Physics Chapter 7 Answers 6d221s

Chapter 7: Work and Kinetic Energy Answers to Even-Numbered Conceptual Questions 2.

False. Any force acting on an object can do work. The work done by different forces may add to produce a greater net work, or they may cancel to some extent. It follows that the net work done on an object can be thought of in the following two equivalent ways: (i) The sum of the works done by each individual force; or (ii) the work done by the net force.

4.

If the net work done on an object is zero, it follows that its change in kinetic energy is also zero. Therefore, its speed remains the same.

6.

Frictional forces do negative work whenever they act in a direction that opposes the motion. For example, friction does negative work when you push a box across the floor, or when you stop your car.

8.

The fact that the ski boat’s velocity is constant means that its kinetic energy is also constant. Therefore, the net work done on the boat is zero. It follows that the net force acting on the boat does no work. (In fact, the net force acting on the boat is zero, since its velocity is constant.)

10.

No. What we can conclude, however, is that the net force acting on the object is zero.

12.

No. Power depends both on the amount of work done by the engine, and the amount of time during which the work is performed. For example, if engine 2 does its work in less than half the time of engine 1, it can produce more power even if the amount of work the engines do is the same.

Solutions to Problems and Conceptual Exercises 1.

Picture the Problem: The International Space Station orbits the Earth in an approximately circular orbit at a height of above the Earth’s surface. Strategy: According to equation 7-3, the work done on an object is positive if the force and the displacement are along the same direction, but zero if the force is perpendicular to the displacement. Solution: The gravitational force exerted by the Earth is radial (toward the center of the circular orbit), but the displacement is tangential, perpendicular to the force. We conclude that the work done by the Earth on the space station is zero. Insight: If the Earth were to do any work on the space station, its energy would change. Of course the rockets that lifted the parts of the space station into space did do work on them, increasing their energy sufficiently to achieve orbit.

2.

Picture the Problem: A pendulum bob swings from point I to point II along the circular arc indicated in the figure at right. Strategy: Apply equation 7-3, which says that the work done on an object is positive if the force and the displacement are along the same direction, but zero if the force is perpendicular to the displacement. Solution: 1. (a) As the pendulum bob swings from point I to point II, the force of gravity points downward and a component of the displacement is downward. Therefore, the work done on the bob by gravity is positive. 2. (b) As the pendulum bob swings, the force exerted by the string is radial (toward the pivot point) but the displacement is tangential, perpendicular to the force. We conclude that the work done on the bob by the string is zero. Insight: The work done by the Earth is negative if the bob swings from point II to point III because a component of the displacement is upward but the force is downward.

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–1

James S. Walker, Physics, 4th Edition

Chapter 7: Work and Kinetic Energy 3.

Picture the Problem: A pendulum bob swings from point II to point III along the circular arc indicated in the figure at right. Strategy: Apply equation 7-3, which says that the work done on an object is positive if the force and the displacement are along the same direction, but zero if the force is perpendicular to the displacement. Solution: 1. (a) As the pendulum bob swings from point II to point III, the force of gravity points downward and a component of the displacement is upward. Therefore, the work done on the bob by gravity is negative. 2. (b) As the pendulum bob swings, the force exerted by the string is radial (toward the pivot point) but the displacement is tangential, perpendicular to the force. We conclude that the work done on the bob by the string is zero. Insight: The work done by the Earth is positive if the bob swings from point I to point II because a component of the displacement is downward and the force is downward.

4.

Picture the Problem: The farmhand pushes the hay horizontally.

88 N

Strategy: Multiply the force by the distance because in this case the two point along the same direction. Solution: Apply equation 7-1 directly:

3.9 m

W = Fd = ( 88 N )( 3.9 m ) = 340 J = 0.34 kJ

Insight: The 26-kg mass is unneeded information unless we needed to know the amount of friction or the acceleration of the bale. 5.

Picture the Problem: The children lift the bucket vertically. Strategy: Multiply the force by the distance because in this case the two point along the same direction. Solution: Apply equation 7-1:

W = Fd = mgd

Now solve for m:

m=

mg

4.70 m

W 201 J = = 4.36 kg gd ( 9.81 m/s 2 ) ( 4.70 m )

Insight: The applied force equals the weight as long as the bucket does not accelerate. 6.

Picture the Problem: The pumpkin is lifted vertically then carried horizontally. Strategy: Multiply the force by the distance because during the lift the two point along the same direction. Solution: 1. (a) Apply equation 7-1 directly:

W = mgd = ( 3.2 kg ) ( 9.81 m/s 2 ) (1.2 m ) = 38 J

2. (b) The force is perpendicular to the displacement so W = 0 . Insight: You can still get tired carrying a pumpkin horizontally even though you’re doing no work! 7.

Picture the Problem: The suitcase is pushed horizontally. Strategy: Determine the applied force and solve for d.

Solution: Solve equation 7-1 for d:

d=

F

d

W W 642 J = = = 3.37 m f k μ k mg ( 0.272 )( 71.5 kg ) ( 9.81 m/s 2 )

Insight: The applied force equals the friction force as long as the suitcase does not accelerate. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–2


Chapter 7: Work and Kinetic Energy 8.

Picture the Problem: The paint can is lifted vertically. Strategy: Multiply the force by the distance because the two vectors point in the same direction in part (a). In part (b) the distance traveled is zero, and in part (c) the force and distance are antiparallel.

mg

1.8 m

W = Fd = mgd

Solution: 1. (a) Apply equation 7-1:

W = ( 3.4 kg ) ( 9.81 m/s 2 ) (1.8 m ) = 60 J

2. (b) Now the force and distance are perpendicular:

W= 0

3. (c) Now the force and distance are antiparallel:

W = −mgd = − 60 J = − 0.060 kJ

Insight: The applied force equals the weight as long as the paint can does not accelerate. The can gains potential energy as it is lifted and loses potential energy as it is lowered. (We will discuss potential energy in Chapter 8.) 9.

Picture the Problem: The water skier is pulled horizontally. Strategy: Multiply the force by the distance because in this case the two point along the same direction. Solution: 1. (a) The work is positive because the force is along the direction of motion (θ = 0°). 2. (b) Apply equation 7-1 directly:

W = Fd = (120 N )( 65 m ) = 7800 J = 7.8 kJ

Insight: While the work done by the rope is positive, the work done by friction is negative, so as long as the skier moves at constant speed she doesn’t gain or lose kinetic energy. 10. Picture the Problem: The water skier is pulled horizontally. Strategy: Take the negative of the force times the distance because in this case the two point in opposite directions. Solution: 1. (a) The work is negative because the force on the boat is opposite the direction of motion (θ = 180°). 2. (b) Apply equation 7-3 directly:

W = Fd cos θ = (120 N )( 65 m ) cos180o = −7800 J = −7.8 kJ

Insight: While the works done on the boat by the rope and by friction are negative, the work done by the motor is positive, so as long as the boat moves at constant speed it doesn’t gain or lose kinetic energy. 11. Picture the Problem: The wagon rolls horizontally but the force pulls upward at an angle.

F

θ

Strategy: Use equation 7-3 keeping in mind the angle between the force and the direction of motion. Solution: Use equation 7-3:

d

W = Fd cos θ = (16 N )(10.0 m ) cos 25o = 150 J = 0.15 kJ

Insight: Only the component of the force along the direction of the motion does any work. The vertical component of the force reduces the normal force a little. 12. Picture the Problem: The packing crate slides horizontally but the force pulls upward at an angle. Strategy: Use equation 7-3 keeping in mind the angle between the force and the direction of motion. Solution: Use equation 7-3:

F

d

W = Fd cos θ = (115 N )( 8.0 m ) cos 43.5° = 670 J = 0.67 kJ

Insight: Only the component of the rope force along the direction of the motion does any work. The vertical component of the force reduces the normal force a little. Friction must do − 0.67 kJ of work on the crate because the crate does not gain or lose energy as it is pulled at constant speed. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–3


Chapter 7: Work and Kinetic Energy 13. Picture the Problem: The mop head is being pushed downward into the floor.

