibrium, Newton’s first law is Fnet = T + FG + Fe = 0. In component form,
( )
( Fnet ) = Tx + ( FG ) x + ( Fe ) x = 0 N
⇒ −T sinθ + 0 N + ⇒ T sin θ =
( Fnet ) y = Ty + ( FG ) + ( Fe ) y = 0 N
Kq 2 =0N d2
T cosθ − mg + 0 N = 0 N
+ Kq 2 Kq 2 = 2 d2 ( 2 L sinθ )
T cosθ = + mg
Dividing the two equations and solving for q, 4 ( sin 2 20° tan 20° ) (1.0 m ) ( 3.0 ×10−3 kg ) ( 9.8 N/kg ) 4sin 2 θ tanθ L2 mg = = 0.75 μ C 9.0 ×109 N m 2 /C 2 K 2
q=
26.61. Model: The electric field is that of a positive point charge located at the origin. Visualize: Please refer to Figure P26.61. Place the 10 nC charge at the origin. Solve: The electric field is 9 2 2 −9 ⎞ G ⎛ 1 q ⎞ ⎛ ( 9.0 ×10 N m /C )(10 ×10 C ) ⎜ E =⎜ q = , away from , away from q ⎟ ⎟ 2 2 ⎜ ⎟ r ⎝ 4πε 0 r ⎠ ⎝ ⎠
⎛ 90.0 N m 2 /C ⎞ =⎜ , away from q ⎟ 2 r ⎝ ⎠ At each of the three points, ⎞ G ⎛ 90.0 N m 2 /C E1 = ⎜ , away from q ⎟ = 1.0 ×105 ˆj N/C 2 ⎜ ( 3.0 ×10−2 m ) ⎟ ⎝ ⎠ ⎞ G ⎛ 90.0 N m 2 /C , away from q ⎟ = ( 3.6 × 104 N/C ) cosθ iˆ + sinθ ˆj E2 = ⎜ 2 ⎜ ( 5.0 ×10−2 m ) ⎟ ⎝ ⎠ 4 3 ˆ 4 ˆ = ( 3.6 ×10 N/C ) 5 i + 5 j = 2.9 × 104iˆ + 2.2 × 104 ˆj N/C
(
(
) (
⎞ G ⎛ 90.0 N m 2 /C ⎟ = 5.6 × 104iˆ N/C , away from q E3 = ⎜ ⎜ ( 4.0 ×10−2 m )2 ⎟ ⎝ ⎠
)
)
26.62. Model: The electric field is that of a positive point charge. Visualize: Please refer to Figure P26.62. Place point 1 at the origin. Solve: The electric field is 9 2 2 −9 ⎞ G ⎛ 1 q ⎞ ⎛ ( 9.0 ×10 N m /C )( 2 × 10 C ) ⎜ E =⎜ q = , toward , toward q ⎟ ⎟ 2 2 ⎜ ⎟ r ⎝ 4πε 0 r ⎠ ⎝ ⎠
⎛ 18.0 N m 2 /C ⎞ =⎜ , toward q ⎟ 2 r ⎝ ⎠ The electric fields at the two points are ⎞ G ⎛ 18.0 N m 2 /C ⎟ E1 = ⎜ q , toward ⎜ (1.0 ×10−2 m )2 ⎟ ⎝ ⎠ = (1.8 × 105 N/C, 60° counter clockwise from the +x-axis or 60° north of east) ⎞ G ⎛ 18.0 N m 2 /C E2 = ⎜ , toward q ⎟ 2 ⎜ (1.0 ×10−2 m ) ⎟ ⎝ ⎠ = (1.8 × 105 N/C, 60° clockwise from the −x-axis or 60° north of west)
26.63. Model: The electric field is that of a positive point charge located at the origin. Visualize: Please refer to Figure P26.63. Place the 5.0 nC charge at the origin. Solve: The electric field is 9 2 2 −9 ⎞ G ⎛ 1 q ⎞ ⎛ ( 9 ×10 N m / C )( 5.0 × 10 C ) ⎜ E =⎜ q = , away from , away from q ⎟ ⎟ 2 2 ⎜ ⎟ r ⎝ 4πε 0 r ⎠ ⎝ ⎠
⎛ 45 N m 2 /C ⎞ =⎜ , away from q ⎟ 2 r ⎝ ⎠ At each of the three points, ⎞ G ⎛ 45 N m 2 /C ⎟ = ( 9.0 × 104 N/C ) cosθ iˆ + sin θ ˆj E1 = ⎜ q , away from ⎜ ( 2.0 ×10−2 m )2 + (1.0 × 10−2 m )2 ⎟ ⎝ ⎠ ⎛ 1 ˆ 2 ˆ⎞ 4 4ˆ = ( 9.0 ×10 N/C ) ⎜ i+ j ⎟ = 4.0 × 10 i + 8.0 × 104 ˆj N/C 5 ⎠ ⎝ 5
(
(
)
⎞ G ⎛ 45 N m 2 /C ⎟ = 4.5 ×105iˆ N/C E2 = ⎜ , away from q ⎜ (1.0 ×10−2 m )2 ⎟ ⎝ ⎠ ⎛ ⎞ 2 G 45 N m /C ⎟ = 4.0 × 104 iˆ − 8.0 × 104 ˆj N/C E3 = ⎜ q , away from ⎜ ( 2.0 × 10−2 m )2 + (1.0 × 10−2 m )2 ⎟ ⎝ ⎠
(
)
)
26.64. Model: The electric field is that of a negative charge at (x, y) = (2.0 cm, 1.0 cm). Visualize:
Solve: (a) The electric field of a negative charge points toward the charge, so we can roughly locate where the field has a particular value by inspecting the signs of Ex and Ey. At point a, the electric field has no y-component and the x-component points to the left, so its location must be to the right of the charge along a horizontal line. Using the equation for the field of a point charge, Ex = E =
Kq ⇒ ra = ra2
( 9.0 ×10
Kq = Ex
9
N m 2 /C 2 )(10.0 ×10−9 C ) 225,000 N/C
= 0.0200 m = 2.00 cm
Thus, point a is at the position (xa, ya) = (4 cm, 1 cm). (b) Point b is above and to the left of the charge. The magnitude of the field at this point is E = Ex2 + E y2 =
(161,000 N/C ) + ( 80,500 N/C ) 2
2
= 180,000 N/C
Using the equation for the field of a point charge, E=
Kq ⇒ rb = rb2
Kq = E
( 9.0 ×10
9
N m 2 /C2 )(10 ×10−9 C ) 180,000 N/C
= 2.236 cm
This gives the total distance but not the horizontal and vertical components. However, we can determine the G angle θ because Eb points straight toward the negative charge. Thus, ⎛ Ey ⎜ E ⎝ x
θ = tan −1 ⎜
⎞ ⎛ 80,500 ⎞ −1 ⎛ 1 ⎞ ⎟ = tan −1 ⎜ ⎟ = tan ⎜ ⎟ = 26.57° ⎟ ⎝ 2⎠ ⎝ 161,000 ⎠ ⎠
The horizontal and vertical distances are then d x = rb cosθ = 2.00 cm and d y = rb sin θ = 1.00 cm. Thus, point b is
at the position (xb, yb) = (0 cm, 2 cm). (c) Point c, which is below and to the left of the charge, is calculated by following a similar procedure. We first find that E = 36,000 N/C . From this we find that the total distance rc = 5.00 cm. The angle φ is ⎛ Ey ⎜ Ex ⎝
φ = tan −1 ⎜
⎞ ⎛ 21,600 ⎞ ⎟ = tan −1 ⎜ ⎟ = 36.87° ⎟ ⎝ 28,000 ⎠ ⎠
which gives the distances d x = rc cos φ = 4.00 cm and d y = rc sin φ = 3.00 cm. Thus point c is at position (xc, yc) = (−2 cm, − 2 cm).
26.65. Model: Visualize:
The electric field is that of a positive charge at (x, y) = (1.0 cm, 2.0 cm).
Solve: (a) The electric field of a positive charge points straight away from the charge, so we can roughly locate the points of interest based simply on whether the signs of Ex and Ey are positive or negative. For point a, the electric field has no y-component and the x-component points to the left, so point a must be to the left of the charge along a horizontal line. Using the field of a point charge,
Ex = E =
Kq ⇒ ra = ra2
Kq = Ex
( 9.0 ×10
9
N m 2 /C 2 )(10.0 ×10−9 C ) 225,000 N/C
= 0.0200 m = 2.00 cm
Thus, (xa, ya) = (−1 cm, 2 cm). (b) Point b is above and to the right of the charge. The magnitude of the field at this point is E = Ex2 + E y2 =
(161,000 N/C ) + (80,500 N/C ) 2
2
= 180,000 N/C
Using the field of a point charge, E=
Kq ⇒ rb = rb2
Kq G = E
( 9.0 ×10
9
N m 2 /C2 )(10.0 ×10−9 C ) 180,000 N/C
= 2.236 cm
This gives the total distance but not the horizontal and vertical components. However, we can determine the G angle θ because Eb points straight away from the positive charge. Thus, ⎛ Ey ⎞ ⎛ 80,500 N/C ⎞ −1 ⎛ 1 ⎞ ⎟ = tan −1 ⎜ ⎟ = tan ⎜ ⎟ = 26.57° ⎜ Ex ⎟ 161,000 N/C ⎠ ⎝ 2⎠ ⎝ ⎝ ⎠ The horizontal and vertical distances are then d x = rb cosθ = ( 2.236 cm ) cos 26.57° = 2.00 cm and d y = rb sin θ = 1.00 cm. Thus, point b is at position (xb, yb) = (3 cm, 3 cm). (c) To calculate point c, which is below and to the right of the charge, a similar procedure is followed. We first find E = 36,000 N/C from which we find the total distance rc = 5.00 cm. The angle φ is
θ = tan −1 ⎜
⎛ Ey ⎞ ⎛ 28,800 ⎞ ⎟ = tan −1 ⎜ ⎟ = 53.13° ⎜ Ex ⎟ ⎝ 21,600 ⎠ ⎝ ⎠ which gives distances d x = rc cos φ = 3.00 cm and d y = rc sin φ = 4.00 cm. Thus, point c is at position (xc, yc) = (4 cm, − 2 cm).
φ = tan −1 ⎜
26.66. Model: The electric field is that of three point charges. Visualize:
Solve:
(a) In the figure, the distances are r1 = r3 =
(1 cm ) + ( 3 cm ) 2
2
= 3.162 cm and the angle is
θ = tan −1 (1/ 3) = 18.43°. Using the equation for the field of a point charge, E1 = E3 =
9 2 2 −9 K q1 ( 9.0 ×10 N m /C )(1.0 ×10 C ) = = 9.0 kN/C 2 r12 ( 0.03162 m )
We now use the angle θ to find the components of the field vectors:
( ) N/C = (8.5iˆ − 2.8 ˆj ) kN/C = E cosθ iˆ + E sinθ ˆj = ( 8540iˆ + 2840 ˆj ) N/C = ( 8.5iˆ + 2.8 ˆj ) kN/C
E1 = E1 cosθ iˆ − E1 sin θ ˆj = 8540iˆ − 2840 ˆj E3
3
3
E2 is easier since it has only an x-component. Its magnitude is E2 =
9 2 2 −9 K q2 ( 9.0 ×10 N m /C )(1.0 ×10 C ) = = 10,000 N/C ⇒ E2 = E2 iˆ = 10.0 iˆ kN/C 2 r22 ( 0.0300 m )
(b) The electric field is defined in terms of an electric force acting on charge q: E = F q . Since forces obey a principle of superposition ( Fnet = F1 + F2 + … ) it follows that the electric field due to several charges also obeys a
principle of superposition. (c) The net electric field at a point 3 cm to the right of q2 is Enet = E1 + E2 + E3 = 27 iˆ kN/C. The y-components of
E1 and E2 cancel, giving a net field pointing along the x-axis.
26.67. Model: The charged ball attached to the string is a point charge. Visualize:
The ball is in static equilibrium in the external electric field when the string makes an angle θ = 20° with the vertical. The three forces acting on the charged ball are the electric force due to the field, the gravitational force on the ball, and the tension force. G G G G G Solve: In static equilibrium, Newton’s second law for the ball is Fnet = T + FG + Fe = 0. In component form,
( Fnet ) x = Tx + 0 N + qE = 0 N
( Fnet ) y = Ty − mg + 0 N = 0 N
The above two equations simplify to T sin θ = qE Dividing both equations, we get
tan θ =
T cosθ = mg
−3 qE mg tan θ ( 5.0 × 10 kg ) ( 9.8 N/kg ) tan 20° ⇒q= = = 1.78 × 10−7 C = 178 nC 100,000 N/C mg E
26.68. Model: The charged ball attached to the string is the point charge. Visualize:
The charged ball is in static equilibrium in the external electric field when the string makes an angle θ with the vertical. The three forces acting on the charge are the electric force due to the electric field, the gravitational force on the ball, and the tension force. G G G G G Solve: In static equilibrium, Newton’s second law for the charged ball is Fnet = T + FG + F3 = 0. In component form,
( Fnet ) x = Tx + 0 N + qE = 0 N
( Fnet ) y = Ty − mg + 0 N = 0 N
These two equations become T sin θ = qE and T cosθ = mg . Dividing the equations gives qE ( 25 ×10 C ) ( 200,000 N/C ) = = 0.255 ⇒ θ = 14.3° mg ( 2.0 ×10−3 kg ) ( 9.8 N/kg ) −9
tan θ =
26.69. Solve: (a) How many excess electrons on a dust particle produce an electric field of magnitude 1.0 ×
106 N/C a distance of 1.0 μm from the dust particle? (b) The number of electrons is
N=
(1.5 ×10
6
N/C )(1.0 × 10−6 m )
2
(9.0 ×109 N m 2/C2 )(1.60 ×10−19 C )
= 1.04 ×103
26.70. Solve: (a) Two equal charges separated by 1.50 cm exert repulsive forces of 0.020 N on each other. What is the magnitude of the charge? (b) The charge is q=
( 0.020 N )( 0.0150 m ) 9.0 ×109 N m 2 /C 2
2
= 22 nC
The problem does not give the direction of the force. So, the charges could be both positive or both negative.
26.71. Solve: (a) At what distance from a 15 nC charge is the electric field strength 54,000 N/C? (b) The distance is r=
( 9.0 ×10
9
N m 2 /C 2 )(15 ×10−9 C ) 54,000 N/C
= 0.050 m = 5.0 cm
26.72. Solve: (a) A 1.0 nC charge is placed at (0 cm, 2 cm) and another 1.0 nC charge is placed at (0 cm, –2 cm). A third charge +q is placed along a line halfway between q1 and q2 such that the angle between the forces on q due to each of the other two charges is 60°. The resultant force on q is 5.0 × 10−5 N, iˆ . What is the magnitude
(
)
of the charge q? (b)
We have ⎛ ( 9.0 ×109 N m 2 /C 2 )(1.0 ×10−9 C ) q ⎞ G , away from q3 ⎟ = ( 5625 N/C ) q cos30°iˆ − sin 30° ˆj F1 on 3 = ⎜ 2 ⎜ ⎟ ( 0.020 m sin 30°) ⎝ ⎠ G F2 on 3 = ( 5625 N/C ) q cos30°iˆ + sin 30 ˆj
(
(
)
G ⇒ Fon 3 = 2 × ( 5625 N/C ) q cos30°iˆ 2 ( 5625 N/C ) q cos30° = 5.0 ×10−5 N ⇒ q =
( 5.0 ×10
−5
N)
2 ( 5625 N/C ) cos30°
= 5.1 nC
)
26.73. Model: Use the charge model. Solve:
The mass of copper in a 2.0-mm-diameter copper ball is 3⎤ ⎛ 4π 3 ⎞ ⎡ 4π M = ρV = ρ ⎜ r ⎟ = ( 8920 kg/m3 ) ⎢ (1.0 ×10−3 m ) ⎥ = 3.736 × 10−5 kg = 0.03736 g 3 3 ⎝ ⎠ ⎣ ⎦
The number of moles in the ball is n=
M 0.03736 g = = 5.884 ×10−4 mol A 63.5 g/mol
The number of copper atoms in the ball is
N = nN A = ( 5.884 ×10−4 mol )( 6.02 ×1023 mol−1 ) = 3.542 ×1020 We note that the number of electrons per atom is the atomic number, and both the atomic number (29) and the average atomic mass (63.5 g) are taken from the periodic table in the textbook. The number of electrons in the copper ball is thus 29 × 3.542 × 1020 = 1.03 × 1022. The number of electrons removed from the copper ball is 50 ×10−9 C = 3.13 ×1011 1.60 ×10−19 C So, the fraction of electrons removed from the copper ball is 3.13 × 1011 = 3.0 ×10−11 1.03 × 1022 Assess:
This is indeed a very small fraction of the available number of electrons in the copper ball.
26.74. Model: The charged balls are point charges. Visualize:
Because of symmetry and the fact that the three balls have the same charge, the magnitude of the electric force on each ball is the same. The other forces acting on each ball are the gravitational force on the ball and the tension force. Solve: The force on ball 3 is the sum of the force from ball 1 and ball 2. We have
G ⎛ K q1 q3 ⎞ Kq 2 F1 on 3 = ⎜ , away from q ⎟ = 2 sin 30°iˆ + cos30° ˆj 1 2 r ⎝ r13 ⎠
(
G Kq 2 F2 on 3 = 2 − sin 30°iˆ + cos30 ˆj r
(
)
)
G 2 Kq 2 2 Kq 2 ⇒ Fon 3 = 2 cos30° ˆj ⇒ Fon 3 = 2 cos30° = Fon 2 = Fon 1 = Fe r r
⇒ Fe =
2 ( 9.0 ×109 N m 2 /C2 ) q 2 cos30°
( 0.20 m )
2
= ( 3.897 ×1011 q 2 ) N/C 2
The distance l, between one of the balls and the center of the equilateral triangle, is l cos30° =
r 0.10 m = 0.10 m ⇒ l = = 0.1155 m 2 cos30°
Thus, the angle made by the string with the plane containing the three balls is cosθ =
l 0.1155 m = ⇒ θ = 81.70° L 0.80 m
From the free-body diagram, we have T sin θ − mg = 0 N −T cosθ + Fe = 0 N
⇒ tanθ =
⇒ q=
mg = 11 3.897 10 N/C 2 ) tanθ × (
mg mg = Fe ( 3.897 ×1011 q 2 ) N/C2
( 3.0 ×10
−3
kg ) ( 9.8 N/kg )
( 3.897 ×10 N/C2 ) tan81.70° 11
= 1.05 ×10−7 C = 105 nC
26.75. Model: The charged spheres are point charges. Visualize:
The figure shows the free-body diagram of the forces on the sphere with the negative charge that is shown in Figure CP26.75. The force FE is due to the external electric field. The force Fe is the attractive force between the positive and the negative spheres. The tension in the string and the gravitational force are the remaining two forces on the spheres. Solve: The two electrical forces are calculated as follows:
FE = q E = (100 ×10−9 C )(105 N/C ) = 1.00 × 10−2 N −9 9 2 2 K q1 q2 ( 9.0 ×10 N m /C )(100 ×10 C ) 9.0 ×10−5 N m 2 /C Fe = = = 2 2 r r r2 2
From the geometry of Figure CP26.75,
r = 2 ( L sin10° ) = 2 ( 0.50 m ) sin10° = 0.174 m ⇒ Fe =
9.0 ×10−5 N m 2 /C
( 0.174 m )
2
= 3.0 ×10−3 N
From the free-body diagram, T cos10° = mg
T sin10° + Fe = FE
Rearranging and dividing the two equations, sin10° F − Fe = tan10° = E mg cos10° m=
FE − Fe 1.00 ×10−2 N − 0.30 ×10−2 N = = 0.41× 10−2 kg = 4.1 g g tan10° ( 9.8 N/kg ) tan10°
26.76.
Model: Visualize:
The charges are point charges.
Solve: The forces on the –1.0 nC charge lie along the line connecting the pairs of charges. Since the 10 nC G G charge is positive, F10 points as shown. The angle formed by the dashed lines where they meet is 90°, so Fq G G G must point towards q in order for F = F10 + Fq to hold. Therefore F=
F10 cos30°
The distance between the 10 nC and –1.0 nC charges is r = 5.0sin 30° cm = 2.5 cm. Thus
F10 = ( 9.0 ×109 N m 2 /C2 )
(10 ×10 C )(1.0 ×10 ( 2.5 ×10 m ) −9
−2
−9
C)
2
Hence F=
1.44 ×10−4 N = 1.66 ×10−4 N cos30°
= 1.44 ×10−4 N
26.77. Model: Charge Q and the dipole charges (q and −q) are point charges. Visualize: Please refer to Figure CP26.77. Solve: (a) The force on the dipole is the vector sum of the force on q and –q. We have ⎛ KQ q ⎞ KQq , away from Q ⎟ = −iˆ FQ on + q = ⎜ ⎜ ( r + s / 2 )2 ⎟ ( r + s / 2 )2 ⎝ ⎠
( )
FQ on −q =
KQq
( r − s / 2)
2
( +iˆ ) ⇒ F
net
⎛ ⎞ 1 1 = KQq ⎜ − ⎟ iˆ ⎜ ( r − s / 2 )2 ( r + s / 2 )2 ⎟ ⎝ ⎠
(b) The net force Fnet is toward the charge Q, because the attractive force due to Q on the negative charge of the dipole is more than the repulsive force of Q on the positive charge. (c) Using the binomial approximation, we get
( r ± s / 2)
−2
−2
s ⎞ ⎛ ⎛ 2s = r −2 ⎜1 ± ⎟ ≅ r −2 ⎜1 ∓ + ⎝ 2r ⎠ ⎝ 2r
KQq ⎛ ⎛ 2s ⎞ ⎛ 2s ⎞ ⎞ 2 KQqs ⎞ ⎟ ⇒ Fnet = 2 ⎜ ⎜1 + ⎟ − ⎜ 1 − ⎟ ⎟ = r ⎝ ⎝ 2r ⎠ ⎝ 2r ⎠ ⎠ r3 ⎠
(d) Coulomb’s law applies only to the force between two point charges. A dipole is not a point charge, so there’s no reason that the force between a dipole and a point charge should be an inverse-square force. Assess: Note that when s → 0 m, Fnet → 0 N. In this limit the dipole is a point with zero charge.
26-1
27.1.
Model: The electric field is that of the two charges placed on the y-axis. Visualize: Please refer to Figure EX27.1. We denote the upper charge by q1 and the lower charge by q2. Because both the charges are positive, their electric fields at P are directed away from the charges. Solve: The electric field from q1 is 9 2 2 −9 G ⎛ 1 q1 ⎞ ( 9.0 × 10 N m /C )( 3.0 × 10 C ) = cosθ iˆ − sin θ ˆj θ + E1 = ⎜ , below x -axis ⎟ 2 2 2 ( 0.050 m ) + ( 0.050 m ) ⎝ 4πε 0 r1 ⎠
(
)
Because tan θ = 5 cm 5 cm = 1, the angle θ = 45°. Hence, G ⎛ 1 ˆ 1 E1 = ( 5400 N/C ) ⎜ i− 2 ⎝ 2
ˆj ⎞ ⎟ ⎠
Similarly, the electric field from q2 is
G ⎛ 1 q2 ⎞ ⎛ 1 ˆ 1 ˆ⎞ E2 = ⎜ i+ j⎟ , θ above + x -axis ⎟ = ( 5400 N/C ) ⎜ 2 r πε 4 2 ⎠ ⎝ 2 0 2 ⎝ ⎠
G G G ⎛ 1 ⎞ˆ 3ˆ ⇒ Enet at P = E1 + E2 = 2 ( 5400 N/C ) ⎜ ⎟ i = 7.6 ×10 i N/C ⎝ 2⎠ Thus, the strength of the electric field is 7.6×103 N/C and its direction is horizontal. Assess: Because the charges are located symmetrically on either side of the x-axis and are of equal value, the ycomponents of their fields will cancel when added.
27.2. Model: The electric field at the point is found by superposition of the fields due to the two charges located on the y-axis. Visualize: Please refer to Figure EX27.2. The electric field due to the positive charge q1 at the point is away from q1. On the other hand, the electric field due to the negative charge q2 at the point is toward q2. These two electric fields are then added vertically to obtain the net electric field at the point. Solve: The electric field from q1 is 9 2 2 −9 G ⎛ 1 q1 ⎞ ⎛ ( 9.0 × 10 N m /C )( 3.0 × 10 C ) ⎞ ⎜ ⎟ cosθ iˆ − sin θ ˆj = E1 = ⎜ θ x , below + -axis ⎟ 2 2 2 ⎟ ( 0.050 m ) + ( 0.050 m ) ⎝ 4πε 0 r1 ⎠ ⎜⎝ ⎠
(
)
Because tan θ = 5 cm 5 cm, θ = 45°. So, G ⎛ 1 ˆ 1 ˆ⎞ E1 = ( 5400 N/C ) ⎜ + i− j⎟ 2 2 ⎠ ⎝ Similarly, the electric field from q2 is
G ⎛ 1 q2 ⎞ ⎛ 1 ˆ 1 i− E2 = ⎜ , θ below − x -axis ⎟ = ( 5400 N/C ) ⎜ − 2 2 2 ⎝ ⎝ 4πε 0 r2 ⎠
ˆj ⎞ ⎟ ⎠
G G G ⎛ 1 ⎞ˆ 3 ˆ ⇒ Enet = E1 + E2 = 2 ( 5400 N/C ) ⎜ − ⎟ j = −7637 j N/C = −7.6 × 10 N/C 2 ⎝ ⎠ Thus, the strength of the electric field is 7.6×103 N/C and its direction is vertically downward. Assess: A quick visualization of the components of the two electric fields shows that the horizontal components cancel.
27.3. Model: The electric field is that due to superposition of the fields of the two 3.0 nC charges located on the y-axis. Visualize: Please refer to Figure EX27.3. We denote the top 3.0 nC charge by q1 and the bottom 3.0 nC charge G G by q2. The electric fields ( E1 and E2 ) of both the positive charges are directed away from their respective G charges. With vector addition, they yield the net electric field Enet at the point P indicated by the dot. Solve: The electric fields from q1 and q2 are 9 2 2 −9 G ⎛ 1 q1 ⎞ ( 9.0 ×10 N m /C )( 3.0 ×10 C ) ˆ i = 10,800iˆ N/C , along + x -axis ⎟ = E1 = ⎜ 2 2 ( 0.05 m ) ⎝ 4πε 0 r1 ⎠
G ⎛ 1 q2 ⎞ E2 = ⎜ , θ above + x -axis ⎟ 2 ⎝ 4πε 0 r2 ⎠ Because tan θ = 10 cm 5 cm, θ = tan −1 ( 2 ) = 63.43°. So, G ( 9.0 × 109 N m 2 /C2 )( 3.0 × 10−9 C ) cos63.43° iˆ + sin 63.43° ˆj = 966iˆ + 1127 ˆj N/C E2 = 2 2 ( 0.10 m ) + ( 0.050 m )
(
) (
)
The net electric field is thus
G G G Enet at P = E1 + E2 = 11,766iˆ + 1127 ˆj N/C
(
)
To find the angle this net vector makes with the x-axis, we calculate
tan φ =
1127 N/C ⇒ φ = 5.5° 11,766 N/C
Thus, the strength of the electric field at P is Enet =
(11,766 N/C )
2
+ (1127 N/C ) = 11,820 N/C = 1.18×103 N/C 2
G and Enet makes an angle of 5.5° above the +x-axis. Assess: Because of the inverse square dependence on distance, E2 < E1. Additionally, because the point P has no special symmetry relative to the charges, we expected the net field to be at an angle relative to the x-axis.
27.4. Model: The electric field at the point is the superposition of the fields due to the two charges located on the y-axis. Visualize: Please refer to Figure EX27.4. We denote the positive charge by q1 and the negative charge by q2. G G The electric field E1 of the positive charge q1 is directed away from q1, but the field E2 is toward the negative G G charge q2. We will add E1 and E2 vectorially to find the strength and the direction of the net electric field vector. Solve: The electric fields from q1 and q2 are 9 2 2 −9 G ⎛ 1 q1 ⎞ ( 9.0 × 10 N m /C )( 3.0 × 10 C ) ˆ E1 = ⎜ , away from q along + x -axis = i = 10,800iˆ N/C ⎟ 1 2 2 ( 0.050 m ) ⎝ 4πε 0 r1 ⎠
G ⎛ 1 q2 ⎞ , θ below − x-axis ⎟ E2 = ⎜ 2 πε 4 r 0 2 ⎝ ⎠ From the geometry of the figure, tan θ =
10 cm ⇒ θ = 63.43° 5 cm
G ( 9.0 × 109 N m 2 /C 2 )( 3.0 × 10−9 C ) ⇒ E2 = − cos 63.43°iˆ − sin 63.43° ˆj = − 966iˆ + 1127 ˆj N/C 2 2 ( 0.10 m ) + ( 0.050 m )
(
) (
G G G ⇒ Enet = E1 + E2 = 9834iˆ − 1127 ˆj N/C ⇒ Enet =
(
)
( 9834 N/C ) + ( −1127 N/C ) 2
)
2
= 9.9×103 N/C
To find the angle this net vector makes with the horizontal, we calculate
tan φ =
( Enet ) y ( Enet ) x
=
1127 N/C ⇒ φ = 6.5° 9834 N/C
G Thus, the strength of the net electric field at the point is 9.9×103 N/C and Enet makes an angle of 6.5° below the +x-axis.
27.5. Model: The distances to the observation points are large compared to the size of the dipole, so model the field as that of a dipole moment. Visualize:
The dipole consists of charges ±q along the y-axis. The electric field in (a) points down. The field in (b) points up. Solve: (a) The dipole moment is G p = (qs, from – to +) = (1.0 × 10−9 C) ( 0.0020 m ) ˆj = 2.0 × 10−12 ˆj C m
The electric field at (10 cm, 0 cm), which is at distance r = 0.10 m in the plane perpendicular to the electric dipole, is G G 1 p 2.0 × 10−12 ˆj C m E=− = −(9.0 × 109 N m 2 /C 2 ) = −18.0 ˆj N/C 3 4π ε 0 r (0.10 m)3
The field strength, which is all we’re asked for, is 18.0 N/C. (b) The electric field at (0 cm, 10 cm), which is at r = 0.10 m along the axis of the dipole, is G −12 ˆ G 1 2p j C m) 9 2 2 2(2.0 × 10 = × = 36 ˆj N/C E= (9.0 10 N m /C ) 3 4π ε 0 r (0.10 m)3 The field strength at this point is 36 N/C.
27.6. Model: The distances to the observation points are large compared to the size of the dipole, so model the field as that of a dipole moment. Visualize:
The dipole consists of charges ±q along the y-axis. The electric field in (a) points up. The field in (b) points down. Solve: (a) The electric field at (0 cm, 10 cm), which is at r = 0.10 m along the axis of the dipole, is G G 1 2p E = 360 ˆj N/C = 4π ε 0 r 3 G r 3E (0.10 m)3 (360 ˆj N/C ) G ⇒ p= = = (2.0 × 10−11 ˆj C m) 2 (1/4π ε 0 ) 2(9.0 × 109 N m 2 /C 2 )
G By definition, the dipole moment is p = 2.0 × 10−11 ˆj C m = (qs, from – to +) = q (0.010 m) ˆj. Thus q=
2.0 × 10−11 C m = 2.0 ×10−9 C = 2.0 nC 0.010 m
G G (b) Point (10 cm, 0 cm) is in the plane perpendicular to the dipole. The electric field E = –x(1/4πε0) p /r3 is half the strength of the field at an equal distance r on the axis of the dipole. Hence the field strength at this point is 180 N/C.
27.7. Model: We will assume that the wire is thin and that the charge lies on the wire along a line. Solve: From Equation 27.15, the electric field strength of an infinitely long line of charge having linear charge density λ is
Eline = ⇒ Eline ( r = 5.0 cm ) =
2λ 4πε 0 5.0 ×10−2 m 1
1 2λ 4πε 0 r Eline ( r = 10.0 cm ) =
2λ 4πε 0 10.0 ×10−2 m 1
Dividing the above two equations gives
⎛ 5.0 ×10−2 m ⎞ 1 3 Eline ( r = 10.0 cm ) = ⎜ ⎟ Eline ( r = 5.0 cm ) = ( 2000 N/C ) = 1.00 × 10 N/C = 1000 N/C −2 10.0 10 m × 2 ⎝ ⎠
27.8. Model: The rods are thin. Assume that the charge lies along a line. Visualize:
The electric field of the positively charged glass rod points away from the glass rod, whereas the electric field of the negatively charged plastic rod points toward the plastic rod. The electric field strength is the magnitude of the electric field and is always positive. Solve: Example 27.3 shows that the electric field strength in the plane that bisects a charged rod is
Erod =
Q 1 4πε 0 r r 2 + ( L / 2 )2
The electric field from the glass rod at r = 1 cm from the glass rod is Eglass = ( 9.0 ×109 N m 2 /C2 )
10 ×10−9 C
( 0.01 m ) ( 0.01 m ) + ( 0.05 m ) 2
2
= 1.765 ×105 N/C
The electric fields from the glass rod at r = 2 cm and r = 3 cm are 0.835 × 105 N/C and 0.514 × 105 N/C. The electric field from the plastic rod at distances 1 cm, 2 cm, and 3 cm from the plastic rod are the same as for the glass rod. Point P1 is 1.0 cm from the glass rod and is 3.0 cm from the plastic rod, point P2 is 2 cm from both rods, and point P3 is 3 cm from the glass rod and 1 cm from the plastic rod. Because the direction of the electric fields at P1 is the same, the net electric field strength 1 cm from the glass rod is the sum of the fields from the glass rod at 1 cm and the plastic rod at 3 cm. Thus At 1.0 cm
E = 1.765 ×105 N/C + 0.514 ×105 N/C = 2.3 × 105 N/C
At 2.0 cm
E = 0.835 × 105 N/C + 0.835 ×105 N/C = 1.67 × 105 N/C
At 3.0 cm
E = 0.514 × 105 N/C + 1.765 × 105 N/C = 2.3 × 105 N/C
Assess: The electric field strength in the space between the two rods goes through a minimum. This point is exactly in the middle of the line connecting the two rods. Also, note that the arrows shown in the figure are not to scale.
27.9. Model: The rods are thin. Assume that the charge lies along a line. Visualize:
Because both the rods are positively charged, the electric field from each rod points away from the rod. Because the electric fields from the two rods are in opposite directions at P1, P2, and P3, the net field strength at each point is the difference of the field strengths from the two rods. Solve: Example 27.3 gives the electric field strength in the plane that bisects a charged rod:
Erod =
1
Q
4πε 0 r r 2 + ( L / 2 )2
The electric field from the rod on the right at a distance of 1 cm from the rod is
Eright = ( 9.0 ×109 N m 2 /C2 )
10 × 10−9 C
( 0.01 m ) ( 0.01 m ) + ( 0.05 m ) 2
2
= 1.765 ×105 N/C
The electric field from the rod on the right at distances 2 cm and 3 cm from the rod are 0.835 × 105 N/C and 0.514 × 105 N/C. The electric fields produced by the rod on the left at the same distances are the same. Point P1 is 1.0 cm from the rod on the left and is 3.0 cm from the rod on the right. Because the electric fields at P1 have opposite directions, the net electric field strengths are At 1.0 cm
E = 1.765 ×105 N/C − 0.514 ×105 N/C = 1.25 × 105 N/C
At 2.0 cm
E = 0.835 × 105 N/C − 0.835 × 105 N/C = 0 N/C
At 3.0 cm
E = 1.765 × 105 N/C − 0.514 × 105 N/C = 1.25 × 105 N/C
27.10. Model: The rod is thin, so assume the charge lies along a line. Visualize:
G G Solve: The force on charge q is F = qErod . From Example 27.3, the electric field a distance r from the center of a charged rod is G ( 9.0 ×109 N m 2/C2 )( 40 ×10−9 C ) iˆ = 1.406 ×105 iˆ N/C 1 Q Erod = iˆ = 2 5 4πε 0 r r 2 + ( L / 2 )2 ( 0.04 m ) ( 0.04 m ) + ( 0.05 m ) Thus, the force is
G F = ( 6.0 ×10−9 C )(1.406 × 105 N/C ) iˆ = 8.4 × 10−4 iˆ N G More generally, F = ( 8.4 ×10−4 N, away from the rod ) .
27.11. Model: Assume that the rings are thin and that the charge lies along circle of radius R. Visualize:
The rings are centered on the z-axis. Solve: (a) According to Example 27.5, the field of the left (negative) ring at z = 10 cm is
( E1 ) z =
zQ1
4πε 0 ( z + R 2
)
2 3/ 2
=
( 9.0 ×10
9
N m 2 /C 2 ) ( 0.10 m ) ( −20 × 10−9 C ) 2 3/ 2
⎡( 0.10 m ) + ( 0.050 m ) ⎤ ⎣ ⎦ 2
= −1.288 × 104 N/C
G That is, the field is E1 = (1.288 × 104 N/C, left ) . Ring 2 has the same quantity of charge and is at the same distance, G so it will produce a field of the same strength. Because Q2 is positive, E2 will also point to the left. The net field at the midpoint is G G G E = E1 + E2 = ( 2.6 × 104 N/C, left ) (b) The force is
G G F = qE = ( −1.0 × 10−9 C )( 2.6 × 104 N/C, left ) = ( 2.6 × 10−5 N, right )
27.12. Model: Assume that the rings are thin and that the charge lies along circle of radius R. Visualize:
Solve: cm is
(a) Let the rings be centered on the z-axis. According to Example 27.5, the field of the left ring at z = 10
( E1 ) z =
zQ1
4πε 0 ( z + R 2
)
2 3/ 2
=
( 9.0 ×10
9
N m 2 /C 2 ) ( 0.10 m ) ( 20 ×10−9 C ) 2 3/ 2
⎡( 0.10 m ) + ( 0.050 m ) ⎤ ⎣ ⎦ 2
= 1.29 ×104 N/C
G That is, E1 = (1.29 × 104 N/C, right ) . Ring 2 has the same quantity of charge and is at the same distance, so it will G produce a field of the same strength. Because Q2 is positive, E2 will point to the left. The net field at the G G G midpoint between the two rings is E = E1 + E2 = 0 N/C. (b) The field of the left ring at z = 0 cm is ( E1 ) z = 0 N/C. The field of the right ring at z = 20 cm to its left is
( E2 ) z =
( 9.0 ×10
9
N m 2 /C2 ) ( 0.20 m ) ( 20 ×10−9 C ) 2 32
= 4.1×103 N/C
⎡( 0.20 m ) + ( 0.050 ) ⎤ ⎣ ⎦ G G G ⇒ E = E1 + E2 = 0 N/C + (4.1×103 N/C, left)
So the electric field strength is 4.1×103 N/C.
