Student Solutions Manual for Physics for Scientists and Engineers: A Strategic Approach Vol 2 (Chs 20-43) (v. 2, Chapters 20-43)

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 T