F

θ

Strategy: Use equation 7-3 keeping in mind the angle between the force and the direction of motion. Solution: 1. (a) Use equation 7-3:

d

W = Fd cos θ = ( 50.0 N )( 0.50 m ) cos 55o = 14 J

2. (b) If the angle is increased to 65°, a smaller component of the force will be along the direction of motion and therefore the work done by the janitor will decrease. Insight: Only the component of the force along the direction of the motion does any work. The vertical component of the force increases the normal force. 14. Picture the Problem: The plane and glider must be at different altitudes. Since the altitudes are constant, both are moving horizontally. Strategy: Use equation 7-3, solving for the angle between the force and the direction of motion.

airplane

F

θ

glider

Solution: 1. Solve equation 7-3 for the angle:

W = Fd cos θ or cos θ =

2. Calculate the angle:

θ = cos −1 ⎜

d

W Fd

⎡ 2.00 × 105 J ⎤ ⎛W ⎞ −1 o ⎥ = 57.4 ⎟ = cos ⎢ Fd 2560 N 145 m ( )( ) ⎝ ⎠ ⎣⎢ ⎦⎥

Insight: Only the component of the force along the direction of the motion does any work. The vertical component of the force helps to lift the glider a little. 17 N

15. Picture the Problem: The force and distance vectors for the woman are depicted at right. Strategy: Multiply the distance by the component of the force that is parallel to the distance.

skateboard

G G d = vt

Solution: 1. (a) First find the distance traveled:

G G d = vt = ( 4.1 m/s )( 25 s ) xˆ = 102.5 m xˆ

2. Now multiply only the x-components:

W = Fx d x = (17 N )(102.5 m ) = 1700 J

3. (b) The force on the bicycle is opposite that on the skateboard by Newton’s third law. 4. Multiply only the x-components:

bicycle

G F

12 N

G Fbike = ( −17 N ) xˆ + ( −12 N ) yˆ W = Fbike,x d x = ( −17 N )(102.5 m ) = −1700 J

Insight: The bicyclist must do work while pedaling or she will lose kinetic energy and come to a stop. The force of friction on the skateboard must do −1700 J of work because her velocity (and therefore kinetic energy) is constant. 16. Picture the Problem: The force and distance vectors have varying relationships to each other in parts (a), (b), and (c). The appropriate vectors are drawn at right. Strategy: Multiply the distance by the component of the force that is parallel to the distance. The perpendicular components contribute no work to the total. Solution: 1. (a) Multiply only the x-components:

W = Fx d x = ( 2.2 N )( 0.25 m )

2. (b) Multiply only the y-components:

W = Fy d y = (1.1 N )( 0.25 m )

W = 0.55 J

W = 0.28 J

2.2 N

(a)

1.1 N 0.25 m

G d

(b)

G F G d G F

0.25 m

1.1 N 2.2 N


7–4


Chapter 7: Work and Kinetic Energy 3. (c) Multiply the parallel components and add:

W = Fx d x + Fy d y

G F

(c)

= ( 2.2 N )( − 0.5 m ) + (1.1 N )( − 0.25 m )

−0.50 m

W = −1.4 J

−0.25 m

G d

1.1 N

2.2 N

Insight: Only the component of the force along the direction of the motion does any work. Multiplying components in G G this manner is equivalent to calculating the dot product of two vectors W = F ⋅ d . (See Appendix A.)

17. Picture the Problem: The boat and skier are both moving toward the left but the rope is pulling at an angle. Strategy: Solve equation 7-3 for the angle between the force and the direction of motion. Solution: 1. Solve eq. 7-3 for the angle:

W = Fd cos θ ⇒

2. Calculate the angle:

θ = cos −1 ⎜

W Fd

cos θ =

⎡ ⎤ 3500 J ⎛W ⎞ −1 o ⎥ = 21 ⎟ = cos ⎢ ⎝ Fd ⎠ ⎢⎣ ( 75 N )( 50 m ) ⎥⎦

Insight: Only the component of the force along the direction of the motion does any work. The work the boat does on the skier is balanced by the negative work friction does on the skier, so that the kinetic energy of the skier is constant.

18. Picture the Problem: A pitcher throws a ball at 90 mi/h and the catcher stops it in her glove. Strategy: Use the work-energy theorem to answer the conceptual question. Solution: 1. (a) The ball gains kinetic energy from the action of the pitcher, so we conclude that the work done on the ball by the pitcher is positive. 2. (b) The ball loses kinetic energy from the action of the catcher, so we conclude that the work done on the ball by the catcher is negative. Insight: The answers are consistent with the previously described definition of work: The pitcher does positive work on the ball by exerting a force in the same direction as the motion of the ball. The catcher does negative work on the ball because the force exerted by the catcher is opposite in direction to the motion of the ball.

19. Picture the Problem: The runner accelerates horizontally and runs in a straight line. Strategy: The work done equals the change in kinetic energy. Solution: Find the change in kinetic energy: W = ΔK = 12 mvf2 − 12 mvi2 =

1 2

( 73 kg ) ⎡⎣( 7.7 m/s )

2

− 0 ⎤ = 2200 J = 2.2 kJ ⎦

Insight: The runner’s kinetic energy comes from the forces his muscles exert on his center of mass over the distance which his center of mass moves.

20. Picture the Problem: The fragment moves at high speed in a straight line. Strategy: Calculate the kinetic energy using equation 7-6. Solution: Apply equation 7-6 directly:

K = 12 mv 2 =

1 2

(1770 kg )(120 m/s )

2

= 1.27 × 107 J = 12.7 MJ

Insight: The energy came from the work the rocket motor did in order to place Skylab into orbit.


7–5


Chapter 7: Work and Kinetic Energy

21. Picture the Problem: The bullet moves at high speed in a straight line. Strategy: Calculate the kinetic energy using equation 7-6, then use ratios to find the new kinetic energies in (b) and (c).

( 0.00950 kg )(1300 m/s )

Solution: 1. (a) Apply equation 7-6 directly:

K = 12 mv 2 =

2. (b) Use a ratio to predict the new kinetic energy:

2 (1 v ) 1 K new 12 mvnew so that = 1 2 = 2 2old = K old 4 vold 2 mvold

1 2

2

= 8030 J = 8.03 kJ

2

K new = 14 K old =

1 4

(8.03 kJ ) =

2.01 kJ

2 K new 12 mvnew ( 2vold ) = 1 2 = = 4 so that 2 K old vold 2 mvold 2

3. (c) Use a ratio to predict the new kinetic energy:

K new = 4 K old = 4 ( 8030 J ) = 32.1 kJ

Insight: The kinetic energy is proportional to the velocity squared, a useful thing to when calculating ratios.

22. Picture the Problem: The work W0 accelerates a car from 0 to 50 km/h. Strategy: Use the work-energy theorem to answer the conceptual question. Solution: 1. (a) The work done on the car is equal to its change in kinetic energy, which in turn is proportional to the square of its velocity. Therefore, if the kinetic energy of the car is W0 when its speed is 50 km/h, its kinetic energy will be 9W0 when its speed is 150 km/h. We conclude that the work required to accelerate the car from 50 km/h to 150 km/h is 9W0 − W0 = 8W0 . 2. (b) The best explanation is II. The final speed is three times the speed that was produced by the work W0. Statement I is false because the kinetic energy depends on the speed squared, and the work is proportional to the change in kinetic energy. Statement III is true but irrelevant. Insight: By using a similar analysis you can show that a car that is traveling 35 mph has twice the kinetic energy as a car that is traveling 25 mph. In other words, it takes just as much work to increase the car’s speed from 25 to 35 mph as it did to accelerate from 0 to 25 mph.

23. Picture the Problem: Four joggers have a variety of masses and speeds. Strategy: Use the definition of kinetic energy to determine the relative magnitudes of the kinetic energies. Solution: 1. Calculate the kinetic energies of each jogger. 2. K A =

1 2

( m )( v )

4. K C =

1 2

( 3m ) ( 12 v )

2

= 12 mv 2 2

=

3 4

(

1 2

mv 2

)

3. K B =

1 2

( 12 m ) ( 3v )

5. K D =

1 2

( 4m ) ( 12 v )

2

2

=

9 2

(

1 2

mv 2

)

= 12 mv 2

6. By comparing the magnitudes of the kinetic energies we arrive at the ranking C < A = D < B. Insight: Even with three times the mass of jogger A, jogger C has only three-fourths the kinetic energy because the kinetic energy is proportional to the square of the speed.

24. Picture the Problem: The pine cone falls straight down under the influence of gravity. Strategy: The work done by gravity equals the change in kinetic energy according to equation 7-7. The work done by gravity is always W = mgh as indicated in Example 7-2 and Conceptual Checkpoint 7-1. Solution: 1. (a) The work done by gravity on the pine cone equals the increase in its kinetic energy. Set the energies equal and solve for v:

W = ΔK = mgh = 12 mv 2 v = 2 gh = 2 ( 9.81 m/s 2 ) (16 m ) = 18 m/s

2. (b) Air resistance did negative work because the speed and therefore the kinetic energy of the pine cone when it landed was reduced. Air resistance removed energy from the pine cone. Insight: Kinetic friction always does negative work because the force is always opposite to the direction of motion.


7–6



25. Picture the Problem: The pine cone falls straight down for 16 m under the influence of gravity. Strategy: The work done by air resistance is the difference in kinetic energies between the air resistance and no air resistance cases. The work done by gravity is always W = mgh (see Example 7-2 and Conceptual Checkpoint 7-1). Solution: 1. (a) Find the difference in kinetic energies between the air resistance and no air resistance cases: 2. (b) The work done by air resistance equals the average force of air resistance times the distance the pine cone falls. It is negative because the upward force is opposite to the downward distance traveled.

W = ΔK = K f − K i = 12 mvf2 − mgh 2 = ( 0.140 kg ) ⎡ 12 (13 m/s ) − ( 9.81 m/s 2 ) (16 m ) ⎤ = −10 J ⎣ ⎦

W = − Fd so that F = −

( −10 J ) W =− = 0.63 N upward d 16 m

Insight: Kinetic friction always does negative work because the force is always opposite to the direction of motion.