2
27.13. Model: Each disk is a uniformly charged disk. When the disk is charged negatively, the on-axis electric field of the disk points toward the disk. The electric field points away from the disk for a positively charged disk. Visualize:
Solve:
(a) The surface charge density on the disk is Q Q 50 ×10−9 C η = = = = 6.366 × 10−6 C/m 2 2 2 A πR π ( 0.050 m )
From Equation 27.23, the electric field of the left disk at z = 0.10 m is
η ⎡ 1 −6.366 × 10−6 C/m 2 ⎡ 1 ⎤ ⎤ 1− = 1− = −38,000 N/C ⎢ ⎥ −12 2 2 ⎢ 2 2 2 ⎥ 2ε 0 ⎣ 2 8.85 × 10 C /N m ( ) ⎣⎢ 1 + ( 0.050 m /0.10 m ) ⎦⎥ 1 + R /z ⎦ G In other words, E1 = (38,000 N/C, left). Similarly, the electric field of the right disk at z = 0.10 m (to its left) is G G G G E2 = (38,000 N/C, left). The net field at the midpoint between the two rings is E = E1 + E2 = (7.6×104 N/C, left).
( E1 ) z =
(b) The force on the charge is G G F = qE = ( −1.0 ×10−9 C )( 7.6 ×104 N/C, left ) = ( 7.6 × 10−5 N, right ) Assess:
Note that the force on the negative charge is to the right because the electric field is to the left.
27.14. Model: Model each disk as a uniformly charged disk. When the disk is positively charged, the onaxis electric field of the disk points away from the disk. Visualize:
Solve:
(a) The surface charge density on the disk is
η=
50 ×10−9 C Q Q = = = 6.366 ×10−6 C/m 2 A π R 2 π ( 0.050 m )2
From Equation 27.23, the electric field of the left disk at z = 0.10 m is
( E1 ) z =
η ⎡ 1 6.366 × 10−6 C/m 2 1 ⎤ ⎡ ⎤ 1 − = 1− = 38,000 N/C −12 2 2 ⎢ 2 2 ⎥ 2 ⎥ 2ε 0 ⎢ 2 8.85 10 C /N m × ( ) R z 1 + 1 + ( 0.050 m /0.10 m ) ⎥⎦ ⎣ ⎦ ⎢⎣
G Hence, E1 = (38,000 N/C, right) . Similarly, the electric field of the right disk at z = 0.10 m (to its left) is G G G G E2 = (38,000 N/C, left) . The net field at the midpoint between the two disks is E = E1 + E2 = 0 N/C . (b) The electric field of the left disk at z = 0.050 m is G 6.366 × 10−6 C/m 2 1 ⎡ ⎤ 1− = 1.05 × 105 N/C ⇒ E1 = (1.05 × 105 N/C, right ) ⎥ 2 2 ⎢ −12 2 2 ( 8.85 × 10 C /N m ) ⎢ 1 + ( 0.050 m /0.10 m ) ⎥⎦ ⎣ G Similarly, the electric field of the right disk at z = 0.15 m (to its left) is E2 = (1.85 ×104 N/C, left ) . The net field
( E1 ) z =
is thus The field strength is 8.7 ×104 N/C.
G G G E = E1 + E2 = ( 8.7 × 104 N/C, right )
27.15. Model: The distance 2.0 mm is very small in comparison to the size of the electrode, so we can model the electrode as a plane of charge. Solve: From Equation 27.26, the electric field of a plane of charge is Eplane =
η 80 × 10−9 C Q = = = 1.13 × 105 N/C 2ε 0 2ε 0 A 2 ( 8.85 × 10−12 C 2 /N m 2 ) ( 0.20 × 0.20 m )
27.16. Model: A spherical shell of charge Q and radius R has an electric field outside the sphere that is exactly the same as that of a point charge Q located at the center of the sphere. Visualize: In the case of a metal ball, the charge resides on its surface. This can then be visualized as a charged spherical shell of radius R. Solve: Equation 27.28 gives the electric field of a charged spherical shell at a distance r > R: G Q Eball = rˆ 4πε 0 r 2 In the present case, Eball = 50,000 N/C at r = Q = 4πε 0 r 2 Eball =
1 2
(10 cm ) + 2.0 cm = 7.0 cm = 0.070 m.
( 0.070 m )
2
So,
50,000 N/C = 2.7 × 10−8 C = 27 nC 9.0 ×10 N m 2 /C 2 9
27.17. Model: The electric field is uniform in a region of space between closely spaced capacitor plates. Solve: The electric field inside a capacitor is E = Q ε 0 A . Thus, the charge needed to produce a field of strength E is Q = ε 0 AE = ( 8.85 ×10−12 C 2 /N m 2 ) ( 0.04 m × 0.04 m ) (1.0 ×106 N/C ) = 14.2 nC Thus, one plate has a charge of 14.2 nC and the other has a charge of −14.2 nC. Assess: Note that the capacitor as a whole has no net charge.
27.18. Model: The electric field in a region of space between two charged circular disks is uniform. Solve: The electric field strength inside the capacitor is E = Q ε 0 A . Thus, the area is A=
( 3.0 ×109 )(1.6 ×10−19 C ) Q D2 = = 2.71× 10−4 m 2 = π −12 2 2 5 ε 0 E ( 8.85 ×10 C /N m )( 2.0 ×10 N/C ) 4 ⇒D=
4A
π
= 1.86 cm
Assess: As long as the spacing is much less than the plate dimensions, the electric field is independent of the spacing and depends only on the diameter of the plates.
27.19. Model: The electric field in a region of space between the plates of a parallel-plate capacitor is uniform. Solve: The electric field inside a capacitor is E = Q ε 0 A. Thus, the charge needed to produce a field of strength E is 2 Q = ε 0 AE = ( 8.85 ×10−12 C 2 /N m 2 ) ⎡π ( 0.020 m ) ⎤ ( 3.0 ×106 N/C ) = 33.4 nC ⎣ ⎦
The number of electrons transferred from one plate to the other is 33.4 ×10−9 C = 2.1×1011 1.60 ×10−19 C
27.20. Model: The charged plastic bead is a point charge. Visualize:
The bead hangs suspended in the air when the net force acting on the bead is zero. The two forces that act on the bead are the electric force and the gravitational force. Because the bead is negatively charged, the electric field must be pointed downward to cause an upward force, which will balance the gravitational force. Solve: For the bead to be in static equilibrium,
( Fnet ) y = qE − mg = 0 N ⇒ E =
−3 mg ( 0.10 ×10 kg ) ( 9.8 N/kg ) = = 6.1× 105 N/C q (1.0 ×1010 )(1.60 ×10−19 C )
G Thus the required field is E = ( 6.1×105 N/C, down ).
27.21. Model: The disks form a parallel-plate capacitor. The electric field inside a parallel-plate capacitor is a uniform field, so the proton will have a constant acceleration. Visualize:
Solve:
(a) The two disks form a parallel-plate capacitor with surface charge density
η=
Q Q 10 × 10−9 C = = = 3.18 × 10−5 C/m 2 2 A πR π (0.010 m)2
From Equation 27.29, the field strength inside a capacitor is
E=
η 3.18 ×10−5 C/m 2 = = 3.6 ×106 N/C ε 0 8.85 ×10−12 C2 / N m 2
(b) G The electric field points toward the Gnegative Gplate,G so in the coordinate system of the figure E = −3.6 × 106 ˆj N/C. The field exerts forces F = qproton E = eE on the proton, causing an acceleration with a y-
component that is ay =
F eE y (1.6 × 10−19 C)( −3.6 × 106 N/C) = = = −3.45 × 1014 m/s 2 1.67 × 10−27 kg m m
After the proton is launched, this acceleration will cause it to lose speed. To just barely reach the positive plate, it should reach v1 = 0 m/s at y1 = 1 mm. The kinematic equation of motion is v12 = 0 m 2 / s 2 = v02 + 2a y Δy
⇒ v0 = −2a y Δy = −2(−3.45 × 1014 m/s 2 )(0.0010 m) = 8.3 × 105 m/s
Assess: The acceleration of the proton in the electric field is enormous in comparison to the gravitation acceleration g. That is why we did not explicitly consider g in our calculations.
27.22. Model: A uniform electric field causes a charge to undergo constant acceleration. Solve:
Kinematics yields the acceleration of the electron.
v12 = v0 2 + 2aΔx ⇒ a =
v12 − v0 2 (4.0 × 107 m/s) 2 − (2.0 × 107 m/s) 2 = = 5.0 × 1016 m/s 2 2Δx 2 ( 0.012 m )
The magnitude of the electric field required to obtain this acceleration is
E=
Fnet me a (9.11 × 10−31 kg)(5.0 × 1016 m/s 2 ) = = = 2.8 × 105 N/C. e e 1.6 × 10−19 C
27.23. Model: The infinite charged plane produces a uniform electric field. Solve: (a) The electric field of a plane of charge with surface charge density η is E=
η 2ε 0
⎛F⎞ ⎛ ma ⎞ ⇒ η = 2ε 0 E = 2ε 0 ⎜ ⎟ = 2ε 0 ⎜ ⎟ q ⎝ ⎠ ⎝ q ⎠
where m is the mass, q is the charge, and a is the acceleration of the electron. To obtain η we must first find a. From the kinematic of motion equation v12 = v02 + 2aΔx, a=
⇒η =
v12 − v02 (1.0 × 107 m/s) 2 − (0 m/s) 2 = = 2.5 × 1015 m/s 2 2Δx 2(2.0 × 10−2 m)
2(8.85 × 10−12 C 2 /N m 2 )(9.11 × 10−31 kg)(2.5 × 1015 m/s 2 ) = 2.5 × 10−7 C/m 2 1.60 × 10−19 C
(b) Using the kinematic equation of motion v1 = v0 + aΔt ,
Δt =
v1 − v0 1.0 ×107 m/s − 0 m/s = = 4.0 ×10−9 s a 2.5 ×1015 m/s 2
27.24. Model: The infinite negatively charged plane produces a uniform electric field that is directed toward the plane. Visualize:
Solve:
From the kinematic equation of motion v12 = 0 = v02 + 2aΔx and F = qE = ma, a=
qE −v02 − mv02 = ⇒ Δx = 2qE m 2Δx
Furthermore, the electric field of a plane of charge with surface charge density η is E = η 2ε 0 . Thus, −27 −12 6 2 2 −mv02ε 0 − (1.67 ×10 kg )( 2.0 ×10 m/s ) ( 8.85 ×10 C /N m ) Δx = = = 0.185 m qη (1.60 ×10−19 C )( −2.0 ×10−6 C/m2 ) 2
27.25. Model: The external electric field exerts a torque on the dipole moment of the water molecule. Solve:
From Equation 27.34, the torque exerted on a dipole moment by an electric field is τ = pE sin θ . The
maximum torque is exerted when sinθ = 1 or θ = 90°. Thus,
τ max = pE = ( 6.2 ×10−30 C m )( 5.0 ×108 N/C ) = 3.1×10−21 N m
27.26. Model: Because r >> s, we can use Equation 27.12 for the electric field in the plane perpendicular to the dipole. Visualize:
Solve: (a) From Newton’s third law, the force of Q on the dipole is equal and opposite to the force of the G G G dipole on Q. You can see from the diagram that Fdipole on Q is down and FQ on dipole is up, in the direction of p. The magnitude of the dipole field at the position of Q is
Edipole = G The magnitude of Fdipole on Q is
1 p 1 qs = 4πε 0 r 3 4πε 0 r 3
Fdipole on Q = QEdipole = G FQ on dipole has the same magnitude, so
1 qQs 4πε 0 r 3
G ⎛ 1 qQs →⎞ FQ on dipole = ⎜ , direction of p ⎟ 3 πε 4 r 0 ⎝ ⎠ G G (b) The electric field of charge Q exerts a torque on the dipole. The field E is perpendicular to the dipole p, so
θ = 90°. The torque is given by Equation 27.34: ⎛ 1 Q⎞ 1 qQs sin 90° = 2 ⎟ 4πε 0 r 2 ⎝ 4πε 0 r ⎠
τ = pE sin θ = (qs ) ⎜
27.27. Model: The size of a molecule is ≈ 0.1 nm. The proton is 2.0 nm away, so r >> s and we can use Equation 27.12 for the electric field in the plane that bisects the dipole. Visualize:
Solve:
G G You can see from the diagram that Fdipole on proton is opposite to the direction of p. The magnitude of the
dipole field at the position of the proton is Edipole =
p 5.0 ×10−30 C m = (9.0 × 109 N m 2 /C2 ) = 5.624 × 105 N/C 3 4πε 0 r (2.0 ×10−9 m)3 1
G The magnitude of Fdipole on proton is
Fdipole on proton = eEdipole = (1.60 × 10−19 C)(5.624 × 105 N/C) = 9.0 × 10−13 N G G –13 Including the direction, the force is Fdipole on proton = (9.0 × 10 N, direction opposite p ).
27.28. Model: The electric field is that of three point charges q1, q2, and q3. Visualize: Please refer to Figure P27.28. Assume the charges are in the x-y plane. The 5.0 nC charge is q1, the G G G G 10 nC charge is q3, and the −5.0 nC charge is q2. The net electric field at the dot is Enet = E1 + E2 + E3 . The procedure will be to find the magnitudes of the electric fields, to write them in component form, and to add the components. Solve: (a) The electric field produced by q1 is 9 2 2 −9 q1 ( 9.0 ×10 N m /C )( 5.0 ×10 C ) = = 112,500 N/C 2 4πε 0 r12 ( 0.020 m ) G G E1 points away from q1, so in component form E1 = 112,500iˆ N/C. The electric field produced by q2 is E2 = G G 28,120 N/C. E2 points toward q2, so E2 = 28,120ˆj N/C . Finally, the electric field produced by q3 is
E1 =
E3 =
1
9 2 2 −9 q3 ( 9.0 ×10 N m /C )(10 ×10 C ) = = 45,000 N/C 2 2 4πε 0 r32 ( 0.020 m ) + ( 0.040 m )
1
G E3 points away from q3 and makes an angle φ = tan −1 ( 4 / 2 ) = 63.43° with the x-axis. So, G E3 = E3 cos φ iˆ − E3 sin φ ˆj = 20,130iˆ − 40,250ˆj N/C
(
)
Adding these three vectors gives G G G G Enet = E1 + E2 + E3 = 132,600iˆ − 12,130ˆj N/C = 1.33 × 105iˆ − 1.21×104 ˆj N/C
(
)
(
)
This is in component form. (b) The magnitude of the field is
Enet = Ex2 + E y2 =
(132,600 N/C ) + ( −12,130 N/C ) 2
(
and its angle from the x-axis is θ = tan −1 Ex E y + x-axis).
)
2
= 133,200 N/C = 1.33 × 105 N/C G = 5.2°. We can also write Enet = (1.33 ×105 N/C, 5.2° CW from the
27.29. Model: The electric field is that of three point charges q1, q2, and q3. Visualize: Please refer to Figure P27.29. Assume the charges are in the x-y plane. The −5.0 nC charge is q1, the G G G G bottom 10 nC charge is q3, and the top 10 nC charge is q2. The net electric field at the dot is Enet = E1 + E2 + E3 . The procedure will be to find the magnitudes of the electric fields, to write them in component form, and to add the components. Solve: (a) The electric field produced by q1 is 9 2 2 −9 q1 ( 9.0 ×10 N m /C )( 5.0 ×10 C ) = = 112,500 N/C 2 4πε 0 r12 ( 0.020 m ) G G E1 points toward q1, so in component form E1 = −112,500 ˆj N/C. The electric field produced by q2 G G is E2 = 56,250 N/C. E2 points away from q2, so E2 = −56,250iˆ N/C. Finally, the electric field produced by q3 is
1
E1 =
E3 =
9 2 2 −9 q3 ( 9.0 ×10 N m /C )(10 ×10 C ) = = 45,000 N/C 2 2 4πε 0 r32 ( 0.020 m ) + ( 0.040 m )
1
G E3 points away from q3 and makes an angle φ = tan −1 ( 2 / 4 ) = 26.6° with the x-axis. So, G E3 = − E3 cos φ iˆ + E3 sin φ ˆj = −40,250iˆ + 20,130 ˆj N/C
(
)
Adding these three vectors gives G G G G Enet = E1 + E2 + E3 = −96,500iˆ − 92,400 ˆj N/C = −9.7 × 104 iˆ − 9.2 × 104 ˆj N/C
(
)
(
)
This is in component form. (b) The magnitude of the field is Enet = Ex2 + E y2 =
( 96,500 N/C ) + ( 92,400 N/C ) 2
(
and its angle below the –x-axis is θ = tan −1 Ex E y from the + x -axis ) .
)
2
= 133,600 N/C = 1.34 × 105 N/C G = 44°. We can also write Enet = (1.34 ×105 N/C, 136° CW
27.30. Model: The electric field is that of three point charges q1, q2, and q3. Visualize: Please refer to Figure P27.30. Assume the charges are in the x-y plane. The 10 nC charge is q1, the G G G G −10 nC charge is q3, and the –5.0 nC charge is q2. The net electric field at the dot is Enet = E1 + E2 + E3 . The procedure will be to find the magnitudes of the electric fields, to write them in component form, and to add the components. Solve: (a) The electric field produced by q1 is 9 2 2 −9 q1 ( 9.0 ×10 N m /C )(10 ×10 C ) = = 100,000 N/C 2 4πε 0 r12 ( 0.030 m ) G G E1 points away from q1, so in component form E1 = −100,000 ˆj N/C . The electric field produced by q2 is G G E2 = 18,000 N/C . E2 points towards q2, so E2 = 18,000iˆ N/C. Finally, the electric field produced by q3 is
E1 =
E3 =
1
9 2 2 −9 q3 ( 9.0 ×10 N m /C )(10 ×10 C ) = = 26,471 N/C 2 2 2 4πε 0 r3 ( 0.030 m ) + ( 0.050 m )
1
G E3 points toward q3 and makes an angle φ = tan −1 ( 3/ 5 ) = 31° with the x-axis. So, G E3 = E3 cos φ iˆ + E3 sin φ ˆj = 22,700iˆ + 13,634 ˆj N/C
(
)
Adding these three vectors gives G G G G Enet = E1 + E2 + E3 = 40,700iˆ − 86,366 ˆj N/C = 4.1× 104 iˆ − 8.6 × 104 ˆj N/C
(
)
(
)
This is in component form. (b) The magnitude of the field is
Enet = Ex2 + E y2 =
( 40,700 N/C ) + ( 86,366 N/C ) 2
(
and its angle from the x-axis is θ = tan −1 Ex E y the + x-axis).
)
2
= 95,476 N/C = 9.5 × 104 N/C G = 25° . We can also write Enet = (9.5 × 104 N/C, 25° CW from
27.31. Model: The electric field is that of three point charges q1 = −Q, q2 = −Q, and q3 G= +4Q.G G G Visualize: Assume the charges are in the x-y plane. The net electric field at point P is Enet = E1 + E2 + E3 . The procedure will be to find the magnitudes of the electric fields, to write them in component form, and to add the components.
Solve:
(a) The electric field produced by q1 points toward q1 and is given by
G ⎛ 1 Q ⎞ˆ E1 = − ⎜ i 2 ⎟ ⎝ 4πε 0 L ⎠ The electric field produced by q2 points toward q2 and is given by G ⎛ 1 Q⎞ˆ E2 = − ⎜ j. 2 ⎟ ⎝ 4πε 0 L ⎠ The electric field produced by q3 is
E3 =
1 ⎛ 4Q ⎞ 1 ⎛ 2Q ⎞ ⎜ 2 2⎟= ⎜ ⎟ 4πε 0 ⎝ L + L ⎠ 4πε 0 ⎝ L2 ⎠
G E3 points away from q3 and makes an angle φ = tan −1 ( L L ) = 45° with the x-axis. So G 1 ⎛ 2Q ⎞ 1 2Q ⎛ 1 ˆ E3 = i + ⎜ ⎟ cos φ iˆ + sin φ ˆj = 4πε 0 ⎝ L2 ⎠ 4πε 0 L2 ⎜⎝ 2
(
)
1 2
ˆj ⎞ ⎟ ⎠
Adding these three vectors gives
G G G G 1 Q ⎡⎛ 2 1 Q ˆ ˆ ⎞ ⎛ 2 ⎞ ⎤ − 1⎟ iˆ + ⎜ − 1⎟ ˆj ⎥ = Enet = E1 + E2 + E3 = i + j 2 ⎢⎜ 4πε 0 L ⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎦ 4πε 0 L2
(
(b) The force on the charge is F = ma = qEnet . Therefore,
a=
q 1 Q m 4πε 0
(
)
2 −1 2
L
2
=
1 ⎛ 0.586 qQ ⎞ ⎜ ⎟ 4πε 0 ⎝ mL2 ⎠
)(
)
2 −1
27.32. Model: Visualize:
Solve:
The electric field is that of two charges −q and +2q located at x = ±a.
(a) At point 1, the electric field from −q is
−q
1
E− q =
=
1
q 4πε 0 5a 2
4πε 0 ( a ) + ( 2a ) G E− q points toward −q and makes an angle φ1 = tan −1 ( 2a a ) = 63.43° below the −x-axis, hence 2
2
G 1 ⎛ q ⎞ 1 ⎛ q ⎞ ⎛ 1 ˆ 2 ˆ⎞ 1 ⎛ q ⎞ ˆ ˆ E− q = i− j⎟ = ⎜ ⎟ − cos φ1iˆ − sin φ1 ˆj = ⎜ ⎟ − ⎜ ⎟ − i −2j 4πε 0 ⎝ 5a 2 ⎠ 4πε 0 ⎝ 5a 2 ⎠ ⎜⎝ 5 5 ⎠ 4πε 0 ⎝ 5 5a 2 ⎠
(
)
(
The electric field from the +2q is
E+2 q =
1 2q 1 2q = 4πε 0 a 2 + ( 2a )2 4πε 0 5a 2
G E+2q points away from +2q and makes an angle φ1 = tan −1 ( 2a a ) = 63.43° above the −x-axis. So, G 1 ⎛ 2q ⎞ 1 ⎛ q ⎞ E+2 q = −2iˆ + 4ˆj ⎜ ⎟ − cos φ2iˆ + sin φ2 ˆj = 4πε 0 ⎝ 5a 2 ⎠ 4πε 0 ⎜⎝ 5 5a 2 ⎟⎠
(
)
(
Adding these two vectors, G G G 1 ⎛ q ⎞ E1 net = E− q + E+2 q = −3iˆ + 2ˆj 4πε 0 ⎜⎝ 5 5a 2 ⎟⎠
(
At point 2, the electric field from −q points toward −q, so
G 1 ⎛ q ⎞ˆ E− q = ⎜ ⎟i 4πε 0 ⎝ a 2 ⎠ The electric field from +2q points away from +2q, so G 1 ⎛ 2q ⎞ ˆ E+2 q = − ⎜ ⎟i 4πε 0 ⎝ 9a 2 ⎠ Adding these two vectors,
)
)
)
G G G 1 ⎛ 7q ⎞ ˆ E2 net = E− q + E+2 q = ⎜ ⎟i 4πε 0 ⎝ 9a 2 ⎠ At point 3, the electric field from −q points toward −q, so G 1 ⎛ q ⎞ˆ E− q = − ⎜ ⎟i 4πε 0 ⎝ 9a 2 ⎠ The electric field from +2q points away from +2q, so G 1 ⎛ 2q ⎞ ˆ E+2 q = ⎜ ⎟i 4πε 0 ⎝ a 2 ⎠ Adding these vectors, G G G 1 ⎛ 17q ⎞ ˆ E3 net = E− q + E+2 q = ⎜ ⎟i 4πε 0 ⎝ 9a 2 ⎠ G G Point 4 is a mirror image of point 1. Since E1 net points to the left and up, E4 net has a reversed y-component and points to the left and down. Thus, G 1 ⎛ q ⎞ E4 net = −3iˆ − 2ˆj 4πε 0 ⎜⎝ 5 5a 2 ⎟⎠
(
(b) The four electric field vectors at points 1-4 are shown in the figure.
)
27.33. Model: The electric field is that of two positive charges. Visualize:
G G G G G The figure shows E1 and E2 due to the individual charges. The total field is E = E1 + E2 . Solve: (a) From symmetry, the y-components of the two electric fields cancel out. The x-components are equal and add. Thus, G E = 2 ( E1 ) x iˆ = 2E1 cosθ iˆ The field strength and angle due to q1 are
E1 =
q q = 4πε 0 r 2 4πε 0 ( x 2 + s 2 /4 )
G Thus the magnitude of the net field is E =
cosθ =
2qx
4πε 0 ( x 2 + s 2 /4 )
3/ 2
x = r
x x + s 2 /4 2
.
(b) The electric field strength at position x is E=
( 9.0 ×10
9
N m 2 /C 2 ) ( 2 ) (1.0 × 10−9 C ) x
⎡ x 2 + ( 0.003 m )2 ⎤ ⎣ ⎦
3/ 2
=
18 x ⎡ x 2 + ( 0.003 m )2 ⎤ ⎣ ⎦
3/ 2
N/C
where x has to be in meters. We can now evaluate E for different values of x: x (mm) 0 2 4 6 10
x (m) 0.000 0.002 0.004 0.006 0.010
E (N/C) 0 768,000 576,000 358,000 158,000
(c) We can use the values of part (b) as a start in drawing the graph. Also note that E → 0 N/C as x → 0 m and as x → ∞. The graph is shown in the figure above.
27.34. Model: The electric field of a dipole is that of two opposite charges ±q that make the dipole. Visualize: Please refer to Figure 27.7. The figure shows a dipole aligned on the y-axis, so the x-axis is the G G G bisecting axis. The field at a point on the x-axis is Edipole = E+ + E− . Solve: From the symmetry of the situation we can see that the x-components of the two contributions to the G electric field will cancel, that they have equal y-components, and that Edipole points in the −y-direction. Thus,
(E
)
dipole x
(E
= 0 N/C
)
dipole y
= −2 E+ sin θ
G where θ is the angle E+ makes with the x-axis. From the geometry of the figure, sin θ =
s/2 1/ 2
⎡ x + ( s / 2 )2 ⎤ ⎣ ⎦ 2
For a point charge +q, the field is
E+ =
1 q 4πε 0 x 2 + ( s / 2 )2
Combining these pieces gives the dipole field at distance x along the bisecting axis: G 1 q s/2 qs ˆj = − ˆj Edipole = − ( 2 ) 3/ 2 2 4πε 0 x 2 + ( s / 2 )2 ⎡ x 2 + ( s / 2 )2 ⎤1/ 2 4πε 0 ( x + s 2 / 4 ) ⎣ ⎦
If x >> s, then ( x 2 + s 2 / 4 )
3/ 2
≈ x3 . Thus G 1 qsjˆ Edipole ≈ − 4πε 0 x 3
G If we note that p = qsjˆ and if we replace x with a more general variable r to denote the distance from the dipole, then G G 1 p Edipole ≈ − 4πε 0 r 3 This is Equation 27.12.
27.35. Model: The electric field is that of three point charges. Visualize:
G G G G The field at points on the x-axis is Enet = E1 + E2 + E3 .
G G G Solve: (a) We note from the symmetry that the y-component of E1 and E3 cancel. Since E2 has no yG G component, the net field will have only an x-component. The x-components of E1 and E3 are equal, so
( Enet ) x = E2 − 2E1 cosθ
( Enet ) y = ( Enet ) z = 0 N/C
Note that the signs of q1 and q3 were used in writing this equation. The electric field strength due to q2 is E2 =
1 q2 1 2q = 4πε 0 r22 4πε 0 x 2
The electric field strength due to q1 is E1 =
1 q1 1 q = 4πε 0 r12 4πε 0 x 2 + d 2
From geometry, cosθ =
x = r1
x x2 + d 2
Assembling these pieces, the net field is
( Enet ) x =
1 2q 1 2q − 4πε 0 x 2 4πε 0 x 2 + d 2
x x2 + d 2
=
2q ⎡ 1 x ⎤ − 4πε 0 ⎢ x 2 ( x 2 + d 2 )3 / 2 ⎥ ⎢⎣ ⎥⎦
G 2q ⎡ 1 x ⎤ˆ ⇒ Enet = − i 4πε 0 ⎢ x 2 ( x 2 + d 2 )3 / 2 ⎥ ⎢⎣ ⎥⎦
G G (b) For x << d, the observation point is very close to q2 = +2q. Furthermore, at x ≈ 0 m the fields E1 and E3 are nearly opposite to each other and will nearly cancel. So for x << d we expect the field to be that of a point charge +2q at the origin. To test this prediction, we note that for x << d
(x
x 2
+d
)
2 3/ 2
≈
G x 1 2q ˆ ≈ 0 ⇒ Enet = i 3 d 4πε 0 x 2
This is, indeed, the field on the x-axis of point charge 2q at the origin. For x >> d, the three charges appear as a G single charge of value qnet = q1 + q2 + q3 = 0. So we expect E ≈ 0 when x >> d. In this limit,
1 x 1 x 1 1 − ≈ − = − =0 x 2 ( x 2 + d 2 )3/ 2 x 2 x 3 x 2 x 2
so the field does rapidly become zero, as expected. (c) A graph of Ex is shown in the figure above.
27.36. Model: The electric field is that of two infinite lines of charge extending out of the page. Visualize: Please refer to Figure P27.36. The line charges lie on the x-axis. Solve: (a) From Equation 27.15, the electric field strength due to an infinite line of charge at a distance r from the line charge is E=
1 2λ 4πε 0 r
For the left and right line charges, the electric fields are Eleft = Eright =
1
2λ
4πε 0
y + ( d / 2) 2
2
=
1
4λ
4πε 0
4 y2 + d 2
G G Eleft makes an angle φ above the +x-axis and Eright makes the same angle φ above the −x-axis. From the geometry of the figure,
cos φ = G ⇒ Eleft =
d /2
=
y +d /4 2
2
1 4πε 0
d 4y + d 2
2
4λ
sin φ =
d ⎛ iˆ + ⎜ 2 4y + d ⎝ 4y + d2 2
2
2y 4 y2 + d 2
ˆj ⎞ ⎟ 4y + d ⎠ 2y 2
2
G 1 4λ d 2y ⎛ ˆj ⎞ − Eright = iˆ + ⎟ 2 2 2 2 4πε 0 4 y 2 + d 2 ⎜⎝ 4y + d 4y + d ⎠ G G G We now add these two vectors to find Enet = Eleft + Eright . The x-components cancel to give
G 1 ⎛ 4λ 2y ⎞ ⎛ Enet = ( 2) ⎜ ⎜ ⎟ 2 2 4πε 0 ⎝ 4 y + d ⎠ ⎝ 4 y 2 + d 2
1 16λ y ˆ ⎞ˆ j ⎟ j = 4πε 2 2 0 (4y + d ) ⎠
Thus the field strength is
E=
1 16λ y 4πε 0 4 y 2 + d 2
(b) To draw a graph of Enet versus y, we calculated Enet at a few selected values of y.
y (in units of d)
0 ± 0.5 ±1 ±2 ±3 ±4
⎛ 1 λ⎞ Enet ⎜ in units of ⎟ 4πε 0 d ⎠ ⎝ 0 ±4 ± 3.2 ± 1.88 ± 1.30 ± 0.98
27.37. Model: The electric field is that of two infinite lines of charge extending out of the page. Visualize: Please refer to Figure P27.37. The line charges lie on the x-axis. Solve: (a) From Equation 27.15, the electric field strength due to an infinite line of charge at a distance r from the line charge is 1 2λ 4πε 0 r
E=
For the left and right line charges, the electric fields are
Eleft = Eright =
cos φ =
1
2λ
4πε 0
y + ( d / 2)
d /2 y + d /4 2
2
G 1 ⇒ Eleft = 4πε 0 G 1 Eright = 4πε 0
=
2
2
=
d 4y + d 2
2
1
4λ
4πε 0
4 y2 + d 2
4λ d ⎛ iˆ + ⎜ 4 y2 + d 2 ⎝ 4 y2 + d 2 4λ
d ⎛ iˆ − ⎜ 2 4y + d ⎝ 4y + d2 2
2
2y
sin φ =
4 y2 + d 2
2y ˆj ⎞ ⎟ 4 y2 + d 2 ⎠ ˆj ⎞ ⎟ 4y + d ⎠
G G G 1 ⎛ 4λ d ⎞ ⎛ ⇒ Enet = Eleft + Eright = ( 2) 4πε 0 ⎜⎝ 4 y 2 + d 2 ⎟⎠ ⎜⎝ 4 y 2 + d 2
2y 2
2
1 8λ d ⎞ˆ iˆ ⎟ i = 4πε 2 + d2) y 4 ( 0 ⎠
Thus the field strength is
E=
1 8λ d 4πε 0 4 y 2 + d 2
(b) To draw a graph of Enet versus y, we calculated Enet at a few selected values of y.
y (in units of d) 0 0.5 1.0 2.0 3.0 4.0
⎛ 1 λ⎞ Enet ⎜ in units of ⎟ 4πε 0 d ⎠ ⎝ 8 4 1.6 0.47 0.22 0.12
27.38. Model: The fields are those of two infinite lines of charge with linear charge density λ. Visualize:
Solve:
The field at distance r from an infinite line of charge is
G 1 2λ E= rˆ 4πε 0 r It points straight away from the line. With two perpendicular lines, the field due to the line along the x-axis points in the y-direction and depends inversely on distance y. Similarly, the field due to the line along the y-axis points in the x-direction and depends inversely on distance x. That is Ex =
1 2λ 4πε 0 x
Ey =
1 2λ 4πε 0 y
G 2λ ⎛ 1 ˆ 1 ˆ ⎞ ⇒ E = Exiˆ + E y ˆj = ⎜ i + j⎟ y ⎠ 4πε 0 ⎝ x The field strength at this point in space is E = Ex2 + E y2 =
2λ 1/ x 2 + 1/ y 2 4πε 0
27.39. Model: Assume that the wire is thin and that the charge lies on the wire along a line. Solve:
From Equation 27.15, the electric field for an infinite uniformly charged line is Eline =
1 2λ 4πε 0 r
where r is the distance from the line in the plane that bisects the line. Solving for the linear charge density,
λ =
rEline ( 0.050 m )( 2000 N/C ) = 5.56 ×10−9 C/m = 2 (1 4πε 0 ) 2 ( 9.0 ×109 Nm 2 / C2 )
The charge in 1.0 cm is
Q = L λ = (1.0 × 10−2 m )( 5.56 ×10−9 C/m ) = 5.56 × 10−11 C = 0.056 nC
Because the electric field is directed toward the line, Q is negative. Thus
Q = −0.056 nC.
27.40. Model: The rods are lines of charge with a uniform linear charge density. Visualize:
Solve:
From Example 27.3, the electric field strength E at distance d from the center of a charged rod is
E=
Q
1
4πε 0 d d 2 + ( L / 2 )2
Because point P is equidistant from the center for each of the three rods, the electric field strength is the same for each rod. We have 10 ×10−9 C E = E1 = E2 = E3 = ( 9.0 ×109 N m 2 /C2 ) 2 d d 2 + ( 0.050 m ) From the geometry in the figure,
d ⎛ L⎞ = tan 30° ⇒ d = ⎜ ⎟ tan 30° = ( 0.05 m ) tan 30° = 0.02887 m L ⎝2⎠
1 2
⇒E=
90 N m 2 /C
( 0.02887 m ) ( 0.02887 m ) + ( 0.050 m ) 2
2
= 54,000 N/C
G G G E1 is along the −y-direction, E2 makes an angle of 30° below the −x-axis, and E3 makes an angle of 30° below the +x-axis. In component form, G E1 = − ( 54,000 N/C ) ˆj
G E2 = ( 54,000 N/C ) − cos30°iˆ − sin 30° ˆj
(
G E3 = ( 54,000 N/C ) cos30°iˆ − sin 30° ˆj
(
)
)
Adding these three vectors, G Enet = −2 ( 54,000 N/C ) sin 30° ˆj − ( 54,000 N/C ) ˆj = −1.08 × 105 ˆj N/C Thus, the electric field strength at the center of the triangle is 1.08 × 105 N/C.
27.41. Model: The electric field is that of an infinite line charge. Visualize:
Solve: The wire must be negative to attract the proton. From Equation 27.15, the electric field strength due to an infinite line of charge at a distance r from the line charge is
E=
1 2λ 4πε 0 r
The force on the proton due to this electric field causes the centripetal acceleration: F = qE =
1 2q λ mv 2 mv 2 = ⇒ λ = ( 4πε 0 ) 4πε 0 r r 2q
Using v = 2π r T = 2π rf ,
λ = ( 4πε 0 )
m ( 4π 2 r 2 f 2 ) 2q
(1.67 ×10 kg ) 4π (1.0 ×10 m ) (1.0 ×10 = ( 9.0 ×10 N m /C ) 2 (1.60 ×10 C ) −27
−2
2
9
Because the wire has to be negative, λ = −2.3 nC/m .
2
2
2
−19
6
s −1 )
2
= 2.3 × 10−9 C/m
27.42. Model: Visualize:
The electric field is that of a line of charge of length L.