26. Picture the Problem: The object falls straight down under the influence of gravity. Strategy: Use the dependence of kinetic energy upon mass and speed to answer parts (a) and (b). The work done by gravity can be found from the change in the kinetic energy.

( 0.40 kg )( 6.0 m/s )

2

Solution: 1. (a) Apply equation 7-6 directly:

K = 12 mv 2 =

2. (b) Solve equation 7-6 for speed:

v=

3. (c) Calculate W = ΔK :

W = ΔK = K f − K i = 25 J − 7.2 J = 18 J

1 2

2 ( 25 J )

2K = m

0.40 kg

= 7.2 J

= 11 m/s

Insight: As an object falls, the work done by gravity increases the kinetic energy of the object.

27. Picture the Problem: The runner slides horizontally on level ground over a distance of 3.40 m and comes to rest. Strategy: The work done by friction equals the negative of the kinetic energy the runner had just before the slide. It also equals the force exerted by friction times the distance of the slide. Solution: 1. (a) Calculate W = ΔK :

W = ΔK = K f − K i = 0 − 12 mvi2 W = − 12 ( 62.0 kg )( 4.35 m/s ) = −587 J 2

W = − Fd = − ( μ k mg ) d so that μ k = −

2. (b) The work done by friction equals the average force of friction times the distance the player slid:

μk = −

( −587 J )

( 62.0 kg ) ( 9.81 m/s 2 ) ( 3.40 m )

W mgd

= 0.284


28. Picture the Problem: The car slows down as it rolls horizontally a distance of 32.0 m through the sand. Strategy: The kinetic energy of the car is reduced by the amount of work done by friction. The work done is the force times the distance, so once we know the work done and the distance, we can find the force. Solution: 1. (a) The net work done on the car must have been negative because the kinetic energy decreased. 2. (b) The work done by friction equals the average force of friction times the distance the car traveled. Apply equations 7-1 and 7-7, including a minus sign to indicate the force and distance are in opposite directions:

W = − Fd = ΔK = mv − mv 1 2

F =−

1 2

2 f

1 2

(1100 kg ) (122 − 192 32 m

2 i

so that F = −

m 2 /s 2 )

1 2

m ( vf2 − vi2 ) d

= 3700 N = 3.7 kN

Insight: Kinetic friction always does negative work because the force is always opposite to the direction of motion. The actual force exerted on the car would be −3.7 kN if the displacement is taken to be +30.0 m. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–7



29. Picture the Problem: The car slows down as it rolls horizontally a distance of 16.0 m through the sand. Strategy: The kinetic energy of the car is reduced by the amount of work done by friction. The change in kinetic energy is related to the initial and final speeds. We can use the speed information from the previous problem to set up a ratio and determine the change in speed for this new case of a shorter sandy section. Solution: 1. (a) Since ΔK ∝ v 2 ∝ Fd , halving the distance would decrease the speed by 1 decrease by less than 3.5 m/s.

2 so the speed will

2 2 ΔK new 12 mvf, new − 12 mvi = 1 2 2 1 ΔK old 2 mvf, old − 2 mvi

2. (b) Create a ratio of the kinetic energies:

2 2 ΔK new vf, new − vi 1 = 2 = 2 ΔK old 2 vf, old − vi

3. We know ΔK new = 12 ΔK old because the

force is the same but the distance is cut in half: 4. Solve for vf, new :

vf,2 new = 12 vf,2 old − 12 vi2 + vi2 ⇒ vf, new =

5. Substitute the speeds from the previous question:

vf, new =

1 2

(12 m/s )

2

1 2

(v

2 f, old

+ vi2 )

+ 12 (19 m/s ) = 15.9 m/s thus 2

Δv′ = 15.9 − 19 m/s = −3.1 m/s


30. Picture the Problem: The bicycle rolls horizontally on level ground, slows down, and comes to rest. Strategy: The work done by the brakes equals the change in the kinetic energy of the bicycle (equation 7-7). Use equation 2-10 to find the distance traveled, and equation 7-1 to find the magnitude of the braking force. W = ΔK = 12 mvf2 − 12 mvi2 = 0 − 12 ( 65 + 8.8 kg )(14 m/s )

Solution: 1. (a) Calculate W = ΔK :

2

= −7200 J = −7.2 kJ

2. (b) Use equation 2-10 to find the distance:

Δx =

3. (c) The force can be found from equation 7-1:

F=

1 2

( v0 + v ) t = 12 (14 + 0 m/s )( 4.0 s ) =

W d

=

W Δx

=

28 m

7200 J = 260 N ≅ 58 lb 28 m

Insight: Kinetic friction always does negative work because the force is always opposite to the direction of motion. The average velocity is half the initial velocity as long as the acceleration is constant.

31. Picture the Problem: A block of mass m and speed v collides with a spring, compressing it a distance Δx. Strategy: Answer the conceptual question using the expression that describes the work required to compress a spring. Solution: The work done in compressing a spring a distance Δx is

1 2

k ( Δ x ) , where k is the force constant of the 2

spring. If the force constant is increased by a factor of four, the compression distance must be halved to produce the same amount of work; that is,

1 2

( 4k ) ( 12 Δ x )

2

= 12 k ( Δ x ) . Thus, the compression of the spring is 2

1 2

Δx .

Insight: The force constant would have to be increased by a factor of nine in order to reduce the compression distance by a factor of three.

32. Picture the Problem: The spring is stretched to x = 0.050 m by pulling to the right or compressed to x = −0.050 m by pushing to the left. Strategy: The work to stretch or compress a spring a distance x is 12 kx 2 . Solution: 1. (a) Apply equation 7-8:

W = 12 kx 2 =

1 2

( 3.5 ×10

4

N/m ) ( 0.050 m ) = 44 J

2. (b) Repeat:

W = 12 kx 2 =

1 2

( 3.5 ×10

4

N/m ) ( − 0.050 m ) = 44 J

2

2

Insight: The work done on the spring is the same in either case because the spring force is opposite the applied force in each case. The applied force and distance vectors are parallel in each case, so the works are positive. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–8



33. Picture the Problem: The compressed spring pushes the block from rest horizontally on a frictionless surface. The block slides to the left as indicated in the figure. Strategy: The work to stretch or compress a spring a distance x is 12 kx 2 . The work done by the spring equals the kinetic energy gained

by the block. Solution: 1. Apply equations 7-7 and 7-8:

W = 12 kx 2 = ΔK = 12 mvf2 − 0

2. Now solve for vf :

⎛ 1.0 ×104 N/m ⎞ ⎛ k ⎞ ⎟ ( 0.15 m ) = 14 m/s vf = ⎜⎜ ⎟⎟ x = ⎜⎜ ⎟ 1.2 kg ⎝ m⎠ ⎝ ⎠

Insight: The work done on the spring in order to compress it becomes stored potential energy. That stored energy becomes the kinetic energy of the block as the spring accelerates it.

34. Picture the Problem: The block slides toward the right and into the spring. After compressing the spring the block comes to rest. Strategy: The work to stretch or compress a spring a distance x is 12 kx 2 .

The work done on the block by the spring equals the kinetic energy lost by the block. The work done on the block is negative because the force on the block is toward the left while the motion is toward the right. Solution: 1. Apply equations 7-7 and 7-8:

Won block = ΔK block = 0 − 12 mvi2 = − 12 kx 2

2. Now solve for k:

k =m

( 2.2 m/s ) vi2 = (1.8 kg ) = 91 N/m 2 2 x ( 0.31 m ) 2

Insight: The kinetic energy of the block is transformed into the energy stored in the spring as it is compressed.

35. Picture the Problem: The work done by the force is the area under the force versus position graph. Strategy: The total work done by the force is the total area under the graph from zero to 0.75 m. The work done from 0.15 m to 0.60 m is the area shaded in gray in the figure at right. Add the works done in each of the three segments to find the total work. Solution: 1. (a) The total area under the graph from 0 to 0.75 m:

Wtotal = A = ( 0.25 m )( 0.6 + 0.4 + 0.8 N )

2. (b) The area of the shaded graph portion:

W = ( 0.25 − 0.15 m )( 0.6 N ) + ( 0.25 m )( 0.4 N ) + ( 0.60 − 0.50 m )( 0.8 N )

= 0.45 J

W = 0.24 J

Insight: The work is positive as long as the object moves from left to right (from small x to large x). Therefore the object gains energy as it moves from left to right.