The origin of the coordinate system is at the center of the rod. Divide the rod into many small segments of charge Δ q and length Δ x′. G G Solve: (a) Segment i creates a small electric field Ei at point P that points to the right. The net field E will point to the right and have Ey = Ez = 0 N/C. The distance to segment i is x′, so
Ei = ( Ei ) x =
Δq 4πε 0 ( x − xi′ )
2
⇒ Ex = ∑ ( Ei ) x =
1 Δq ∑ 4πε 0 ( x − xi′ )2
Δ q is not a coordinate, so before converting the sum to an integral we must relate the charge Δ q to length Δ x′. This is done through the linear charge density λ = Q/L, from which Δq = λΔx′ =
Q Δx′ L
With this charge, the sum becomes
Ex =
(Q / L) 4πε 0
Δx′
∑ ( x − x′ ) i
2
i
Now we let Δ x′ → dx′ and replace the sum by an integral from x′ = − 12 L to x′ = + 12 L. Thus, Ex =
(Q / L ) L / 2 4πε 0
dx′
∫ ( x − x′)
−L / 2
2
=
(Q / L ) ⎡ 4πε 0
L/2 G 1 ⎤ 1 Q (Q / L ) L = ⇒E= iˆ 2 2 2 ⎢⎣ x − x′ ⎦⎥ − 4 x L 4 4 πε πε x − L2 4 −L / 2 0 0
The electric field strength at x is
E=
1
Q 4πε 0 x − L2 4 2
(b) For x >> L ,
E=
1 Q 4πε 0 x 2
That is, the line charge behaves like a point charge. (c) Substituting into the above formula 3.0 ×10−9 C E = ( 9.0 × 109 N m 2 /C2 ) = 9.8 × 104 N/C 2 2 ( 3.0 ×10−2 m ) − ( 12 × 5.0 ×10−2 m )
27.43. Model: The electric field is that of a line charge of length L. Visualize: Please refer to Figure P27.43. Let the bottom end of the rod be the origin of the coordinate system. Divide the rod into many small segments of charge Δ q and length Δ y′. Segment i creates a small electric field at the point P that makes an angle θ with the horizontal. The field has both x and y components, but Ez = 0 N/C. The distance to segment i from point P is ( x 2 + y′2 ) . 12
Solve:
The electric field created by segment i at point P is
G Ei =
x y′ Δq Δq ⎛ ˆj ⎞ cosθ iˆ − sin θ ˆj = iˆ − ⎜ ⎟ 4πε 0 ( x 2 + y′2 ) 4πε 0 ( x 2 + y′2 ) ⎝ x 2 + y′2 x 2 + y′2 ⎠ G G G The net field is the sum of all the Ei , which gives E = ∑ Ei . Δ q is not a coordinate, so before converting the
(
)
i
sum to an integral we must relate charge Δ q to length Δ y′. This is done through the linear charge density λ = Q/L, from which we have the relationship Δq = λΔy′ =
Q Δy′ L
With this charge, the sum becomes G Q/L ⎡ xΔy′ y′Δy′ E= iˆ − ∑ 3/ 2 2 4πε 0 i ⎢ ( x 2 + y′2 )3 / 2 ( x + y′2 ) ⎢⎣
ˆj ⎤ ⎥ ⎥⎦
Now we let Δ y′ → dy′ and replace the sum by an integral from y′ = 0 m to y′ = L . Thus, L G (Q / L) ⎛ L xdy′ y′dy′ E= iˆ − ∫ ⎜∫ 2 3 / 2 2 2 3/ 2 4πε 0 ⎜ 0 ( x + y′2 ) 0 ( x + y′ ) ⎝
=
L L ⎛ ⎞ y′ −1 ⎤ ⎤ ˆ ⎡ ˆj ⎟ = ( Q / L ) ⎜ x ⎡ − i ⎢ 2 2 ⎥ ⎟ 4πε 0 ⎜ ⎢⎣ x 2 x 2 + y′2 ⎥⎦ ⎣ x + y′ ⎦ 0 0 ⎝ ⎠
Q 1 1 ⎛ Q ⎞⎛ x iˆ − ⎜ ⎟ 1− 2 4πε 0 x x 2 + L2 4πε 0 ⎝ Lx ⎠ ⎜⎝ x + L2
⎞ˆ ⎟j ⎠
⎞ ˆj ⎟ ⎟ ⎠
27.44. Model: Assume that the ring of charge is thin and that the charge lies along circle of radius R. Visualize: Please refer to Figure 27.15. Solve: From Example 27.5, the on-axis field of a ring of charge Q and radius R is Ez =
zQ
4πε 0 ( z 2 + R 2 )
3/ 2
When z << R, this means we are near the center of the ring. At that point, segments of charge i and j that are G G 180° apart create fields Ei and E j that cancel each other. When the fields of all segments of charge around the ring are added, the net result is zero. This is indicated by the above expression because when z = 0 m, the electric field is zero. When z >> R, then
(z
2
+ R2 )
3/ 2
= ( z2 )
3/ 2
= z3
If this is used in the expression for Ez, we get
Ez ≈
zQ 1 Q = 4πε 0 z 3 4πε 0 z 2
This is the field of a point charge Q as seen along the z-axis.
27.45. Model: Assume that the ring of charge is thin and that the charge lies along circle of radius R. Solve:
(a) From Example 27.5, the on-axis field of a ring of charge Q and radius R is
Ez =
1 zQ 4πε 0 ( z 2 + R 2 )3/ 2
For the field to be maximum at a particular value of z, dE dz = 0. Taking the derivative, z ( 3/ 2 )( 2 z ) ⎤ dE Q ⎡ 1 1 3z 2 = − =0⇒ = ⎢ 2 ⎥ 3 / 2 5 / 2 3 2 5/ 2 dz 4πε 0 ⎢ ( z + R 2 ) ( z 2 + R 2 ) ⎥⎦ ( z 2 + R2 ) ( z 2 + R2 ) ⎣
⇒1= (b) The field strength at the point z = R
( Ez )max =
Q 4πε 0
(R 2) ⎡ R 2 +R ⎤ ) ⎥⎦ ⎢⎣( 2
2
3/ 2
=
2
2 is Q
2 3 3 4πε 0 R
3z 2 R ⇒z= 2 2 z +R 2
27.46. Model: Assume that the semicircular rod is thin and that the charge lies along the semicircle of radius R. Visualize:
The origin of the coordinate system is at the center of the circle. Divide the rod into many small segments of G charge Δ q and arc length Δ s. Segment i creates a small electric field Ei at the origin. The line from the origin to segment i makes an angle θ with the x-axis. Solve: Because every segment i at an angle θ above the axis is matched by segment j at angle θ below the axis, the y-components of the electric fields will cancel when the field is summed over all segments. This leads to a net field pointing to the right with
Ex = ∑ ( Ei ) x = ∑ Ei cosθ i i
E y = 0 N/C
i
Note that angle θ i depends on the location of segment i. Now all segments are at the same distance ri = R from the origin, so
Ei =
Δq Δq = 4πε 0 ri 2 4πε 0 R 2
The linear charge density on the rod is λ = Q/L, where L is the rod’s length. This allows us to relate charge Δ q to the arc length Δ s through Δ q = λ Δ s = (Q/ L)Δ s Thus, the net field at the origin is Ex = ∑ i
( Q / L ) Δ s cosθ 4πε 0 R
2
i
=
Q ∑ cosθi Δ s 4πε 0 LR 2 i
The sum is over all the segments on the rim of a semicircle, so it will be easier to use polar coordinates and integrate over θ rather than do a two-dimensional integral in x and y. We note that the arc length Δ s is related to the small angle Δθ by Δ s = RΔθ , so Ex =
Q ∑ cosθ i Δθ 4πε 0 LR i
With Δ θ → dθ , the sum becomes an integral over all angles forming the rod. θ varies from Δ θ = −π/2 to θ = +π/2. So we finally arrive at
Ex =
π /2 Q Q cosθ dθ = sin θ ∫ − π / 2 4πε 0 LR 4πε 0 LR
π /2 −π / 2
=
2Q 4πε 0 LR
Since we’re given the rod’s length L and not its radius R, it will be convenient to let R = L/π. So our final G expression for E , now including the vector information, is G 1 ⎛ 2π Q ⎞ ˆ E= ⎜ ⎟i 4πε 0 ⎝ L2 ⎠
(b) Substituting into the above expression,
E=
( 9.0 ×10
9
N m 2 /C2 ) 2π ( 30 ×10−9 C )
( 0.10 m )
2
= 1.70 ×105 N/C
27.47. Model: Assume that the quarter-circle plastic rod is thin and that the charge lies along the quartercircle of radius R. Visualize:
The origin of the coordinate system is at the center of the circle. Divide the rod into many small segments of charge Δ q and arc length Δ s. G Solve: (a) Segment i creates a small electric field Ei at the origin with two components:
( Ei ) x = Ei cosθ i
( Ei ) y = Ei sinθ i
Note that the angle θ i depends on the location of the segment i. Now all segments are at distance ri = R from the origin, so
Ei =
1 Δq 1 Δq = 2 4πε 0 ri 4πε 0 R 2
The linear charge density of the rod is λ = Q/L, where L is the rod’s length (L = quarter-circumference = π R/2). This allows us to relate charge Δ q to the arc length Δ s through ⎛Q⎞ ⎛ 2Q ⎞ Δq = λΔs = ⎜ ⎟ Δs = ⎜ ⎟ Δs ⎝L⎠ ⎝πR ⎠ Using Δ s = RΔ θ , the components of the electric field at the origin are
( Ei ) x =
1 Δq 1 1 ⎛ 2Q ⎞ 1 ⎛ 2Q ⎞ cosθi = ⎜ ⎟ RΔθ cosθ i = ⎜ ⎟ Δθ cosθ i 4πε 0 R 2 4π 0 R 2 ⎝ π R ⎠ 4πε 0 ⎝ π R 2 ⎠
( Ei ) y =
1 Δq 1 1 ⎛ 2Q ⎞ 1 ⎛ 2Q ⎞ sin θi = ⎟ RΔθ sin θi = ⎜ ⎟ Δθ sin θ i 2 2 ⎜ 4πε 0 R 4πε 0 R ⎝ π R ⎠ 4πε 0 ⎝ π R 2 ⎠
(b) The x- and y-components of the electric field for the entire rod are the integrals of the expressions in part (a) from θ = 0 rad to θ = π/2. We have
Ex =
1 ⎛ 2Q ⎞ π / 2 cosθ dθ ⎜ ⎟ 4πε 0 ⎝ π R 2 ⎠ ∫ 0
1 ⎛ 2Q ⎞ π / 2 sin θ dθ ⎜ ⎟ 4πε 0 ⎝ π R 2 ⎠ ∫ 0
Ey =
(c) The integrals are
∫
π /2 0
sin θ dθ = [ − cosθ ]0
π /2
π ⎛ ⎞ = − ⎜ cos − cos0 ⎟ = +1 2 ⎝ ⎠
∫
π /2 0
The electric field is
G 1 2Q ˆ ˆ E= i+j 4πε 0 π R 2
(
)
cosθ dθ = [sin θ ]0
π /2
= sin
π 2
− sin 0 = +1
27.48. Model: Assume that the plastic sheets are planes of charge. Visualize: Please refer to Figure P27.48. Solve: At point 1 the electric field due to the left sheet and the right sheet are
G ⎛η ⎞ η Eleft = ⎜ 0 , toward right ⎟ = 0 iˆ ε 2 ⎝ 0 ⎠ 2ε 0
G ⎛ 3η ⎞ 3η Eright = ⎜ 0 , toward left ⎟ = − 0 iˆ ε 2 2ε 0 ⎝ 0 ⎠
G G G η ⇒ Enet = Eleft + Eright = − 0 iˆ
ε0
G G G G At point 2, Eleft = − (η0 2ε 0 ) iˆ, Eright = − ( 3η0 2ε 0 ) iˆ, and Enet = − ( 2η0 ε 0 ) iˆ. At point 3, Enet = + (η0 ε 0 ) iˆ.
27.49. Motion: The very wide charged electrode is a plane of charge. Solve: For a plane of charge with surface charge density η, the electric field is
(E )
plane z
=
η 2ε 0
⇒ η = 2ε 0 ( Eplane ) = z
Q Q = A π R2
⇒ Q = 2π R 2ε 0 ( Eplane ) = 2π ( 0.0050 m ) (8.85 × 10−12 C 2 /N m 2 ) (1000 N/C ) = 1.39 × 10 −12 C = 1.39 × 10−3 nC 2
z
Assess: Note that the field strength of a plane of charge does not depend on z and is a constant. We did not have to use the distance z = 5.0 cm given in the problem.
27.50. Model: An insulating sphere of charge Q and radius R has an electric field outside the sphere that is exactly the same as that of a point charge Q located at the center of the sphere. Visualize:
Solve:
The electric field of a charged sphere at a distance r > R is given by Equation 27.28:
G Esphere =
Q 4πε 0 r 2
rˆ
G G G G G In the present case, E = E1 + E2 , where E1 and E2 are the fields of the individual spheres. The distance from the
center of each sphere to the midpoint between is r1 = r2 = 4 cm. Thus, E1 =
(9.0 ×10
9
( 0.040 m )
( 9.0 × 10 =
9
E2
N m 2 / C2 )(10 × 10−9 C ) 2
N m 2 / C2 )(15 × 10−9 C )
( 0.040 m )
2
= 5.625 × 104 N/C
= 8.438 × 104 N/C
The fields point in the same direction, so G E = ( 5.625 × 104 N/C + 8.438 × 104 N/C, right ) = (1.41×105 N/C, right ) .
The electric field will point left when Q1 and Q2 are interchanged. The electric field strength in both cases is 1.41 × 105 N/C.
27.51. Model: The parallel plates form a parallel-plate capacitor. The electric field inside a parallel-plate capacitor is a uniform field, so the electron and proton will have constant acceleration. Visualize:
The negative plate is at x = 0 m and the positive plate is at x = d = 1 cm. Solve: Both particles accelerate from rest (v0 = 0 m/s), so at time t their positions are
1 1 xe = xe0 + aet 2 = aet 2 2 2
1 1 xp = xp0 + apt 2 = d + apt 2 2 2
At some instant of time t1 the electron and proton have the same position: xe = xp = x1. This is the point where they pass each other. At this instant, 1 x1 = aet12 2
1 x1 = d + apt12 2
These are two equations in the two unknowns x1 and t1. From the first equation, 12 t12 = x1 ae . Using this in the second equation gives
x1 = d +
ap ae
x1 ⇒ x1 =
d 1 + ap /ae
To finish, we need to find the accelerations of the electron and proton. Both particles are in a parallel-plate capacitor with Ecap = Q/ε0A. The field points to the left, so Ex = −Q/ε0A. The proton’s acceleration is
ap =
Fp mp
=
qp E x mp
=
e ( −Q/ε 0 A ) eQ/ε 0 A =− mp mp
The proton’s acceleration is negative, as expected. For the electron, ae =
Fe qe Ex −e ( −Q/ε 0 A ) eQ/ε 0 A = = = me me me me
Consequently, the acceleration ratio is ap ae = me mp . Using this, the point where the two charges pass is x1 =
1 cm d = = 0.9995 cm 1 + me /mp 1 + 9.11 × 10−31 kg/1.67 × 10−27 kg
Assess: This is very close to where the proton starts. Since there is a factor of ~1800 difference in the masses of the proton and electron, the electron accelerates much faster than the proton.
27.52. Model: The electric field is uniform inside the capacitor, so constant-acceleration kinematic equations apply to the motion of the proton. Visualize:
G Solve: From Equation 27.29, the electric field between the parallel plates E = (η ε 0 ) ˆj. The force on the proton is G G G G G qE qη ˆ qη = F = ma = qE ⇒ a = j ⇒ ay = m mε 0 mε 0
Using the kinematic equation y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) , 2
1 ⎛ qη ⎞ 1 2 Δy = y1 − y0 = ( 0 m/s )( t1 − t0 ) + a y ( t1 − t0 ) = ⎜ ⎟ ( t1 − t0 ) 2 ⎝ mε 0 ⎠ 2 To determine t1 − t0, we consider the horizontal motion of the proton. The proton travels a distance of 2.0 cm at a constant speed of 1.0 × 106 m/s. The velocity is constant because the only force acting on the proton is due to the field between the plate along the y-direction. Using the same kinematic equation, Δx = 2.0 × 10−2 m = v0 x ( t1 − t0 ) + 0 m ⇒ ( t1 − t0 ) =
2.0 ×10−2 m = 2.0 ×10−8 s 1.0 ×106 m/s
−19 −6 −8 2 1 (1.60 × 10 C )(1.0 × 10 C/m )( 2.0 × 10 s ) = 2.2 mm 2 (1.67 ×10−27 kg )(8.85 ×10−12 C2/Nm2 ) 2
⇒ Δy =
27.53. Model: Assume that the electric field inside the capacitor is constant, so constant-acceleration kinematic equations apply. Visualize: Please refer to Figure P27.53. Solve: (a) The force on the electron inside the capacitor is G G G G G qE F = ma = qE ⇒ a = m G Because E is directed upward (from the positive plate to the negative plate) and q = −1.60 ×10−19 C , the
acceleration of the electron is downward. We can therefore write the above equation as simply ay = qE/m. To determine E, we must first find ay. From kinematics, 1 2 x1 = x0 + v0 x ( t1 − t0 ) + ax ( t1 − t0 ) ⇒ 0.040 m = 0 m + v0 cos 45° ( t1 − t0 ) + 0 m 2
⇒ ( t1 − t0 ) =
( 0.040 m )
( 5.0 ×10
6
m/s ) cos 45°
= 1.1314 × 10−8 s
Using the kinematic equations for the motion in the y-direction,
⎛ t −t ⎞ ⎛ qE ⎞⎛ t1 − t0 ⎞ v1 y = v0 y + a y ⎜ 1 0 ⎟ ⇒ 0 m/s = v0 sin 45° + ⎜ ⎟⎜ ⎟ 2 ⎝ ⎠ ⎝ m ⎠⎝ 2 ⎠ ⇒E=−
2 ( 9.1× 10−31 kg )( 5.0 ×106 m/s ) sin 45° 2 m v0 sin 45° =− = 3550 N/C = 3.6 × 103 N/C −19 −8 q ( t1 − t0 ) − × × 1.60 10 C 1.1314 10 s ( )( )
(b) To determine the separation between the two plates, we note that y0 = 0 m and v0 y = (5.0 ×106 m/s)sin 45°, but at
y = y1, the electron’s highest point, v1y = 0 m/s. From kinematics, v12y = v02y + 2a y ( y1 − y0 ) ⇒ 0 m 2 / s 2 = v02 sin 2 45° + 2a y ( y1 − y0 )
⇒ ( y1 − y0 ) = −
v02 sin 2 45° v2 =− 0 2a y 4a y
From part (a), ay =
−19 qE ( −1.60 ×10 C ) ( 3550 N/C ) = = −6.242 ×1014 m/s 2 9.1×10−31 kg m
⇒ y1 − y0 = −
( 5.0 × 10
6
m/s )
2
4 ( −6.242 × 1014 m/s 2 )
= 0.010 m = 1.0 cm
This is the height of the electron’s trajectory, so the minimum spacing is 1.0 cm.
27.54. Model: The parallel plates form a parallel-plate capacitor. The electric field inside a parallel-plate capacitor is a uniform field, so the electrons will have a constant acceleration. Visualize:
Solve: (a) The bottom plate should be positive. The electron needs to be repelled by the top plate, so the top plate must be negative and the bottom plate positive. In other words, the electric field needs to point away from G the bottom plate so the electron’s acceleration a is toward the bottom plate. (b) Choose an xy-coordinate system with the x-axis parallel to the bottom plate and the origin at the point of G entry. Then the electron’s acceleration, which is parallel to the electric field, is a = ajˆ . Consequently, the problem looks just like a projectile problem. The kinetic energy K = 12 mv02 = 3.0 ×10−17 J gives an initial speed
v0 = ( 2 K m )
1/ 2
= 8.115 ×106 m/s . Thus the initial components of the velocity are vx 0 = v0 cos 45° = 5.74 × 106 m/s
v y 0 = v0 sin 45° = 5.74 × 106 m/s
G What acceleration a will cause the electron to pass through the point (x1, y1) = (1.0 cm, 0 cm)? The kinematic equations of motion are 1 x1 = x0 + vx 0t1 + axt12 = vx 0t1 = 0.010 m 2
1 1 y1 = y0 + v y 0t1 + a yt1 = v y 0t1 + at12 = 0 m 2 2
From the x-equation, t1 = x1 vx 0 = 1.742 × 10−9 s. Using this in the y-equation gives
a=−
2v y 0t1 t12
= −6.59 × 1015 m/s 2
But the acceleration of an electron in an electric field is a=
( 9.11×10−31 kg )( −6.59 ×1015 m/s2 ) = 37,500 N/C = 3.8 ×104 N/C Felec qelec E −eE ma = = ⇒E=− =− m m m e 1.60 ×10−19 C
(c) The minimum separation dmin must equal the “height” ymax of the electron’s trajectory above the bottom plate. (If d were less than ymax, the electron would collide with the upper plate.) Maximum height occurs at t = 12 t1 = 8.71 × 10−10 s. At this instant, 1 ymax = v y 0t + at 2 = 0.0025 m = 2.5 mm 2 Thus, dmin = 2.5 mm.
27.55. Model: Assume that the plates form a parallel-plate capacitor. The electric field inside the capacitor is uniform, so that the protons will have a constant acceleration. Visualize:
Solve: (a) The acceleration needed to slow the protons to 2.0 × 105 m/s can be obtained using the kinematic equation of motion v12 = v02 + 2a ( x1 − x0 ) . We have
( 2.0 ×105 m/s ) − ( 2.0 ×106 m/s ) = −9.9 ×1013 m/s2 v2 − v2 a= 1 0 = 2 ( x1 − x0 ) 2 ( 0.020 m ) 2
2
The force on the proton due to the electric field between the plates is F = ma = qEx . The electric field is Ex =
−27 13 2 ma (1.67 ×10 kg )( −9.9 ×10 m/s ) = = −1.033 ×106 N/C +1.6 ×10−19 C q
The minus sign indicates that the field points to the left. Because E = η ε 0 for a parallel-plate capacitor, where E is the field strength, η = ε 0 E = (8.85 ×10−12 C2 /Nm 2 )(1.033 ×106 N/C ) = 9.1× 10−6 C/m 2 It is clear that the electric field must oppose the motion of the protons. This requires that the first plate be negatively charged and the second plate be positively charged. (b) Let us calculate E that makes protons come to rest completely. Combining the equations for E and a in part (a),
E=
m ( v12 − v02 )
2q ( x1 − x0 )
=
(1.67 ×10
2 2 kg ) ⎡( 0 m/s ) − ( 2 ×106 m/s ) ⎤ ⎥⎦ ⎣⎢ = 1.04 ×106 N/C 2 (1.6 ×10−19 C ) ( 0.020 m ) −27
⇒ η = ε 0 E = ( 8.85 ×10−12 C2 /Nm 2 )(1.04 × 106 N/C ) = 9.24 × 10−6 C/m 2 Because the surface charge density of ±1.0 × 10−5 C/m 2 in your device is larger than 9.24 × 10−6 C/m 2 , the protons will not reach the positive plate. The beam will in fact reverse its direction. The protons will not be stopped, so no, the device does not work.
27.56. Model: An orbiting electron experiences a force that causes centripetal acceleration. Visualize:
The electron orbits at a radius r = R + h = 12 ( 2.0 mm ) + 1.0 mm = 2.0 mm. The force that causes the G circular motion is simply the electric force F given by Coulomb’s law:
Solve:
F=
K Qsphere qelec r
2
=
KeQsphere r
2
=
( 9 ×10
9
N m 2 / C2 )(1.6 × 10−19 C )(1.0 × 10−9 C )
( 2.0 ×10
−3
m)
2
For circular motion, F = macentripetal =
mv 2 rF ⇒v= = r m
( 2.0 ×10
−3
m )( 3.6 ×10−13 N )
9.11×10−31 kg
= 2.8 × 107 m/s
= 3.60 × 10−13 N
27.57. Model: An orbiting proton experiences a force that causes centripetal acceleration. Visualize:
Solve: The proton orbits at a radius r = 5.0 mm + 1.0 mm = 6.0 mm = 6.0 × 10−3 m. The force that causes the G circular motion is simply the electric force F given by Coulomb’s law:
F=
1
Qball qproton
4πε 0
r2
= macentripetal =
mv 2 ( 4πε 0 ) mv 2r ⇒ Qball = r qproton
The speed of the proton is v=
−3 2π r 2π ( 6.0 ×10 m ) = = 3.77 ×104 m/s 1.0 ×10−6 s T
(1.67 ×10 kg )( 3.77 ×10 m/s ) ( 6.0 ×10 = ( 9.0 ×10 N m /C )(1.60 ×10 C ) −27
⇒ Qball
4
9
2
2
2
−19
−3
m)
= 9.9 ×10−12 C
The charge on the ball must be negative to hold the proton in its orbit, so Q = −9.9 ×10−12 C.
27.58. Model: Visualize:
Solve:
The electron orbiting the proton experiences a force given by Coulomb’s law.
The force that causes the circular motion is
F=
2 2 2 1 qp qe mev 2 me ( 4π r f ) = = 2 4πε 0 r r r
where we used v = 2π r T = 2π rf . The frequency is
⎛ 1 ⎞ qp qe f = ⎜ ⎟ 2 3 = ⎝ 4πε 0 ⎠ 4π me r
( 9.0 ×10 N m /C )(1.60 ×10 4π ( 9.11×10 kg )( 5.3 × 10 9
2
2
−31
2
−19
−11
C)
m)
2
3
= 6.56 ×1015 Hz
27.59. Model: Visualize:
Solve:
The electron orbiting the proton experiences a force given by Coulomb’s law.
The force that causes the circular motion is F=
2 2 2 1 qp qe mev 2 me ( 4π r f ) = = 2 4πε 0 r r r
where we used v = 2π r T = 2π rf . The radius is 1/ 3
qp qe ⎡ 1 ⎤ r=⎢ 2 2 ⎥ ⎢⎣ 4πε 0 me ( 4π f ) ⎥⎦
⎡ (1.60 ×10−19 C )(1.60 ×10−19 C ) ⎤⎥ = ⎢( 9.0 × 109 N m 2 /C 2 ) 2 ⎢ ( 9.1× 10−31 kg ) 4π 2 (1.0 × 1012 s−1 ) ⎥⎦ ⎣
1/ 3
= 1.86 × 10−8 m = 18.6 nm
27.60. Model: The electric field at the dipole’s location is that of the ion with charge q. Visualize:
Solve:
(a) We have p = α E. The units of α are the units of p/E and are
C m C 2 m C 2s 2 = = N/C N kg (b) The electric field due to the ion at the location of the dipole is
⎛ 1 q ⎞ 1 q ˆ Eat dipole = ⎜ i , away from q ⎟ = 2 2 ⎝ 4πε 0 r ⎠ 4πε 0 r
G G Because p = α E , the induced dipole moment is
⎛ 1 q ⎞ˆ G p =α ⎜ i 2 ⎟ ⎝ 4πε 0 r ⎠ From Equation 27.11, the electric field produced by the dipole at the location of the ion is 2 G G 1 2p 1 ⎛ 2 ⎞ ⎛ 1 q ⎞ ˆ ⎛ 1 ⎞ ⎛ 2qα ⎞ ˆ Edipole = α i = = ⎟ ⎜ ⎟ ⎜ ⎜ ⎟ ⎜ ⎟i 4πε 0 r 3 4πε 0 ⎝ r 3 ⎠ ⎝ 4πε 0 r 2 ⎠ ⎝ 4πε 0 ⎠ ⎝ r 5 ⎠ The force the dipole exerts on the ion is 2
G G ⎛ 1 ⎞ ⎛ 2q 2α ⎞ ˆ Fdipole on ion = qEdipole = ⎜ ⎟ ⎜ 5 ⎟i ⎝ 4πε 0 ⎠ ⎝ r ⎠ G G According to Newton’s third law, Fdipole on ion = − Fion on dipole . Therefore,
⎛ ⎛ 1 ⎞ 2 2q 2α ⎞ G Fion on dipole = ⎜ ⎜ , toward ion ⎟ ⎟ 5 ⎜ ⎝ 4πε 0 ⎠ r ⎟ ⎝ ⎠
27.61. Model: density λ. Visualize:
Solve:
The electric field at the dipole’s location is that of the infinite line of charge of linear charge
The field at distance r from an infinite line of charge is G 1 2λ E= rˆ 4πε 0 r
It points straight away from the line. The dipole consists of charge +q at distance r + s/2 and charge –q at distance r – s/2. The net force on the dipole due to the field of the line is
G G G 2qλ ⎛ 1 1 ⎞ 2qλ ⎛ ( r − s/2 ) − (r + s/2 ) ⎞ F = qE+ − qE− = − ⎜ ⎟ rˆ ⎜ ⎟ rˆ = 4πε 0 ⎝ r + s/2 r − s/2 ⎠ 4πε 0 ⎝ (r + s/2 )(r − s/2 ) ⎠ =−
2qsλ 2 pλ rˆ = − rˆ 4πε 0 ( r 2 − ( s/2) 2 ) 4πε 0 ( r 2 − ( s/2) 2 )
where we used p = qs for the dipole moment. The minus sign shows that this is an attractive force, toward the line of charge. If r >> s, the (s/2)2 term in the denominator is negligible and we find that the line of charge exerts an attractive force of magnitude 2 pλ F= 4πε 0 r 2
27.62. Solve: (a) Charges ±2.0 nC form an electric dipole. The electric field strength 2.5 cm from the dipole in the plane perpendicular to the dipole is 1150 N/C. What is the charge separation? (b) Solving for s, 3 (1150 N/C )( 0.025 m ) = 9.98 × 10−4 m = 1.0 × 10−3 m = 1.0 mm s= ( 9 ×109 N m 2 /C2 )( 2.0 ×10−9 C )
27.63. Solve: (a) A very long charged wire has linear charge density 2.0×10–7 C/m. At what distance from the wire is the field strength 25,000 N/C? (b) Solving for r, ( 9 × 109 N m2 /C2 ) 2 ( 2.0 × 10−7 C/m ) = 0.144 m = 14.4 cm r= 25,000 N/C
27.64. Solve: (a) At what distance along the axis of a charged disk is the electric field strength half its value at the center of the disk? Give your answer in terms of the disk’s radius R. (b) Dividing both sides of the equation by η 2ε 0 , 1−
z z 2 + R2
=
1 R ⇒ 2z = z 2 + R 2 ⇒ 4z 2 = z 2 + R 2 ⇒ z = 2 3
27.65. Solve: (a) A proton is released from the positive plate of a parallel-plate capacitor and accelerates toward the negative plate at 2.0 × 1012 m/s2. If the capacitor plates are 2.0 cm × 2.0 cm squares, what is the magnitude of the charge on each? (b) Solve the first equation for E and substitute into the second equation. The charge is
Q=
(1.67 ×10
−27
kg )( 2.0 ×1012 m/s 2 )( 8.85 ×10−12 C2 /N m 2 ) ( 0.020 m ) 1.60 ×10−19 C
2
= 7.4 ×10−11 C
27.66. Model: The long charged wire is an infinite line of charge. The charges on the wire and the plastic rod are uniform. Visualize: Please refer to Figure P27.66. The plastic stirrer is located on the x-axis. Solve: The electric field of the infinite line of charge at a distance x from its axis is
Ex =
1 2λ 4πε 0 x
Because the electric field is a function of x, the plastic stirrer experiences an electric field that varies along the length of the stirrer. We handled such problems earlier. Take a small segment of the charge on the stirrer and calculate the electric force due to the line charge on this charge segment Δ q. A summation of all such forces on the charge segments Δ q will yield the net force on the stirrer. This procedure is equivalent to integrating the electric force on a small segment Δ q over the length of the stirrer. The force on charge Δ q of length Δ x at position x due to the infinite line of charge is ΔF = E Δq = Eλ ′Δx =
1 ⎛ 2λλ ′Δx ⎞ ⎜ ⎟ 4πε 0 ⎝ x ⎠
In the above expression, λ′ = Q/L is the linear charge density of the stirrer and λ is the linear charge density of the plastic rod. We have also used the relation Δq Δx = λ ′. Changing Δ x → dx and integrating x from x = 0.020 m to x = 0.080 m, the total force on the stirrer is ⎛ 1 ⎞ ⎛ 1 ⎞ 1 2λλ ′ 0.080 m ⎛ 0.080 m ⎞ dx = ⎜ ⎟ ( 2λλ ′ ) ln x 0.020 m = ⎜ ⎟ ( 2λλ ′ ) ln ⎜ ⎟ 4 πε x 4 πε 4 πε ⎝ 0.020 m ⎠ 0 0 ⎠ 0 ⎠ ⎝ ⎝ 0.020 m 0.080 m
F=
∫
⎛ 10 × 10−9 C ⎞ −4 = ( 9.0 × 109 N m 2 /C 2 ) 2 (1.0 × 10−7 C/m ) ⎜ ⎟ ln ( 4 ) = 4.2 × 10 N 0.060 m ⎝ ⎠
27.67. Model: The electric field is that of the ball. This electric field polarizes charge in the aluminum foil. Visualize:
Please refer to Figure CP27.67. Solve: (a) The electric field at the location of the piece of foil is
G ⎛ Q ⎞ Q ˆ Eball = ⎜ j , straight down ⎟ = − 2 4πε 0 h 2 ⎝ 4πε 0 h ⎠ (b) The electric field inside the foil is not G just the field due to the +q Gand −q charges on the surfaces of the foil. It is the superposition of the ball’s field Eball and the polarization field Epolar that arises when the foil is polarized by G G G G G the charged ball. That is, Ein = Eball + Epolar . In electrostatic equilibrium, the internal field must be Ein = 0 N/C.
(An internal field that wasn’t zero would cause the motion of electrons and thus prevent the foil from being in electrostatic the charged ball is brought near, the foil is polarized just to the point where G G When G G equilibrium.) Epolar = − Eball , causing Ein = 0 N/C. G G G G (c) In part (b) we noted that Ein = Eball + Epolar = 0 N/C. We know the ball’s field, from part (a). The polarization field is that of a parallel-plate capacitorGsince the polarized foil, with +q and −q surfaces, is just like a capacitor. The polarization field points upward, Epolar = (η ε 0 ) ˆj , where η is the surface charge density. Adding the two fields and setting the sum to zero gives G G Eball + Epolar = −
G ˆj + η ˆj = 0 N/C ⇒ η = Q ε0 4πε 0 h 4π h 2 Q
2
Once we know η, the actual charge on the foil surface of area A = πR2 is q = η A = ηπ R 2 =
R2 Q 4h 2
(d) The attractive and repulsive forces can each be separately calculated and combined, but it is easier to consider the polarized foil to be a dipole since it has two opposite charges ± q separated by a distance t. Because t << h, we can write the on-axis electric field of the foil as the on-axis field of a dipole. Using Equation 27.11, the electric field produced by the dipole is 2p 2qt = Edipole = 3 4πε 0 r 4πε 0 h3 G G The dipole exerts a force on the charged ball Ffoil on ball = QEdipole . From Newton’s third law,
Fball on foil = Ffoil on ball =
2Qqt 4πε 0 h3
The foil is lifted when Fball on foil ≥ mg = ρVg , where ρ is the mass density of the foil and V = πR2t is the foil’s volume. That is, when 2Qqt 2Qq ≥ ρπ R 2tg ⇒ ≥ ρπ R 2 g 4πε 0 h3 4πε 0 h3 since the foil thickness t cancels. Using the expression for q we found in part (c),
⎡ ⎤ R 2Q 2 Q2 ≥ ρπ R 2 g ⇒ h ≤ ⎢ ⎥ 5 2 ( 4πε 0 ) h 2 4 g π πε ρ 0) ⎣⎢ ( ⎦⎥
1/ 5
The minimum height for which the ball lifts the foil is
⎡ ( 9.0 × 109 N m 2 /C2 )( 50 × 10−9 C )2 ⎤ ⎥ h=⎢ ⎢ 2π ( 2700 kg/m3 )( 9.8 m/s 2 ) ⎥ ⎣ ⎦
1/ 5
= 0.0106 m = 1.06 cm
27.68. Model: The rod is thin and is assumed to be a line of charge of length L. Visualize:
Solve: (a) The λ-versus-y graph over the length of the rod is shown in the figure. (b) Consider a segment of charge Δ q of length Δ y at a distance y from the center of the rod. The amount of charge in this segment is Δq = λΔy = a y Δy Converting Δ q to dq, Δ y to dy, and integrating from y = −L/2 to y = +L/2, the total charge is
Q = ∫ dq =
+L/2
∫
−L / 2
L/2
a y dy = 2 ∫ ay dy = 2a 0
y2 2
L/2
= 0
aL2 4
Thus the constant a is a=
4Q L2
(c) With the origin of the coordinate system at the center of the rod, consider two symmetrically located charge segments Δ q with length Δ y. The electric field at P from the top charge segment makes an angle θ below the xaxis and the electric field of the bottom charge segment makes an angle θ above the x-axis. Because the charge density is symmetric about the origin (i.e., λ (at –y) = λ (at y)), the y-components of the two contributions cancel out. Thus, we have to calculate only the x-component of the electric field at P. Because the electric field strength of the lower half of the rod is the same as that of the upper half, we only need to obtain the electric field strength of half the rod, then multiply by two. The electric field along the x-direction due to a charge segment Δ q is xa y Δy 1 Δq 1 λΔy 1 x ΔE x = cosθ = = 4πε 0 ( x 2 + y 2 ) 4πε 0 ( x 2 + y 2 ) x 2 + y 2 4πε 0 ( x 2 + y 2 )3 / 2
Changing Δ E to dE, Δ y = dy, integrating y from y = 0 m to y = L/2, and multiplying by 2 to take into account the entire rod, the electric field is L/2
Ex = 2
∫ 0
=
ax y dy 2ax = 3/ 2 2 2 4πε 0 ( x + y ) 4πε 0
L/2
∫ 0
(x
L/2
y dy
2
+ y2 )
3/ 2
=
2ax ⎡ −1 ⎤ 4πε 0 ⎢⎣ x 2 + y 2 ⎥⎦ 0
2ax ⎡ 1 1 8Q ⎡ x ⎤ ⎤ − ⎥ = 4πε L2 ⎢1 − 2 ⎥ 2 2 2 4πε 0 ⎢⎣ x x +L 4⎦ x +L 4⎦ 0 ⎣
The last step used the expression for a from part (b).