7–9



36. Picture the Problem: The work done by the force is the area under the force versus position graph. Strategy: The initial position is shown by the dark vertical line. Positive work is done on the object if it moves to the right of the line, negative work if it moves to the left. Use the fact that the work is the area under the graph to determine the final position of the object in each case. The shaded area to the right of the dark line represents the positive work done on the object, and the shaded area to the left of the line represents the negative work. Solution: 1. (a) The work done on the object is sufficient to move it past the 0.75 m mark:

W = ( 0.50 − 0.40 m )( 0.8 N ) + ( 0.25 m )( 0.4 m ) + ( x − 0.75 m )( 0.2 N ) = 0.21 J

2. (b) The negative work done on the object requires that it be moved left of the 0.25 m mark:

W = ( 0.25 − 0.40 m )( 0.8 N ) + ( x − 0.25 m )( 0.6 N ) = − 0.19 J

x=

x=

0.21 − 0.08 − 0.10 J + 0.75 m = 0.90 m 0.2 N −0.19 − ( −0.12 ) J

+ 0.25 m = 0.13 m 0.6 N Insight: The work is positive as long as the object moves from left to right (from small x to large x). Therefore the object gains energy as it moves from left to right, but loses energy as it is moved from right to left.

37. Picture the Problem: A block of mass m and speed v collides with a spring, compressing it a distance Δx. Strategy: Answer the conceptual question using the work-energy theorem. Solution: The work done in compressing a spring a distance Δx is

1 2

k ( Δ x ) , where k is the force constant of the 2

spring. In this case the spring does work to remove the kinetic energy K = 12 mv 2 of the block and bring it to rest. If the mass is halved and the speed is doubled, the amount of kinetic energy that must be removed is 1 2 1 2

( k ( 2 Δ x) = 2(

( 12 m ) ( 2v )

2

2

=2

)

1 2

mv 2 = 2 K . This means that the compression distance must increase to

1 2

k Δ x 2 = 2 K , or twice as much energy is removed than in the first case.

2 ( Δ x ) so that

)

Insight: The force constant would have to be increased by a factor of nine in order to reduce the compression distance by a factor of three.

38. Picture the Problem: The springs are stretched horizontally in each case. Strategy: The works and stretch distances can be used to find the spring constants by applying equation 7-8. Solution: 1. Solve equation 7-8 for k1 :

k1 =

2 (150 J ) 2W1 = 2 2 x1 ( 0.20 m )

= 7500 N/m

2. Solve equation 7-8 for k2 :

k2 =

2 ( 210 J ) 2W2 = = 4700 N/m 2 2 x2 ( 0.30 m )

3. Stiffer springs have larger spring constants so we conclude spring 1 is stiffer. Insight: The work done on the spring in order to stretch it becomes stored potential energy. We will discuss potential energy more in Chapter 8.


7 – 10



39. Picture the Problem: The spring is compressed horizontally. Strategy: The work and stretch distance can be used to find the spring constant by applying equation 7-8. Solution: 1. (a) Solve equation 7-8 for k :

k=

2 (180 J ) 2W = 2 2 x ( 0.15 m )

= 1.6 × 10 4 N/m = 16 kN/m

2. (b) It would take more than 180 J of work because W is proportional to x2:

W = 12 kx22 − 12 kx12 =

1 2

(1.6 ×10

4

2 2 N/m ) ⎡( 0.30 m ) − ( 0.15 m ) ⎤ = 540 J ⎣ ⎦

Insight: The extra work required to stretch the spring an additional 0.15 m can be pictured as the difference in area of two triangles, one with base 0.30 m and one with base 0.15 m, both bounded by the line given by kx (Figure 7-10).

40. Picture the Problem: The work done by the force is the area under the force versus position graph. Strategy: Positive work is done on the object as it moves to the right of the initial position, negative work if it moves to the left. Use the fact that the work is the area under the graph to determine the final velocity of the object in part (a). Use the change in kinetic energy to find the work done on the object in part (b), and use the area under the graph to figure out the final position. The shaded area to the right of the dark line represents the positive work done on the object, and the shaded area to the left of the line represents the negative work. Solution: 1. (a) The work done on the object is the area under the graph between 0.27 and 0.99 m:

W = ( 0.50 − 0.27 m )( 0.8 N ) + ( 0.25 m )( 0.4 m ) + ( 0.99 − 0.75 m )( 0.2 N )

2. The final velocity can be found by setting the work equal to the change in kinetic energy:

W = ΔK = 12 mvf2 − 12 mvi2 now solve for vf :

3. (b) The decrease in speed and kinetic energy means the object must move to the left. Determine the work required:

W = ΔK = 12 mvf2 − 12 mvi2 =

4. Now set the work equal to the area under the graph to find the final x:

W = ( 0.25 − 0.27 m )( 0.8 N ) + ( x − 0.25 m )( 0.6 N ) = − 0.078 J

= 0.332 J

vf =

2W + vi2 = m

2 ( 0.332 J ) 1.7 kg 1 2

+ ( 0.44 m/s ) = 0.76 m/s 2

(1.7 kg ) ⎡⎣( 0.32 m/s )

2

2 − ( 0.44 m/s ) ⎤ ⎦

= − 0.078 J

x=

− 0.078 J − ( − 0.016 J ) + 0.150 J 0.6 N

= 0.15 m

Insight: The work is positive as long as the object moves from left to right (from small x to large x). Therefore the object gains energy as it moves from left to right, but loses energy as it is moved from right to left.


7 – 11



41. Picture the Problem: The work done by the force is the area under the force versus position graph.

Strategy: Determine the geometric area under the graph for the various given starting and ending positions. The area under the graph equals the work done on the block.

Solution: 1. (a) The work done on the object is the area under the graph between 0 and 0.30m:

W=

2. The magnitude of the force at x = 0.10 m can be found from the given formula:

F = kx = ( 2.0 × 10 4 N/m ) ( 0.10 m ) = 2000 N

3. (b) The work done is the area under the graph between 0 and 0.40 m minus the area of the triangle between 0 and 0.10 m:

W=

1 2

1 2

( 0.21 m )( 4200 N ) + ( 0.30 − 0.21 m )( 4200 N ) =

0.82 kJ

( 0.21 m )( 4200 N ) + ( 0.40 − 0.21 m )( 4200 N ) − 12 ( 0.10 m )( 2000 N ) = 1.1 kJ

Insight: When the force varies as a function of x it is often useful to break the area under the graph into simple geometric shapes to aid in the calculation of work.

42. Picture the Problem: Forces of different magnitudes do different amounts of work in different intervals of time. Strategy: Use the definition of power to determine the ranking of the powers. Solution: 1. Calculate the powers produced by each force: 2. P1 =

W1 5J = = 0.50 W Δ t1 10 s

3. P2 =

W2 3 J = = 0.60 W Δ t2 5 s

4. P3 =

W3 6J = = 0.33 W Δ t3 18 s

5. P4 =

W4 25 J = = 0.20 W Δ t4 125 s

6. By comparing the magnitudes of the powers we arrive at the ranking F4 < F3 < F1 < F2 . Insight: If the amount of work and the time interval are each doubled, the power produced remains the same.

43. Picture the Problem: The runner does work against gravity as he elevates his center of gravity. Strategy: The power required is the work required to change the elevation divided by the time. As in Conceptual Checkpoint 7-1 and Example 7-2 the work required to change the elevation of an object is W = mgh. Solution: Divide the work required by the time:

P=

2 W mgh ( 70.0 kg ) ( 9.81 m/s ) (1576 steps × 0.20 m/step ) = = t t ( 9 min × 60 s/min + 33 s )

P = 380 W × 1 hp 746 W = 0.51 hp

Insight: The energy and power required of the runner is much higher than this because muscles aren’t 100% efficient at converting food energy into mechanical energy, and the body requires additional energy to stay warm, keep the heart pumping, etc.

44. Picture the Problem: This is a units conversion problem. Strategy: Use the conversion factors given in the inside front cover of the book. Solution: Convert the units:

W = Pt = (1 kW × 1000 W/kW )(1 h × 3600 s/h ) = 3.6 × 106 J = 3.6 MJ

Insight: Electrical energy is sold in units of kWh, which are really units of energy in multiples of 3.6 million joules.


7 – 12



45. Picture the Problem: The fly does work against gravity as it elevates its center of mass. Strategy: The power required is the force times the velocity, where the force is just the weight of the fly. Solution: Apply equation 7-13:

P = Fv = ( mg ) v = (1.4 × 10-3 kg )( 9.81 m/s 2 ) ( 0.023 m/s ) P = 3.2 × 10−4 W = 0.32 mW

Insight: The energy and power required of the fly is higher than this because it isn’t 100% efficient at converting food energy into mechanical energy.

46. Picture the Problem: The microwave oven delivers energy to the ice cube via electromagnetic waves. Strategy: The power required is the energy delivered divided by the time. Solution: Solve equation 7-10 for t:

t=

W 32200 J = = 307 s = 5.11 min P 105 W

Insight: Power can be regarded as the rate of energy transfer because work is essentially transferred energy. We’ll learn more about melting ice cubes in Chapter 17 and electromagnetic waves in Chapter 25.

47. Picture the Problem: The bucket is lifted vertically upward. Strategy: The power required is the force times the velocity, where the force is just the weight of the bucket. Solution: Solve equation 7-13 for v:

v=

P P 108 W = = = 2.20 m/s F mg ( 5.00 kg ) ( 9.81 m/s 2 )

Insight: Lifting faster than this would require more power. If the rope’s mass were not ignored it would require additional power since its center of mass is also lifted. If the force exceeded the weight, the bucket would accelerate.

48. Picture the Problem: The water is pumped vertically upward. Strategy: The power required is the work required to change the elevation divided by the time. As in Conceptual Checkpoint 7-1 and Example 7-2 the work required to change the elevation of an object is W = mgh. Solution: Divide the work required by the time:

P=

W mgh (12.0 lb × 4.448 N/lb )( 2.00 m ) = = t t (1.00 s )

P = 107 W × 1 hp 746 W = 0.143 hp

Insight: A ¼-hp motor would do the trick. Pumping faster than this would require more power.

49. Picture the Problem: The kayaker paddles horizontally in a straight line at constant speed. Strategy: The kayaker does positive work on the kayak as she paddles, and friction does negative work at the same time. The two works are equal because the kayak does not change its kinetic energy. Therefore the force she exerts must be equivalent to the force of friction. Solution: 1. (a) Solve equation 7-13 for F:

F=

P 50.0 W = = 33.3 N = 7.5 lb v 1.50 m/s

2. (b) Since the speed of the kayak is proportional to P / F, doubling the power would double the speed for the same F. Insight: Newton’s Second Law F = ma states that the net force on the kayak must be zero because it is not accelerating. That’s another way of figuring that the magnitude of the paddling force equals the magnitude of the friction force.


7 – 13



50. Picture the Problem: The aircraft flies horizontally at constant speed. Strategy: The power required is the work required to fly the plane (against friction) divided by the time. Solution: 1. (a) Solve equation 7-10 for W:

W = Pt = ( 0.30 hp × 746 W/hp )( 2 h × 3600 s/h + 49 m × 60 s/m ) W = 2.3 × 106 J = 2.3 MJ

2. (b) Convert W into units of Snickers bars:

2.27 ×106 J ×

1 Cal 1 Snickers bar × = 1.9 bars or about 2 Snickers 4186 J 280 Cal

Insight: The energy required of the pilot is much higher than this because muscles aren’t 100% efficient at converting food energy into mechanical energy, and the body requires additional energy to stay warm, keep the heart pumping, etc.

51. Picture the Problem: The weight slowly descends straight down. Strategy: The power delivered is the force (the weight) times the speed. Solution: 1. (a) Apply equation 7-13:

0.760 m ⎛ ⎞ P = Fv = mgv = ( 4.35 kg ) ( 9.81 m/s 2 ) ⎜ ⎟ ⎝ 3.25 d × 86400 s/d ⎠ P = 1.15 × 10−4 W = 0.115 mW

2. (b) To increase the power delivered you must either increase the force or the velocity. In this case, the time it takes for the mass to descend should be decreased so the velocity will increase and so will the delivered power. Insight: The weight delivers energy to the clock by doing work. The downward force it exerts on the clock is parallel to its displacement, so it is doing positive work on the clock.

52. Picture the Problem: You run up a flight of stairs, gaining elevation. Strategy: The power required is the work necessary to elevate your center of mass divided by the time. As in Conceptual Checkpoint 7-1 and Example 7-2 the work required to change the elevation of your center of mass is W = mgh . Solution: Suppose you climb one flight of stairs (3 m) in 2 seconds (you run up the stairs quickly). Estimate your mass to be 70 kg (about 150 lb at sea level):

P= P=

W mgh = t t ( 70 kg ) ( 9.81 m/s2 ) ( 3.0 m ) 2.0 s

= 1030 W ×

1 hp = 1.4 hp 746 W

Insight: The energy and power required of you is much higher than this because muscles aren’t 100% efficient at converting food energy into mechanical energy, and your body requires additional energy to stay warm, keep the heart pumping, etc.

53. Picture the Problem: The car accelerates horizontally in a straight line. Strategy: The power required is the work required to change the kinetic energy divided by the time. Use ratios to easily find the desired quantities. Solution: 1. (a) Combine equations 7-7 and 7-10: 2. Now divide top and bottom by

1 2

m and

substitute for the velocities:

t=

W ΔK 12 mvf2 − 12 mvi2 = = 1 2 P P 2 mvi T

t=

2 vf2 − vi2 ( 2v ) − v = = 3T vi2 T v2 T 2

3. (b) Again combine equations 7-7 and 7-10:

⎛ 1 mv 2 ⎞ W = ΔK = 12 mvf2 = Pt = ⎜ 2 ⎟ ( 2T ) ⎝ T ⎠

4. Now solve for vf :

1 2

mvf2 = mv 2 so vf2 = 2v 2 and thus vf = v 2

Insight: The assumption that the power remains constant is not realistic because car engines only generate their rated horsepower at high engine rpm, so less power is generated when the car first begins to accelerate. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7 – 14


Chapter 7: Work and Kinetic Energy G 54. Picture the Problem: A force F is exerted on a G sailboat while it drifts a distance d .

Strategy: The work done by the force is Fd cos θ , G G where θ is the angle between the vectors F and d . Solution: No work is done on boat A because θ is 90°. The work done on boat B is positive because θ is less than 90°, but the work done on boat C is negative because θ is greater than 90° (the horizontal components of the vectors point in opposite directions). We therefore arrive at the ranking C < A < B. G Insight: When negative work is done on the boat by force F , the boat loses kinetic energy and slows down.

55. Picture the Problem: A youngster rides on a skateboard with a speed of 2 m/s. After a force acts on the youngster, her speed is 3 m/s. Strategy: The work-energy theorem indicates that when positive work is done on an object, the object gains kinetic energy. Solution: The youngster gains kinetic energy because her speed increases. We conclude that the work done by the force is positive. Insight: When negative work is done on the youngster by a force, the youngster loses kinetic energy and slows down.

56. Picture the Problem: A car is accelerated from rest to final speed v by a constant force F acting over a distance Δx. Strategy: Use the definition of kinetic energy and the work-energy theorem to determine the distance over which the force should act in order to double the car’s speed. Solution: 1. (a) We are given that the force F times the distance Δx produces an increase in kinetic energy from 0 to K = 12 mv 2 , where m is the mass of the car. That is, FΔx = K. To increase the speed from v to 2v implies a change in

kinetic energy from K to K ′ = 12 m ( 2v ) = 4 2

(

1 2

)

mv 2 = 4 K , or an increase of 3K. It follows, then, that the required

distance Δ x′ satisfies the relation F Δ x ′ = 3K = 3 ( F Δ x ) , and hence that Δ x′ = 3 Δ x . 2. (b) The best explanation is III. Work is force times distance, and work depends on the speed squared. Statements I and II are each true, but neither statement is a complete explanation of the answer. Insight: In order to increase the car’s speed from v to 3v, the kinetic energy must change from K to 9K, a change of 8K. In order to accomplish this the force must act over a distance Δ x′ = 8 Δ x .

57. Picture the Problem: Car 1 has four times the mass of car 2, but they both have the same kinetic energy. Strategy: Use the definition of kinetic energy to answer the conceptual question. Solution: Let car 1’s mass be 4m and car 2’s mass be m. If we let car 1’s speed be v, we can set the kinetic energies equal to each other and find car 2’s speed: 12 ( 4m )v 2 = 12 mv22 ⇒ v2 = 2v . Insight: In a mixture of gases the molecules have different masses but the same average kinetic energy if they are in equilibrium. A calculation similar to the one in this problem shows that in such a case the lighter molecules have the higher average speed. There will be more discussion about the kinetic theory of gases in chapter 17.

58. Picture the Problem: The protein exerts a force that is parallel to the displacement. Strategy: The work done by the protein equals the force times the displacement because the two vectors are parallel. Solution: Apply equation 7-1 directly:

W = Fd = ( 7.5 × 10 −12 N )( 8.0 ×10−9 m ) = 6.0 ×10−20 J

Insight: A convenient unit of energy at this scale is the electron volt (eV), where 1.0 eV = 1.60×10-19 J. The work done in this problem by the protein is thus 0.375 eV.


7 – 15



59. Picture the Problem: Your teeth exert a force that is parallel to the displacement. Strategy: The work done by your teeth equals the force times the displacement because the two vectors are parallel. Solution: 1. (a) Solve equation 7-1 for F, estimating your teeth bite over a distance of 3 cm:

F=

W 19 J = = 630 N = 142 lb d 0.030 m

2. (b) The power can be found using equation 7-10, estimating the time as 0.25 s:

P=

W 19 J = = 76 W t 0.25 s

Insight: The energy delivered to the apple goes into the deformation of the material as it gets chewed up as well as some heating of the material.

60. Picture the Problem: The candy bar travels horizontally at constant speed. Strategy: Convert the metabolic energy in nutritional calories (1 Cal = 1000 cal) to mechanical energy in joules (1 cal = 4.186 J) and set it equal to the kinetic energy. Then solve for v. Solution: Solve equation 7-6 for v:

K = 12 mv 2 so that v = v=

2K m

2 ( 210 Cal × 4186 J/Cal ) 0.045 kg

= 6300 m/s = 6.3 km/s = Mach 18!

Insight: 1 nutritional calorie of metabolic energy corresponds to a very large amount of mechanical energy.