27.69. Model: Model the infinitely long sheet of charge with width L as a uniformly charged sheet.
Solve: (a) Divide the sheet into many long strips parallel to the y-axis and of width Δ x. Each strip has a linear charge density λ = ηΔ x and acts like a long, charged wire. At the point of interest, strip i contributes a small field G Ei of strength
Ei =
2λ 2ηΔx = 4πε 0 ri 4πε 0 ri
By symmetry, the x-components of all the strips will add to zero and the net field will point straight away from the sheet along the z-axis. The net field’s z-component will be Ez = ∑ ( Ei ) z = ∑ Ei cosθ i i
The distance ri = ( xi2 + z
)
2 1/ 2
i
and cosθi = z ri = z ( xi2 + z
Ez = ∑ i
2ηΔx
4πε 0 ( x + z 2 i
) (x
2 1/ 2
)
2 1/ 2
2 i
. Thus,
z +z
)
2 1/ 2
=
2η z Δx ∑ 4πε 0 i xi2 + z 2
If we now let Δ x → dx, the sum becomes an integral ranging from x = Δ L/2 to x = L/2. This gives L/2
2η z dx η z ⎡ 1 −1 ⎛ x ⎞ ⎤ η ⎡ −1 ⎛ L ⎞ ⎛ −L ⎞⎤ = tan ⎜ ⎟ ⎥ = tan ⎜ ⎟ − tan −1 ⎜ ⎟⎥ ⎢ ⎢ 2 2 ∫ 4πε 0 − L / 2 x + z 2πε 0 ⎣ z ⎝ z ⎠ ⎦ − L / 2 2πε 0 ⎣ ⎝ 2z ⎠ ⎝ 2z ⎠⎦ L/2
Ez =
From trigonometry, tan −1 ( −φ ) = − tan −1 (φ ) . So finally,
Ez =
G η η ⎛ L⎞ ⎛ L⎞ tan −1 ⎜ ⎟ ⇒ E = tan −1 ⎜ ⎟ kˆ 2 z πε 0 πε ⎝ ⎠ ⎝ 2z ⎠ 0
(b) As z → 0 m,
G η π ˆ η ˆ ⎛ L⎞ π tan −1 ⎜ ⎟ → ⇒E→ k= k 2ε 0 2 πε 0 2 ⎝ 2z ⎠
This is the electric field due to a plane as you can see from Equation 27.26. We obtain this result because in the limit as z → 0 m, the dimension L becomes extremely large. As z → ∞,
G L η L ˆ 1 2λ ˆ ⎛ L⎞ tan −1 ⎜ ⎟ → ⇒E→ k= k 2 2 z z z πε 2 4 πε ⎝ ⎠ 0 0 z
where we have used ηL = λ as the charge per unit length of the sheet. This is the electric field due to a long, charged wire. We obtain this result because for z >> L, the infinitely long sheet “looks” like an infinite line charge. (c) The following table shows the field strength Ez in units of η/ε0 for selected values of z in units of L. A graph of Ez is shown in the figure above. z L 0 0.25 0.50 1.0 2.0 3.0 4.0
Ez
η ε0
0.5 0.35 0.25 0.15 0.08 0.05 0.04
27.70. Model: Model the infinitely long sheet of charge with width L as a uniformly charged sheet. Visualize:
Solve: (a) Consider a point on the x-axis at a distance d from the center of the sheet of charge. (We’ll call this distance d to begin with, rather than x, to avoid confusion with x as the integration variable.) Once again, let the sheet of charge be divided into small strips of width Δ x. Each strip has a linear charge density λ = ηΔ x and acts like a long, charged wire. Strip i is at distance ri = d − xi from the point of interest, so it contributes the small field
G 2λ ˆ 2ηΔx Ei = i = iˆ 4πε 0 ri 4πε 0 ( d − xi )
G In this situation all fields Ei point in the same direction. Their x-components all add to give a net field in the +x-direction:
Ex = ∑ ( Ei ) x = i
2η Δx ∑ 4πε 0 i ( d − xi )
We’ll replace the sum with an integral from x = −L/2 to x = +L/2, giving L/2 2η dx 2η 2η 2η ⎛ 2d + L ⎞ = ⎡ − ln ( d − x ) ⎦⎤ − L / 2 = ⎡ − ln ( d − L 2 ) + ln ( d + L 2 ) ⎦⎤ = ln ⎜ ⎟ 4πε 0 − L∫/ 2 ( d − x ) 4πε 0 ⎣ 4πε 0 ⎣ 4πε 0 ⎝ 2d − L ⎠ L/2
Ex =
Now that the integration is complete, we can note that d really is the x-coordinate of the point of interest. Substituting x for d and changing to vector form, we end up with G 2η ⎛ 2 x + L ⎞ ˆ ln ⎜ Enet = ⎟i 4πε 0 ⎝ 2 x − L ⎠ (b) If the point is very distant compared to width of the sheet of charge (x >> L), then the sheet of charge looks like a line of charge with linear density (charge per unit length) λ = ηL. Write
Ex =
2η ⎛ 1 + L/2 x ⎞ 2η ln ⎜ [ln(1 + L/2 x ) − ln(1 − L/2 x )] ⎟= 4πε 0 ⎝ 1 − L/2 x ⎠ 4πε 0
If x >> L, then L/2x << 1 and we can use the approximation ln(1 + u) ≈ u if u << 1. Thus Ex ≈
2η ⎡ L ⎛ L ⎞ ⎤ 2η L 2λ − ⎜ − ⎟⎥ = = ⎢ 4πε 0 ⎣ 2 x ⎝ 2 x ⎠ ⎦ 4πε 0 x 4πε 0 x
This is the field of a line of charge with λ = ηL, as we expected.
(c) The following table shows the field strength Ex in units of 2η/4πε0 for selected values of z in units of L. A graph of Ex is shown in the figure above.
z L
0.75 1.0 1.5 2.0 3.0 4.0
Ex
2η 4πε 0
1.61 1.10 0.69 0.51 0.34 0.25
27.71. Model: The electric field is uniform inside the capacitor, so constant-acceleration kinematic equations apply to the motion of the electron. Visualize: Please refer to Figure CP27.71. The condition for the electron to not hit the negative plate is that its vertical velocity should just become zero as the electron reaches the plate. Solve: The force on the electron inside the capacitor is G G G G G qE F = ma = qE ⇒ a = m
⇒ ay =
− (1.60 ×10−19 C )(1.0 ×104 N/C ) 9.11×10−31 kg
= −1.756 × 1015 m/s 2
The initial velocity v0 has two components: v0 x = v0 cos 45° and v0 y = v0 sin 45°. Because the electric field inside
the capacitor is along the +y-direction, the electron has a negative acceleration that reduces the vertical velocity. We require v1y = 0 m/s if it is not to hit the plate. Using kinematics,
v12y = v02y + 2a y ( y1 − y0 ) ⇒ ( 0 m/s ) = v02y + 2a y Δy 2
⇒ v0 y = −2a y Δy = −2 ( −1.756 × 1015 m/s 2 ) ( 0.02 m ) = 8.381 × 106 m/s
⇒ v0 =
8.381× 106 m/s = 1.19 ×107 m/s sin 45°
27.72. Model: The two electrodes form a parallel-plate capacitor. The electric field inside the electrodes is uniform, so the electrons have constant acceleration. Visualize:
Solve: (a) The ink drops are deflected up or down as they experience an electric field between the two parallel electrodes. The electric field exerts a force on the ink drops which is F = qE = may ⇒ E = may /q We therefore need to determine the mass m, the charge q, and the acceleration ay of the ink drops. We have 3⎤ ⎛ 4π 3 ⎞ ⎡ 4π m = ρV = ρ ⎜ r ⎟ = ( 800 kg/m3 ) ⎢ (15 ×10−6 m ) ⎥ = 1.131×10−11 kg ⎝ 3 ⎠ ⎣ 3 ⎦
q = ( 8 × 105 )(1.60 × 10−19 C ) = 1.28 × 10−13 C
At maximum deflection, the drop’s angle upon exiting the plates must be tan θ =
vy vx
⎛ 3 mm ⎞ =⎜ ⎟ = ±0.15 ⇒ v y = ( ±0.15) vx = ( ±0.15)( 20 m/s ) = ±3.0 m/s ⎝ 2.0 cm ⎠
From the kinematic equation v1 y = v0 y + a y ( t1 − t0 ) , ay =
v1 y − v0 y t1 − t0
⇒ ay =
±3.0 m/s − 0 m/s t1 − t0
We can obtain t1 − t0 from the x-motion between the plates as follows:
1 2 x1 = x0 + v0 x ( t1 − t0 ) + ax ( t1 − t0 ) ⇒ x1 − x0 = 6.0 mm = v0 x ( t1 − t0 ) + 0 m 2 ⇒ t1 − t0 =
±3.0 m/s 6.0 × 10 −3 m 6.0 × 10−3 m = ±1.0 ×104 m/s 2 = = 3.0 × 10−4 s ⇒ a y = 3.0 ×10−4 s 20 m/s v0 x
We are now in a position to obtain the field strength E from the equation
E=
ma y q
=
(1.131×10
−11
kg )(1.0 ×104 m/s 2 )
1.28 ×10−13 C
= 8.84 × 105 N/C
The electric field strength is 8.8×105 N/C. (b) The surface charge density is
η = ε 0 E = (8.85 ×10−12 C 2 /Nm 2 )( 8.84 ×105 N/C ) = 7.82 × 10−6 C/m 2 The area of each plate is
Aplate = ( 6.0 ×10−3 m )( 4.0 ×10−3 m ) = 24.0 × 10−6 m 2
Thus the charge on the plates is Q = ±η Aplate = ±1.88 × 10−10 C = ±0.188 nC. Assess: Because of the large acceleration due to the electric field, the acceleration due to gravity can be ignored in the calculations.
27.73. Model: The orbital motion of the positron-electron system about their center of mass is due to the electric force between the positron and the electron. Solve: The distance between the charges is twice the radius of their orbits about the center of mass. The force that causes the circular motion is
F=
2 1 qpositron qelectron melectron v 2 mpositron v m 4π 2 f 2 r 2 = = = electron 2 4πε 0 r r r ( 2r )
where we used v = 2πrf. The frequency is
⎛ 1 ⎞ qpositron qelectron f = ⎜ = ⎟ 2 3 ⎝ 4πε 0 ⎠ melectron 16π r
( 9.0 ×10 ( 9.1×10
9
N m 2 /C 2 )(1.60 ×10−19 C )
−31
2
kg )16π 2 ( 0.50 ×10−9 m )
3
= 1.13 × 1014 Hz
27.74. Model: radius R. Visualize:
Solve:
The electric field is that of a positively charged ring. The ring is thin and the charge lies along circle of
(a) From Example 27.5, the electric field on the axis of a ring of radius R is
Ez =
1
zQ
4πε 0 ( z + R 2 )3/ 2 2
The force on negative charge −q is Fz = −qEz = −
1 qQz 4π 0 ( z 2 + R 2 )3/ 2
The meaning of the negative sign is that the force on the electron points left when the displacement is to the right and to the right when the displacement is to the left. That is, the electric force tries to keep the charge at z = 0 m. (b) For small amplitude oscillations, that is, when z << R, the force is Fz = −
1 qQ ⎛ z2 ⎞ + z 1 ⎜ ⎟ 4πε 0 R 3 ⎝ R 2 ⎠
−3/ 2
=−
1 qQ ⎛ 3 z 2 …⎞ 1 qQ z ⎜1 − 2 + ⎟ ≈ − z 3 4πε 0 R ⎝ 2 R 4πε 0 R 3 ⎠
This is simply Hooke’s law Fz = − kz where the “spring constant” is
k=
qQ 4πε 0 R3
A particle that obeys Hooke’s law undergoes simple harmonic motion with a frequency given by f =
1 2π
k 1 = m 2π
qQ 4πε 0 mR 3
(c) Substituting into the above expression,
f =
1 2π
(1.0 ×10
−13
C )(1.6 × 10−19 C )( 9.0 × 109 N/m 2 C2 )
( 9.11×10
−31
kg )(1.0 × 10−6 m )
3
= 2.0 × 1012 Hz
27-1
28.1. Visualize:
As discussed in Section 28.1, the symmetry of the electric field must match the symmetry of the charge distribution. In particular, the electric field of a cylindrically symmetric charge distribution cannot have a component parallel to the cylinder axis. Also, the electric field of a cylindrically symmetric charge distribution cannot have a component tangent to the circular cross section. The only shape for the electric field that matches the symmetry of the charge distribution with respect to (i) translation parallel to the cylinder axis, (ii) rotation by an angle about the cylinder axis, and (iii) reflections in any plane containing or perpendicular to the cylinder axis is the one shown in the figure.
28.2. Visualize:
The object has spherical symmetry, so the electric field is radial.
28.3. Visualize:
Figure 28.6 shows the electric field for an infinite plane of charge. For two parallel planes, this is the only shape of the electric field vectors that matches the symmetry of the charge distribution.
28.4. Model: The electric flux “flows” out of a closed surface around a region of space containing a net positive charge and into a closed surface surrounding a net negative charge. Visualize: Please refer to Figure EX28.4. Let A be the area in m2 of each of the six faces of the cube. G G Solve: The electric flux is defined as Φ e = E ⋅ A = EA cosθ , where θ is the angle between the electric field and a line perpendicular to the plane of the surface. The electric flux out of the closed cube surface is
Φ out = ( 20 N/C + 20 N/C + 10 N/C ) A cos0° = ( 50 A ) N m 2 /C Similarly, the electric flux into the closed cube surface is Φ in = (15 N/C + 15 N/C + 15 N/C ) A cos180° = − ( 45 A ) N m 2 /C The net electric flux is ( 50 A ) N m 2 /C − ( 45 A ) N m 2 /C = ( 5 A ) N m 2 /C. Since the net electric flux is positive (i.e., outward), the closed box contains a positive charge.
28.5. Model: The electric flux “flows” out of a closed surface around a region of space containing a net positive charge and into a closed surface surrounding a net negative charge. Visualize: Please refer to Figure EX28.5. Let A be the area of each of the six faces of the cube. G G Solve: The electric flux is defined as Φ e = E ⋅ A = EA cosθ , where θ is the angle between the electric field and a line perpendicular to the plane of the surface. The electric flux out of the closed cube surface is
Φ out = (10 N/C + 10 N/C + 20 N/C + 5 N/C ) A cos0° = ( 45 A ) N m 2 /C Similarly, the electric flux into the closed cube surface is Φ in = (15 N/C + 20 N/C ) A cos (180° ) = − ( 35 A ) N m 2 /C Hence, Φout + Φin = 10 N m2/C. Since the net electric flux is positive (i.e., outward), the closed box contains a positive charge.
28.6. Model: The electric flux “flows” out of a closed surface around a region of space containing a net positive charge and into a closed surface surrounding a net negative charge. Visualize: Please refer to Figure EX28.6. Let A be the area of each of the six faces of the cube. G G Solve: The electric flux is defined as Φ e = E ⋅ A = EA cosθ , where θ is the angle between the electric field and a line perpendicular to the plane of the surface. The electric flux out of the closed cube surface is
Φ out = ( 20 N/C + 10 N/C + 10 N/C ) A cos0° = ( 40 A ) N m 2 /C Similarly, the electric flux into the closed cube surface is Φ in = ( 20 N/C + 15 N/C ) A cos180° = − ( 35 A ) N m 2 /C
Because the cube contains negative charge, Φout + Φin must be negative. This means Φout + Φin + Φunknown < 0 N m2/C. Therefore,
( 40 A) N m 2 /C + ( −35 A) N m 2 /C + Φ unknown < 0 N m 2 / C ⇒ Φ unknown < ( −5 A ) N m 2 /C That is, the unknown vector points into the front face of the cube and its field strength is greater than 5 N/C.
28.7. Model: The electric flux “flows” out of a closed surface around a region of space containing a net positive charge and into a closed surface surrounding a net negative charge. Visualize: Please refer to Figure EX28.7. Let A be the area of each of the six faces of the cube. G G Solve: The electric flux is defined as Φ e = E ⋅ A = EA cosθ , where θ is the angle between the electric field and a line perpendicular to the plane of the surface. The electric flux out of the closed cube surface is
Φ out = ( 20 N/C + 10 N/C ) A cos0° = ( 30 A ) N m 2 /C Similarly, the electric flux into the closed cube surface is Φ in = ( 20 N/C + 15 N/C + 10 N/C ) A cos180° = − ( 45 A ) N m 2 /C
Because the cube contains positive charge, Φout + Φin must be positive. This means Φout + Φin + Φunknown > 0 N m2/C. Therefore,
( 30 A) N m 2 /C + ( −45 A) N m 2 /C + Φ unknown > 0 N m 2 / C ⇒ Φ unknown > (15 A ) N m 2 /C That is, the unknown vector points out of the front face of the cube and its field strength is greater than 15 N/C.
28.8. Model: The electric flux “flows” out of a closed surface around a region of space containing a net positive charge and into a closed surface surrounding a net negative charge. Visualize: Please refer to Figure EX28.8. Let A be the area of each of the six faces of the cube. G G Solve: The electric flux is defined as Φ e = E ⋅ A = EA cosθ , where θ is the angle between the electric field and a line perpendicular to the plane of the surface. The electric flux out of the closed cube surface is
Φ out = (15 N/C + 15 N/C + 10 N/C ) A cos0° = ( 40 A ) N m 2 /C Similarly, the electric flux into the closed cube surface is Φ in = ( 20 N/C + 15 N/C ) A cos180° = − ( 35 A ) N m 2 /C Because the cube contains no charge, the total flux is zero. This means Φin + Φout + Φunknown = 0 N m2/C ⇒ Φ unknown = − ( 5 A ) N m 2 /C
The unknown electric field vector on the front face points in, and the electric field strength is 5 N/C.
28.9. Model: The electric field is uniform over the entire surface. Visualize: Please refer to Figure EX28.9. The electric field vectors make an angle of 30° with the planar surface. Because the normal nˆ to the planar surface is at an angle of 90° with the surface, the angle between nˆ G and E is θ = 60°. Solve: The electric flux is G G Φ e = E ⋅ A = EA cosθ = ( 200 N/C ) (1.0 ×10−2 m 2 ) cos 60° = 1.0 N m 2 /C
28.10. Model: The electric field is uniform over the entire surface. Visualize: Please refer to Figure EX28.10. The electric field vectors make an angle of 30° below the surface. G Because the normal nˆ to the planar surface is at an angle of 90° relative to the surface, the angle between nˆ and E is θ = 120°. Solve: The electric flux is G G 2 Φ e = E ⋅ A = EA cosθ = ( 200 N/C ) (15 ×10−2 m ) cos120° = −2.3 N m 2 /C
28.11. Model: The electric field is uniform over the entire surface. Visualize: Please refer to Figure EX28.11. The electric field vectors make an angle of 60° above the surface. G Because the normal nˆ to the planar surface is at an angle of 90° relative to the surface, the angle between nˆ and E is θ = 30°. Solve: The electric flux is G G Φ e = E ⋅ A = EA cosθ ⇒ 25 N m 2 /C = E (1.0 ×10−2 m 2 ) cos30° ⇒ E = 2.9×103 N/C
28.12. Model: The electric field is uniform over the rectangle in the xy plane. Solve:
(a) The area vector is perpendicular to the xy plane. Thus G A = ( 2.0 cm × 3.0 cm ) kˆ = ( 6.0 × 10−4 m 2 ) kˆ
The electric flux through the rectangle is G G Φ e = E ⋅ A = 50iˆ + 100kˆ ⋅ 6.0 × 10−4 kˆ N m 2 /C = 6.0 × 10−2 N m 2 /C
(
(b) The electric flux is
)(
)
G G Φ e = E ⋅ A = 50iˆ + 100ˆj ⋅ 6.0 × 10−4 kˆ N m 2 /C = 0 N m 2 / C
(
)(
)
G Assess: In (b), E is in the plane of the rectangle. That is why the flux is zero.
28.13. Model: The electric field over the rectangle in the xz plane is uniform. Solve:
(a) The area vector is perpendicular to the xz plane. Thus G A = ( 2.0 cm × 3.0 cm ) ˆj = ( 6.0 × 10−4 m 2 ) ˆj
The electric flux through the rectangle is G G Φ e = E ⋅ A = 50iˆ + 100kˆ N/C ⋅ ( 6.0 × 10−4 m 2 ) ˆj
(
= ( 300 × 10 (b) The flux is
)
−4
( )
( )
N m /C ) iˆ ⋅ ˆj + ( 600 × 10−4 N m 2 /C ) kˆ ⋅ ˆj = 0 N m 2 /C 2
G G Φ e = E ⋅ A = 50iˆ + 100ˆj N/C ⋅ ( 6.0 × 10−4 m 2 ) ˆj
(
)
( )
= ( 300 ×10−4 N m 2 /C ) iˆ ⋅ ˆj + ( 600 × 10−4 N m 2 /C ) ˆj ⋅ ˆj = 0 N m /C+ ( 6.0 ×10 2
−2
N m 2 /C ) = 6.0 ×10−2 N m 2 /C
28.14. Model: The electric field over the circle in the xy plane is uniform. Solve:
The area vector of the circle is G 2 A = π r 2 kˆ = π ( 0.015 m ) kˆ =
( 7.07 ×10
−4
m 2 ) kˆ
Thus, the flux through the area of the circle is G G Φ e = E ⋅ A = 1500iˆ + 1500 ˆj + 1500 kˆ N/C ⋅ ( 7.07 ×10−4 m 2 ) kˆ
(
)
Using iˆ ⋅ kˆ = ˆj ⋅ kˆ = 0 and kˆ ⋅ kˆ = 1,
Φ e = (1500 N/C ) ( 7.07 ×10−4 m 2 ) = 1.06 N m 2 /C
G
28.15. Model: The electric field over the two faces of the box perpendicular to E is uniform. Visualize:
Solve: The electric field is parallel to four sides of the box, so the electric fluxes through these four surfaces are zero. The field is out of the top surface, so Φtop = EA. The field is into the bottom surface, so Φbottom = –EA. Thus the net flux is Φe = 0 N m2/C2.
28.16. Model: The electric field through the two cylinders is uniform. Visualize: Please refer to Figure EX28.16. Let A = π R2 be the area of the end of the cylinder and let E be the electric field strength. Solve:
→
(a) There’s no flux through the side walls of the cylinder because E is parallel to the wall. On the right →
end, where E points outward, Φright = EA = π R2E. The field points inward on the left, so Φleft = –EA = –π R2E. Altogether, the net flux is Φe = 0 N m2/C2. →
(b) The only difference from part (a) is that E points outward on the left end, making Φleft = EA = π R2E. Thus the net flux through the cylinder is Φe = 2π R2E.
28.17. Visualize:
For any closed surface that encloses a total charge Qin, the net electric flux through the closed surface is Φ e = Qin ε 0 .
28.18. Visualize:
For any closed surface that encloses a total charge Qin, the net electric flux through the closed surface is Φ e =
Qin ε 0 .
28.19. Visualize: Please refer to Figure EX28.19. Solve: For any closed surface that encloses a total charge Qin, the net electric flux through the surface is Φ e = Qin ε 0 . We can write three equations from the three closed surfaces in the figure:
ΦA = −
q
ε0
=
q1 + q3
ε0
⇒ q1 + q3 = −q ΦC =
−2q
ε0
=
q2 + q3
ε0
ΦB =
3q
ε0
q1 + q2
ε0
⇒ q2 + q3 = −2q
Subtracting third equation from the first, q1 – q2 = +q Adding second equation to this equation, 2q1 = +4q ⇒ q1 = 2q That is, q1 = +2q, q2 = +q, and q3 = −3q.
=
⇒ q1 + q2 = 3q
28.20. Visualize: Please refer to Figure EX28.20. For any closed surface that encloses a total charge Qin, the net electric flux through the closed surface is Φ e = Qin ε 0 . For the closed surface of the torus, Qin includes only the −1 nC charge. So, the net flux through the torus is due to this charge: Φe =
This is inward flux.
−1.0 ×10−9 C = −113 N m 2 /C 8.85 ×10−12 C2 /Nm 2
28.21. Visualize: Please refer to Figure EX28.21. For any closed surface that encloses a total charge Qin, the net electric flux through the closed surface is Φ e = Qin ε 0 . The cylinder encloses the +1 nC charge only as both the +100 nC and the −100 nC charges are outside the cylinder. Thus, Φe =
This is outward flux.
1.0 ×10−9 C = 113 N m 2 /C. 8.85 ×10−12 C2 /Nm 2
28.22. Solve: For any closed surface enclosing a total charge Qin, the net electric flux through the surface is Φe =
Qin
ε0
⇒ Qin = ε 0Φ e = ( 8.85 ×10−12 C2 /Nm 2 )( −1000 N m 2 /C ) = −8.85 nC
28.23. Solve: For any closed surface that encloses a total charge Qin, the net electric flux through the surface is
Φe =
Qin
ε0
( 55.3 ×10 )( −1.60 ×10 6
=
8.85 × 10
−12
2
−19
C /N m
2
C)
= −1.00 N m 2 /C
28.24. Model: A copper penny is a conductor. The excess charge on a conductor resides on the surface. Solve:
The electric field at the surface of a charged conductor is
G ⎛η ⎞ Esurface = ⎜ , perpendicular to surface ⎟ ⎝ ε0 ⎠
⇒ η = ε 0 Esurface = ( 8.85 ×10−12 C 2 /Nm 2 ) ( 2000 N/C ) = 17.7 ×10−9 C/m 2
28.25. Model: The excess charge on a conductor resides on the outer surface. Solve:
The electric field at the surface of a charged conductor is
G ⎛η ⎞ Esurface = ⎜ , perpendicular to surface ⎟ ⎝ ε0 ⎠
⇒ η = ε 0 Esurface = ( 8.85 ×10−12 C 2 /Nm 2 )( 3.0 ×106 N/C ) = 2.7 ×10−5 C/m 2 Assess: It is the air molecules just above the surface that “break down” when the E-field becomes strong enough to accelerate stray charges to approximately 15 eV between collisions, thus causing collisional ionization. It does not make any difference whether E points toward or away from the surface.
28.26. Model: The excess charge on a conductor resides on the outer surface. Visualize: Please refer to Figure EX28.26. Solve: Point 1 is at the surface of a charged conductor, hence
G ⎛η ⎞ ( 5.0 ×1010 )(1.60 ×10−19 C m2 ) = 904 N/C Esurface = ⎜ , perpendicular to surface ⎟ ⇒ Esurface = 8.85 ×10−12 C2 /N m 2 ⎝ ε0 ⎠ At point 2 the electric field strength is zero because this point lies inside the conductor. The electric field strength at point 3 is zero because there is no excess charge on the interior surface of the box. This can be quickly seen by considering a Gaussian surface just inside the interior surface of the box as shown in Figure 28.31.
28.27. Model: The copper plate is a conductor. The excess charge resides on the surface of the plate. Solve: (a) One-half of the electrons are located on the top surface and one-half on the bottom surface of the copper plate, so the surface charge density is
( 0.5 ×10 )(1.60 ×10 10
η=
−19
0.1 m × 0.1 m
C)
= 80 ×10−9 C/m 2
Thus, the electric field at the surface of the plate is E=
η 80 ×10−9 C/m 2 = = 9.0 × 103 N/C ε 0 8.85 ×10−12 C2 /N m 2
Because the charge on the plate is negative, the direction of the electric field is toward the plate. (b) E = 0 N/C because the electric field within a conductor is zero. (c) The electric field E = 9.0×103 N/C, toward the plate.
28.28. Visualize: Please refer to Figure EX28.28. Solve: For any closed surface that encloses a total charge Qin, the net electric flux through the closed surface is Φ e = Qin ε 0 . In the present case, the conductor is neutral and there is a point charge Q inside the cavity. Thus Qin = Q and the flux is
Φe =
Q
ε0
28.29. Model: The electric field over the four faces of the cube is uniform. Visualize: Please refer to Figure P28.29. G G G Solve: (a) The electric flux through a surface area A is Φ e = E ⋅ A = EA cosθ , where θ is the angle between G the electric field and the vector A that points outward from the surface. Thus
Φ1 = EA cosθ1 = ( 500 N/C ) ( 9.0 ×10−4 m 2 ) cos150° = −0.39 N m 2 /C Φ 2 = EA cosθ 2 = ( 500 N/C ) ( 9.0 ×10−4 m 2 ) cos60° = 0.23 N m 2 /C
Φ 3 = EA cosθ 3 = ( 500 N/C ) ( 9.0 ×10−4 m 2 ) cos30° = 0.39 N m 2 /C Φ 4 = EA cosθ 4 = ( 500 N/C ) ( 9.0 × 10−4 m 2 ) cos120° = −0.23 N m 2 /C (b) The net flux through these four sides is Φ e = Φ1 + Φ 2 + Φ 3 + Φ 4 = 0 N m 2 /C. Assess: The net flux through the four faces of the cube is zero because there is no enclosed charge.
28.30. Model: The electric field over the five surfaces is uniform. Visualize: Please refer to Figure P28.30. G G G Solve: The electric flux through a surface area A is Φ e = E ⋅ A = EA cosθ where θ is the angle between the electric field and a line perpendicular to the plane of the surface. The electric field is perpendicular to side 1 and is G G parallel to sides 2, 3, and 5. Also the angle between E and A4 is 60°. The electric fluxes through these five surfaces are
Φ1 = E1 A1 cosθ1 = ( 400 N/C )( 2 m × 4 m ) cos (180° ) = −3200 N m 2 /C Φ 2 = E2 A2 cos90° = Φ 3 = Φ 5 = 0 N m 2 /C
Φ 4 = E4 A4 cosθ 4 = ( 400 N/C ) ( ( 2 m sin 30° ) × 4 m ) cos60° = +6400 N m 2 /C Assess: Because the flux into these five faces is equal to the flux out of the five faces, the net flux is zero, as we found.
28.31. Model: Because the tetrahedron contains no charge, the net flux through the tetrahedron is zero. Visualize:
Solve: (a) The area of the base of the tetrahedron is 43 a 2 , where a = 20 cm is the length of one of the sides. Because the base of the tetrahedron is parallel to the ground and the vertical uniform electric field passes upward G G through the base, the angle between E and A is 180°. Thus,
G G 3 2 3 2 Φ base = E ⋅ Abase = EAbase cos180° = − E a = − ( 200 N/C ) ( 0.20 m ) = −3.46 N m 2 /C 4 4 The electric flux through the base is −3.5 N m2/C. G (b) Since E is perpendicular to the base, the other three sides of the tetrahedron share the flux equally. Because the tetrahedron contains no charge,
Φ net = Φ base + 3Φ side = 0 N m 2 / C ⇒ −3.46 N m 2 /C + 3Φ side = 0 N m 2 / C
⇒ Φ side = 1.15 N m 2 /C
28.32. Solve: (a) When centered at the origin the sphere encloses both q1 and q2. For any closed surface that encloses a total charge Qin, the net electric flux through the closed surface is Φe =
Qin
ε0
=
( −4Q + 2Q ) = −2Q ε0
ε0
(b) When centered at x = 2a, the sphere encloses only q2 located at x = +a. The net electric flux through the closed surface is thus
Φe =
Qin
ε0
=
2Q
ε0
28.33. Solve: For any closed surface that encloses a total charge Qin, the net electric flux through the closed surface is Φ e = Qin ε 0 . The flux through the top surface of the cube is one-sixth of the total: Φ e surface =
Qin 10 ×10−9 C = = 188 N m 2 /C 6ε 0 6 ( 8.85 ×10−12 C2 /N m 2 )
28.34. Solve: For any closed surface that encloses a total charge Qin, the net electric flux through the closed surface is Φ e = Qin ε 0 . The total flux through the cube is Φ e = 6 (100 N m 2 /C ) =
Qin
ε0
⇒ Qin = ( 600 N m 2 /C )( 8.85 × 10−12 C 2 /N m 2 ) = 5.31×10−9 C = 5.31 nC
28.35. Solve: (a) The electric field is G 2 E = ( 5000 r 2 ) rˆ N/C = 5000 ( 0.20 ) rˆ N/C = 200rˆ N/C So the electric field strength is 200 N/C. 2 (b) The area of the surface is Asphere = 4π ( 0.20 m ) = 0.5027 m 2 . Thus, the electric flux is
G G Φ e = AE ⋅ dA = EAsphere = ( 200 N/C ) ( 0.5027 m 2 ) = 100.5 N m 2 /C ≈ 101 N m 2 /C (c) Because Φ e = Qin ε 0 , Qin = ε0Φe = (8.85 × 10−12 C2/N m2)(100.5 N m2/C) = 8.9 × 10−10 C
28.36. Solve: (a) The electric field is G ⎛ 200 ⎞ E =⎜ ⎟ rˆ N/C = 2000rˆ N/C ⎝ 0.10 ⎠ So the electric field strength is 2000 N/C. 2 (b) The area of the spherical surface is Asphere = 4π ( 0.10 m ) = 0.1257 m 2 . Hence, the flux is G G Φ e = AE ⋅ dA = EAsphere = ( 2000 N/C ) ( 0.1257 m 2 ) = 250 N m 2 /C (c) Because Φ e = Qin ε 0 , Qin = ε0Φe = (8.85 × 10−12 C2/N m2)(250 N m2/C) = 2.2 × 10−9 C = 2.2 nC
28.37. Model: The excess charge on a conductor resides on the outer surface. Visualize:
Solve:
(a) Consider a Gaussian surface surrounding the cavity just inside the conductor. The electric field →
inside a conductor in electrostatic equilibrium is zero, so E is zero at all points on the Gaussian surface. Thus Φe = 0. Gauss’s law tells us that Φe = Qin/ε0, so the net charge enclosed by this Gaussian surface is Qin = Qpoint + Qwall = 0.
We know that Qpoint = +100 nC, so Qwall = –100 nC. The positive charge in the cavity attracts an equal negative charge to the inside surface. (b) The conductor started out neutral. If there is –100 nC on the wall of the cavity, then the exterior surface of the conductor was initially +100 nC. Transferring –50 nC to the conductor reduces the exterior surface charge by 50 nC, leaving it at +50 nC. Assess: The electric field inside the conductor stays zero.
28.38. Model: The ball is uniformly charged. The charge distribution is spherically symmetric. Visualize:
The figure shows spherical Gaussian surfaces with radii r = 5 cm, 10 cm, and 20 cm. These surfaces match the G symmetry of the charge distribution. So, E is perpendicular to the Gaussian surface and the value of the field strength is the same at all points on the surface. Solve: (a) Because the ball is uniformly charged, its charge density is
ρ=
3 ( 80 ×10−9 C ) Q Q =4 3= = 2.387 ×10−6 C/m3 ≈ 2.4 ×10−6 C/m3 3 V 3πr 4π ( 0.20 m )
(b) Once again, because the ball is uniformly charged,
⎛ 4π 3 ⎞ ⎛ 4π 3 ⎞ Q = ρV = ρ ⎜ r ⎟ = ( 2.387 ×10−6 C/m3 ) ⎜ r ⎟ = (1.00 ×10−5 C/m3 ) r 3 3 3 ⎝ ⎠ ⎝ ⎠ When r = 5 cm, Q5 = (1.00 ×10−5 C/m3 ) ( 0.050 m ) = 1.25 nC. When r = 10 cm and 20 cm, Q10 = 10.0 nC and 3
Q20 = 80 nC.
G G (c) Gauss’s law is Φ e = AE ⋅ dA = EA = Qin ε 0 . For the 5.0-cm-radius Gaussian spherical surface,
Q5
ε0
= E5 A5 ⇒ E5 =
Q5 1.25 × 10−9 C = = 4.5 × 103 N/C ε 0 A5 (8.85 ×10−12 C 2 /Nm 2 ) ⎡ 4π ( 0.050 m )2 ⎤ ⎣ ⎦
Similarly, we can apply Gauss’s law to the 10 cm-radius and 20 cm-radius spherical surfaces and obtain E10 = 9.0 ×103 N/C and E20 = 1.80 × 103 N/C. Assess: We note that E20 = 2E10 = 4E5. This result is consistent with the result of Example 28.4 according to which
Einside =
1
Q r 4πε 0 R3
28.39. Model: The excess charge on a conductor resides on the outer surface. The field inside, outside, and within the hollow metal sphere has spherical symmetry. Visualize:
The figure shows spherical Gaussian surfaces with radii r ≤ a, a < r < b, and r ≥ b. These surfaces match the symmetry of the charge distribution. Solve: (a) For r ≤ a, Gauss’s law is G G Q +Q Φ e = AE ⋅ dA = in =
ε0
ε0
Notice that the electric field is everywhere perpendicular to the spherical surface. Because of the spherical symmetry of the charge, the electric field magnitude E is the same at all points on the Gaussian surface. Thus,
Φ e = EAsphere = E ( 4π r 2 ) =
Q
ε0
⇒ E=
G 1 Q ⇒E= rˆ 4πε 0 r 4πε 0 r 2 Q
2
where we made use of the fact that E is directed radially outward. The field depends only on the enclosed charge, not on the charge on the outer sphere. For a < r < b, Gauss’s law is G G Q Φ e = AE ⋅ dA = EAsphere = E ( 4π r 2 ) = in
ε0
Here Qin = 0 C. It is not +Q, because the charge in the cavity polarizes the metal sphere in such a way that E = 0 in the metal. Thus a charge −Q moves to the inner surface. Because the hollow sphere has a net charge of +2Q, the exterior surface now has a charge of +3Q. Thus, the electric field E = 0 N/C. For r ≥ b, Qin = Qexterior + Qinterior + Qcavity = +3Q + (−Q) + (+Q) = +3Q Gauss’s law applied to the Gaussian surface at r ≥ b yields: G G G Q +3Q 1 ⎛ 3Q ⎞ 1 ⎛ 3Q ⎞ Φ e = AE ⋅ dA = EAsphere = E ( 4π r 2 ) = in = ⇒ E= ⎜ 2 ⎟ ⇒E= ⎜ ⎟ rˆ 4πε 0 ⎝ r ⎠ 4πε 0 ⎝ r 2 ⎠ ε0 ε0 (b) As determined in part (a), the inside surface of the hollow sphere has a charge of −Q, and the exterior surface of the hollow sphere has a charge of +3Q. Assess: The hollow sphere still has the same charge +2Q as given in the problem, although the sphere is polarized.
28.40. Model: The excess charge on a conductor resides on the outer surface. The charge distribution on the two spheres is assumed to have spherical symmetry. Visualize: Please refer to Figure P28.40. The Gaussian surfaces with radii r = 8 cm, 10 cm, and 17 cm match G the symmetry of the charge distribution. So, E is perpendicular to these Gaussian surfaces and the field strength has the same value at all points on the Gaussian surface. G G Solve: (a) Gauss’s law is Φ e = AE ⋅ dA = Qin ε 0 . Applying it to a Gaussian surface of radius 8 cm,
Qin = −ε 0 EAsphere = − ( 8.85 ×10−12 C2 /N m 2 ) (15,000 N/C ) ⎡ 4π ( 0.08 m ) ⎤ = −1.07 × 10−8 C ⎣ ⎦ Because the excess charge on a conductor resides on its outer surface and because we have a solid metal sphere inside our Gaussian surface, Qin is the charge that is located on the exterior surface of the inner sphere. (b) In electrostatics, the electric field within a conductor is zero. Applying Gauss’s law to a Gaussian surface just inside the inside surface of the hollow sphere at r = 10 cm, G G Q Φ e = AE ⋅ dA = in ⇒ Qin = 0 C 2
ε0
That is, there is no net charge. Because the inner sphere has a charge of −1.07 × 10−8 C, the inside surface of the hollow sphere must have a charge of +1.07 × 10−8 C. (c) Applying Gauss’s law to a Gaussian surface at r = 17 cm, G G 2 Qin = ε 0 AE ⋅ dA = ε 0 EAsphere = ( 8.85 ×10−12 C 2 /Nm 2 ) (15,000 N/C ) 4π ( 0.17 m ) = 4.82 ×10−8 C This value includes the charge on the inner sphere, the charge on the inside surface of the hollow sphere, and the charge on the exterior surface of the hollow sphere due to polarization. Thus, Qexterior hollow + (1.07 × 10−8 C ) + ( −1.07 ×10−8 C ) = 4.82 × 10−8 C ⇒ Qexterior hollow = 4.82 ×10−8 C
28.41. Model: The charge distribution at the surface of the earth is assumed to be uniform and to have spherical symmetry. Visualize:
G Due to the symmetry of the charge distribution, E is perpendicular to the Gaussian surface and the field strength has the same value at all points on the surface. G G Solve: Gauss’s law is Φ e = AE ⋅ dA = Qin ε 0 . The electric field points inward (negative flux), hence Qin = −ε 0 EAsphere = − ( 8.85 × 10−12 C 2 /N m 2 ) (100 N/C ) 4π ( 6.37 ×106 m ) = −4.51× 105 C 2
G
28.42. Model: The electric field inside the box is 0. Visualize: Please refer to Figure 28.32. Solve: The charge on the capacitor plates produces a uniform electric field pointing to the right. When the metal G G box is added, the field inside the box is Enet = 0 because there is no charge enclosed within the box. The principle G G G G of superposition tells us that Enet = Ecapacitor + Ebox , where Ebox is the electric field due to the charge on the
exterior surface of the box. The net surface charge is zero, because the box as a whole is neutral, but the box is polarized by the capacitor’s electric field to where the left end of the box is negative and the right end is positive. G G G Since Enet = 0, we’re led to conclude that the field due to the charge on the surface of the box is Ebox = – G Ecapacitor = a uniform field pointing to the left. Note that this is the electric field inside the box due to the surface
charge on the box. The field outside the box is much more complex. If we freeze the surface charges and remove the G G box from the capacitor, we’re left with Enet = Ebox = a uniform field pointing to the left. This is shown in the picture below.