61. Picture the Problem: The bricks are lifted vertically upward. Strategy: The power required is the work required to change the elevation divided by the time. As in Conceptual Checkpoint 7-1 and Example 7-2 the work required to change the elevation of an object is W = mgh . Solution: Divide the work required by the time:

P=

W mgh ( 836 N )(10.7 m ) = = = 386 W = 0.517 hp t t ( 23.2 s )

Insight: A ¾ hp motor would do the trick. Lifting faster than this would require more power.

62. Picture the Problem: The box slides horizontally in a straight line at constant speed. Strategy: You do positive work on the box as it slides, and friction does negative work at the same time. The two works are equal because the kinetic energy of the box is constant. Therefore the force you must generate equals the friction force. Solution: 1. (a) Substitute for F in equation 7-13:

P = Fv = ( μk mg ) v = ( 0.55 )( 67 kg ) ( 9.81 m/s 2 ) ( 0.50 m/s ) P = 180 W = 0.18 kW

2. (b) Solve equation 7-10 for W:

W = Pt = (180 W )( 35 s ) = 6300 J = 6.3 kJ

Insight: Newton’s Second Law F = ma states that the net force on the box must be zero because it is not accelerating. That’s another way of figuring that the magnitude of the pushing force equals the magnitude of the friction force.


7 – 16



63. Picture the Problem: The heart’s produces power in order to push the blood through your veins and arteries. Strategy: Power is work (or energy) per time. The work necessary to elevate your center of mass is W = mgh , as in Conceptual Checkpoint 7-1 and Example 7-2. Solution: 1. (a) Solve equation 7-10 for W:

W = Pt = (1.33 W )( 24 h × 3600 s/h ) = 1.15 × 105 J = 115 kJ

2. (b) Convert W into elevation:

h=

1.15 × 105 J W = = 170 m = 550 ft! mg ( 70 kg ) ( 9.81 m/s 2 )

Insight: You wouldn’t be able to climb quite this high because muscles aren’t 100% efficient at converting food energy into mechanical energy. Besides, you still need that energy to keep your heart pumping!

64. Picture the Problem: The energy required to run the clock is compared to the energy consumed by a light bulb. Strategy: Find the energy consumed by the light bulb and divide by the given ratio. Solution: 1. Solve equation 7-10 for Wbulb :

Wbulb = Pt = ( 60 W )( 24 h × 3600 s/h ) = 5.18 × 106 J

2. Use a ratio to find WAtmos :

WAtmos =

Wbulb 5.18 × 106 J = = 0.0216 J = 21.6 mJ 240 × 106 240 × 106

Insight: Light bulbs actually aren’t very efficient, producing more heat than visible light. The Atmos clock had to be designed and built very carefully in order to run on such a small amount of energy.

65. Picture the Problem: The work W0 is required to accelerate a car from rest to the speed v 0 . Strategy: Use the work-energy theorem and the definition of kinetic energy to answer the conceptual questions. Solution: 1. (a) A car with a speed of v0 2 has a kinetic energy that is 1/4 the kinetic energy it has when its speed is v0 . Therefore, the work required to accelerate this car from rest to v0 2 is W0 4 .

2. (b) Suppose the car has a kinetic energy K 0 when its speed is v0 . As we saw in part (a), its kinetic energy when the speed is v0 2 is K 0 4. It follows that the increase in kinetic energy in going from a speed of v0 2 to a speed of v0 is 3K 0 4. As a result, the work required for this increase in speed is 3W0 4 .

Insight: The work required to accelerate the car from v0 to 2v0 is 3W0 because the kinetic energy must change from K0 to 4K0, a change of 3K0.

66. Picture the Problem: The work W0 is required to stretch a certain spring a distance 2 cm from equilibrium. Strategy: Use the work-energy theorem and the expression for the work done on a spring to answer the conceptual questions. Solution: 1. (a) The work required to stretch a spring depends on the square of the amount of stretch. Therefore, to stretch a spring by the amount x requires only 1/4 the work required to stretch it by the amount 2x. In this case, the work required is W0 4 . 2. (b) To stretch this spring by 3 cm from equilibrium requires 32 = 9 times the work to stretch it 1 cm. Therefore, stretching to 3 cm requires the work 9W0 4. Subtracting W0 , the work required to stretch to 2 cm, we find that an

additional work of 5W0 4 is required to stretch from 2 cm to 3 cm. Thus, it takes 5 times as much work to produce a 1-cm stretch from 2 cm to 3 cm as it does to produce a 1 cm stretch from 0 to 1 cm. Insight: The reason the work becomes so much greater in the second case can be pictured in of the area of the force vs. distance graph (such as Figure 7-10). The larger force exerted by the spring at large stretch distances produces a larger area under the graph and therefore requires a larger amount of work. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7 – 17



67. Picture the Problem: The straw moves horizontally at high speed and is then brought to rest by the tree’s force. Strategy: Set the work done by the tree’s force equal to the initial kinetic energy of the straw and solve for vi . Solution: Solve equations 7-1 and 7-7 for vi :

W = − Fd = ΔK = 0 − 12 mvi2 so that vi2 = 2 Fd m or vi =

2 Fd = m

2 ( 65 N )( 0.023 m ) 0.55 × 10 −3 kg

= 74 m/s = 165 mph!

Insight: The tree does negative work on the straw because the force is opposite to the direction of motion.

68. Picture the Problem: The glove rises vertically to height h, stops, and returns to the ground. Strategy: Gravity does negative work on the glove, reducing its kinetic energy to zero when it reaches height h. Set the work done by gravity equal to the change in kinetic energy between the ground and h/2. Solution: 1. Combine equations 7-1 and 7-7:

W = Fd = − ( mg ) ( 12 h ) = K f − K i K f = K i − 12 mgh W = Fd = − ( mg ) h = K f − K i = 0 − K i

2. Now set the work done by gravity when the glove rises to height h equal to the initial kinetic energy:

K i = mgh

3. Substitute the result into the first equation:

K f = K i − mgh 2 = mgh − mgh 2 = mgh 2 = K i 2 = K 2

Insight: The work done by gravity converts the glove’s kinetic energy into gravitational potential energy, a topic discussed in Chapter 8.

69. Picture the Problem: The boat and skier are both moving toward the left but the rope is pulling at an angle. Strategy: Use equation 7-3 to calculate the work done by the rope on the skier. The rope does positive work on the skier as it skis, and friction does negative work at the same time. The two works are equal because the kinetic energy of the skier is constant. Therefore the towing force equals the friction force. Solution: 1. (a) Apply equation 7-3:

W = Fd cos θ = F ( vt ) cos θ = ( 90.0 N )(14 m/s )(10.0 s ) cos 35o = 10,300 J = 10 kJ

2. (b) The work done by friction must be opposite the work done by the rope: Wfriction = −10 kJ Insight: Newton’s Second Law F = ma states that the net force on the skier must be zero because it is not accelerating. That’s another way of figuring that the magnitude of the towing force equals the magnitude of the friction force.

70. Picture the Problem: The sled slides horizontally but the force pulls upward at an angle. Strategy: Use equation 7-1 keeping in mind that only the component of G the force parallel to d does any work. Solution: 1. (a) Use equation 7-1:

W = Fd = ( 2.89 N )( 4.55 m ) = 13.1 J

2. (b) Since the parallel component of the force remains unchanged, so does the work:

W = 13.1 J

F

d

3. (c) Since the work depends only upon force and distance, the work done in parts (a) and (b) would stay the same if the mass of the sled were increased. Both the mass and the upward component of the force would affect the motion of the sled, but the total work done would remain unchanged. If the mass were increased, the amount of friction would increase and the speed would be less at the end of the 4.55 m. Insight: Only the component of the force along the direction of the motion does any work. The vertical component of the force would reduce the normal force, and increasing the mass would increase the normal force.


7 – 18



71. Picture the Problem: The apple falls straight down. Strategy: The power delivered is the force (the weight) times the speed. Find the velocity using equation 2-12, then use the velocity in equation 7-13 to find the power. Repeat the procedure for y = 1.5 m. Solution: 1. (a) The power delivered to the apple by gravity will increase as the apple falls because the force stays the same while the velocity increases. 2. (b) First find the velocity :

v 2 = v02 − 2 g ( y − y0 ) but v0 = 0 so v = −2 g ( y − y0 ) v = −2 ( 9.81 m/s 2 ) ( 2.5 m − 3.5 m ) = 4.4 m/s

3. Now apply equation 7-13:

P = Fv = mgv = ( 0.19 kg ) ( 9.81 m/s 2 ) ( 4.4 m/s ) = 8.2 W

4. (c) Repeat for y = 1.5 m:

v = −2 g ( y − y0 ) = −2 ( 9.81 m/s 2 ) (1.5 m − 3.5 m ) = 6.3 m/s P = Fv = mgv = ( 0.19 kg ) ( 9.81 m/s 2 ) ( 6.3 m/s ) = 12 W

Insight: The downward force gravity exerts on the apple is parallel to its displacement so it is doing positive work on the apple. The quantity v2 and thus the kinetic energy increases linearly with the distance the apple has fallen.