28.43. Model: The hollow metal sphere is charged such that the charge distribution is spherically symmetric. Visualize:
The figure shows spherical Gaussian surfaces at r = 4 cm, at r = 8 cm, and at r = 12 cm. These surfaces match the G symmetry of the spherical charge distribution. So E is perpendicular to the Gaussian surface and the field strength has the same value at all points on the Gaussian surface. Solve: The charge on the inside surface is
Qinside = (−100 nC/m2) 4π (0.06 m)2 = −4.524 nC This charge is caused by polarization. That is, the inside surface can be charged only if there is a charge of +4.524 nC at the center which polarizes the metal sphere. Applying Gauss’s law to the 4.0-cm-radius Gaussian surface, which encloses the +4.524 nC charge,
G G Q Φ e = AE ⋅ dA = EAsphere = in
ε0
⇒E=
−9 9 2 2 Qin 4.524 ×10−9 C 4.524 ×10 C ( 9.0 × 10 N m /C ) = = = 2.54 × 104 N/C 2 Aε 0 4πε 0 ( 0.04 m )2 ( 0.04 m )
G At r = 4 cm, E = ( 2.54 ×104 N/C, outward ).
G G There is no electric field inside a conductor in electrostatic equilibrium. So at r = 8 cm, E = 0. Applying Gauss’s law to a 12.0-cm-radius sphere,
G G Q Φ e = AE ⋅ dA = EAsphere = in
ε0
The charge on the outside surface is
Qoutside = (+100 nC/m2)4π (0.10 m)2 = 1.257 × 10−8 C ⇒ Qin = Qoutside + Qinside + Qcenter = 1.257 × 10−8 C − 0.452 ×10−8 C + 0.452 × 10−8 C = 1.257 × 10−8 C ⇒E=
At r = 12 cm,
1.257 × 10−8 C Qin = = 7.86 × 103 N/C ε 0 A (8.85 ×10−12 C 2 /Nm 2 ) 4π ( 0.12 m )2
G E = ( 7.86 ×103 N/C, outward ) .
28.44. Model: The charged hollow spherical shell is assumed to have a spherically symmetric charge distribution. Visualize:
The figure shows two spherical Gaussian surfaces at r < R and r > R. These surfaces match the symmetry of the G spherical charge distribution. So, E is perpendicular to the Gaussian surface and the field strength has the same value at all points on the surface. Solve: (a) Gauss’s law applied to the Gaussian surface inside the sphere (r < R) is
G G Q +q 1 +q Φ e = AE ⋅ dA = EAsphere = in ⇒ E ( 4π r 2 ) = ⇒ E= 4πε 0 r 2 ε0 ε0
G ⎛ 1 +q ⎞ 1 q rˆ , outward ⎟ = ⇒E =⎜ 2 2 πε πε r 4 4 0 0 r ⎝ ⎠ (b) Gauss’s law applied to the Gaussian surface outside the sphere (r > R) is
G G Q −2q + q − q 1 +q ⇒ E= Φ e = AE ⋅ dA = − EAsphere = in = = 4πε 0 r 2 ε0 ε0 ε0 G ⎛ 1 q ⎞ 1 q ⇒E =⎜ rˆ , inward ⎟ = − 2 2 πε πε r 4 4 0 0 r ⎝ ⎠ Assess: A uniform spherical shell of charge has the same electric field at r > R as a point charge placed at the center of the shell. Additionally, the electric field for r < R is zero, meaning that the shell of charge exerts no electric force on a charged particle inside the shell.
28.45. Model: The hollow plastic ball has a charge uniformly distributed on its outer surface. This distribution leads to a spherically symmetric electric field. Visualize:
The figure shows Gaussian surfaces at r < R and r > R. Solve: (a) Gauss’s law for the Gaussian surface for r < R where Qin = 0 is G G Q Φ e = AE ⋅ dA = in = 0 N m2/C ⇒ E = 0 N/C
ε0
(b) Gauss’s law for the Gaussian surface for r > R is G G Q Q Q Φ e = AE ⋅ dA = EAsphere = in = ⇒ EAsphere = ⇒E=
ε0
ε0
ε0
Q
ε 0 Asphere
=
1 Q 4πε 0 r 2
Assess: A uniform spherical shell of charge has the same electric field at r > R as a point charge placed at the center of the sphere. Additionally, the shell of charge exerts no electric force on a charged particle inside the shell.
28.46. Model: The charge distributions of the ball and the metal shell are assumed to have spherical symmetry. Visualize:
The spherical symmetry of the charge distribution tells us that the electric field points radially inward or outward. We will therefore choose Gaussian surfaces to match the spherical symmetry of the charge distribution and the field. The figure shows four Gaussian surfaces in the four regions: r ≤ a, a < r < b, b ≤ r ≤ c and r > c. G G Solve: (a) Gauss’s law is Φ e = AE ⋅ dA = Qin ε 0 . Applying it to the region r ≤ a, where the charge is negative G so E points inward, we get
− EAsphere = − E ( 4π r 2 ) = Here ρ = −Q
4π 3
+ ρ ( 43π r 3 )
ε0
⇒E=
−ρr 3ε 0
a 3 is the charge density (C/m3). Thus
G ⎛ 1 Qr ⎞ 1 Qr E =⎜ rˆ , inward ⎟ = − 3 4πε 0 a 3 ⎝ 4πε 0 a ⎠ Applying Gauss’s law to the region a < r < b, G −Q 1 Q 1 Q − EAsphere = ⇒E= ⇒E=− rˆ 2 ε0 4πε 0 r 4πε 0 r 2 G G E = 0 in the region b ≤ r ≤ c because this is a conductor in electrostatic equilibrium. To apply Gauss’s law to the region r > c, we use Qin = −Q + 2Q = +Q. Thus,
EAsphere = (b)
Q
ε0
⇒E=
G 1 Q 1 Q ⇒E= rˆ 2 4πε 0 r 4πε 0 r 2
28.47. Model: The three planes of charge are infinite planes. Visualize:
From planar symmetry the electric field can point straight toward or away from the plane. The three planes are labeled as P (top), P′, and P″(bottom). Solve: From Example 28.6, the electric field of an infinite charged plane of charge density η is
Eplane =
η 2ε 0
⇒ EP = EP′′ =
η 4ε 0
=
EP′ 2
In region 1 the three electric fields are G G η ˆ η ˆ EP = − j EP ′ = j 4ε 0 2ε 0 G G Adding the three contributions, we get Enet = 0 N/C. In region 2 the three electric fields are
G Thus, Enet = (η 2ε 0 ) ˆj.
G η ˆ EP = j 4ε 0
G η ˆ EP ′ = j 2ε 0
G η ˆ EP′′ = − j 4ε 0
G −η ˆ EP′′ = j 4ε 0
In region 3,
G Thus, Enet = − (η 2ε 0 ) ˆj.
G η ˆ EP = j 4ε 0
G η ˆ EP ′ = − j 2ε 0
G η ˆ EP′′ = − j 4ε 0
In region 4,
G η ˆ EP = j 4ε 0 G G Thus Enet = 0 N/C.
G η ˆ EP ′ = − j 2ε 0
G η ˆ EP′′ = j 4ε 0
28.48. Model: The charge has planar symmetry, so the electric field must point toward or away from the slab. Furthermore, the field strength must be the same at equal distances on either side of the center of the slab. Visualize:
Choose Gaussian surfaces to be cylinders of length 2z centered on the z = 0 plane. The ends of the cylinders have area A. Solve: (a) For the Gaussian cylinder inside the slab, with z < z0, Gauss’s law is
G G G G G G G G Qin AE ⋅ dA = ∫ top E ⋅ dA + ∫ bottom E ⋅ dA + ∫ sides E ⋅ dA =
ε0
The field is parallel to the sides, so the third integral is zero. The field emerges from both ends, so the first two integrals are the same. The charge enclosed is the volume of the cylinder multiplied by the charge density, or Qin = ρV = ρ (2zA). Thus
Qin
ε0
=
G G G G ρ (2 zA) ρz E ⋅ dA + 0 = 2 EA ⇒ E = = ∫ E ⋅ dA + ∫ top bottom ε0 ε0
The field increases linearly with distance from the center. (b) The analysis is the same for the cylinder that extends outside the slab, with z > z0, except that the enclosed charge Q = ρ (2z0A) is that within a cylinder of length 2z0 rather than 2z. Thus Qin
ε0
=
G G G G ρ (2 z0 A) ρz E ⋅ dA + 0 = 2 EA ⇒ E = 0 = ∫ E ⋅ dA + ∫ top bottom ε0 ε0
The field strength outside the slab is constant, and it matches the result of part (a) at the boundary. (c)
28.49. Model: The infinitely wide plane of charge with surface charge density η polarizes the infinitely wide conductor. Visualize:
G G Because E = 0 in the metal there will be an induced charge polarization. The face of the conductor adjacent to the plane of charge is negatively charged. This makes the other face of the conductor positively charged. We thus have three infinite planes of charge. These are P (top conducting face), P′ (bottom conducting face), and P″(plane of charge). Solve: Let η1, η2, and η3 be the surface charge densities of the three surfaces with η2 a negative number. The electric field due to a plane of charge with surface charge density η is E = η 2ε 0 . Because the electric field inside a conductor is zero (region 2), G G G G G η η η EP + EP′ + EP′′ = 0 N/C ⇒ − 1 ˆj + 2 ˆj + 3 ˆj = 0 N/C ⇒ −η1 + η2 + η = 0 C/m 2 2ε 0 2ε 0 2ε 0
We have made the substitution η3 = η. Also note that the field inside the conductor is downward from planes P and P′ and upward from P″. Because η1 + η2 = 0 C/m2, because the conductor is neutral, η2 = −η1. The above equation becomes
−η1 − η1 + η = 0 C/m 2 ⇒ η1 = 12 η ⇒ η2 = − 12 η We are now in a position to find electric field in regions 1–4. For region 1, G G η ˆ η ˆ EP = j EP′ = − j 4ε 0 4ε 0 G G G G The electric field is Enet = EP + EP′ + EP′′ = (η 2ε 0 ) ˆj. G G In region 2, Enet = 0 N/C. In region 3, G G η ˆ η ˆ EP = − j EP ′ = j 4ε 0 4ε 0 G The electric field is Enet = (η 2ε 0 ) ˆj. In region 4, G The electric field is Enet
G η ˆ EP = − j 4ε 0 = − (η 2ε 0 ) ˆj.
G η ˆ EP ′ = j 4ε 0
G η ˆ EP′′ = j 2ε 0
G η ˆ EP′′ = j 2ε 0
G η ˆ EP′′ = − j 2ε 0
28.50. Model: Assume the metal slabs are large enough to model as infinite planes. Then the electric field has planar symmetry, pointing either toward or away from the planes. Visualize:
One Gaussian surface is a cylinder with end-areas a extending past the two metal slabs. A second Gaussian surface ends in region 3, the space between the slabs. G G G Solve: (a) Because these are metals, we immediately know that E2 = E4 = 0. To find the fields outside the slabs, consider the Gaussian surface on the left. The electric field everywhere points up or down, so there is no flux through the sides of this cylinder. Because both slabs are positive, the electric fields in regions 1 and 5 point outward. Gauss’s law is G G G G G G G G Qin AE ⋅ dA = ∫ top E ⋅ dA + ∫ bottom E ⋅ dA + ∫ sides E ⋅ dA = E1a + E5a + 0 =
ε0
The Gaussian surface encloses both the upper and lower surfaces of the top slab. The total charge per unit area on both surfaces of this slab is Q1/A = Q/A, so the charge enclosed within the cylinder is Qa/A. (For this calculation we don’t need to know how the charge is distributed between the two surfaces.) Similarly, the enclosed charge on the lower slab is Q2a/A = 2Qa/A. Thus E1a + E5 a =
Qa 2Qa 3Q + ⇒ E1 + E5 = ε0 A ε0 A ε0 A
G G Fields E1 and E5 are both a superposition of the fields of four sheets of surface charge. Because the field of a plane of charge is independent of distance from the plane, the superposition at points above the top plane must be the same magnitude, but opposite direction, as the superposition at points below the bottom plane. Consequently, E1 = E5 (same field strengths). This is a rather subtle point in the reasoning and one worth thinking about. Thus E1 = E5 = 3Q/2ε0A. Now consider the Gaussian surface on the right. The lower slab is more positive than the upper slab, so the electric field in region 3 must point upward, into the lower face of this cylinder. Thus G G G G G G G G Qin AE ⋅ dA = ∫ top E ⋅ dA + ∫ bottom E ⋅ dA + ∫ sides E ⋅ dA = E1a − E3a + 0 =
ε0
where the minus sign with E3 is because of the direction. We know E1, and we’ve already determined that Qin = Qa/A. Thus E3 = E1 −
Q 3Q Q Q = − = ε 0 A 2ε 0 A ε 0 A 2ε 0 A
Summarizing, E1 = 3Q/2ε0A, E2 = 0, E3 = Q/2ε0A, E4 = 0, and E5 = 3Q/2ε0A. (b) The electric field at the surface of a conductor is E = η/ε0. We can use the known fields and η = ε0E to find G the four surface charge densities. At surface a, E1 points away from the surface. Thus
ηa = +ε 0 E1 = +ε 0 G At surface b, E3 points toward the surface. Thus
3Q 3Q =+ 2ε 0 A 2A
ηb = −ε 0 E3 = −ε 0
Q 1Q =− 2ε 0 A 2A
Surface c is opposite to surface b, because the field points away from the surface, so ηc = +Q/2A. Finally, at G surface d, the field points away from the surface and has the same strength as E1 , hence ηd = +3Q/2A. Assess: Notice that ηa + ηb = Q/A and ηc + ηd = 2Q/A. This is the expected “net” surface charge density for slabs with total charge Q and 2Q.
28.51. Model: A long, charged wire can be modeled as an infinitely long line of charge. Visualize:
The figure shows an infinitely long line of charge that is surrounded by a hollow metal cylinder of radius R. The symmetry of the situation indicates that the only possible shape of the electric field is to point straight in or out from the wire. The shape of the field suggests that we choose our Gaussian surface to be a cylinder of radius r and length L, centered on the wire. Solve: (a) For the region r < R, Gauss’s law is
G G Q Φ e = AE ⋅ dA = in ⇒
ε0
∫
top
G G E ⋅ dA + ∫
bottom
G G E ⋅ dA + ∫
side
G G λL E ⋅ dA =
ε0
G G λL λL ⇒ 0 N m 2 / C + 0 N m 2 / C + E ⋅ Aside = ⇒ E ( 2π r ) L =
ε0
⇒E=
ε0
G ⎛ λ 1 ⎞ λ 1 λ rˆ ⇒E =⎜ , outward ⎟ = 2πε 0 r ⎝ 2πε 0 r ⎠ 2πε 0 r
(b) Applying Gauss’s law to the Gaussian surface at r > R,
G G G G G G G G Qin AE ⋅ dA = ∫ top E ⋅ dA + ∫ bottom E ⋅ dA + ∫ side E ⋅ dA =
ε0
G G Q ⇒ 0 N m 2 /C + 0 N m 2 /C + E ⋅ Awall = in
ε0
⇒ E ( 2π rL ) =
Qline + Qcylinder
ε0
=
G λ L + 2λ L 3λ 1 3λ rˆ ⇒E= ⇒E= ε0 2πε 0 r 2πε 0 r
28.52. Model: A long, charged cylinder is assumed to be infinite and to have linear symmetry. Visualize:
The cylindrical symmetry of the situation indicates that the only possible shape of the electric field is to point straight in or out from the cylinder. The shape of the field suggests that we choose our Gaussian surface to be a cylinder of radius r and length L, which is concentric with the charged cylinder. Solve: (a) For r ≥ R, Gauss’s law is G G G G G G G G Qin AE ⋅ dA = ∫ left E ⋅ dA + ∫ right E ⋅ dA + ∫ wall E ⋅ dA =
ε0
⇒ 0 N m 2 / C + 0 N m 2 / C + EA =
Qin
ε0
⇒E=
(b) For r ≤ R, Gauss’s law is
EA =
Qin
ε0
⇒ E=
G ⎛ λ 1⎞ λL λ = ⇒E =⎜ ⎟ rˆ ( 2π rL ) ε 0 2πε 0r ⎝ 2πε 0 r ⎠
ρ (π r 2 L ) Qin = Aε 0 ( 2π rL ) ε 0
An expression for the volume charge density ρ in terms of the linear charge density can be calculated by considering the charge on a cylinder of length d and radius R:
λ d = ρ (π R 2 d ) ⇒ ρ =
λ π R2
G 1 λr 1 λr ⎛ λ ⎞ πr L rˆ ⇒E =⎜ = ⇒E= 2 ⎟ 2 2πε 0 R 2 ⎝ π R ⎠ ( 2π rL ) ε 0 2πε 0 R 2
(c) At r = R, 1 λR 1 λ Er = R = rˆ = rˆ 2πε 0 R 2 2πε 0 R
28.53. Model: The charge distribution in the shell has spherical symmetry. Visualize:
The spherical surfaces of radii r ≥ Rout, r ≤ Rin, and Rin ≤ r ≤ Rout, concentric with the spherical shell, are Gaussian surfaces. Solve: (a) Gauss’s law for the Gaussian surface r ≥ Rout is
G G Qin G 1 Q 1 Q Q ⇒ E ( 4π r 2 ) = ⇒ E = ⇒E= rˆ AE ⋅ dA = 2 ε0 ε0 4πε 0 r 4πε 0 r 2 The vector form comes from the fact that the field is directed radially outward. (b) For r ≤ Rin, Gauss’s law is
G G Qin 0 C G G = ⇒ E =0 AE ⋅ dA =
ε0
ε0
(c) For Rin ≤ r ≤ Rout, Gauss’s law is
G G Qin Q ⇒ E ( 4π r 2 ) = in AE ⋅ dA =
ε0
Q in =
4π 3 ( r − Rin3 ) ρ = 43π 3
ε0
( r − R )Q = Q ⎛ r ( R − R ) ⎜⎝ R 3
4π 3
3 in
3 out
3 in
3
− Rin3 ⎞ ⎟ − Rin3 ⎠
3 out
G 1 Q r 3 − Rin3 1 Q ⎛ r 3 − Rin3 ⎞ E ⇒ = ⎜ 3 ⎟ rˆ 3 − Rin3 − Rin3 ⎠ 4π r 2 ε 0 Rout 4πε 0 r 2 ⎝ Rout G G (d) The result obtained in part (c) for the electric field simplifies to E = 0, when r = Rin which is the result obtained in part (b). Furthermore, at r = Rout, the electric field obtained in part (c) becomes ⇒E=
G 1 Q E= rˆ 2 4πε 0 Rout
which is the same as the electric field obtained in part (a). (e)
28.54. Model: Assume that the negative charge uniformly distributed in the atom has spherical symmetry. Visualize:
The nucleus is a positive point charge +Ze at the center of a sphere of radius R. The spherical symmetry of the charge distribution tells us that the electric field must be radial. We choose a spherical Gaussian surface to match the spherical symmetry of the charge distribution and the field. The Gaussian surface is at r < R, which means that we will calculate the amount of charge contained in this surface. G G Solve: (a) Gauss’s law is AE ⋅ dA = Qin ε 0 . The amount of charge inside is
⎡ r3 ⎤ ( − Ze ) r3 ⎛ 4π 3 ⎞ Qin = ρ ⎜ r ⎟ + Ze = 4π 3 ( 43π r 3 ) + Ze = − ( Ze ) 3 + Ze = Ze ⎢1 − 3 ⎥ R ⎝ 3 ⎠ (3R) ⎣ R ⎦
⇒ Ein ( 4π r 2 ) =
Ze ⎡ 1 r ⎤ Ze ⎡ r 3 ⎤ ⎢1 − 3 ⎥ ⇒ Ein = 4πε ⎢ r 2 − R 3 ⎥ ε0 ⎣ R ⎦ ⎦ 0 ⎣
(b) At the surface of the atom, r = R. Thus,
Ein =
Ze ⎡ 1 R⎤ − = 0 N/C 4πε 0 ⎢⎣ R 2 R 3 ⎥⎦
This is an expected result, which can be quickly obtained from Gauss’s law. Applying Gauss’s law to a Gaussian surface just outside r = R. Because the atom is electrically neutral, Qin = 0. Thus G G Qin = 0 ⇒ E = 0 N/C AE ⋅ dA =
ε0
(c) The electric field strength at r = 12 R = 0.050 nm is
⎡ 1 ( 0.050 nm ) ⎤ = 4.6 ×1013 N/C − Ein = 92 (1.60 ×10−19 C )( 9.0 ×109 C 2 /Nm 2 ) ⎢ 2 3 ⎥ ⎢⎣ ( 0.050 nm ) ( 0.10 nm ) ⎥⎦
28.55. Model: The long thin wire is assumed to be an infinite line of charge. Visualize: Please refer to Figure CP28.55. The cube of edge length L is centered on the line charge with a linear charge density λ. Although the line charge has cylindrical symmetry, we will take the cube as our Gaussian surface. G G G G Solve: (a) The electric flux through an area dA in the yz plane is d Φ = E ⋅ dA. The electric field E due to an infinite line of charge at a distance s from the line charge is
G λ 1 λ E= rˆ = 2πε 0 r 2π 0
1
y 2 + ( L 2)
2
rˆ
G Also dA = L dyiˆ. Thus,
dΦ = =
λ 2πε 0
1 y + ( L 2) 2
2
λ L dy L dy rˆ ⋅ iˆ = 2πε 0
( )
λ L dy
L2
2πε 0 y 2 + ( L 2 )
2
y2 + ( L 2)
2
=
1 y + ( L 2) 2
2
cosθ
λ L2 dy 2 4πε 0 ⎡ y 2 + ( L 2 ) ⎤ ⎣ ⎦
(b) The expression for the flux dΦ can now be integrated to obtain the total flux through this face as follows: Φ = ∫ dΦ = =
L2
∫
−L 2
λ L2 dy 2 4πε 0 ⎡ y 2 + ( L 2 ) ⎤ ⎣ ⎦
L2
=
λ L2 ⎡ 2 −1 ⎛ 2 y ⎞ ⎤ tan ⎜ ⎟ ⎥ 4πε 0 ⎢⎣ L ⎝ L ⎠⎦ −L 2
λL λ L ⎡π ⎛ π ⎞⎤ λ L ⎡⎣ tan −1 (1) − tan −1 ( −1) ⎤⎦ = −⎜− ⎟ = 2πε 0 2πε 0 ⎢⎣ 4 ⎝ 4 ⎠ ⎥⎦ 4ε 0
(c) Because there are four faces through which the flux flows, the net flux through the cube is Φ e = 4Φ =
λ L ( Qin L ) L Qin = = ε0 ε0 ε0
28.56.
Model: Visualize:
Solve:
The field has cylindrical symmetry.
(a) The volume charge density ρ ( r ) = r ρ0 R is linearly proportional to r. The graph is shown above.
(b) Consider the cylindrical shell of length L, radius r, and thickness dr shown in the diagram. The charge within a small volume dV is
dq = ρ dV =
ρ0r R
( 2π r ) drL =
2πρ 0 L 2 r dr R
Integrating this expression to obtain the total charge in the cylinder: R
R 2πρ 0 L 2 2πρ 0 L ⎡ r 3 ⎤ 2πρ 0 LR 3 3λ = λ L ⇒ ρ0 = Q = ∫ dq = r dr = ⎢ ⎥ = ∫ 3R 2π R 2 R 0 R ⎣ 3 ⎦0
(c) Consider the cylindrical Gaussian surface of length l at r < R shown in the figure. Gauss’s law is G G Φ e = AE ⋅ dA = Qin / ε 0 . The charge on the inside of the Gaussian surface is
⎛ r3 ⎞ 2πρ 0l 2 2πρ0lr 3 2π ⎛ 3λ ⎞ 3 = r dr = lr = λl ⎜ 3 ⎟ ⎜ 2 ⎟ 3R 3R ⎝ 2π R ⎠ R ⎝R ⎠ 0 r
Qin = ∫ dq = ∫ ρ dV = ∫
G G λl ⎛ r 3 ⎞ λ ⎛ r2 ⎞ Q ⇒ AE ⋅ dA = E ( 2π rl ) = in = ⎜ 3 ⎟ ⇒ E = ⎜ ⎟ 2πε 0 ⎝ R 3 ⎠ ε0 ε0 ⎝ R ⎠ The last step is justified because the electric field has radial dependence. (d) The expression for the electric field strength simplifies at r = R to E=
λ R2 λ = 2πε 0 R 3 2πε 0 R
This is the same result as obtained in example 28.5 for a long, charged wire.
28.57.
Visualize:
Solve: (a) Consider the spherical shell of radius r and thickness dr shown in the figure. The charge dq within a small volume dV is C dq = ρ dV = 2 ( 4π r 2 ) dr = 4π Cdr r Integrating this expression to obtain the total charge in the sphere: R
Q = ∫ dq = ∫ 4π Cdr = 4π CR ⇒ C = 0
Q 4π R
(b) Consider the spherical Gaussian surface at r < R shown in the figure. Gauss’s law applied to this surface is G G Q Qin Q Φ e = AE ⋅ dA = in ⇒ E ( 4π r 2 ) = in ⇒ E = ε0 ε0 4πε 0 r 2
Using the results from part (a), r
r
C Q ⎛ Q ⎞ 4π r 2 dr = ∫ ⎜ ⎟ 4π dr = r 2 r R R 4 π ⎝ ⎠ 0 0
Qin = ∫ dq = ∫ ρ dV = ∫
G 1 Q ⎛Q ⎞ 1 rˆ ⇒ E = ⎜ r⎟ ⇒E= 2 4πε 0 Rr ⎝ R ⎠ 4πε 0 r (c) At r = R, the equation for the electric field obtained in part (b) simplifies to G 1 Q E= rˆ 4πε 0 R 2
This is the same result as obtained in Example 28.3. The result was expected because a spherical charge behaves, for r ≥ R, as if the entire charge were at the center.
28.58. Model: The charge density within the sphere is nonuniform. We will assume that the distribution has spherical symmetry. Visualize:
Solve: (a) Consider the thin shell of width dr at a distance r from the center shown in the figure. The volume of this thin shell is dV = ( 4π r 2 ) dr. The charge contained in this volume is R R R⎛ r⎞ r3 ⎞ ⎛ Q = ∫ dq = ∫ 4π r 2 ρ dr = ∫ 4π r 2 ρ0 ⎜1 − ⎟ dr = 4πρ 0 ∫ ⎜ r 2 − ⎟ dr 0 R⎠ ⎝ R⎠ ⎝ 0 0 R
⎡ r3 r 4 ⎤ 3Q R3 = 4πρ 0 ⎢ − ⇒ ρ0 = ⎥ = 4πρ0 π R3 12 ⎣ 3 4R ⎦0 (b) Gauss’s law for the Gaussian surface shown in the figure at r < R is G G Qin Q ⇒ E ( 4π r 2 ) = in AE ⋅ dA =
ε0
ε0
The charge inside the Gaussian surface is r r r ⎛ ⎡ r3 r 4 ⎤ r⎞ r3 ⎞ ⎛ Qin = ∫ dq = ∫ 4π r 2 ρ dr = ∫ 4π r 2 ρ0 ⎜1 − ⎟ dr = 4πρ 0 ∫ ⎜ r 2 − ⎟ dr = 4πρ 0 ⎢ − ⎥ R⎠ ⎝ R⎠ ⎣ 3 4R ⎦ 0 0 0⎝
Gauss’s law becomes E ( 4π r 2 ) =
4πρ0 ⎡ r 3 r 4 ⎤ ρ0 ⎡ r r 2 ⎤ ⎛ 3Q ⎞ 1 = − = − E ⇒ ε 0 ⎢⎣ 3 4 R ⎥⎦ ⎜⎝ π R3 ⎟⎠ ε 0 ε 0 ⎢⎣ 3 4 R ⎥⎦
r ⎡ 3r ⎤ Qr ⎛ 3r ⎞ = 1− 4− ⎟ 3⎜ ⎢ ⎥ 3 ⎣ 4 R ⎦ 4πε 0 R ⎝ R⎠
The direction of the field is radially outward. (c) At r = R, the above expression for the electric field reduces to E=
QR ⎛ 3R ⎞ 1 Q ⎜4− ⎟= 4πε 0 R 3 ⎝ R ⎠ 4πε 0 R 2
This is an expected result, because all charge Q is inside R and the field looks like that of a point charge.
28.59.
Visualize:
Solve:
(a) It is clear that E ( r ) ∝ r 4 up to r = R. That is, the maximum value of E(r) occurs at r = R. At this point,
E=
1
Q 4πε 0 R 2
because all the charge is inside R and the charge is spherically distributed about a point at the center. Thus, E=
⎛ R4 ⎞ Q Q = Emax ⎜ 4 ⎟ ⇒ Emax = 2 2 4 πε 4πε 0 R R ⎝ ⎠ 0R 1
(b) Applying Gauss’s law to the surface at r < R shown in the figure,
G G Qin ⎛ Q ⎞ r4 ⎛ r6 ⎞ ⎛ r4 ⎞ Q ⇒ ⎜ Emax 4 ⎟ 4π r 2 = in ⇒ Qin = ⎜ ε 4 4π r 2 = Q ⎜ 6 ⎟ AE ⋅ dA = 2 ⎟ 0 ε0 R ⎠ ε0 ⎝R ⎠ ⎝ ⎝ 4πε 0 R ⎠ R To find Qin, we consider the thin spherical shell of width dr and charge dq at a distance r from the center shown in the figure. Thus,
dq = ρ dV = ρ ( 4π r 2 ) dr ⇒ ∫ dq = Qin = ∫ ρ ( r ) ( 4π r 2 ) dr Using this form of Qin in the equation obtained from Gauss’s law, Qin = Q
r6 = ρ ( r )4π r 2 dr ⇒ R6 ∫
Q r6
∫ ρ ( r ) r dr = 4π R 2
6
Taking the position derivative on both sides 3Q r 3 d Q d 6 Q 2 2 5 r r r dr r ρ = ⇒ ρ r r = 6 r ⇒ ρ = ( ) ( ) ( ) ( ) ( ) 2π R 6 dr ∫ 4π R 6 dr 4π R 6 (c) Consider a spherical shell of width dr and charge dq at a distance r from the center. Then
dq = ρ ( r ) dV =
R
3Qr 3 6Q 5 6Q R 6 r5 2 4 r dr π ⇒ dq = 6 Q dr ⇒ = = = ⋅ =Q Q dq r dr ∫ 2π R 6 R6 R 6 ∫0 R6 6
28-1
29.1. Model: The mechanical energy of the proton is conserved. A parallel-plate capacitor has a uniform electric field. Visualize:
The figure shows the before-and-after pictorial representation. The proton has an initial speed vi = 0 m/s and a final speed vf after traveling a distance d = 2.0 mm. Solve: The proton loses potential energy and gains kinetic energy as it moves toward the negative plate. The potential energy U is defined as U = U0 + qEx, where x is the distance from the negative plate and U0 is the potential energy at the negative plate (at x = 0 m). Thus, the change in the potential energy of the proton is ΔUp = Uf – Ui = (U0 + 0 J) – (U0 + qEd) = −qEd The change in the kinetic energy of the proton is
ΔK = Kf − Ki = 12 mvf2 − 12 mvi2 = 12 mvf2 The law of conservation of energy is ΔK + ΔUp = 0 J. This means 1 2
⇒ vf =
mvf2 + ( −qEd ) = 0 J
2 ( +1.60 ×10−19 C ) ( 50,000 N/C ) ( 2.0 ×10−3 m ) 2qEd = = 1.38 ×105 m/s m (1.67 ×10−27 kg )
Assess: As described in Section 29.1, the potential energy for a charge q in an electric field E is U = U0 + qEx, where x is the distance measured from the negative plate. Having U = U0 at the negative plate (with x = 0 m) is completely arbitrary. We could have taken it to be zero. Note that only ΔU, and not U, has physical consequences.
29.2. Model: The mechanical energy of the electron is conserved. A parallel-plate capacitor has a uniform electric field. Visualize:
The figure shows the before-and-after pictorial representation. The electron has an initial speed vi = 0 m/s and a final speed vf after traveling a distance d = 1.0 mm. Solve: The electron loses potential energy and gains kinetic energy as it moves toward the positive plate. The potential energy U is defined as U = U0 + qEx, where x is the distance from the negative plate and U0 is the potential energy at the negative plate (at x = 0 m). Thus, the change in the potential energy of the electron is ΔUe = Uf – Ui = (U0 + qEd) – (U0 + 0 J) = qEd The change in the kinetic energy of the electron is
ΔK = Kf − Ki = 12 mvf2 − 12 mvi2 = 12 mvf2 Now, the law of conservation of mechanical energy gives ΔK + ΔU = 0 J. This means 1 2
⇒ vf =
−2qEd = m
mvf2 + qEd = 0 J
( −2 ) ( −1.60 ×10−19 C ) ( 20,000 N/C ) (1.0 ×10−3 m ) 9.11×10−31 kg
= 2.7 × 106 m/s
Assess: Note that ΔUe = qEd is the change in the potential energy of the electron. It is negative because q = − e for the electron. Thus, the potential energy becomes more negative as d increases, that is, the potential energy of the electron decreases with an increase in d (or x).
29.3. Model: The mechanical energy of the proton is conserved. A parallel-plate capacitor has a uniform electric field. Visualize:
The figure shows the before-and-after pictorial representation. Solve: The proton loses potential energy and gains kinetic energy as it moves toward the negative plate. The potential energy is defined as U = U0 + qEx, where x is the distance from the negative plate and U0 is the potential energy at the negative plate (at x = 0 m). Thus, the change in the potential energy of the proton is
ΔU p = U f − U i = (U 0 + 0 J ) − (U 0 + qEd ) = −qEd The change in the kinetic energy of the proton is
ΔK = Kf − Ki = 12 mvf2 − 12 mvi2 = 12 mvf2 Applying the law of conservation of energy ΔK + ΔUp = 0 J, we have 1 2
mvf2 + ( − qEd ) = 0 J ⇒ vf2 =
2qEd m
When the amount of charge on each plate is doubled, then the final velocity of the proton is
vf′2 =
2qE′d m
Dividing these equations, vf′2 E ′ = ⇒ vf′ = vf2 E
E′ vf E
For a parallel-plate capacitor E = η ε 0 = Q Aε 0 . Therefore,
vf′ =
Q′ vf = 2 ( 50,000 m/s ) = 70,711 m/s Q
Assess: The proton’s velocity is expected to increase because an increased charge on the capacitor plates leads to a higher electric field between the plates and hence to an increased force on the proton.
29.4. Model: The mechanical energy of the charged particles is conserved. A parallel-plate capacitor has a uniform electric field. Visualize:
The figure shows the before-and-after pictorial representation. Solve: The potential energy is defined as U = U0 + qEx, where x is the distance from the negative plate and U0 is the potential energy at the negative plate (at x = 0 m). Thus, the change in the potential energy of the proton as it moves from the positive plate to the negative plate is ΔU p = U f − U i = (U 0 + 0 J ) − (U 0 + eEd ) = −eEd
This decrease in potential energy appears as an increase in the proton’s kinetic energy: ΔK = K f − K i = 12 mpvf2 p − 12 mpvi2p = 12 mpvf2 p
Applying the law of conservation of mechanical energy ΔK + ΔUp = 0 J, we have 1 2
mvf2p + ( − eEd ) = 0 J ⇒ vf2 p =
2eEd mp
When the proton is replaced with a helium ion and the same experiment is repeated,
vf2ion =
2eEd mion
Dividing the two equations,
vf ion =
mp mion
vf p =
1 4
( 50,000 m/s ) = 25,000 m/s
Assess: Being a heavier particle, the helium ion’s velocity is expected to be smaller compared to the proton’s velocity.
29.5. Model: The charges are point charges. Visualize: Please refer to Figure EX29.5. Solve: The electric potential energy of the electron is
U electron = U13 + U 23 ⎡ −19 −19 ⎢ (1.60 ×10 C )( −1.60 ×10 C ) = ( 9.0 ×109 N m 2 /C2 ) ⎢ + 2 2 −9 −9 ⎢⎣ ( 2.0 ×10 m ) + ( 0.50 ×10 m ) = −1.12 ×10 −19 J − 1.12 ×10−19 J = −2.24 ×10−19 J
(1.60 ×10 ( 2.0 ×10
⎤ C )( −1.60 × 10−19 C ) ⎥ 2 2 ⎥ m ) + ( 0.50 × 10−9 m ) ⎥ ⎦
−19
−9
29.6. Model: The charges are point charges. Visualize: Please refer to Figure EX29.6. Solve: For a system of point charges, the potential energy is the sum of the potential energies due to all pairs of charges:
U elec = ∑ i,j
Kqi q j rij
= U12 + U13 + U 23 = K
q1q2 qq qq +K 1 3 +K 2 3 r12 r13 r23
⎡ ( 2.0 nC )( −1.0 nC ) ( 2.0 nC )( −1.0 nC ) ( −1.0 nC )( −1.0 nC ) ⎤ = ( 9.0 × 109 N m 2 /C2 ) ⎢ + + ⎥ 0.030 m 0.030 m 0.030 m ⎣ ⎦ −6 −6 −6 −6 = −0.60 × 10 J − 0.60 × 10 J + 0.30 × 10 J = −0.90 × 10 J Assess:
Note that U12 = U21, U13 = U31, and U23 = U32.
29.7. Model: The charges are point charges. Visualize: Please refer to Figure EX29.7. Solve: For a system of point charges, the potential energy is the sum of the potential energies due to all distinct pairs of charges:
U elec = ∑ i,j
Kqi q j rij
= U12 + U13 + U 23
⎡ ( 3.0 ×10−9 C )( 2.0 × 10−9 C ) ( 3.0 × 10−9 C )( 2.0 ×10−9 C ) ( 2.0 ×10−9 C )( 2.0 × 10−9 C ) ⎤ ⎥ = ( 9.0 ×109 N m 2 /C2 ) ⎢ + + 2 2 ⎥ ⎢ 0.030 m 0.040 m 0.030 m 0.040 m + ( ) ( ) ⎦ ⎣ −6 −6 −6 −6 = 1.80 ×10 J + 1.35 × 10 J + 0.72 × 10 J = 3.87 ×10 J Assess:
Note that U12 = U21, U13 = U31, and U23 = U32.
29.8. Model: A water molecule is at the point of minimum potential energy when it is aligned with an electric field. However, an external force can rotate the water molecule causing its dipole moment to make an angle with the field. G G Solve: The potential energy of an electric dipole moment in a uniform electric field is U dipole = − p ⋅ E = − pE cosθ . This means
U dipole parallel = − pE cos0° = − pE
U dipole perpendicular = − pE cos90° = 0 J
⇒ Udipole perpendicular – Udipole parallel = 1.0 × 10−21 J = 0 J – (–pE) ⇒E=
Assess:
1.0 ×10−21 J 1.0 × 10−21 J = = 1.61× 108 N/C p 6.2 ×10−30 C m
Note that the units with E are J/C m. Because 1 J/C m = 1 N m/C m = 1 N/C, the units of E are N/C.
29.9. Model: An external electric field supplies energy to a dipole. Visualize: On an energy diagram, the oscillation occurs between the points where the potential-energy curve crosses the total energy line.
Solve: (a) The potential energy of an electric dipole moment in a uniform electric field is G G U = − p ⋅ E = − pE cos φ . This means
Uφ = 0° = − pE = −2 μ J
U φ = 60° = − pE cos60° = − 12 pE = − 12 ( 2 μ J ) = −1 μ J
The mechanical energy Emech = U + K. We know that at φ = 60°, K φ = 60° = 0 J. So, Emech = U φ = 60° + Kφ = 60° = −1 μ J + 0 J = − 1 μ J
(b) Conservation of mechanical energy gives
Uφ = 60° + Kφ = 60° = U φ = 0° + Kι = 0° ⇒ −1 μ J + 0 J = −2 μ J + Kφ = 0° ⇒ Kφ = 0° = 1 μ J
29.10. Model: Mechanical energy is conserved. The potential energy is determined by the electric potential. Visualize:
The figure shows a before-and-after pictorial representation of an electron moving through a potential difference of 1000 V. A negative charge speeds up as it moves into a region of higher potential (U → K). Solve: The potential energy of charge q is U = qV. Conservation of energy, expressed in terms of the electric potential V, is Kf + qVf = Ki + qVi ⇒ Kf = Ki + q(Vi – Vf) = Ki – q(Vf – Vi)
⇒ 12 mvf2 = 0 J − ( −e )( ΔV ) ⇒ vf =
2 (1.60 ×10−19 C ) (1000 V ) 2eΔV = = 1.874 × 107 m/s m 9.11×10−31 kg
Assess: Note that the electric potential difference of 1000 V already existed in space due to other charges or sources. The electron of our problem has nothing to do with creating the potential.
29.11. Model: Energy is conserved. The potential energy is determined by the electric potential. Visualize:
The figure shows a before-and-after pictorial representation of a proton moving through a potential difference of −1000 V. A positive charge speeds up as it moves into a region of lower potential (U → K). Solve: The potential energy of charge q is U = qV. Conservation of energy, expressed in terms of the electric potential V, is Kf + qVf = Ki + qVi, ⇒ Kf = Ki + q(Vi − Vf) = 0 J − qΔV
⇒ 12 mvf2 = − q ( −1000 V ) ⇒ vf = Assess: V.
2 (1.60 × 10−19 C ) (1000 V ) 1.67 ×10−27 kg
= 4.38 × 105 m/s
Note that the proton of our problem has nothing to do with creating the potential difference of −1000
29.12. Model: Energy is conserved. The potential energy is determined by the electric potential. Visualize:
The figure shows a before-and-after pictorial representation of a He+ ion moving through a potential difference. The ion’s initial speed is zero and its final speed is 2.0 × 106 m/s. A positive charge speeds up as it moves into a region of lower potential (U → K). Solve: The potential energy of charge q is U = qV. Conservation of energy, expressed in terms of the electric potential V, is Kf + qVf = Ki + qVi ⇒ q(Vf – Vi) = Ki − Kf
4 (1.67 ×10−27 kg )( 2.0 ×106 m/s ) K − K f 0 J − 12 mvf2 ⇒ ΔV = i = =− = −8.4 ×104 V q q 2 (1.60 ×10−19 C ) 2
Assess:
This result implies that the helium ion moves from a higher potential toward a lower potential.
29.13. Model: Energy is conserved. The potential energy is determined by the electric potential. Visualize:
The figure shows a before-and-after pictorial representation of an electron moving through a potential difference. Because the negative electron gains speed as it travels, it moves into a region of higher potential (U → K). Solve: The potential energy of charge q is U = qV. Using the conservation of energy equation, Kf + qVf = Ki + qVi ⇒ Vf − Vi = ΔV = 1 ( K i − K f ) = 1 ( 0 J − 12 mvf2 ) q ( −e ) −31 6 mv 2 ( 9.11×10 kg )( 2.0 ×10 m/s ) ⇒ ΔV = f = = 11.4 V 2e 2 (1.60 ×10−19 C ) 2
Assess: A positive value of ΔV shows that the electron moved from a region of lower potential to a region of higher potential.
29.14. Model: Energy is conserved. The potential energy is determined by the electric potential. Visualize:
The figure shows a before-and-after pictorial representation of an electron moving through a potential difference. Solve: (a) Because the electron is a negative charge and it slows down as it travels, it must be moving from a region of higher potential to a region of lower potential. (b) Using the conservation of energy equation, Kf + Uf = Ki + Ui ⇒ Kf + qVf = Ki + qVi 1 1 1 2 ⇒ Vf − Vi = ( Ki − K f ) = ( mv − 0 J ) q ( −e ) 2 i
(9.11×10−31 kg ) ( 500,000 m/s ) = −0.712 V mvi2 =− 2e 2 (1.60 ×10−19 C ) 2
⇒ ΔV = −
Assess: The negative sign with ΔV verifies that the electron moves from a higher potential region to a lower potential region.
29.15. Model: Energy is conserved. The potential energy is determined by the electric potential. Visualize:
The figure shows a before-and-after pictorial representation of a proton moving through a potential difference. Solve: (a) Because the proton is a positive charge and it slows down as it travels, it must be moving from a region of lower potential to a region of higher potential. (b) Using the conservation of energy equation, Kf + Uf = Ki + Ui ⇒ Kf + qVf = Ki + qVi
⇒ Vf − Vi =
1 1 ( Ki − K f ) = ( 12 mvi2 − 0 J ) q (e)
−27 mvi2 (1.67 ×10 kg ) ( 800,000 m/s ) = = 3340 V 2e 2 (1.60 ×10−19 C ) 2
⇒ ΔV = Assess:
A positive ΔV confirms that the proton moves into a higher potential region.
29.16. Model: Energy is conserved. Solve:
(a) The conservation of energy equation Kf + Uf = Ki + Ui is Kf + qVf = 0 J + qVi ⇒ Kf = q(Vi – Vf)
The above equation can be rewritten separately for the proton and the electron as K fp = 12 mpvfp2 = ( +e ) ( ΔVp )
K fe = 12 mevfe2 = ( −e )( −ΔVe )
Note that we have written Vi – Vf as ΔVp for the proton, but Vi – Vf = –ΔVe for the electron. This is because Vi > Vf to speed up a proton, and Vf > Vi to speed up an electron. With vfp = vfe, the above two equations give ΔVp ΔVe
=
mp me
=
1.67 ×10−27 kg = 1833 9.11× 10−31 kg
(b) The first two equations of part (a) can be divided to give
K fp K fe
=
ΔVp ΔVe
For the same final kinetic energy of the proton and the electron ΔVp ΔVe = 1.
29.17. Solve: By definition 1 V = 1 J/C and 1 J = 1 N m. Consequently, 1
V J Nm N =1 =1 =1 m Cm Cm C
29.18. Model: The electric potential difference between the plates is determined by the uniform electric field in the parallel-plate capacitor. Solve: (a) The potential of an ordinary AA or AAA battery is 1.5 V. Actually, this is the potential difference between the two terminals of the battery. If the electric potential of the negative terminal is taken to be zero, then the positive terminal is at a potential of 1.5 V. (b) If a battery with a potential difference of 1.5 V is connected to a parallel-plate capacitor, the potential difference between the two capacitor plates is also 1.5 V. Thus, ΔVC = 1.5 V = V+ − V− = Ed where d is the separation between the two plates. The electric field inside a parallel-plate capacitor is
E= ⇒Q=
⎛ Q ⎞ η Q = ⇒ 1.5 V = ⎜ ⎟d ε 0 Aε 0 ⎝ Aε 0 ⎠
(1.5 V )( Aε 0 ) = (1.5 V )π ( 2.0 ×10−2 m ) d
2
(8.85 ×10
2.0 ×10−3 m
−12
C2 /Nm 2 )
= 8.3 ×10−12 C
Thus, the battery moves 8.3 × 10−12 C of electron charge from the positive to the negative plate of the capacitor.
29.19. Model: The electric potential difference between the plates is determined by the uniform electric field in the parallel-plate capacitor. Solve: (a) The potential difference ΔVC across a capacitor of spacing d is related to the electric field inside by
E=
ΔVC ⇒ ΔVC = Ed = (1.0 ×105 V/m ) ( 0.0020 m ) = 200 V d
(b) The electric field of a capacitor is related to the surface charge density by
E=
η Q A = ε0 ε0
⇒ Q = ε 0 AE = ( 8.85 × 10−12 C 2 /N m 2 )( 4.0 × 10−4 m 2 )(1.0 × 105 V/m ) = 3.5 × 10−10 C
29.20. Model: The electric potential between the plates of a parallel-plate capacitor is determined by the uniform electric field between the plates. Solve: (a) The potential difference across the plates of a capacitor is
⎛η ⎞ ( Q A) d = Qd = ( 0.708 ×10−9 C )(1.00 ×10−3 m ) = 200 V ΔVC = Ed = ⎜ ⎟ d = ε0 Aε 0 ( 4.00 ×10−4 m 2 )(8.85 ×10−12 C 2 /N m 2 ) ⎝ ε0 ⎠ (b) For d = 2.00 mm, ΔVC = 400 V. Assess: Note that the units in part (a) are N m/C. But Exercise 29.17 showed that 1 N/C = 1 V/m, so 1 N m/C = 1 V. We also see that the potential difference across a parallel-plate capacitor is directly proportional to the plate separation.
29.21. Model: Energy is conserved. The electron’s potential energy inside the capacitor can be found from the capacitor’s electric potential. Solve: (a) The voltage across the capacitor is ΔVC = Ed = ( 5.0 ×105 V/m )( 2.0 × 10−3 m ) = 1000 V (b) Because E = η ε 0 for a parallel-plate capacitor, with η = Q A , the charge on each plate is
Q = π R 2 Eε 0 = π (1.0 ×10−2 m ) ( 5.0 ×105 V/m )( 8.85 ×10−12 C2 /N m 2 ) = 1.39 ×10−9 C 2
(c) The electron has charge q = −e, and its potential energy at a point where the capacitor’s potential is V is U = −eV. Since the electron is launched from the negative (lower potential) plate toward the positive (higher potential) plate, its potential energy becomes more negative (because of the negative sign of the electron charge). That is, the potential energy decreases, which must lead to an increase in the kinetic energy. Conversely, the electron’s speed as it is launched is smaller than 2.0 × 107 m/s. The conservation of energy equation is
Kf + qVf = Ki + qVi ⇒
⇒ vi2 = vf2 + ⇒ vi =
( 2.0 ×10
7
m/s ) − 2
1 2
mvi2 = 12 mvf2 + q (Vf − Vi )
2 ( −e )(1000 V ) m
2 (1.60 ×10−19 C ) (1000 V ) 9.11×10−31 kg
= 7.0 ×106 m/s
29.22. Model: The charge is a point charge. Solve:
(a) The electric potential of a charge q is V=
1 q 1 q 25 × 10−9 C 225 N m 2 /C ⇒r= = ( 9.0 ×109 N m 2 /C 2 ) = 4πε 0 r 4πε 0 V V V
For V = 1000 V,
r1000 =
225 N m 2 /C = 0.23 m 1000 V
For V = 2000 V, r2000 = 0.113 m ; for V = 3000 V, r3000 = 0.075 m ; for V = 4000 V, r4000 = 0.056 m . (b)
29.23. Model: The charge is a point charge. Visualize: Please refer to Figure EX29.23. Solve: (a) The electric potential of the point charge q is
V=
⎛ 2.0 ×10−9 C ⎞ 18.0 N m 2 /C 1 q = ( 9.0 ×109 N m 2 /C2 ) ⎜ ⎟= 4πε 0 r r r ⎝ ⎠
For points A and B, r = 0.010 m. Thus,
VA = VB =
18.0 N m 2 /C Nm ⎛V⎞ = 1800 = 1800 ⎜ ⎟ m = 1.80 kV 0.010 m C ⎝m⎠
For point C, r = 0.020 m and VC = 900 V. (b) The potential differences are
ΔVAB = VB − VA = 1.80 kV −1.80 kV = 0 V
ΔVBC = VC − VB = 0.90 kV − 1.80 kV = −0.90 kV
29.24. Solve: Outside a charged sphere of radius R, the electric potential is identical to that of a point charge Q at the center. That is, V=
1 Q 4πε 0 r
r≥R
The potential of the ball bearing is the potential right on the surface of the ball bearing. Thus, ( 9.0 ×109 N m2/C2 )( 2.0 ×109 )( −1.60 ×10−19 C ) = −5.8 kV V= 0.50 ×10−3 m
29.25. Model: Outside a charged sphere the electric potential is identical to that of a point charge at the center. Solve: (a) For a proton, assumed to be a point charge, the electric potential is V=
1 4πε 0
( +e ) = r
( 9.0 ×10
9
N m 2 /C 2 )
1.60 × 10−19 C = 27 V 0.053 ×10−9 m
(b) The potential energy of a charge q at a point where the potential is V is U = qV. The potential energy of the electron in the proton’s potential is U = (−1.60 × 10−19 C) (27 V) = −4.3 × 10−18 J
29.26. Model: The net potential is the sum of the potentials due to each charge. Visualize: Please refer to Figure EX29.26. Solve: The potential at the dot is V=
1 q1 1 q2 1 q3 + + 4πε 0 r1 4πε 0 r2 4πε 0 r3
⎡ 2.0 × 10−9 C 2.0 × 10−9 C 2.0 × 10−9 C ⎤ 3 = ( 9.0 ×109 N m 2 /C 2 ) ⎢ + + ⎥ = + 1.41× 10 V 0.040 m 0.050 m 0.030 m ⎣ ⎦
Assess:
Potential is a scalar quantity, so we found the net potential by adding three scalar quantities.
29.27. Model: The net potential is the sum of the potentials due to each charge. Visualize: Please refer to Figure EX29.27. Solve: From the geometry in the figure,
1.5 cm 1.5 cm 1.5 cm 1.5 cm = = = cos30° ⇒ r1 = r2 = r3 = = 1.732 cm cos30° r1 r2 r3 The potential at the dot is
V=
1 q1 1 q2 1 q3 + + 4πε 0 r1 4πε 0 r2 4πε 0 r3
⎡ 2.0 × 10−9 C 1.0 ×10−9 C 1.0 × 10−9 C ⎤ = ( 9.0 × 109 N m 2 /C 2 ) ⎢ − − ⎥= 0V ⎣ 0.01732 m 0.01732 m 0.01732 m ⎦ Assess:
Potential is a scalar quantity, so we found the net potential by adding three scalar quantities.
29.28. Model: The electric potential at the dot is the sum of the potentials due to each charge. Visualize: Please refer to Figure EX29.28. Solve: The electric potential at the dot is
V=
1 q1 1 q2 1 q3 + + 4πε 0 r1 4πε 0 r2 4πε 0 r3
⎡ −5.0 ×10−9 C 5.0 ×10−9 C ⎤ q = ( 9.0 ×109 Nm 2 /C2 ) ⎢ + + ⎥ = 3140 V 2 2 0.040 m 0.020 m ⎦⎥ ⎣⎢ ( 0.020 m ) + ( 0.040 m ) Solving yields q = 1.00×10–8 C = 10.0 nC. Assess: Potential is a scalar quantity, so we found the net potential by adding three scalar quantities.
29.29. Model: The net potential is the sum of the scalar potentials due to each charge. Visualize:
Solve: Let the point on the x-axis where the electric potential is zero be at a distance x from the origin. At this point, V1 + V2 = 0 V. This means
1 ⎧⎪ 3.0 ×10−9 C −1.0 ×10−9 C ⎫⎪ + ⎨ ⎬ = 0 V ⇒ − x + 3 x − 4.0 cm = 0 cm 4πε 0 ⎪⎩ x x − 4.0 cm ⎪⎭ Either − x + 3( x − 4.0 cm ) = 0 cm, or − x + 3( 4.0 cm − x ) = 0 cm. In the first case, x = 6.0 cm and, in the second case, x 3 cm. That is, we have two points on the x-axis where the potential is zero.
=
29.30. Model: The net potential is the sum of the scalar potentials due to each charge. Visualize:
Solve: Let the point on the y-axis where the electric potential is zero be at a distance y from the origin. At this point, V1 + V2 = 0 V. This means 1 ⎧ q1 q2 ⎫ ⎨ + ⎬=0 V⇒ 4πε 0 ⎩ r2 r2 ⎭
⇒ 3 (16 cm ) + y 2 = 4 2
−3.0 × 10−9 C
( −9.0 cm )
( −9 cm )
2
2
+ y2
+
4.0 × 10−9 C
(16.0 cm )
2
+ y2
=0
+ y 2 ⇒ 9 ( 256 cm2 + y 2 ) = 16 ( 81 cm2 + y 2 )
⇒ 7y2 = 1008 cm2 ⇒ y = ±12 cm.
29.31. Model: While the potential is the sum of the scalar potentials due to each charge, the electric field is the vector sum of the electric fields due to each charge. Visualize: Please refer to Figure EX29.31. Solve: (a) Ex points left for x > b because Ex is negative. So, the charge at x = b is negative. For x < a, Ex points left, so the charge at x = a is positive. For a < x < b, Ex is positive which is consistent with the charge choices. (b) By the symmetry of the drawing about the middle, it appears that the magnitudes of the charges are the same. q = 1. Thus a qb (c)
The graph of the electric potential shown in the figure is consistent with the electric field as well as the charges.
29.32. Model: While the potential is the sum of the scalar potentials due to each charge, the electric field is the vector sum of the electric fields due to each charge. Visualize: Please refer to Figure EX29.32. Solve: (a) As V is always positive, both charges must be positive. The positive signs for the two equal charges at x = a and at x = b are also consistent with the behavior of the potential in the range a < x < b. (b) By the symmetry of the drawing about the middle, we infer that the magnitudes of the charges are the same. Thus qa = 1. qb (c)
The graph of Ex, the x-component of the electric field, as a function of x is shown in the figure.
29.33. Model: Consider the rod to be a line of charge. Visualize:
Divide the rod into small segments, each with charge ±Δq.
Solve: Consider two segments, one positive and one negative, equally distant from the center of the rod. These segments are the same distance r from the dot. Thus the contribution of this pair of segments to the potential at the dot is 1 +Δq 1 −Δq V+ + V– = + =0 V 4πε 0 r 4πε 0 r
Since we can divide the entire rod into pairs of symmetrically placed segments, the net result of adding the potentials due to each pair is V = 0 V. Assess: This conclusion depends on the dot being directly outward from the midpoint of the rod. The potential is not zero at other points.
29.34. Solve: Let the unknown charges be Q1 and Q2. Then Q1 + Q2 = 30 × 10−9 C and U=
1 Q1Q2 1 Q1Q2 = = −180 × 10−6 J 4πε 0 r12 4πε 0 2.0 × 10−2 m
− (180 ×10−6 J )( 2.0 ×10−2 m )
= −4.0 × 10−16 C 2 9.0 ×109 N m 2 /C2 Solving the first equation for Q2 and substituting into the second equation, ⇒ Q1Q2 =
Q1 ( 30 ×10−9 C − Q1 ) = −4.0 ×10−16 C 2 ⇒ Q12 − ( 30 ×10−9 C ) Q1 − ( 4.0 ×10−16 C2 ) = 0
( 30 ×10 ⇒Q = 1
−9
C) ±
( −30 ×10
−9
2
C ) + 4 ( 4.0 ×10−16 C 2 ) 2
⇒ Q1 = +40 nC and −10 nC
That is, the two charges are −10 nC and 40 nC. Assess: As they must, the two charges when added yield a total charge of 30 nC, and when substituted into the potential energy equation yield U = −180 × 10−6 J .
29.35. Solve: Let the two unknown, positive charges be Q1 and Q2. They are separated by a distance r12 = 5.0 × 10−2 m. Their electric potential energy is U=
( 72 ×10−6 J )( 5.0 ×10−2 m ) = 4.0 ×10−16 C2 1 Q1Q2 = 72 ×10−6 J ⇒ Q1Q2 = 4πε 0 r12 9.0 ×109 N m 2 /C2
The electric force between these two charges is F= Assess:
1 Q1Q2 4.0 ×10−16 C 2 9 2 2 = 9.0 × 10 N m /C = 1.44 × 10−3 N ( ) 2 4πε 0 r122 ( 5.0 ×10−2 m )
As far as the magnitudes are concerned, F=
U 72 ×10−6 J = = 1.44 ×10−3 N r 5.0 ×10−2 m
29.36. Model: While the net potential is the sum of the scalar potentials due to each charge, the net electric field is the vector sum of the electric fields. Visualize:
Solve: (a) Let the two charges be labeled as Q1 = −2.0 nC and Q2 = +2.0 nC. As pictorially shown in the figure, the net electric field will not become zero anywhere except at infinity. We will now show this result mathematically. Let the point of zero electric field be at a distance x from the origin. Then, G G G Enet = E1 − E2 = 0 N/C ⇒ E1 = E2 ⇒
1
2.0 ×10−9 C
4πε 0 ( x + 1.0 cm )
2
=
1
2.0 × 10−9 C
4πε 0 ( x − 1.0 cm )2
⇒ (x + 1.0 cm)2 = (x – 1.0 cm)2 Obviously, this equation can only be satisfied if x = ∞. That is, the electric field is zero at +∞ and at −∞. (b) Because potential is a scalar quantity, the potential of charge Q1 is always negative and the potential of charge Q2 is always positive. These two scalar numbers will add to give zero potential at x = 0 m and at ±∞. Mathematically this can be shown as follows. When −1.0 cm ≤ x ≤ 1.0 cm, let the point of zero potential be at x. Then, the condition V1 + V2 = 0 V/m is
1
( −2 ×10
−9
C)
4πε 0 1.0 cm + x
+
1
( 2 ×10
−9
C)
4πε 0 (1.0 cm − x )
= 0 ⇒ −(1.0 cm + x) + (1.0 cm – x) = 0 ⇒ x = 0 cm.
When x > 1.0 cm, 1
( −2 ×10
−9
C)
4πε 0 ( x + 1.0 cm )
+
1
( 2 ×10
−9
C)
4πε 0 ( x − 1.0 cm )
This equation can only be satisfied at x = ±∞. (c) The graphs are shown in the figure above.
= 0 ⇒ (x + 1.0 cm) = (x – 1.0 cm)
29.37. Model: While the net potential is the sum of the potentials due to each charge, the net electric field is the vector sum of the electric fields. Visualize:
The charge Q1 = 20.0 nC is at the origin. The charge Q2 = −10.0 nC is 15.0 cm to the right of the charge Q1 on the x-axis. Solve: (a) As the pictorial representation shows, the point P on the x-axis where the electric field is zero can only be on the right side of the charge Q2, that is, at x ≥ 15.0 cm. At this point E1 = E2 , so we have
1 20.0 ×10−9 C 1 10.0 ×10−9 C = ⇒ x2 = 2(x – 15.0 cm)2 4πε 0 x2 4πε 0 ( x − 15.0 cm )2 ⇒ x2 − (60.0 cm)x + 450 cm2 = 0 ⇒ x =
( 60.0 cm ) ±
3600 cm 2 − 1800 cm 2 2
⇒ x = 51.2 cm and 8.8 cm. The root x = 8.8 cm is not possible physically. So, the electric fields cancel out at x = 51.2 cm. The electric potential at this point is
V=
⎡ 20.0 × 10−9 C 1 Q1 1 Q2 −10.0 ×10−9 C ⎤ + = ( 9.0 ×109 N m 2 /C 2 ) ⎢ + ⎥ = +103 V 4πε 0 r1 4πε 0 r2 ( 0.512 m − 0.150 m ) ⎥⎦ ⎢⎣ 0.512 m
(b) The point on the x-axis where the potential is zero can be obtained from the condition V1 + V2 = 0 V, which is
1 Q1 1 Q2 20.0 × 10−9 C −10.0 ×10−9 C + =0 V⇒ + =0 4πε 0 r1 4πε 0 r2 x (15.0 cm − x )
⇒ 2(15.0 cm – x) – x = 0 ⇒ x = 10.0 cm The electric field 10.0 cm away from charge Q1 is −9 G G G 1 20.0 ×10−9 C ˆ 1 (10.0 ×10 C ) ˆ Enet = E1 + E2 = i + i = 5.40 ×104iˆ N/C 2 4πε 0 ( 0.100 m ) 4πε 0 ( 0.050 m )2
29.38. Model: Mechanical energy is conserved. Visualize:
Solve:
Before the charges are released, the total mechanical energy is the potential energy of the three charges. 1 ⎡ q1q2 q1q3 q2 q3 ⎤ Ei = U12 + U13 + U 23 = + + ⎢ ⎥ r13 r23 ⎦ 4πε 0 ⎣ r12
⎡ ( 5.0 × 10−9 C )( −1.0 × 10−9 C ) ( 5.0 × 10−9 C )( 5.0 × 10−9 C ) ( −1.0 × 10−9 C )( 5.0 × 10−9 C ) ⎤ ⎥ = ( 9.0 × 109 Nm 2 /C2 ) ⎢ + + 0.010 m 0.020 m 0.010 m ⎢⎣ ⎥⎦ −6 = 2.25 × 10 J
After the charges are released and move very far (infinitely) apart, the total potential energy is zero. By the symmetry of the original configuration, the middle charge does not move when released, and the two 5.0 nC charges are moving at the same speed. The final energy is thus 1 1 1 ⎛1 ⎞ Ef = m1v12 + m2v2 2 + m3v32 = 2 ⎜ m1v 2 ⎟ = Ei = 2.25 × 10−6 J 2 2 2 ⎝2 ⎠ ⇒v=
2.25 ×10−6 J = 4.7 × 10−2 m/s = 4.7 cm/s 1.0 ×10−3 kg
29.39. Model: The charged beads are point charges. Visualize:
Solve: To find the maximum speeds vfA and vfB it is easier to use the conservation of momentum and conservation of energy equations than kinematics, as suggested in the hint. The beads achieve their maximum speeds when they are infinitely separated. The momentum conservation equation along x-direction pafter = pbefore = 0 kg m/s means −mAvfA + mBvfB = 0 kg m/s ⇒ −(0.015 kg)vfA + (0.025 kg)vfB = 0 kg m/s = 0 kg m/s ⇒ vfB = 53 vfA
The conservation of energy conservation is Uf + Kf = Ui + Ki. Noting that Uf → 0 J as the masses are infinitely separated and Ui =
9 2 2 −9 −9 1 QAQB ( 9.0 ×10 N m /C )( 5.0 ×10 C )(10.0 ×10 C ) = = 3.75 ×10−6 J 4πε 0 rAB 0.12 m
⇒0J +
⇒
1 2
(
1 2
2 2 mA vfA + 12 mBvfB ) = 3.75 × 10−6 J + 0 J
( 0.015 kg ) vfA2 + 12 ( 0.025 kg ) ( 53 vfA )
2
2 = 3.75 ×10−6 J ⇒ 0.012vfA = 3.75 ×10−6 J
Solving these equations yields vfA = 1.77 cm/s and vfB = 53 (1.77 cm/s ) = 1.06 cm/s. These are the maximum speeds of the two beads and occur when they are infinitely separated from each other.
29.40. Model: Mechanical energy is conserved. Metal spheres are point particles and they have point charges. Visualize:
Solve: (a) The system could have both kinetic and potential energy, although here K = 0 J. The energy of the system is
E0 = K 0 + U 0 = 0 J +
( 9.0 ×109 N m2 /C2 )( 2.0 ×10−6 C )( 2.0 ×10−6 C ) = 0.72 J qA qB = 4πε 0 r0 0.050 m
(b) In static equilibrium, the net force on sphere A is zero. Thus T = FB on A =
9 2 2 −6 −6 qA qB ( 9.0 ×10 N m /C )( 2.0 × 10 C )( 2.0 × 10 C ) = = 14.4 N 2 2 4πε 0 r0 ( 0.050 m )
(c) The spheres move apart due to the repulsive electric force between them. The surface is frictionless, so they continue to slide without stopping. When they are very far apart (r1 → ∞), their potential energy U1 → 0 J. Energy is conserved, so we have 2 2 E1 = K1 + U1 = 12 mAvA1 + 12 mBvB1 + 0 J = E0
Momentum is also conserved: Pafter = mAvA1 + mBvB1 = Pbefore = 0 kg m/s. Note that these are velocities and that vA1 is a negative number. From the momentum equation, mv vA1 = − B B1 mA Substituting this into the energy equation, 2 ⎛ mv ⎞ ⎛m ⎞ 2 2 E0 = 12 mA ⎜ − B B1 ⎟ + 12 mBvB1 = 12 mB ⎜ B + 1⎟ vB1 ⎝ mA ⎠ ⎝ mA ⎠
⇒ vB1 =
2 E0 = mB ( mB mA + 1)
2 ( 0.720 J ) = 10.95 m/s ( 0.004 kg )( 4 g 2 g + 1)
Using this result, we can then find vA1 = − mBvB1 mA = −21.9 m/s. These are the velocities, so the final speeds are 21.9 m/s for the 2 g sphere and 11.0 m/s for the 4 g sphere.
29.41. Model: Energy is conserved. The charged spheres are point charges. Visualize: Please refer to Figure P29.41. Label the spheres 1, 2, 3, and 4 in a clockwise manner, with the sphere in the upper left-hand corner being sphere 1. Solve: The total potential energy of the four spheres before they are allowed to move is Ui = (U12 + U23 + U34 + U41) + (U13 + U24) All the charges are identical, r12 = r23 = r34 = r41 = 1.0 cm, and r13 = r24 = 2.0 cm. The initial potential energy is
⎛ 1 10 × 10−9 C)(10 × 10−9 C) ⎞ ⎛ 1 10 × 10−9 C)(10 × 10−9 C) ⎞ −5 Ui = 4⎜ ⎟ + 2⎜ ⎟ = 48.73 × 10 J 0.010 m 0.01414 m ⎝ 4π e0 ⎠ ⎝ 4π e0 ⎠ Since all charges are at rest, Ki = 0 J. As the spheres are allowed to move away from one another and they are far apart, Uf = 0 J. The final kinetic energy is K f = 4( 12 mvf2 ) = 2(1.0 × 10−3 kg)vf2
From the energy conservation equation Uf + Kf = Ui + Ki,
2 (1.0 × 10−3 kg ) vf2 = 48.73 × 10−5 J ⇒ vf = 0.49 m/s
29.42. Model: Energy is conserved. The potential energy is determined by the electric potential. Visualize: Please refer to Figure P29.42. Solve: The proton at point A is at a potential of 30 V and its speed is 50,000 m/s. At point B, the proton is at a potential of −10 V and we are asked to find its speed. Clearly, the proton moves into a lower potential region, so its speed will increase. The conservation of energy equation Kf + Uf = Ki + Ui is 1 2
⇒ vf =
Assess:
mvf2 + ( +e )( −10 V ) = 12 m ( 50,000 m/s ) + ( + e )( 30 V ) 2
( 50,000 m/s )
2
+
2 (1.60 × 10−19 C ) ( 40 V ) 1.67 ×10−27 kg
The speed of the proton is higher, as expected.
= 1.01× 105 m/s
29.43. Model: Energy is conserved. Solve:
(a)
(b) The potential energy of the positive charge as a function of x is U(x) = 5000 qx2. This is analogous to the potential energy of a mass on a spring. Thus the motion is simple harmonic motion. (c) Turning points at ±8.0 cm mean that xmax = ±8.0 cm. At the turning points, the energy is all potential energy and the kinetic energy is zero. The mechanical energy of the charged particle is
E = U + K = U max + 0 J = 5000 q ( xmax ) = 5000 (10.0 ×10 −9 C ) ( 0.080 m ) = 3.2 ×10−7 J 2
2
(d) The conservation of energy equation K1 + U1 = K2 + U2 is 2 = 3.20 ×10−7 J ⇒ vmax = Kmax + 0 J = 0 J + Umax ⇒ 12 mvmax
2 ( 3.20 ×10−7 J )
(1.0 ×10
−3
kg )
= 2.5 cm/s
29.44. Model: Energy is conserved. Solve:
(a)
(b) The potential energy of the positively charged proton as a function of x is
U ( x ) = eV ( x ) = 6000ex 2 = 6000 (1.60 ×10−19 C ) x 2 = 9.6 × 10−16 x 2 J. This is analogous to the potential energy of a mass on a spring. Thus the motion is simple harmonic motion. 1 (c) The form for the potential energy of a mass on a spring is U ( x ) = kx 2 . Comparing that to the expression 2 above yields 1 keff = 9.6 ×10−16 J/m 2 = 9.6 × 10−16 N/m ⇒ keff = 1.92 × 10−15 N/m 2 Assess: The constant in the given expression for V carry units of V/m2. This must be carefully accounted for in later expressions.
29.45. Model: Energy is conserved. The proton’s potential energy inside the capacitor can be found from the capacitor’s potential difference. Visualize: Please refer to Figure P29.45. Solve: (a) The electric potential at the midpoint of the capacitor is 250 V. This is because the potential inside a parallel-plate capacitor is V = Es where s is the distance from the negative electron. The proton has charge q = e and its potential energy at a point where the capacitor’s potential is V is U = eV. The proton will gain potential energy ΔU = eΔV = e(250 V) = 1.60 × 10−19 C (250 V) = 4.00 × 10−17 J if it moves all the way to the positive plate. This increase in potential energy comes at the expense of kinetic energy which is
K = 12 mv 2 =
1 2
(1.67 ×10
−27
kg ) ( 200,000 m/s ) = 3.34 × 10−17 J 2
This available kinetic energy is not enough to provide for the increase in potential energy if the proton is to reach the positive plate. Thus the proton does not reach the plate because K < ΔU. (b) The energy-conservation equation Kf + Uf = Ki + Ui is 1 2
⇒ vf = vi2 +
mvf2 + qVf = 12 mvi2 + qVi ⇒ 12 mvf2 = 12 mvi2 + q (Vi − Vf )
2q (Vi − Vf ) = m
( 2.0 × 10
5
m/s ) + 2
2 (1.60 × 10−19 C ) ( 250 V − 0 V ) 1.67 × 10−27 kg
= 2.96 × 105 m/s
29.46.
Model:
Energy is conserved.
Visualize:
Solve:
(a) The electric field inside a parallel-plate capacitor is constant with strength
E=
3 ΔV ( 25 ×10 V ) = = 2.1×106 V/m. 0.012 m d
(b) Assuming the initial velocity is zero, energy conservation yields
Ui = Kf + U f 1 0 = mevf 2 + ( −e )( Ed ) 2 ⇒ vf = Assess:
2 (1.60 ×10−19 C )( 2.1× 106 V/m ) ( 0.012 m ) 9.11×10−31 kg
= 9.4 ×107 m/s
This speed is about 31% the speed of light. At that speed, relativity must be taken into account.
29.47. Model: Mechanical energy is conserved. Visualize:
Solve: (a) Both gravitational and electric potential energy are involved. The electric potential between the plates is V = Es, where s is measured from the more negative plate. The electric field between the plates is
E=
ΔV 3.0 ×106 V = = 1.00 ×106 V/m. 3.0 m d
Conservation of energy is used to find the height. K i + U gi + U ei = K f + U gf + U ef 1 mvi 2 + 0 J + q ( 0 V ) = 0 J + mgh + q ( Eh ) 2
1 1 2 mvi 2 1.0 ×10−3 kg ) ( 5.0 m/s ) ( 2 ⇒h= 2 = = 0.85 m mg + qE (1.0 ×10−3 kg )( 9.8 m/s 2 ) + ( 4.0 × 10−9 C )(1.00 × 106 V/m ) (b) The top plate is at a negative potential, so the charge is attracted to it. The electric potential is now zero at the top plate and positive at the bottom plate. Does the charge make it all the way to the top plate? If we assume it does not and obtain a height more than 3.0 m then we will know that it does. Conservation of energy for the charge requires 1 1 mvi 2 + 0 J + qE ( 3.0 m ) = mgh + qE ( 3.0 m − h ) ⇒ mvi 2 = ( mg − qE ) 2 2
1 mvi 2 ⇒h= 2 = 2.6 m mg − qE Assess: Since h < 3.0 m the charge did not hit the top plate.
29.48.
Model: Visualize:
Solve:
Mechanical energy is conserved.
The initial energy of the electron is the same as the energy at the turning point, when its speed is zero.
1 Ei = Ef ⇒ mevi 2 + U1i + U 2i = U1f + U 2f 2 ⎛ 1 1 ( e )( −e ) ⎞ = 2 ⎛ 1 ( e )( −e ) ⎞ ⇒ mevi 2 + 2 ⎜ ⎟ ⎜ ⎟ −11 r 2 ⎝ 4πε 0 5.5 ×10 m ⎠ ⎝ 4πε 0 ⎠ 2 (1.6 ×10 C ) 1 ⇒ ( 9.11×10−31 kg )(1.5 × 106 m/s ) − 2 ( 9.0 ×109 Nm 2 /C2 ) 2 5.5 ×10−11 m −19
= −2 ( 9.0 ×109 Nm 2 /C2 )
(1.6 ×10
−19
C)
2
2
r
⇒ 1.025 ×10−18 J − 8.38 ×10−18 J = −
4.61×10−28 Nm 2 r
⇒ r = 6.27 ×10−11 m = 0.0627 nm Since r 2 = ( 0.055 nm ) + y 2 , y = 2
( 0.0627 nm ) − ( 0.055 nm ) 2
2
= 0.030 nm.
Assess: The electron moves a distance outward that is less than the distance to each of the protons.
29.49. Model: Energy is conserved. The electron ends up so far away from the glass sphere that we can consider its potential energy to be zero. Visualize:
The minimum speed to escape is the speed that allows the electron to reach rf = ∞ when vf = 0 m/s. Solve: The conservation of energy equation Kf + Uf = Ki + Ui is
⎛ 1 q⎞ 0 J + 0 J = 12 mvi2 + qVi ⇒ 0 J = 12 mvi2 + ( −e ) ⎜ ⎟ ⎝ 4πε 0 R ⎠ ⇒ vi =
2 (1.60 ×10−19 C ) ⎛ 10 ×10−9 C ⎞ 2e 1 q 7 = 9.0 × 109 N m 2 /C2 ) ⎜ ( ⎟ = 8.0 × 10 m/s −31 −2 m 4πε 0 R 9.11× 10 kg 0.50 × 10 m ⎝ ⎠
29.50. Model: Energy is conserved. The potential energy is determined by the electric potential. Solve:
The conservation of energy Kf + Uf = Ki + Ui is
Kf + qVf = Ki + qVi ⇒ Kf = Ki + q(Vi – Vf) In this equation, K i = mv = 0 J , the final proton potential is zero, and the initial proton potential is V. We have 1 2
2 i
Kf = eV. That is, the maximum kinetic energy or maximum speed is directly proportional to the potential of the surface the proton is launched from. From Equation 29.32, Example 29.11, and Example 29.12, when z → 0 for the ring and disk:
Vsphere =
1 Q 4πε 0 R
Vring =
1 Q 4πε 0 R
Vdisk =
1 2Q 4πε 0 R
Because the potential on the disk is a factor of 2 larger than the potential on the sphere or the ring, the proton’s kinetic energy or the speed will be maximum in the case of the disk.