72. Picture the Problem: The ball rises vertically to height h, stops, and returns to the ground. Strategy: As the ball rises, gravity does negative work on the ball, reducing its kinetic energy to zero when it reaches height hmax. Set the work done by gravity equal to the change in kinetic energy between h and hmax. As the ball falls gravity does positive work on the ball, increasing its kinetic energy once again. Solution: 1. (a) Combine equations 7-1 and 7-7:

W = Fd = − ( mg )( hmax − h ) = mg ( h − hmax )

2. (b) Repeat for when the ball reaches h = 0:

W = Fd = − ( mg )( 0 − h ) = mgh

3. (c) Apply equation 7-7 again:

K f = Ki + W =

1 2

mv02 + mgh

Insight: The work done by gravity converts the ball’s kinetic energy into gravitational potential energy, a topic discussed in Chapter 8. The gravitational potential energy becomes kinetic energy once again as the ball falls.

73. Picture the Problem: The work done by the force is the area under the force versus position graph. Strategy: Determine the F-values that correspond to the x-values given in the problem and then add or subtract the areas of triangles in order to determine the areas bounded by the given x-values.

The first line is:

F = ( 0.8 1.5 ) x

The second line is:

F = 1.28 − ( 0.8 2.5 ) x

Therefore the F values that correspond to x = 1.0, 1.2, 2.0, 3.5, and 4.0 m are 0.53, 0.65, 0.65, 0.16, and 0 N, respectively. Positive work is done when the object moves in the positive x-direction, negative work otherwise. The area of the entire triangle is 12 ( 4.0 m )( 0.80 N ) = 1.6 J . Solution: 1. (a) The work done on the object is the area under the graph between 0 and 2.0 m:

W = 1.6 J − 12 ( 2.0 m )( 0.64 N ) = 0.96 J

2. (b) The work done on the object is the area under the graph between 1.0 and 4.0 m:

W = 1.6 J − 12 (1.0 m )( 0.53 N ) = 1.3 J

3. (c) The work done on the object is the area under the graph between 3.5 and 1.2 m:

W = − ⎡⎣1.6 J − 12 (1.2 m )( 0.64 N ) − 12 ( 0.5 m )( 0.16 N ) ⎤⎦ = −1.2 J

Insight: The work is positive as long as the object moves from left to right (from small x to large x). Therefore the object gains energy as it moves from left to right, and loses energy as it is moved from right to left. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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74. Picture the Problem: The spider crawls up the window at an angle as pictured. Strategy: Use equation 7-3 keeping in mind the angle between the force exerted by the spider (vertically upward) and the displacement (upward at 25° from the vertical). Solution: 1. Apply equation 7-3:

W = Fd cos θ = mg y cos θ = mg ( vt ) cos θ

2. Now use equation 7-10:

P=

W mg ( vt ) cos θ = = mgv cos θ t t P = (1.8 × 10−3 kg )( 9.81 m/s 2 ) ( 0.023 m/s ) cos 25o P = 3.7 × 10−4 W = 0.37 mW

Insight: Only the component of the force along the direction of the motion does any work. The spider is gaining gravitational potential energy (see Chapter 8) as it elevates its center of mass. Yes, that spider does look curiously like a beetle…

75. Picture the Problem: The ski boat pulls the skier horizontally at constant speed. Strategy: The extra power required to pull the skier at constant speed is the force of friction times the velocity. Write equations for the power required to propel the boat alone and to pull the skier and solve for the tension in the rope. Solution: 1. For the boat alone:

Pb = f b v

2. For the boat and the skier:

Pbs = ( fb + F ) v

3. Solve for the tension F:

F=

Pbs P P P − Pb 37,800 W − 36, 600 W − f = bs − b = bs = = 86 N v v v v 14.0 m/s

Insight: The tension in the rope must also equal the friction force acting on the skier because the skier is not accelerating, so by Newton’s Second Law the net force on the skier must be zero.

76. Picture the Problem: The cookie dough is stirred with a spoon. Strategy: The power required to stir the dough is the force times the velocity. Solution: 1. (a) Apply equation 7-13:

P = F v = ( 21 N )( 0.23 m/s ) = 4.8 W

2. (b) Solve equation 7-10 for W:

W = Pt = ( 4.8 W )(1.5 min × 60 s/min ) = 430 J = 0.43 kJ

Insight: The work you do with all that stirring is converted into heat by viscous frictional forces within the dough.

77. Picture the Problem: The baseball is accelerated along a straight line. Strategy: The work done on the ball by the pitcher changes the ball’s kinetic energy. The power required to accelerate the ball is the work divided by the time. Solution: 1. (a) Apply equation 7-7:

W = 12 mvf2 − 12 mvi2 =

2. (b) Apply equation 7-10:

P=

1 2

( 0.14 kg )( 42.5 m/s )

2

− 0 = 130 J

W 130 J = = 2200 W = 2.9 hp t 0.060 s

3. (c) The power required to accelerate the ball to the same speed in less time would be more than the previous case because the same amount of work would have to be done in a smaller amount of time. Insight: The power generated by the pitcher is the rate at which the ball gains kinetic energy.


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78. Picture the Problem: The jet is accelerated horizontally by the catapult. Strategy: The work done by the catapult on the jet equals the change in kinetic energy of the jet. Since the kinetic energy depends upon the mass, we can find the mass of the jet if we know the amount of work that was done. The power is the work divided by the time. 7 2W 2 ( 7.6 × 10 J ) 2 1 W = 2 mvf − 0 ⇒ m = 2 = = 2.9 ×104 kg = 32 tons! Solution: 1. (a) Solve equation 7-7 for m: 2 vf ( 72 m/s ) 2. (b) Apply equation 7-10 to find P:

P=

W 7.6 × 107 J = = 3.8 × 107 W = 38 MW = 51, 000 hp! t 2.0 s

Insight: The work done by the catapult goes into the kinetic energy of the jet. We’re ignoring any energy removed by frictional forces.

79. Picture the Problem: The brain converts chemical energy into electrical energy and finally into heat energy. Strategy: The power consumed by the brain equals the rate at which it converts chemical energy into other forms. Solution: 1. (a) Solve equation 7-10 for t:

t=

W 280 Cal × 4186 J/Cal 1h = = 53,300 s × = 15 h P 22 W 3600 s

2. (b) Solve equation 7-13 for v:

v=

P P 22 W = = = 0.62 m/s F mg ( 3.6 kg ) ( 9.81 m/s 2 )

3. (c) Solve d = vt for t:

t=

d 1.0 m = = 1.6 s v 0.62 m/s

Insight: The power required for doing mechanical work is somewhat larger than that consumed by the brain. In this case it probably takes you a little less than a second to lift the milk 1.0 m, so you expend about 40 W or twice the power lifting the milk than your brain consumes.

80. Picture the Problem: A car accelerates while traveling on a level road. Strategy: The acceleration can be found from the force and the mass using Newton’s Second Law. The force can be determined from the power and the speed. F P v ( 49 hp × 746 W/hp ) Solution: 1. (a) Combine Newton’s a= = = = 2.0 m/s 2 Second Law with equation 7-13: m m (1300 kg )(14 m/s ) 2. (b) For the same power output the car can exert a smaller force at higher speeds. Therefore the acceleration will decrease if the speed is doubled. 3. (c) Recalculate using the new speed:

a=

P ( 49 hp × 746 W/hp ) = = 1.0 m/s 2 mv (1300 kg )( 28 m/s )

Insight: The question isn’t entirely realistic because car engines generate more power at higher rpm than they do at low speeds, so a real engine would be able to accelerate at a higher rate than that calculated in part (c).

81. Picture the Problem: The meteorite is slowed to a stop while traveling in a straight line. Strategy: The work done on the meteorite by the car equals the negative of the initial kinetic energy of the meteorite. The work can then be used to find the force exerted on the meteorite by the car. Solution: Solve equation 7-7 for the force. The W = Fd = 12 mvf2 − 12 mvi2 = 0 − 12 mvi2 average force is negative if we take the initial 2 ( 27 lb × 1 kg/2.2 lb )( 550 m/s ) mvi2 velocity of the meteorite to be in the positive F =− =− direction: 2d 2 ( 0.22 m ) F = −8.4 × 106 N = −948 tons!

Insight: The kinetic energy of the meteorite is used to deform the metal trunk of the car as well as heat up everything through frictional forces. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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82. Picture the Problem: The pigeon flies horizontally in a straight line at constant speed. Strategy: The pigeon does positive work on its center of mass as it flies, and friction does negative work at the same time. The two works are equal because the pigeon does not change its kinetic energy. Therefore the force the pigeon must generate equals the friction force. The power required to do this depends upon the force and the speed. Solution: 1. (a) Substitute for F in equation 7-13.