29.51. Model: Energy is conserved. Visualize:
The dipole initially is at the higher potential energy. As it moves to its minimum energy state, the decrease in the potential energy appears as rotational kinetic energy. G G Solve: The potential energy of a dipole is U dipole = − p ⋅ E = − pE cosθ , where θ is the angle between the dipole
and the electric field. The energy conservation equation Kf + Uf = Ki + Ui is 1 2
I ω 2 + U f = 0 J + U i ⇒ 12 I ω 2 = U i − U f = − pE cos90° − ( − pE cos 0° ) ⇒ 12 I ω 2 = pE ⇒ ω =
2 pE I
The moment of inertia of the dipole is I dipole = 2m ( 12 s ) . Substituting into the above equation, 2
ω=
2 ( qs ) E
2m ( 12 s )
2
=
4 ( 2.0 × 10−9 C ) ( 0.10 m )(1000 V/m )
(1.0 × 10
−3
kg ) ( 0.10 m )
2
= 0.28 rad/s
29.52. Model: The electrons and the proton are point charges. Visualize:
Solve:
We are given that r12 = r23 = r13 = 1.0 ×10−9 m . From the geometry of the figure,
r r23 = cos30° ⇒ r24 = = 0.5774 × 10−9 m = r14 = r34 r24 2cos30°
1 2 23
The contributions to the total potential energy are
U12 = U13 = U 23 = U14 = U 24 = U 34 =
( 9.0 ×10
9
N m 2 /C2 )( −1.60 ×10−19 C )( −1.60 ×10−19 C ) 1.0 ×10−9 m
( 9.0 ×10
9
N m 2 /C2 )( −1.60 ×10−19 C )(1.60 ×10−19 C ) 0.5774 ×10−9 m
= 2.304 ×10−19 J
= −3.990 ×10−19 J
Summing all of the contributions,
U elec = U12 + U13 + U 23 + U14 + U 24 + U 34
= 3( 2.304 ×10−19 J ) + 3( −3.990 ×10−19 J ) = −5.1×10−19 J
Assess:
Note that U12 = U21, U13 = U31, and U23 = U32, U14 = U41, U24 = U42, and U34 = U43.
29.53. Model: Outside a charged sphere the electric potential is identical to that of a point charge at the center. Visualize:
Solve:
For r ≥ R, V = Q 4πε 0 r . The potentials at the two points are
V2 mm =
( 9.0 ×10
9
N m 2 /C2 ) Q
(1 mm + 2 mm )
( 9.0 ×10
9
=
N m 2 /C2 ) Q −3
3.0 ×10 m
( 9.0 ×10
9
V4 mm =
N m 2 /C2 ) Q
5.0 ×10−3 m
1 1 ⎛ ⎞ ⇒ V2 mm − V4 mm = +500 V = ( 9.0 ×109 N m 2 /C2 ) Q ⎜ − ⎟ −3 −3 × × 3.0 10 m 5.0 10 m ⎝ ⎠ ⇒Q= Assess:
⎛ 15.0 ×10−6 m 2 ⎞ 500 V = 4.2 ×10−10 C ( 9.0 ×109 N m2/C2 ) ⎜⎝ 2.0 ×10−3 m ⎟⎠
Do not forget to include the radius of the glass bead in r.
29.54. Model: Energy is conserved. Because the iron nucleus is very large compared to the proton, we will assume that the nucleus does not move (no recoil) and that the proton is essentially a point particle with no diameter. Visualize:
The proton is fired from a distance much greater than the nuclear diameter, so ri ≈ ∞ and Ui ≈ 0 J. Because the nucleus is so small, a proton that is even a few atoms away is, for all practical purposes, at infinity. As the proton approaches the nucleus, it is slowed by the repulsive electric force. At the end point, the proton has just reached the surface of the nucleus (rf = nuclear diameter) with vf = 0 m/s. (The proton won’t remain at this point but will be pushed back out again, but the subsequent motion is not part of this problem.) Solve: Initially, the proton has kinetic energy but no potential energy. At the point of closest approach, where vf = 0 m/s, the proton has potential energy but no kinetic energy. Energy is conserved, so Kf + Uf = Ki + Ui. This equation is
0 J+
( e )( 26e ) = 1 m 4πε 0 rf
2
v +0 J
2 proton i
where rf is half the nuclear diameter. The initial speed of the proton is
vi =
2 (1.6 ×10−19 C )( 26 ×1.6 ×10−19 C )( 9.0 ×109 N m 2 / C 2 ) 2 ( e )( 26e ) = = 4.0 ×107 m/s 4πε 0 rf mproton (1.67 ×10−27 kg )( 4.5 ×10−15 m )
29.55. Model: Energy is conserved. Because the mercury nucleus is very large compared to the proton, we will assume that the nucleus does not move (no recoil) and that the proton is essentially a point particle with no diameter. Visualize:
The proton is fired from a distance much greater than the nuclear diameter, so ri = ∞ and Ui = 0 J. Because the nucleus is so small, a proton that is even a few atoms away is, for all practical purposes, at infinity. As the proton approaches the nucleus, it is slowed by the repulsive electric force. At the end point of the problem, the proton reaches the distance of closest approach (d) with vf = 0 m/s. (The proton won’t remain at this point but will be pushed back out again, but the subsequent motion is not part of the problem.) Solve: Initially, the proton has kinetic energy but no potential energy. At the point of closest approach, where vf = 0 m/s, the proton has potential energy but no kinetic energy. Energy is conserved, so Kf + Uf = Ki + Ui . This equation is
0 J+
e ( 80e ) 1 = mproton vi2 + 0 J 4πε 0 d 2
( 9.0 ×109 N m2 / C2 ) (160 ) (1.6 × 10−19 C ) = 1.38 × 10−14 m = 13.8 fm 160e2 ⇒d = = 2 2 4πε 0 mproton vi (1.67 × 10−27 kg )( 4.0 ×107 m/s ) 2
The radius of the nucleus is 7.0 fm, so the proton’s closest approach to the surface is 6.8 fm.
29.56. Model: Energy is conserved. Visualize:
The alpha particle is initially at rest (vi alpha = 0 m/s) at the surface of the thorium nucleus. The potential energy of the alpha particle is U i alpha . After the decay, the alpha particle is far away from the thorium nucleus, Uf alpha = 0 J,
and moving with speed vf alpha . Solve: Initially, the alpha particle has potential energy and no kinetic energy. As the alpha particle is detected in the laboratory, the alpha particle has kinetic energy but no potential energy. Energy is conserved, so Kf alpha + Uf = alpha Ki alpha + Ui alpha. This equation is 1 2
⇒ vf alpha =
1 4πε 0
mvf2alpha + 0 J = 0 J +
1
( 2e )( 90e )
4πε 0
ri
( 360e ) = ( 9.0 ×10 N m /C ) 360 (1.60 ×10 C ) mr 4 (1.67 ×10 kg )( 7.5 ×10 m ) 2
9
2
−27
i
−19
2
−15
2
= 4.1× 107 m/s
29.57. Model: Energy is conserved. Visualize:
Solve: (a) The neutron has zero charge. In the decay n → p+ + e– + v, where the neutrino v is neutral, the net charge of the final state is e + (−e) + 0 = 0. So charge is conserved. (b) As shown in the figure, the tritium nucleus has one proton and two neutrons. It is the single proton that makes this an isotope of hydrogen, the Z = 1 element in the periodic table. One of the neutrons undergoes the decay n → p + e + v. The electron and neutrino are ejected from the nucleus as beta radiation, but the new proton remains in the nucleus. The nucleus still has three nucleons, but now two are protons and only one is a neutron. An atom whose nucleus contains two protons is the Z = 2 element of the periodic table, namely helium. (c) As shown in the figure, the electron is “launched” from the surface of the 3He nucleus with speed vi. The initial energy is
Ei = K i + U i = 12 mevi2 +
qelec qnuc 1 ( −e )( 2e ) = 2 mevi2 + 4πε 0 ri 4πε 0 ri
If the electron just barely escapes, it slows to vf → 0 m/s as rf → ∞. Its final energy is Ef = Kf + Uf = 0 J + 0 J = 0 J. Energy is conserved, Ef = Ei, so 1 2
⇒ vi = −
2 ( −e )( 2e ) = me 4πε 0 ri
mevi2 +
( −e )( 2e ) = 0 J 4πε 0 ri
( 2 ) ( 9.0 ×109 N m 2 /C2 ) ( 2 ) (1.60 ×10−19 C )
( 9.11×10
−31
kg )(1.5 ×10−15 m )
2
= 8.21× 108 m/s
Assess: Our “classical” analysis gives an answer that exceeds the speed of light (3 × 108 m/s), which is forbidden by Einstein’s theory of relativity. We would need to perform a relativistic analysis, which is beyond the scope of this chapter, to get the correct minimum escape speed.
29.58.
Model: Energy is conserved in the collision. Solve: (a) The centers of the protons are 2.4 fm apart when they touch and fusion occurs. Assume they start very far apart so that the initial electric potential energy between them may be ignored. The minimum temperature required is found if it is assumed they have zero velocity just as they touch. Energy conservation requires
K1i + K 2i + U12i = K1f + K 2f + U12f 1 1 1 e2 mp v1i 2 + mp v2i 2 + 0 J = 0 J + 0 J + 2 2 4πε 0 2.4 × 10−15 m Since the protons both have exactly the same average kinetic energy of 3/2 kBT,
1 e2 ⎛3 ⎞ 2 ⎜ k BT ⎟ = −15 ⎝2 ⎠ 4πε 0 2.4 ×10 m
( 9.0 ×10 Nm /C ) (1.6 ×10 3(1.38 ×10 J/K ) 2.4 ×10 9
⇒T =
2
2
−23
−19 −15
C) m
2
= 2.3 ×109 K
(b) The ratio of the kinetic energy of the protons in part (a) to the average kinetic energy of protons with a temperature of 15 million K is 3
2
3
k B ( 2.3 ×109 k )
2
k B (15 ×106 k )
= 153.
The energy of the proton undergoing fusion exceeds the average proton energy by a factor of 153. Assess: Then how can fusion take place in the sun? The answer lies in the fact that there is a distribution of proton velocities. Only protons with velocities in the long high-velocity tail of the Maxwell-Boltzman velocity distribution participate in fusion in the sun’s core.
29.59. Model: The electric field inside a capacitor is uniform. Solve: (a) Because the parallel-plate capacitor was connected to the terminals of a 15 V battery for a long time, the potential difference across the capacitor right after the battery is disconnected is ΔV= 15 V. The electric field strength inside the capacitor is
E=
ΔVC 15 V = = 3000 V/m = 3.0 kV/m d 0.50 ×10−2 m
Because E = η ε 0 for a parallel-plate capacitor and η = Q A , the total charge on each plate is Q = EAε 0 = ( 3000 V/m ) π ( 0.050 m ) (8.85 ×10 −12 C 2 /Nm 2 ) = 2.1× 10−10 C 2
(b) After the electrodes are pulled away to a separation of d ′ = 1.0 cm, the charges on the plates are unchanged. That is, Q′= Q. Because A′ = A, the electric field inside the capacitor is also unchanged. So, E′ = E. The potential difference across the capacitor is ΔVC′ = E ′d ′. Because d increases from 0.50 cm to 1.0 cm (d′ = 1.0
cm), the potential difference ΔVC′ increases from 15 V to 30 V. (c) When the electrodes are expanded, the new area is A′ = π(r′)2 = π(2r)2 = 4A. The charge Q′ on the capacitor plates, however, stays the same as before (Q′ = Q). The electric field is
E′ =
η ′ Q′ Q E 3000 V/m = = = = = 750 V/m = 0.75 kV/m ε 0 A′ε 0 4 Aε 0 4 4
The potential difference across the capacitor plates ΔVC′ = E′d ′ = ( 750 V )( 0.050 m ) = 3.8 V.
29.60. Model: The electric field inside a capacitor is uniform. Solve: (a) While the capacitor is attached to the battery, the plates are at the same potentials as the terminals of the battery. Thus, the potential difference across the capacitor is ΔVC = 15 V. The electric field strength inside the capacitor is
E=
ΔVC 15 V = = 3000 V/m = 3.0 kV/m d 0.50 ×10−2 m
Because E = η ε 0 = Q Aε 0 , the charge on each plate is
Q = EAε 0 = (3000 V/m) π (0.050 m)2 (8.85 × 10−12 C2/N m2) = 2.1 × 10−10 C (b) After the electrodes are pulled away to a new separation of d′ = 1.0 cm, the potential difference across the capacitor stays the same as before. That is, ΔVC′ = ΔVC = 15 V. The electric field strength inside the capacitor is
E′ =
ΔV ′ 15 V = = 1500 V/m = 1.50 kV/m ′ d 0.010 m
The charge on each plate is
Q′ = E ′Aε 0 = (1500 V)π (0.05 m) 2 (8.85 × 10−12 C2 / N m 2 ) = 1.04 × 10−10 C (c) The potential difference ΔVC′ = ΔVC = 15 V is unchanged when the electrodes are expanded to area A′ = 4A . The electric field between the plates is
E′ = E =
ΔVC′ 15 V = = 3000 V/m = 3.0 kV/m d′ 0.50 ×10−2 m
The new charge is Q′ = E ′A′ε 0 = 4 EAε 0 = 4Q = 8.3 × 10−12 C
29.61. Solve: (a) Outside a uniformly charged sphere, the electric field is identical to that of a point charge Q at the center. For r ≥ R, E=
1 Q 4πε 0 r 2
Evaluating this at r = R and using Equation 29.32, E0 =
1 Q V0 = 4πε 0 R 2 R
(b) The field strength is E0 =
500 V = 100,000 V/m = 100 kV/m 0.50 ×10−2 m
29.62. Visualize:
Solve: A sphere of radius R and charge Q has a surface potential V0 = Q 4πε 0 R . When the two drops A and B merge to form drop C, two quantities are conserved: the total charge and the total volume of mercury. The conservation of charge yields
QC = Qtotal = QA + QB = 0.10 nC + 0.10 nC = 0.20 nC Since the quantity of mercury does not change, the volumes of drops A, B, and C are related by ⎡⎣ VolC = 34 π RC3 ⎤⎦ = ⎡⎣ VolA + Vol B = 34 π RA3 + 43 π RB2 ⎤⎦ ⇒ RC = ( RA3 + RB3 )
1/ 3
We’re not given the radii of drops A and B, but we know their surface potentials (V0)A and (V0)B. Thus,
(V0 )A =
( 0.10 ×10 QA QA = ⇒ RA = 4πε 0 (V0 )A 4πε 0 RA
−9
C )( 9.0 ×109 N m 2 /C 2 ) 300 V
= 3.0 × 10−3 m = RB
Combining these radii gives RC = 3.8 × 10−3 m, and
(V0 )C =
( 0.20 ×10−9 C )( 9.0 ×109 N m 2/C2 ) = 0.47 kV QC = 4πε 0 RC 3.8 × 10−3 m
29.63. Solve: (a) Because the excess charge resides on the outer surface of a conductor, the charge placed on the inside surface of a hollow metal sphere will move rapidly to the outside surface. (b) Because a spherical shell of charge has the same electric potential as a point charge Q at the center, the potential on the surface is V=
1 Q ⎛ Q ⎞ ⇒ 500,000 V = ( 9.0 × 109 N m 2 /C 2 ) ⎜ ⎟ ⇒ Q = 8.3 μ C 4πε 0 R ⎝ 0.15 m ⎠
(c) The electric field inside a conductor is zero, so the electric field just inside the sphere is zero. From Gauss’s Law, the electric field outside a charged conductor is
E=
η Q 8.33 ×10−6 C = = = 3.3 ×106 V/m ε 0 Aε 0 4π ( 0.15 m )2 ( 8.85 ×10−12 C 2 /Nm 2 )
29.64. Solve: A charged particle placed inside a uniformly charged spherical shell experiences no electric force. That is, E = 0 V/m inside the shell. We know from Section 29.5 that a difference in potential between two points or two plates is the source of an electric field. Since the potential on the surface of the shell is V = Q 4πε 0 R , the potential inside must be the same. This ensures that the potential difference is zero and hence the electric field is zero inside the shell. The potential at the center of the spherical shell is thus the same as at the surface. That is, Vcenter = Vsurface = Q 4πε 0 R .
29.65. Model: The potential at any point is the superposition of the potentials due to all charges. Outside a uniformly charged sphere, the electric potential is identical to that of a point charge Q at the center. Visualize: Please refer to Figure P29.65. Sphere A is the sphere on the left and sphere B is the one on the right. Solve: The potential at point a is the sum of the potentials due to the spheres A and B:
Va = VA at a + VB at a =
1 QA 1 QB + 4πε 0 RA 4πε 0 0.70 m
100 ×10−9 C 25 ×10−9 C + ( 9.0 × 109 N m 2 /C2 ) 0.30 m 0.70 m = 3000 V + 321 V = 3321 V = ( 9.0 ×109 N m 2 /C 2 )
Similarly, the potential at point b is the sum of the potentials due to the spheres A and B: Vb = VB at b + VA at b =
1 QB 1 QA + 4πε 0 RB 4πε 0 0.95 m
⎛ 25 ×10−9 C 100 ×10−9 C ⎞ = ( 9.0 ×109 N m 2 /C2 ) ⎜ + ⎟ 0.95 m ⎠ ⎝ 0.05 m = 4500 V + 947 V = 5447 V Thus, the potential at point b is higher than the potential at a. The difference in potential is Vb – Va = 5447 V – 3321 V = 2126 V = 2.1 kV. Assess: VA at a = 3000 V and the sphere B has a potential of 225 V at point a. The spherical symmetry dictates that the potential on a sphere’s surface be the same everywhere. So, in calculating the potential at point a due to the sphere B we used the center-to-center separation of 1.0 m rather than a separation of 100 cm – 30 cm = 70 cm from the center of sphere B to the point a. The former choice leads to the same potential everywhere on the surface whereas the latter choice will lead to a distribution of potentials depending upon the location of the point a. Similar reasoning also applies to the potential at point b.
29.66. Model: The charges making the dipole are point charges. The potential is the sum of potentials from each charge. Visualize:
Point P is far away compared to the separation s of the dipole charges, that is, y >> s. Solve: (a) The potentials at P due to the positive and negative charges of the dipole are
V+ =
⇒ Vdipole = V+ + V− =
1 q 4πε 0 y − s 2
V− =
1 ( −q ) 4πε 0 y + s 2
⎞ q ⎛ 1 1 ⎞ q ⎛ s 1 p 1 p − ≅ ⎜ ⎟= ⎜ 2 2 ⎟= 2 2 4πε 0 ⎝ y − s 2 y + s 2 ⎠ 4πε 0 ⎝ y − s 4 ⎠ 4πε 0 y − s 4 4πε 0 y 2
(b) The potential due to the dipole moment of water is
Vdipole = ( 9.0 ×109 N m 2 /C 2 )
6.2 ×10−30 C m
(1.0 ×10
−9
m)
2
= 0.056 V
29.67. Model: The potential is the sum of potentials from each charge. Visualize:
Point P is located on the x-axis and the two positive charges are located on the y-axis. The separation between the two charges is s. Solve: (a) The potentials at P due to the two positive charges are
Vtop =
1 4πε 0
( +q ) x +s 4 2
⇒ Vq + q = Vtop + Vbottom =
2
Vbottom =
1 4πε 0
2q x +s 4 2
2
1 4πε 0 =
2q
( +q ) x2 + s2 4 1
4πε 0 x 1 + s 2 4 x 2
(b) The V-versus-x graph is shown in the figure. The potential due to a charge 2q located at the origin is V2 q = 2q 4πε 0 x , and is also plotted. This expression differs from the expression obtained in part (a) by the factor
(1 + s
2
4 x 2 ) 2 . For x >> s , this factor becomes 1 and the two potentials become the same. −1
29.68. Model: The net potential is the sum of potentials from all the charges. Visualize: Please refer to Figure P29.68. Point P at which we want the net potential due to the linear electric quadrupole is far away compared to the separation s, that is, y >> s. Solve: The net potential at P is
Vnet = V1 + V2 + V3 = =
1
q1 1 q2 1 q3 + + 4πε 0 y − s 4πε 0 y 4πε 0 y + s
1 ⎡ ( + q ) ( −2q ) ( + q ) ⎤ q ⎡ y 2 + ys − 2 y 2 + 2 s 2 + y 2 − ys ⎤ q 2s 2 ⎢ ⎥ + + = = ⎢ ⎥ 2 2 4πε 0 ⎣ y − s y y + s ⎦ 4πε 0 ⎢ y ( y2 − s2 ) ⎥⎦ 4πε 0 y ( y − s ) ⎣
At distances y >> s , Vnet =
1 4πε 0
( 2qs ) = 2
y3
1 Q 4πε 0 y 3
where Q = 2qs2 is called the electric quadrupole moment of the charge distribution. Assess: This charge distribution is in fact a combination of two dipoles. As seen above their effects do not completely cancel.
29.69. Model: Assume the thin rod is a line of charge with uniform linear charge density. Visualize: Please refer to Figure P29.69. The point P is a distance d from the origin. Divide the charged rod into N small segments, each of length Δx and with charge Δq. Segment i, located at position xi, contributes a small amount of potential Vi at point P. Solve: The contribution of the ith segment is
Vi =
Δq Δq Q Δx L = = 4πε 0 ri 4πε 0 ( d − xi ) 4πε 0 ( d − xi )
where Δq = λΔx and the linear charge density is λ = Q L . We are placing the point P at a distance d rather than x from the origin to avoid confusion with xi. The Vi are now summed and the sum is converted to an integral giving V=
L/2 ⎛d +L 2⎞ dx Q Q = ⎡⎣ − ln ( d − x ) ⎤⎦ − L / 2 = ln ⎜ ⎟ 4πε 0 L − L/ 2 d − x 4πε 0 L 4πε 0 L ⎝ d − L 2 ⎠
Q
L/ 2
∫
Replacing d with x, the potential due to a line charge of length L at a distance x along the axis is V=
⎛ x+L 2⎞ Q ln ⎜ ⎟ 4πε 0 L ⎝ x − L 2 ⎠
29.70. Model: Assume the thin rod is a line of charge with uniform linear charge density. Visualize: Please refer to Figure P29.69. The point P is located a distance z from the origin. Divide the charged rod into N small segments, each of length Δx and with charge ΔQ. Segment i, located at position xi, contributes a small amount of potential Vi at point P. Solve: The contribution of the ith segment is
Vi =
ΔQ ΔQ = 4πε 0ri 4πε 0 xi2 + z 2
Charge ΔQ is related to length Δx through the linear charge density: ΔQ = λΔx = ( Q L ) Δx . Thus
Vi =
QΔx 4πε 0 L xi2 + z 2
Potential is a scalar, so we can find the net potential at P by summing all the Vi: V = ∑Vi = i
Q Δx ∑ 4πε 0 L i xi2 + z 2
If we now let Δx → dx, the sum becomes an integral over the length of the rod, from xmin = −L/2 to xmax = +L/2:
V=
Q
4πε L ∫ 0
L/2
−L / 2
dx x2 + z 2
=
Q
⎡ln 4πε 0 L ⎣⎢
(
⎡ z 2 + ( L 2 )2 + L L/2 Q x2 + z 2 + x ⎤ ln ⎢ = ⎦⎥ − L / 2 4πε 0 L ⎢ z 2 + ( L 2 )2 − L ⎣
)
2⎤ ⎥ 2 ⎥⎦
We can also get a simpler expression by using symmetry to integrate from 0 to L/2 and multiplying by 2. This gives
V=
2 ⎡⎛ L ⎞ ⎛ L⎞ ⎤ ln ⎢⎜ ⎟ + 1 + ⎜ ⎟ ⎥ 2πε 0 L ⎢⎝ 2z ⎠ ⎝ 2 z ⎠ ⎥⎦ ⎣
Q
This expression can be shown to be equivalent to the one obtained earlier.
29.71. Model: Because the rod is thin, assume the charge lies along the semicircle of radius R. Visualize: Please refer to Figure P29.71. The bent rod lies in the xy-plane with point P as the center of the semicircle. Divide the semicircle into N small segments of length Δs and of charge ΔQ = ( Q π R ) Δs, each of which can be modeled as a point charge. The potential V at P is the sum of the potentials due to each segment of charge. Solve: The total potential is
V = ∑Vi = ∑ i
i
1 ΔQ 1 1 Q 1 Q ⎛ Q ⎞ = R Δθ = ∑ ∑ Δθ . ⎜ ⎟ Δs = 2 ∑ 4πε 0 R 4πε 0 R i ⎝ π R ⎠ 4πε 0 π R i 4πε 0 π R i
All of the terms come to the front of the summation because these quantities did not change as far as the summation is concerned. The summation does not have to convert to an integral because the sum of all the Δθ around the semicircle is π. Hence, the potential at the center of a charged semicircle is
Vcenter =
1 Qπ 1 Q = 4πε 0 π R 4πε 0 R
2 29.72. Model: The disk has a uniform surface charge density η = Q A = Q π ( Rout − Rin2 ) .
Visualize: Please refer to Figure 29.31. Orient the disk in the xy-plane, with point P at distance z. Divide the disk into rings of equal width Δr. Ring i has radius ri and charge ΔQi. Solve: Using the result of Example 29.11, we write the potential at distance z of ring i as
Vi =
1 4πε 0
ΔQi ri + z 2
2
⇒ V = ∑Vi = i
1 ΔQi ∑ 4πε 0 i ri 2 + z 2
Noting that ΔQi = ηΔAi = η2πriΔr, V= =
1 4πε 0
η 2π ∑ i
ri Δr ri + z 2
2
=
η 2ε 0
Rout
∫
Rin
r dr r +z 2
2
=
η ⎡ 2 2 ⎤R η ⎡ 2 r +z Rout + z 2 − Rin2 + z 2 ⎤ = ⎣ ⎦ ⎦ R 2ε 0 2ε 0 ⎣
Q ⎡ R2 + z 2 − R2 + z 2 ⎤ out in 2 2 ⎣ ⎦ 2πε 0 ( Rout − Rin )
In the limit Rin → 0 m,
V=
Q ⎡ 2 Rout + z 2 − z ⎤ 2 ⎣ ⎦ 2πε 0 Rout
This is the same result obtained for a disk of charge in Example 29.12.
out
in
29.73. Solve: (a) A charge of 40 nC is separated into two charges which are placed 3.0 cm apart. The potential energy of the charge arrangement is 90 μJ. What are the two charges?
(b) We have q1q2 = 3.0 × 10−16 C2 and q1 + q2 = 40 × 10−9 C. Substituting into the first equation an expression for q1 from the second equation,
( 40 ×10
−9
C − q2 ) q2 = 3.0 × 10−16 C 2 ⇒ q22 − ( 40 × 10−9 C ) q2 + 3.0 × 10−16 C 2 = 0
The solutions to this equation are q2 = 30 nC and 10 nC. This means the two charges are 10 nC and 30 nC.
29.74. Solve: (a) A proton is shot toward a 2.0 nC point charge with a speed of 2.5 × 106 m/s. What is the proton’s speed when it is 1.0 mm from the charge? (b) Solving the equation yields vi = 1.67 × 106 m/s.
29.75. Solve: (a) Two 3.0 nC charges are distance d apart. At a point 3.0 cm from one charge, on the side opposite the other charge, the potential is 1200 V. Find the separation d between the charges. (b) The given equation simplifies to 27 N m 2 /C 27 N m 2 /C 900 N m/C + = 1200 V ⇒ = 300 V ⇒ d = 0.060 m = 6.0 cm 0.030 m + d 0.030 m + d
29.76. Model: The charges are point charges. The potential at a point is the sum of the potentials from all the charges. Visualize:
The two charges are in the xy-plane. Solve: Let the potential be zero at the point P(x, y). Then, VP =
1 4πε 0
Q1 x2 + y2
+
1 4πε 0
⇒ 3 x2 + y 2 =
Q2
( 4.0 cm − x )
(4 − x)
2
2
+ y2
=0 V=
1.0 ×10−9 C x2 + y2
+
−3.0 × 10 −9 C
(4 − x)
2
+ y2
+ y 2 ⇒ 9x2 + 9y2 = 16 + x2 + y2 – 8x
⇒ 8x2 + 8x – 16 + 8y2 = 0 ⇒ 8 ( x + 12 ) − 2 − 16 + 8 y 2 = 0 ⇒ ( x + 12 ) + y 2 = 188 = 2
2
9 4
This is the equation of a circle, x′2 + y′2 = R2, whose center is at ( x = − 12 cm, y = 0 cm ) and whose radius is
R=
9 4
= 32 = 1.5 cm. All values of (x, y) that satisfy this equation are the points with zero potential. This zero
potential contour map is shown as a dotted line in the figure. Assess: It is clear from the contour map that two zero-potential points lying on the x-axis are at x = 1.0 cm and x = −2.0 cm. Each position is nearer to the smaller charge and away from the bigger charge, as it must be.
29.77. Model: Apply the principles of conservation of energy and conservation of momentum. Visualize:
The figure shows a before-and-after pictorial representation of the charged particles moving toward each other. The proton’s physical quantities are denoted by the subscript p and that of the alpha particle by the subscript a. Solve: The conservation of energy equation is Kf + Uf = Ki + Ui. Initially when the charges are far away, Ui = 0 J and K i = 12 mp vip2 + 12 ma via2 . At the distance of closest approach, Uf =
1 ( e )( 2e ) 4πε 0 d
K f = 12 ma vfa2 + 12 mpvfp2
The particles are not at rest at closest approach. That would violate conservation of momentum. Instead, the particles are instantaneously moving with the same velocity vfa = vfp = vf so that the distance between them is (instantaneously) not changing. The conservation of energy equation simplifies to 1 2
(m
a
+ mp ) vf2 +
1 2e 2 = 4πε 0 d
1 2
(m
p
+ ma ) ( 0.010c )
2
The conservation of momentum equation Pafter = Pbefore is mavf + mpvf = ma(0.010c) – mp(0.01c)
⇒ vf =
( 0.010c ) ( ma − mp ) ( 0.010 ) ( 3 × 108 m/s ) ( 4 u − 1 u ) ma + mp
=
4 u +1 u
= 1.80 × 106 m/s
Substituting into the energy-conservation equation, 1 2
( 4 u + 1 u ) (1.80 ×106 m/s )
2
+ ( 9.0 × 109 N m 2 /C2 )
⇒
⇒d =
2 (1.60 × 10−19 C ) d
2
= 12 ( 4 u + 1 u ) ( 3.0 × 106 m/s )
4.608 × 10−28 N m 2 = 1.44 × 1013 u m 2 /s 2 d
4.608 ×10−28 N m 2 = 1.93 ×10−14 m (1.44 ) (1.661×10−27 kg ) ×1013 ( m2 /s2 )
2
29.78. Model: Energy and momentum are conserved. Visualize: Please refer to Figure CP29.78. Let the two spheres have masses mA and mB, and speeds vA and vB when they are very far apart. Solve: The energy-conservation equation Kf + Uf = Ki + Ui is
( ⇒
1 2
1 2
mA vA2 + 12 mBvB2 ) + 0 J = 0 J +
1 qA qB 4πε 0 rAB
( 0.002 kg ) vA2 + 12 ( 0.001 kg ) vB2 = ( 9.0 ×109 N m 2 /C2 )
( 2.0 ×10
−9
C )(1.0 × 10−9 C )
2.0 ×10−3 m
⇒ vA2 + 0.5 vB2 = 0.0090 m 2 /s 2 To solve for vA and vB, we need another equation relating vA and vB. From the momentum conservation equation Pafter = Pbefore we get mAvA + mBvB = 0 kg m/s ⇒ (2.0 g)vA + (1.0 g)vB = 0 kg m/s ⇒ vB = −2vA Substituting this expression into the energy-conservation equation,
vA2 + 0.5 ( 2vA ) = 0.009 m 2 /s 2 ⇒ 3vA2 = 0.009 m2 /s 2 2
Solving these equations, we get vA = –0.0548 m/s and vB = +0.110 m/s. Thus the speeds are 0.055 m/s for A and 0.110 m/s for B.
29.79. Model: Energy and momentum are conserved. Visualize: Please refer to Figure CP29.79. Let the two spheres have masses mC and mD, and speeds vC and vD when the two spheres collide. Solve: The energy-conservation equation Kf + Uf = Ki + Ui is
(
1 2
⎛ 1 qC qD ⎞ qC qD 1 mCvC2 + 12 mDvD2 ) + ⎜ ⎟=0 J+ 4πε 0 0.010 m ⎝ 4πε 0 0.002 m ⎠
The momentum conservation equation Pafter = Pbefore is
⎛ 2.0 g ⎞ mCvC + mDvD = 0 kg m/s ⇒ vC = − ⎜ ⎟ vD = −2vD ⎝ 1.0 g ⎠ Substituting this expression in the energy equation, 1 2
( 0.001 kg ) 4vD2 + 12 ( 0.002 kg ) vD2 = ( 9.0 ×109 N m2 /C2 )( 2.0 ×10−9 C )( −1.0 ×10−9 C ) ⎡⎢
1 1 ⎤ − ⎥⎦ 0.010 m 0.002 m ⎣
⇒ ( 0.0030 kg ) vD2 = 7.2 ×10−6 N m Solving these equations gives vD = 0.049 m/s = 4.9 cm/s and hence vC = −9.8 cm/s. The speed of C is +9.8 cm/s.
29.80. Model: The charges making the dipole are point charges. The potential is the sum of potentials from each charge. Visualize:
Point P is located at a distance r from the center of the dipole and it makes an angle θ with the dipole axis. Solve: The potential at P due to the dipole is
⎡ ⎤ 1 Q 1 ( −Q ) Q ⎢ 1 1 ⎥ + = − 2 ⎥ 2 4πε 0 r+ 4πε 0 r− 4πε 0 ⎢ x 2 + ( y − s 2 )2 + + x y s 2 ( ) ⎣ ⎦ ⎡ ⎤ Q ⎢ 1 1 ⎥. = − 2 ⎥ 2 2 4πε 0 ⎢ x 2 + y 2 − ys + ( s 2 )2 x y ys s 2 + + + ( ) ⎣ ⎦
V=
Because x, y >> s , (s/2)2 can be ignored compared with x2, y2, and ys. The potential is
V=
−1/ 2 −1/ 2 ⎤ 1 1 1 ⎛ ys ⎞ ⎤ Q ⎡ Q ⎡ 1 ⎛ ys ⎞ − = − − + 1 1 ⎢ 2 ⎥ ⎢ ⎜ ⎜ ⎟ ⎥. 2 ⎟ 4πε 0 ⎣⎢ r − ys r ⎝ r 2 ⎠ ⎦⎥ r 2 + ys ⎦⎥ 4πε 0 ⎣⎢ r ⎝ r ⎠
Using the binomial expansion, V= =
Q 4πε 0 1 4πε 0
1 ⎡⎛ ys … ⎞ ⎛ ys Q 1 2 ys ⎞⎤ + ⎟ − ⎜1 − 2 + … ⎟ ⎥ = ⎜1 + 2 πε r ⎢⎣⎝ 2r 2 2 r 4 ⎠ ⎝ ⎠⎦ 0 r 2r
( Qs ) y = r
3
1 4πε 0
where p = Qs is the dipole moment of the dipole.
( Qs ) r cosθ r
3
=
p cosθ 4πε 0 r 2 1
29.81. Model: The potential energy is determined by the electric potential. Solve:
(a) The electric field in the parallel-plate capacitor when the plates have charges ±q is η q qd E= = ⇒ ΔVC = Ed = ε 0 Aε 0 Aε 0
The increase in the potential energy dU due to moving charge dq through potential difference ΔVC is
dU = dq ΔVC =
qd dq Aε 0
(b) Integrating the expression for dU from q = 0 (uncharged) to q = Q (fully charged), Q
U = ∫ dU = ∫ 0
Q
Q qd d d ⎡ q2 ⎤ 1 ⎛ Qd ⎞ 1 dq = q dq = ⎟ Q = QΔVC ⎢ ⎥ = ⎜ ∫ Aε 0 Aε 0 0 Aε 0 ⎣ 2 ⎦ 0 2 ⎝ Aε 0 ⎠ 2
where we have used the expression for ΔVC from part (a).
29.82. Model: The potential energy is determined by the electric potential. Solve: (a) A sphere of radius R that is uniformly charged with a charge q is at a potential V0 (see Equation 29.32): 1 q V0 = 4πε 0 R The increase in the potential energy dU when an infinitesimal amount of charge dq is moved from infinity (where the potential is zero) to the surface of the sphere (where the potential is V0) is
dU = dqV0 =
1 q dq 4πε 0 R
(b) Integrating the expression for U from q = 0 (uncharged) to q = Q (fully charged), Q
Q
q 1 1 Q2 dq = qdq = ∫ 4πε 0 R 4πε 0 R 0 4πε 0 2 R 0
U = ∫ dU = ∫
1
(c) The self-energy of a proton is
(1.60 ×10−19 C) = 2.3 ×10−13 J 1 Q2 U= = ( 9.0 ×109 N m 2 /C2 ) 4πε 0 2 R ( 2 ) ( 0.50 ×10−15 m ) 2
29.83. Model: Assume the wire is a line of charge with uniform linear charge density. The electric potential at the point is the sum of contributions from all charges present. Visualize: Please refer to Figure CP29.83. Let the origin be the center of our coordinate system. Divide the charged wire into two straight sections and the central semicircle. Divide each section into N small segments, each of length Δx and with charge Δq. Segment i, located at position xi, contributes a small amount of potential Vi at the semicircle center. Solve: For the right hand straight section, the contribution of the ith segment is
Vi =
Δq 4πε 0 ri
=
Δq 4πε 0 xi
=
λΔx 4πε 0 xi
where Δq = λΔx. The Vi are now summed and the sum is converted to an integral giving
Vstraight section =
3R λ 3 R dx λ λ = ln ( 3) ⎡⎣ln ( x ) ⎤⎦ R = ∫ 4πε 0 R x 4πε 0 R 4πε 0 R
The integration limits are set by the physical location of the straight section. By symmetry, the left hand straight section adds the same amount to the total potential at the semicircle center. The potential due to the ith segment of the semicircular section is Vi =
λ ( R Δθ ) Δq Δq λ = = = Δθ 4πε 0 ri 4πε 0 R 4πε 0 R 4πε 0 R
where we have used the arc length Δx = R Δθ . The Vi are now summed and the sum converted to an integral giving
Vsemicircle =
λ
π
4πε 0 R ∫ 0
dθ =
λπ 4πε 0 R
=
λ 4ε 0 R
The total potential is the sum of the potentials due to the three sections: ⎛ λ ⎞ λ λ ⎛ 2ln(3) ⎞ V = Vsemicircle + 2Vstraight section = ln ( 3) ⎟ = + 2⎜ ⎜1 + 4ε 0 R 4 R 4 πε ε π ⎟⎠ 0 0R ⎝ ⎝ ⎠
29.84.