P = Fv = ( bv 2 ) v = bv3

2. Now solve for b and determine the units:

b=

3. (b) Solve the above equation for v:

v=

P W N ⋅ m s3 kg ⋅ m m s3 kg = = = = 3 3 3 3 2 3 v m s s m s s m m 3

P b

vnew = vold

4. (c) Use a ratio to calculate the speed increase:

3

Pnew b

3

Pold b

=

3

2P 3 = 2 = 1.26 P

Insight: Air resistance effectively limits the maximum speed of birds. Here we find that doubling the power only increases the maximum speed by 26%. Since most birds have a similar aerodynamic shape, we expect the b constant to be similar for most birds so that their maximum flight speeds will be similar.

83. Picture the Problem: The two springs are connected in series and are stretched vertically. Strategy: Because of the arrangement of the springs the forces applied to the two springs must be the same. The total stretch distance is the sum of the individual stretch distances. In order to stretch the springs we must do work against both springs. Add the required works to find the total work. ⎡1 1 ⎤ F F + =F⎢ + ⎥ k1 k2 ⎣ k1 k2 ⎦

Solution: 1. Add the stretch distances and solve for the force:

x = x1 + x2 =

2. Solve for F:

F=

3. Now add the works:

W = W1 + W2 = 12 k1 x12 + 12 k2 x22

4. Substitute for x1 and x2 :

⎛F⎞ ⎛F⎞ F2 ⎛ 1 1 ⎞ W = k1 ⎜ ⎟ + 12 k2 ⎜ ⎟ = ⎜ + ⎟ 2 ⎝ k1 k2 ⎠ ⎝ k1 ⎠ ⎝ k2 ⎠

5. Substitute for F:

W=

x 1 k1 + 1 k2

2

2

1 2

1 2 ⎤⎛ 1 1 ⎞ x2 1⎡ 2 x + ⎟= ⎢ ⎥ 2 ⎜ 2 ⎢⎣ (1 k1 + 1 k2 ) ⎥⎦ ⎝ k1 k2 ⎠ 1 k1 + 1 k2

Insight: Two springs in series act like a single spring with a force constant that is smaller than either of the individual spring constants, where 1 keffective = 1 k1 + 1 k2 .

84. Picture the Problem: The two springs are connected in parallel and are stretched vertically together. Strategy: Because of the arrangement of the springs the stretch distances must be the same. In order to stretch the springs a distance x we must do work against both springs. Add the required works to find the total work. Solution: Add the works using equation 7-8:

W = W1 + W2 = 12 k1 x 2 + 12 k2 x 2 =

1 2

( k1 + k2 ) x 2

Insight: Two springs in parallel act like a single spring with a force constant equal to the sum of the individual force constants. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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85. Picture the Problem: The block slides horizontally but the pulling force is inclined at an angle. Strategy: Use the definition of work to find the angle between the force and displacement vectors. Then set the work done by the pulling force equal to the change in kinetic energy of the block. Solution: 1. (a) Solve equation 7-3 for θ:

⎡ ⎤ 50.0 J ⎛W ⎞ −1 o W = Fd cos θ so θ = cos −1 ⎜ ⎥ = 42.2 ⎟ = cos ⎢ ⎝ Fd ⎠ ⎢⎣ ( 45.0 N )(1.50 m ) ⎥⎦

2. (b) Solve equation 7-7 for m:

W = ΔK = 12 mvf2 − 0 so m =

2 ( 50.0 J ) 2W = = 14.8 kg 2 2 vf ( 2.60 m/s )

Insight: Only the component of the force along the direction of the motion does any work. The vertical component of the force reduces the normal force a little.

86. Picture the Problem: The power required for Microraptor gui to fly, as a function of speed, is depicted in the graph at right. Strategy: Examine the lower horizontal line that shows the estimated 9.8-W power output of Microraptor. The two locations where it intersects the graph of the power required for flight indicate the minimum and maximum flight speeds the creature could achieve. Solution: The lower horizontal line at 9.8 W power output intersects the graph of the power required for flight at 7.7 m/s and 15 m/s, indicating the creature could fly at speeds between 7.7−15 m/s. Insight: If the Microraptor could only generate 7.5 W of power, it would not have been able to fly at all! A minimum of 8.1 W is required, as can be seen from the minimum of the graph of the power required for flight.

87. Picture the Problem The power required for Microraptor gui to fly, as a function of speed, is depicted in the graph at right. Strategy: Examine the upper horizontal line that shows the hypothetical 20-W power output of Microraptor. The two locations where it intersects the graph of the power required for flight indicate the minimum and maximum flight speeds the creature could achieve if it could generate that much power. Solution: The upper horizontal line at 20-W power output intersects the graph of the power required for flight at 2.5 m/s and 25 m/s, indicating the creature could fly at speeds between 2.5 − 25 m/s. Insight: The Microraptor would be able to hover (fly at 0 m/s) if it could generate at least 25 W of power.


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88. Picture the Problem: The power required for Microraptor gui to fly, as a function of speed, is depicted in the graph at right. Strategy: A minimum of 8.1 W of power would be required for the Microraptor to fly at 10 m/s. Use this power value to determine the energy expended if the creature flew at 10 m/s for 1.0 minute. Solution: Multiply the power by the time to find the work:

W = Pt = ( 8.1 W )( 60 s ) W = 490 J

Insight: In chapter 16 we will learn about thermal energy and food calories. If the Microraptor could somehow use the energy of 100 food calories (4.2×105 J) solely for flying at 10 m/s, it could fly for over 14 hours!

89. Picture the Problem: The power required for Microraptor gui to fly, as a function of speed, is depicted in the graph at right. Strategy: Note the two dashed, straight lines labeled 1 and 2. These lines represent constant ratios of power to speed; that is, a constant value for P/v. Referring to Equation 7-13, we see that P/v = Fv/v = F, so the lines 1 and 2 correspond to lines of constant force. Line 2 has the smallest slope that still touches the power-versus-speed curve. The slope of line 2 is therefore the minimum amount of force that the Microraptor had to generate in order to fly. Solution: Determine the slope of line 2:

slope = Fmin =

9.8 W − 0 = 0.65 N 15 m/s − 0

Insight: The Microraptor had to generate maximum force at its minimum flying speed (dashed line 1), a force of about 9.8 W / 7.7 m/s = 1.3 N.

90. Picture the Problem: The block slides into the spring and compresses it. Strategy: The work to stretch or compress a spring a distance x is 12 kx 2 .

The work done on the block by the spring equals the kinetic energy lost by the block. The work done on the block is negative because the force on the block is toward the left while the motion is toward the right. Solution: 1. (a) Apply equations 7-7 and 7-8:

Won block = ΔK block = 0 − 12 mv02 = − 12 kx 2

2. Now solve for k:

k =m

3. (b) Use a ratio to find the new compression:

xnew = xold

( 0.62 m/s ) v02 = (1.4 kg ) = 930 N/m = 0.93 kN/m 2 x2 ( 0.024 m ) 2

xnew =

2 mnew vnew k 2 old old

m v xold 2

=

k

2.4 cm 2

=

( 2mold ) ( 12 vold ) mold

2 old

v

2

=

1 2

= 1.7 cm

Insight: The kinetic energy of the block is transformed into the energy stored in the spring as it is compressed.


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91. Picture the Problem: The compressed spring pushes the block horizontally on a frictionless surface. Strategy: The work to stretch or compress a spring a distance x is 12 kx 2 .

The work done by the spring equals the kinetic energy gained by the block. Solution: 1. (a) Apply equations 7-7 and 7-8:

W = 12 kx 2 = ΔK = 12 mvf2 − 0

2. Now solve for m:

m=

kx 2 ( 750 N/m )( 0.041 m ) = = 1.6 kg 2 vf2 ( 0.88 m/s ) 2

3. (b) Since vf ∝ 1 m , doubling the mass will decrease the speed by a factor of will be greater than 0.44 m/s. vf =

4. (c) Solve the above equation for vf :

kx 2 = m

2 , not 2. Therefore the final speed

( 750 N/m )( 0.041 m ) 2 × 1.6 kg

2

= 0.63 m/s

Insight: The work done on the spring in order to compress it becomes stored potential energy. That stored energy becomes the kinetic energy of the block as the spring accelerates it. We will learn more about potential energy in Chapter 8.

92. Picture the Problem: The car accelerates horizontally in a straight line. Strategy: The power generated by the car’s engine equals the rate at which it can change the kinetic energy of the car. Solution: 1. (a) Combine equations 7-7 and 7-10:

P=

2. Now solve for t:

t=

W ΔK 12 mvf2 − 12 mvi2 = = t t t 1 2

(1200 kg ) ( 252 − 152 m 2 /s 2 ) = ( 48 hp × 746 W/hp )

7.8 s

3. (b) Since t ∝ vf2 − vi2 the time required to increase the speed from 5.0 m/s to 15 m/s will be less than the time required to change the speed over the next interval of 10 m/s as in part (a). 4. (c) Repeat the above calculation:

t=

1 2

(1200 kg ) (152 − 5.02 m 2 /s 2 ) = ( 48 hp × 746 W/hp )

3.9 s

Insight: The problem isn’t entirely realistic because the power output of a car’s engine is not constant. Automobile engines generate more power at higher engine rpm than they do at lower speeds.


7 – 25




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