Model: The electric potential at a point is the sum of contributions from all charges present. Visualize: Please refer to Figure 29.32 in Example 29.12. Divide the disk into rings of equal width Δr. Ring i has radius ri and charge Qi. We can use the result of Example 29.11 to write the potential at distance z of ring i as
Vi =
ΔQi
1 4πε 0
ri 2 + z 2
The potential at the point is the sum of all rings. V = ΣVi =
1
N
4πε 0
∑ i =1
ΔQi ri 2 + z 2
The charge on each ring is related to the area of each ring by
ΔQi = ηΔAi = cri 2π rΔr = 2π cri 2Δr With this substitution and converting the sum to an integral, the potential is
2π c r 2 dr ∫ 4πε 0 0 r 2 + z 2 R
V= Using an integral table, V=
)
(
R
⎛ z 1 c ⎛1 c ⎛ 2 2 2 2 ⎞ 2 2 ⎜ R R + z + ln ⎜ ⎜ r r + z − ln r + r + z ⎟ = 2 2 2ε 0 ⎝ 2 2 ⎠ 0 4ε 0 ⎜⎝ ⎝R+ R +z
The total charge on the disk is R
Q = ∫η dA = ∫ ( cr )2π rdr = 0
3Q Thus c = . With this substitution, 2π R3 z ⎛ 3Q ⎛ V= R R 2 + z 2 + ln ⎜ 3⎜ ⎜ 2 2 8πε 0 R ⎝ ⎝R+ R +z
⎞⎞ ⎟ ⎟⎟ ⎠⎠
2π c 3 R 3
⎞⎞ ⎟ ⎟⎟ ⎠⎠
29.85. Model: Assume the cylinder is uniformly charged. Visualize:
Divide the cylinder into narrow rings of width dz, radius R, and charge dq.
Solve: The potential at the center is the sum of the potential of all the rings. The potential at distance z on the axis of a ring of charge was found in Example 29.11 to be 1 dq Vring = 4πε 0 R 2 + z 2
The surface area of the ring is dA = (2πR)dz, so the amount of charge is dq = η dA = (2πRη)dz. The surface area of the cylinder is A = (2πR)L, so the surface charge density is η = Q/2πRL. Thus dq = (Q/L)dz and the potential of the ring is Vring =
Q/L 4πε 0
dz R2 + z 2
The potential at the center of the cylinder is found by summing (i.e., integrating) the potential due to all rings. That is,
Vring = =
Q/L L / 2 dz 2 Q/L = 4πε 0 ∫− L / 2 R 2 + z 2 4πε 0
∫
2 2 Q/L ⎛⎜ L ⎛ L ⎞ ⎞⎟ + 1+ ⎜ ln ⎟ 4πε 0 ⎜ 2 R ⎝ 2 R ⎠ ⎟⎠ ⎝
L/2 0
dz R2 + z 2
=
(
)
L/2 2 Q/L ⎡ ln z + R 2 + z 2 ⎤ ⎦⎥ 0 4πε 0 ⎣⎢
30.1. Solve: The potential difference ΔV between two points in space is xf
ΔV = V ( xf ) − V ( xi ) = − ∫ Ex dx xi
where x is the position along a line from point i to point f. When the electric field is uniform, xf
ΔV = − Ex ∫ dx = − Ex Δx = − (1000 V/m )( 0.30 m − 0.10 m ) = −200 V xi
30.2. Visualize:
Solve:
The potential difference ΔV between two points on the y-axis is yf
ΔV = − ∫ E y dy yi
When the electric field is uniform, the above result simplifies to ΔV = − E y Δy. In the present problem,
Δy = yf – yi = 0.05 m – (−0.05 m) = 0.10 m The y-component of the electric field is
E y = −50,000 V/m ⇒ ΔV = – (–50,000 V/m)(0.10 m) = +5.0 kV Assess: Vf – Vi = 5.0 kV shows that the potential at point f is higher than at point i. This is because the electric field is directed from the point at the higher potential to the point at the lower potential.
30.3. Model: The potential difference is the negative of the area under the Ex vs x curve. Visualize: Please refer to Figure EX30.3. Solve: The potential difference between x = 1.0 m and x = 3.0 m is 1 ⎛ ⎞ ΔV = − ⎜ ( 200 V/m )( 2.0 m − 1.0 m ) + ( 200 V )( 3.0 m − 2.0 m ) ⎟ = −300 V. 2 ⎝ ⎠ Assess: The potential difference is negative since the electric field points in the direction of decreasing potential.
30.4. Model: The potential difference is the negative of the area under the Ex vs x curve. Visualize: Please refer to Figure EX30.4. Solve: The potential difference between the origin and x = 3.0 m is 1 ⎛1 ⎞ ΔV = V ( x = 3.0 m ) − V ( x = 0 m ) = − ⎜ ( −100 V )(1.0 m − 0 m ) + ( 200 V )( 3.0 m − 1.0 m ) ⎟ 2 2 ⎝ ⎠ = −150 V Thus V ( 3.0 m ) = V ( 0 m ) − 150 V = −50 V − 150 V = −200 V Assess:
The potential decreases from the origin to x = 3.0 m.
30.5. Solve: The work done is exactly equal to the increase in the potential energy of the charge. That is, W = ΔU = qΔV = q (Vf − Vi ) = (1.0 ×10−6 C ) (1.5 V ) = 1.5 × 10−6 J Assess:
The work done by the escalator on the charge is stored as electric potential energy of the charge.
30.6. Solve: The Van de Graaff generator or the motor that runs the belt does work in lifting a positive ion (q = e) against the downward force on the positive charge that is moving up the belt. The work done is
W = ΔU = qΔV = q (1.0 × 106 V − 0 V ) = (1.60 × 10−19 C )(1.0 × 106 V ) = 1.6 × 10−13 J Assess:
The work done by the generator in lifting the charge is stored as electric potential energy of the charge.
30.7. Solve: The emf is defined as the work done per unit charge by the charge escalator or the battery. That is, ε=
Wchem 0.60 J = = 12 V q 0.050 C
30.8.
Solve: The work done is the potential energy gained by the electron.
W = U = qΔV = −eΔV
( 2.4 ×10 − (1.6 ×10
−19
⇒ ΔV =
−19
J)
C)
= −1.5 V
The potential difference from the positive to the negative terminal is –1.5 V, so the emf of the solar cell is 1.5 V.
30.9. Model: The electric field points in the direction of decreasing potential and is perpendicular to the equipotential lines. Visualize: Please refer to Figure EX30.9. The three equipotential surfaces correspond to potentials of 0 V, 100 V, and 200 V. Solve: The electric field component along a direction of constant potential is Es = − dV ds = 0 V/m. But, the electric field component perpendicular to the equipotential surface is G ΔV 100 V E = = = 10 kV/m Δ s 0.01 m The direction of the electric field vector is “downhill,” perpendicular to the equipotential surfaces, and directed to the left. That is, the electric field is 10 kV/m to the left.
30.10. Model: The electric field is perpendicular to the equipotential lines and points “downhill.”
Visualize: Please refer to Figure EX30.10. Three equipotential surfaces at potentials of −200 V, 0 V, and 200 V are shown. Solve: The electric field component perpendicular to the equipotential surface is
E=−
ΔV 200 V =− = −20 kV/m Δs 0.01 m
The electric field vector is in the third quadrant, 45° below the negative x-axis. That is, G E = ( 20 kV/m, 45° below − x -axis )
30.11. Model: The electric field is the negative of the slope of the graph of the potential function. Visualize: Please refer to Figure EX30.11. Solve: There are three regions of different slope. For 0 cm < x < 10 cm and 20 cm < x < 30 cm,
ΔV = 0 V/m ⇒ Ex = 0 V/m Δx For 10 cm < x < 20 cm, ΔV −100 V − (100 V ) = = −2000 V/m ⇒ Ex = +2000 V/m Δx 0.20 m − 0.10 m
Assess:
Because Es = − dV ds , the electric field is zero where the potential is not changing.
30.12. Model: The electric field is the negative of the slope of the graph of the potential function. Visualize: Please refer to Figure EX30.12. Solve: There are three regions of different slope. For 0 cm < x < 1 cm,
ΔV 50 V − 0 V = = 5000 V/m ⇒ Ex = −5000 V/m Δ x 0.01 m − 0 m For 1 cm < x < 2 cm, ΔV −50 V − ( 50 V ) = = −10,000 V/m ⇒ Ex = 10,000 V/m dx 0.02 m − 0.01 m For 2 cm < x < 3 cm, 0 V − ( −50 V ) ΔV = = 5000 V/m ⇒ Ex = −5000 V/m Δ x 0.03 m − 0.02 m
Assess:
ΔV Δ x and Ex are the negative of each other.
30.13. Visualize:
Solve:
The electric potential difference ΔV between two points in a uniform electric field is
V ( x f ) − V ( xi ) = − ∫ Ex dx = − Ex ( x f − xi ) Choosing xi = −1.0 m and xf = +1.0 m, +1000 V − (−1000 V) = −Ex[1.0 m − (−1.0 m)] ⇒ Ex = −1.0 kV/m Alternatively, xi = 1.0 m and xf = −1.0 m. For this choice,
−1000 V − (+1000 V) = −Ex [−1.0 m − (1.0 m)] ⇒ Ex = −1.0 kV/m Assess: The choice of initial and final positions does not change the physical nature of the electric field or the potential difference.
30.14. Model: The electric field is the negative of the derivative of the potential function. Solve: (a) From Equation 30.11, the component of the electric field in the s-direction is Es = − dV ds . For the given potential,
dV d V = (100 x 2 V ) = 200 x ⇒ Ex = −200 x V m dx dx m At x = 0 m, Ex = 0 V/m. (b) At x = 1 m, Ex = −200 (1) V/m = −200 V/m. Assess: The potential increases with x, so the electric field must point in the −x-direction.
30.15.
Solve:
(a) Since Ex = −
dV , we have dx
Ex = −
d ( 50 x − 100 / x ) V/m = − ( 50 + 100 / x 2 ) V/m dx
At x = 1.0 m,
(
Ex = − 50 + 100 / (1.0 )
2
) V/m = −150 V/m
(b) At x = 2.0 m, ⎛ 100 ⎞ Ex = −⎜ 50 + ⎟ V/m = −75 V 2 ⎜ ( 2.0 ) ⎟⎠ ⎝
30.16. Model: Use Kirchhoff ’s loop law. Visualize: Please refer to Figure EX30.16. Solve: For any path that starts and ends at the same point, ΔVloop = ∑ ( ΔV )i = 0 V. Therefore, i
ΔV12 + ΔV23 + ΔV34 + ΔV41 = 30 V + 50 V + ΔV34 – 60 V = 0 V ⇒ ΔV34 = −20 V
That is, the potential at point 4 is less than the potential at point 3.
30.17. Model: Assume that the battery is ideal and that the capacitor is a parallel-plate capacitor. Solve:
(a) From Equation 30.17, the capacitance is
C=
ε0 A d
=
(8.85 × 10
−12
C 2 / N m 2 ) ( 0.020 m × 0.020 m ) 0.00050 m
= 7.1 × 10−12 F = 7.1 pF
(b) The charges on the electrodes are
Q = ±C ΔVC = ± ( 7.1×10−12 F ) (100 V ) = ±7.1× 10−10 C = ±0.71 nC
30.18. Solve: From Equation 30.17, the capacitance is C= Assess:
ε0 A d
=
ε 0 L2 d
⇒L=
Cd
ε0
=
(100 × 10
−12
F )( 0.20 × 10−3 m )
8.85 × 10−12 C 2 / N m 2
The capacitance depends only on the geometry of the capacitor.
= 0.048 m = 4.8 cm
30.19. Model: Assume that the battery is ideal and the connecting wires have zero resistance. Solve:
The capacitance of a capacitor is the ratio of the charge and the potential difference ΔVC:
C=
Q Q 30 ×10−6 C ⇒ ΔVC = = = 3.0 V ΔVC C 10 ×10−6 F
For an ideal battery, E = ΔVbat = ΔVC . So, E = 3.0 V.
30.20. Model: Assume that the battery is ideal and the wires connecting the battery with the capacitor have zero resistance. Solve: Using Equation 30.19,
E = ΔVbat = ΔVC =
48 ×10−6 C Q = = 24 V 2.0 ×10−6 F C
30.21. Model: Assume that the battery is ideal. Solve:
Equation 30.19 is ΔVC =
Q Q 45 ×10−9 C =E ⇒ C = = = 5.0 ×10−9 F C ΔVC 9.0 V
30.22. Solve: According to Equation 30.21, Ceq = C1 + C2 + C3 = 6 μF + 10 μF + 16 μF = 32 μF
30.23. Solve: According to Equation 30.23, −1
−1
⎛ 1 ⎛ 1 1 1 ⎞ 1 1 ⎞ −6 Ceq = ⎜ + + ⎟ =⎜ + + ⎟ = 3.0 × 10 F = 3.0 μ F μ μ μ C C C 6 F 10 F 16 F ⎝ ⎠ 2 3 ⎠ ⎝ 1
30.24. Model: Two capacitors in parallel combine to give greater capacitance. Solve: Since we want a capacitance of 50 μF and we have a 30 μF capacitor, we must connect the second capacitor in parallel with the 30 μF capacitor. That is, C + 30 μF = 50 μF ⇒ C = 50 μF − 30 μF = 20 μF
30.25. Model: Two capacitors in series combine to give less capacitance. Solve: Since we have a 75 μF capacitor and we want a 50 μF capacitance, we must connect the second capacitor in series with the 75 μF capacitor. The capacitance of the second capacitor is calculated as follows: 1 1 1 + = ⇒ C = 150 μF 75 μ F C 50 μ F
30.26. Visualize: Please refer to Figure EX30.26. Solve: Any two electrodes, regardless of their shape, form a capacitor whose capacitance is defined as C = Q ΔVC . The capacitance is C=
20 ×10−9 C = 2.0 ×10−10 F = 200 pF 100 V
30.27. Model: Assume the battery is ideal. Visualize: Please refer to Figure EX30.27. Solve: For an ideal battery, the potential difference across the capacitor is the same as the emf of the battery. Thus,
E = ΔVC =
Q ⇒ Q = CE = (10 × 10−6 F)(1.5 V) = 15.0 × 10−6 C C
30.28. Solve: From Equation 30.26, the energy stored in a capacitor is U C = 12 C ( ΔVC ) ⇒ ΔVC = 2
2 (1.0 J ) 2U C = = 1.41 kV C 1.0 ×10−6 F
30.29. Solve: The graph of Figure EX30.29 shows that the capacitor is fully charged at 3.0 s and that its charge is 200 μC. From t = 0 s to t = 3.0 s, the charge on the capacitor follows the equation Q = ( 200 μ C 3.0 s ) t. Using Equation 30.26, the energy stored in the capacitor as a function of time is 2
UC =
1 Q 2 1 ⎛ 200 μ C ⎞ t2 = ⎜ = (1.111×10−3 C2 /s F ) t 2 ⎟ 2 C 2 ⎝ 3.0 s ⎠ 2.0 μ F
30.30. Solve: Using Equation 30.26, U C1 = 12 C1 ( ΔVC1 )
2
U C2 = 12 C2 ( ΔVC2 )
2
Because C2 = 12 C1 and ΔVC2 = 2ΔVC1, we have C1 ( ΔVC1 ) U C1 12 C1 ( ΔVC1 ) 1 = = = U C2 12 C2 ( ΔVC2 )2 ( 12 C1 ) 4 ( ΔVC1 )2 2 2
2
30.31. Solve: The energy density is uE = ( 5.0 ×10−12 J ) ( 0.020 m × 0.020 m × 0.020 m ) = 6.25 ×10−7 J m3 . uE = 12 ε 0 E 2 ⇒ E =
2uE
ε0
=
2 ( 6.25 ×10−7 J/m3 ) 8.85 ×10−12 C2 /N m 2
= 1.19 kV/m
30.32. Solve: (a) From Equation 30.26, the energy stored in the charged capacitor is 1 1 ⎛ Aε ⎞ 2 2 U C = C ( ΔVC ) = ⎜ 0 ⎟ ( ΔVC ) 2 2⎝ d ⎠
2 2 −12 1 π ( 0.010 m ) (8.85 ×10 C /N m ) 2 ( 200 V ) = 1.11×10−7 J 2 0.50 ×10−3 m 2
=
(b) From Equation 30.28, the energy density in the electric field is 2
2
1 1 ⎛ ΔV ⎞ 1 200 V ⎛ ⎞ 3 uE = ε 0 E 2 = ε 0 ⎜ C ⎟ = ( 8.85 ×10−12 C2 /N m 2 ) ⎜ ⎟ = 0.71 J/m −3 2 2 ⎝ d ⎠ 2 0.50 × 10 m ⎝ ⎠
30.33.
Model: Assume the capacitor is a parallel-plate capacitor. Solve: (a) The capacitance is increased by a factor of the dielectric constant κ over the equivalent vacuumfilled capacitor.
C = κ C0 = κ
ε0 A d
(8.85 ×10 = ( 3.7 )
−12
C2 / N m 2 ) π ( 0.050 m )
0.20 ×10−3 m
2
= 1.29 × 10−9 F = 1.29 nF
We have used κ = 3.7 for paper from Table 30.1. (b) The dielectric strength is the maximum possible electric field in the capacitor before breakdown. In this capacitor,
E=
ΔV ΔV ⇒ 16 ×106 V/m = ⇒ ΔV = 3.2 kV d ( 0.20 ×10−3 m )
30.34. Solve:
Model: The electrodes form a parallel-plate capacitor. (a) The capacitance of the equivalent vacuum-filled capacitor is
C0 =
ε0 A
=
d
(8.85 ×10
−12
C 2 /Nm 2 )( 5.0 ×10−3 m )
( 0.10 ×10−3 m )
2
= 2.21×10−12 F
From Table 30.1, the dielectric constant of mylar is κ = 3.1. With the battery attached, the potential difference across the plates is ΔVC = ΔVbatt = 9.0 V . The charge on the plates is
Q = C ΔV = κ C0 ΔV = ( 3.1) ( 2.21 × 10−12 F ) ( 9.0 V ) = 6.17 × 10−11 C = 62 pC The electric field inside the capacitor is
E=
E0
κ
=
1 ΔVC
κ d
9.0 V ⎛ 1 ⎞⎛ ⎞ =⎜ ⎟⎜ ⎟ = 20 kV/m −3 3.1 0.10 × 10 m ⎝ ⎠⎝ ⎠
(b) With the battery connected, ΔVC = 9.0 V. The capacitor is now vacuum-insulated. E = E0 =
ΔVC 9.0 V = = 90 kV/m d 0.10 ×10−3 m
Q = C0 ΔVC = ( 2.21×10−12 F ) ( 9.0 V ) = 20 pC Assess: Since the battery remains connected as the mylar is withdrawn, the potential difference across the plates does not change.
30.35. Solve:
Model: The disks form a parallel-plate capacitor. (a) The charge on each of the metal disks is
Q = C ΔVC = κ Pyrex C0 ΔVC = κ Pyrex
ε0 A d
ΔVC
The charge density on a disk is
η=
(8.85 × 10−12 C2 /N m2 ) (1000 V ) Q ε ΔV = κ Pyrex 0 C = ( 4.7 ) A d ( 0.50 × 10−3 m )
= 8.3 × 10−5 C/m 2 = 83 μ C/m 2 (b) In the absence of the dielectric, the surface charge density on the disks is
η0 =
η κ Pyrex
= 1.77 × 10−5 C/m 2 .
The surface charge density on the Pyrex is ⎛ 1 ⎞ 1 ⎞ ⎛ = 1.77 ×10−5 C/m 2 ) ⎜1 − ⎟ ⎜ κ Pyrex ⎟⎟ ( 4.7 ⎝ ⎠ ⎝ ⎠ −5 2 2 = 1.39 × 10 C/m = 13.9 μ C/m
ηinduced = η0 ⎜1 −
Assess:
The induced charge density on the dielectric is less than the charge density on the plates.
30.36. Visualize: Please refer to Figure P30.36. Solve: (a) The electric field points “downhill.” So, point A is at a higher potential than point B. (b) In a region that has a uniform electric field, Equation 30.3 gives the potential difference between two points: ΔV = − ∫ Ex dx = − Ex ( xf − xi ) ⇒ VB – VA = −(1000 V/m)(0.07 m) = −70 V That is, the potential at point A is 70 V higher than the potential at point B.
30.37. Solve: (a)
(b) Equation 30.3 gives the potential difference between two points in space: xf
30 cm
xi
−20 cm
ΔV = − ∫ Ex dx = −
∫ ( −1000 x V/m ) dx
0.30 m
⎡ x2 ⎤ 1000 ⎡ 2 2 = + ⎢1000 ⎥ V=+ ( 0.30 m ) − ( −0.20 m ) ⎦⎤ V = +65 V ⎣ 2 2 ⎣ ⎦ −0.20 m Assess: E is positive for negative x but negative for positive x, so the potential difference depends on the square of the positions.
30.38. Solve: (a)
(b) Equation 30.3 gives the potential difference between two points in space: x
x
xf
f f ⎡ x2 ⎤ ΔV = V ( xf ) − V ( xi ) = − ∫ Ex dx = − ∫ ( 5000 x V/m ) dx = −5000 ⎢ ⎥ V = −2500 ( xf2 − xi2 ) V ⎣ 2 ⎦ xi xi xi
Taking V(xi) = 0 V at xi = 0 m, and replacing xf with simply x, V(x) = −(2500 x2) V. (c) A graph of V versus x over the region −1 m ≤ x ≤ 1 m is shown in part (a). Assess: As it must be, we have
−
dV d = − ( −2500 x 2 ) V = 5000x = Ex dx dx
30.39. Visualize:
Solve:
Equation 30.3 gives the potential difference between two points in space rf
ΔV = V ( rf ) − V ( ri ) = − ∫ Er dr ri
r
⇒ V ( r ) − V ( R ) = −∫
λ
2πε 0 r R
dr = −
λ 2πε 0
[ln r ]R = −
⇒ V ( r ) = V0 − Assess:
r
λ 2πε 0
[ln r − ln R ] = −
λ 2πε
ln
r R
r λ ln 2πε 0 R
At r = R, ln ( R R ) = 0 and V(R) = V0, as it must be due to the assumed constant of integration.
30.40. Model: The electric field is the negative of the slope of the graph of the potential function. Visualize: Please refer to Figure P30.40. Solve: We have
Ex = −
dV ⇒ dV = −Exdx dx
For x ≥ 1 cm, Ex = 1000 V/m. Integrating the above expression, V = − (1000 V/m)x + C Because V = 0 V at x = 0.03 m, C = (1000 V/m)(0.03 m) = 30 V. Thus, V = − (1000 V/m)x + 30 V (x ≥ 1 cm) This means V at x = 0.02 m is 10 V and V at x = 0.01 m is 20 V. For x < 1 cm, Ex = (105 V/m)x. Therefore, dV = − (105 V/m) x dx. Integrating again, 5 5 ∫ dV = − ∫ (10 V/m ) x dx + C ⇒ V = − (10 V/m )
Because V = 20 V at x = 0.01 m, C = 25 V. At x = 0 m, V = 25 V.
x2 + C (x < 1 cm) 2
30.41. Model: Assume the electrodes form parallel-plate capacitors with a uniform electric field between the plates. Visualize:
Please refer to Figure P30.41. The three metal electrodes serve as plates for two capacitors. On the middle electrode, half the charge is located on the left face and half on the right face, thus forming two capacitors. Each plate of the two capacitors carries a charge of ±50 nC. Solve: (a) In the space 0 cm < x < 1 cm, the electric field points to the left and its magnitude is
E=
η Q 50 × 10−9 C = = = 1.41 × 107 V/m ε 0 Aε 0 ( 0.02 m )2 ( 8.85 × 10−12 C2 / N m 2 )
G In the region 1 cm ≤ x ≤ 2 cm, E = 0 because in electrostatics the inside of a conductor has no free charge. The electric field in the region 2 cm < x < 3 cm points to the right and has the same magnitude as the electric field in the region 0 cm < x < 1 cm. (b) The potential difference between two points in space with a uniform electric field is
ΔV = Vf − Vi = E ( xf − xi ) Assuming that the negative plate at x = 0 m is at zero potential (Vi = 0 V at xi = 0 cm), Vf = xf E, or simply V = xE. Thus, the potential increases linearly with distance x from the negative plate in the region 0 ≤ x ≤ 1. At x = 1 cm, the potential is
V = xE = (1.0 × 10−2 m)(1.41 × 107 V/m) = 1.41 × 10−5 V The potential must be the same throughout the region 1 cm ≤ x ≤ 2 cm. If this were not the case, we would not have an electrostatic situation with the electric field E = 0 V/m. Using the previous reasoning, the potential decreases linearly in the region 2 cm < x < 3 cm.
30.42. Model: The electric field is the negative of the slope of the graph of the potential function. Visualize:
G Solve: (a) The x-component of E is Ex = −dV/dx. From Figure P30.42, the slope of the graph can be measured in five different regions. For x < −3 m and x > 3 m, the slope is 0 V/m. For −3 m < x < −1 m and 1 m < x < 3 m the slope is −5 V/m. For −1 m < x < 1 m the slope is +10 V/m. (b) The potential is independent of y, so the equipotential lines are parallel to the y-axis. The x-values of various equipotentials can be read from Figure P30.42. These give the potential map shown above. (c) The field vectors point in the direction of decreasing potential. Furthermore, the results of part (a) show that the field vectors in the inner region (−1 m < x < 1 m) are twice as long as the field vectors in the outer region. As you can see by comparing the vectors to the graph of the electric field, this information was used to add the field vectors to the potential map.
30.43. Model: The electric field is the negative of the slope of the potential function. Solve:
The on-axis potential of a charged disk was obtained in Chapter 29 to be
Vdisk =
Q ⎡ 2 2 R + z − z⎤ ⎦ 2πε 0 R 2 ⎣
where the charge on the disk of radius R is Q and the point is a distance z away from the center of the disk. Because E = − dV dz , we have Edisk = − Assess:
⎤ ⎤ dVdisk Q ⎡ 12 ( 2 z ) Q ⎡ z =− − 1⎥ = ⎢ ⎢1 − 2 2 ⎥ dz 2πε 0 R 2 ⎣ R 2 + z 2 ⎦ 2πε 0 R 2 ⎣ R +z ⎦
Using binomial expansion with z >> R, we have
Edisk =
⎡ ⎤ −1/ 2 z ⎢1 − ⎥ = Q ⎡1 − (1 + R 2 z 2 ) ⎤ = Q ⎡1 − 1 + 12 R 2 z 2 + "⎤ = Q 1/ 2 2 2 ⎢ ⎦ 4πε z 2 ⎥⎦ 2πε R 2 ⎣ 2 2 2πε 0 R ⎢ z (1 + R z ) ⎥ 2πε 0 R ⎣ 0 0 ⎣ ⎦ Q
That is, the disk behaves like a point charge. This is an expected result.
30.44. Model: Assume the charged rod is a line of charge of length L. Visualize: Please refer to Figure P30.44. Solve: (a) Divide the charged rod into N small segments, each of length Δx and with charge Δq. The segment i located at position xi contributes a small amount of potential Vi at point P:
Vi =
Δq 4πε 0 ri
=
Δq ( QΔ x L ) = 4πε 0 ( x0 − xi ) 4πε 0 ( x0 − xi )
Point P is at a distance x0 from the origin. This is done to avoid confusion with xi. The Vi are now summed and the sum is converted to an integral giving
V=
⎛ x +L 2⎞ L/2 dx Q Q ln ⎜ 0 = ⎡ − ln ( x0 − x ) ⎤⎦ − L / 2 = ⎟ 4πε 0 L − L∫/ 2 x0 − x 4πε 0 L ⎣ 4πε 0 L ⎝ x0 − L 2 ⎠ Q
L/2
Replacing x0 with x, the potential due to a line charge of length L at a distance x from the center is V=
Q ⎛ x + L/2 ⎞ ln ⎜ ⎟ 4πε 0 L ⎝ x − L/2 ⎠
(b) Because Ex = − dV dx ,
Ex = −
⎤ Q d Q ⎡ 1 1 Q 1 ⎡ln ( x + L/2 ) − ln ( x − L/2 ) ⎦⎤ = − − ⎢ ⎥= 4πε 0 L dx ⎣ 4πε 0 L ⎣⎢ ( x + L/2 ) ( x − L/2 ) ⎦⎥ 4πε 0 x 2 − L2 /4
Assess: When L = 0 m, Ex = Q 4πε 0 x 2. This is the electric field of a point charge Q a distance x away from a point charge, as expected.
30.45. Model: The electric field is the negative of the slope of the potential graph. Visualize: Please refer to Figure P30.45. Solve: Since the contours are uniformly spaced along the y-axis above and below the origin, the slope method is the easiest to apply. Point 1 is in the center of a 75 V change (25 V to 100 V) over a distance of 2 cm, so the slope ΔV/Δ s is 37.5 V/cm or 3750 V/m. Point 2 has the same potential difference in half the distance. Thus the slope at point 2 is 7500 V/m. The magnitudes of the electric fields at points 1 and 2 are 3750 V/m and 7500 V/m. The directions of the electric fields are downward at point 1 and upward at point 2, that is, from the higher potential to the lower potential. That is, G G E1 = ( 3750 V m, down ) E2 = ( 7000 V m, up )
30.46. Model: The electric field is the negative of the slope of the graph of the potential function. Visualize:
→
Solve: (a) At each point, E is perpendicular to the equipotential lines, points “downhill” toward lower potential, and has strength E = dV/ds ≈ ΔV/Δ s. We can read ΔV from the contour lines and use the 8-cm-scale at the bottom of the figure to estimate Δ s. Point 1, on the 100 V contour, is between the 200 V and the 0 V contours. The actual distance on the figure between the 200 V and the 0 V contours can be measured to be ≈6 mm. The length of the 8-cm-scale is ≈12 mm. Thus
200 V ⎛ 6 mm ⎞ Δ s1 ≈ ⎜ = 5000 V/m ⎟ (8 cm) ≈ 4 cm ⇒ E1 ≈ 12 mm 0.040 m ⎝ ⎠ From the symmetry of the figure, the field strengths are the same at points 2, 4, and 5. That is, E1 = E2 = E4 = E5 ≈ 5000 V/m. The contour lines are closer at point 3, so the field strength E3 is larger. The measured spacing between the 100 V and –100 V lines, which bracket the point, is ≈1.5 mm. Thus 200 V ⎛ 1.5 mm ⎞ Δ s3 ≈ ⎜ = 20,000 V/m ⎟ (8 cm) ≈ 1 cm ⇒ E3 ≈ 0.010 m ⎝ 12 mm ⎠
The field strength at 3 is four times that at the other points. (b) The field vectors are shown on the figure. Each is drawn perpendicular to the equipotential at that point.
30.47. Model: The electric field is the negative of the slope of the graph of the potential function.
Solve: The electric potential in a region of space is V = (150 x2 – 200 y2) V where x and y are in meters. The xand y-components are dV dV Ey = − = + ( 400 y ) V/m Ex = − = − ( 300 x ) V/m dy dx At (x,y) = (2.0 m, 2.0 m), Ex = −600 V/m and Ey = 800 V/m. The magnitude and direction of the electric field are E = E x2 + E y2 =
tan θ =
Ey Ex
=
( −600 V/m ) + ( 800 V/m ) 2
2
= 1000 V/m
800 V/m 4 = ⇒ θ = 53.1° above the −x-axis 600 V/m 3
The electric field points 180° − 53° = 127° counterclockwise (ccw) from the +x-axis.
30.48.
Solve:
The electric field is related to the electric potential by G G ⎛ ∂V ˆ ∂V ˆ ⎞ E ( x, y ) = −∇V ( x, y ) = − ⎜ i+ j⎟ dy ⎠ ⎝ dx Compute the partial derivatives. −3 200 x ∂V 200 y ∂V ∂ ⎛ 200 ⎞ ⎛ 1⎞ ⎟ = 200 ⎜ − ⎟ ( 2 x ) ( x 2 + y 2 ) 2 = − =− = ⎜ 3 3 2 2 2 2 2 2 ∂ ∂x ∂x ⎝⎜ x + y ⎟⎠ y ⎝ 2⎠ (x + y ) ( x + y2 ) 2 G G 200 Thus E = xiˆ + yjˆ . At (x,y) = (2.0 m, 1.0 m), E = 17.9 2.0iˆ + 1.0 ˆj V/m. The magnitude 3 ( x2 + y 2 ) 2
(
(
)
E = 17.9
( 2.0
2
+ 1.02 ) V/m = 40 V/m
⎛E ⎞ ⎛1⎞ The direction is θ = tan −1 ⎜ y ⎟ = tan −1 ⎜ ⎟ = 27D ccw from the +x-axis. E ⎝2⎠ ⎝ x⎠
)
30.49. Model: The field strength is the negative of the slope of the graph of the potential function. Solve:
(a)
EquipotentialsG are drawn by connecting points of equal potential. (b) To find E , first determine the direction perpendicular to the equipotential and pointing toward a lower potential. Call this direction the s-axis. The field strength is E = dV ds ≈ ΔV Δ s , where Δ s is the distance between equipotential lines that have potential difference ΔV. The electric field at point A points straight to the left, toward the 50 V equipotential. The distance between the 50 V line and the 75 V line is one grid spacing, or Δ s = 5 cm. Thus, 25 V EA ≈ = 500 V/m 0.05 m G As a vector, the field is EA = ( 500 V/m, left ) . Using the 50 V and 100 V equipotentials to calculate the field at point G B, EB ≈ ( 50 V ) ( 0.05 m ) = 1000 V/m. Thus EB = (1000 V/m, right ) . The field at point C points at a 45° angle toward the 50 V equipotential. The distance Δs between the 50 V and 100 V equipotentials is 5 cm × 2 = 7.07 cm, so G EC ≈ ( 50 V ) ( 0.0707 m ) = 707 V. As a vector, EC = ( 707 V/m, 45° above straight left ) . The field at point D is G symmetrical with point C, so ED = ( 707 V/m, 45° below straight left ) .
(c) The electric field vectors are shown in the figure.
30.50. Model: Conductors connected by a conducting wire are at the same potential. Visualize:
Charge flows from sphere 1 to sphere 2 when the wire is connected. Once connected, the entire system is one large conductor in electrostatic equilibrium. Consequently, the entire system must be an equipotential. That is, the two spheres must have equal potentials V1 = V2 after being connected by the wire. Solve: The potential of a sphere is V = Q 4πε 0 R , so the final charges Q1f and Q2f on the two spheres must be such that Q1f Q2f Q2f ⇒ Q1f = 12 Q2f = = 4πε 0 R1 4πε 0 R2 4πε 0 ( 2 R1 )
where we used the fact that R2 = 2R1. It is also true that charge must be conserved, so Q1f + Q2f = Q1i + Q2i = 6 nC Using Q2f = 2Q1f, we find Q1f + 2Q1f = 3Q1f = 6 nC from which we finally get Q1f = 2 nC and Q2f = 4 nC.
30.51.
Solve:
Because the spheres have equal and opposite charges and they have the same diameter, their potential is zero on a line that bisects the two spheres. On the left and right of this 0 V equipotential line, the equipotential lines show the curvature of the sphere the line is nearer to. This is because the potential nearer to one sphere is dominated by the potential of this sphere. The equipotential lines at 0 V, ±100 V and ±200 V are shown in the figure.
30.52. Model: The potential inside a conductor is the same as the potential on the surface. Solve:
The potential on the surface of the copper sphere must be 500 V. That is,
500 V =
1 Q ( 0.020 m )( 500 V ) = 1.11 nC ⇒ Q = ( 4πε 0 ) R ( 500 V ) = 4πε 0 R 9.0 ×109 N m 2 /C
30.53. Solve:
Model: The charged metal spheres are isolated and far from each other and anything else. (a) The charge on a sphere of radius r, charged to a potential V0 is given by equation 29.33:
Q1 = 4πε 0 rV 1 0
Q2 = 4πε 0 r2V0
The ratio of charge densities is ⎛ Q1 ⎞ ⎜ ⎟ η1 ⎝ A1 ⎠ Q1 A2 ⎛ r1 ⎞⎛ 4π r2 2 ⎞ r2 2 R = = = ⎜ ⎟⎜ =2 ⎟= = η 2 ⎛ Q2 ⎞ Q2 A1 ⎝ r2 ⎠⎝ 4π r12 ⎠ r1 R ⎜ ⎟ ⎝ A2 ⎠
(b) The electric field strengths at the surface of the conductors is E1 =
E η1 η and E2 = 2 , thus 1 = 2. E2 ε0 ε0
Assess: Halving the radius doubles the electric field strength as well as the charge density. Sharp corners (i.e., parts of conductors with a small radius of curvature, like a small sphere) have strong electric fields. This characteristic is why lightning rods work.
30.54. Model: The battery is assumed to be ideal. Visualize:
The pictorial representation shows the capacitor plates connected to a battery, the battery removed from the plates, and the plates moved apart with insulating handles. Solve: (a) The battery charges the plates through the wires. Once the wires are disconnected, the charge is trapped on the plates and will not change as their spacing is increased. The initial capacitance of the plates is Ci =
ε0 A di
=
(8.85 ×10
−12
C 2 /N m 2 ) ( 0.020 m × 0.020 m ) 0.0010 m
= 3.54 × 10−12 F
Consequently, the initial voltage ΔVi = 9 V charges the plates to
Q = ±C ΔV = ± ( 3.54 × 10−12 F ) ( 9.0 V ) = ±3.2 × 10−11 C
(b) The charge is still Q = ±3.2 × 10−11 C after the spacing is increased, but the final capacitance is
Cf =
ε0 A df
=
ε0 A 2d i
=
Ci 2
As a result, the final potential difference is
ΔVf =
Q Q Q = = 2 = 2ΔVi = 2 × 9 V = 18 V Cf Ci /2 Ci
30.55. Model: Assume the battery is ideal. Visualize:
The pictorial representation shows the capacitor plates connected to a battery, and the capacitor plates moved apart with insulating handles while they are connected to the battery. Solve: (a) The initial capacitance of the plates is Ci =
ε0 A di
=
(8.85 ×10
−12
C 2 /N m 2 ) ( 0.020 m )
1.0 × 10−3 m
2
= 3.54 × 10−12 F
Consequently, an initial voltage ΔVi = 9.0 V charges the plates to Q = ±Ci ΔVi = ± ( 3.54 × 10−12 F ) ( 9.0 V ) = ±31.9 × 10−12 C = ±32 pC (b) The new capacitance is Cf = ε 0 A 2di = 12 Ci . The potential difference across the plates is determined by the
battery and is unchanged: ΔVf = ΔVi = 9.0. Thus, the new charge on the plates is
Q = ±Ci ΔVf = ± 12 ( 3.54 ×10−12 F ) ( 9.0 V ) = 16.0 × 10−12 C = 16.0 pC
30.56. Model: Capacitance is a geometric property of two electrodes. Visualize:
