11.1. Visualize: Please refer to Figure Ex11.1.Solve: (a)

r rA B AB⋅ = = ° =cos ( )( )cos . .α 4 5 40 15 3

(b) r rC D CD⋅ = = ° = −cos ( )( )cos . .α 2 4 120 4 0

(c) r rE F EF⋅ = = ° =cos ( )( )cos .α 3 4 90 0

11.2. Visualize: Please refer to Figure Ex11.2.Solve: (a)

r rA B AB⋅ = = ° = −cos ( )( )cos . .α 3 4 110 4 1

(b) r rC D CD⋅ = = ° = −cos ( )( )cos .α 4 5 180 20

(c) r rE F EF⋅ = = ° =cos ( )( )cos . .α 4 3 30 10 4

11.3. Solve: (a) r rA B A B A Bx x y y⋅ = + = − + − = −( ( ( )( ) .3) )2 4 6 30

(b) r rA B A B A Bx x y y⋅ = + = + − =( ( ( )( )2) )6 3 4 0 .

11.4. Solve: (a) r rA B A B A Bx x y y⋅ = + + − − == 5) )( ( ( )( ) .2 3 2 16

(b) r rA B A B A Bx x y y⋅ = + = − + − = −( ( ( )( ) .4) )3 2 6 24

11.5. Solve: (a) W F r i j i i i j i= ⋅ = − ⋅ ⋅ = ⋅ − ⋅ =r r

∆ ( . ˆ . ˆ) ( . ˆ) ( . ˆ ˆ . ˆ ˆ) .6 0 3 0 2 0 12 0 3 0 12 0 N m J J.

(b) W F r i j j i j j j= ⋅ = − ⋅ ⋅ = ⋅ − ⋅ = −r r

∆ ( . ˆ . ˆ) ( . ˆ) ( . ˆ ˆ . ˆ ˆ) .6 0 3 0 2 0 12 0 6 0 6 0 N m J J.

11.6. Solve: (a) r r r

W F r i j i i i j i= ⋅ = − + ⋅ = − ⋅ + ⋅ −∆ ( . ˆ . ˆ) ( . ˆ) ( . ˆ ˆ . ˆ ˆ)5 0 4 0 3 0 15 0 12 0 N m J = 15.0 J.

(b) r r r

W F r i j j i j j j= ⋅ = − + ⋅ − = ⋅ − ⋅ −∆ ( . ˆ . ˆ) ( . ˆ) ( . ˆ ˆ . ˆ ˆ)5 0 4 0 3 0 15 0 12 0 N m J = 12.0 J.

11.7. Model: Use the work-kinetic energy theorem to find the work done on the particle.Visualize:

Solve: From the work-kinetic energy theorem,

W K mv mv m v v= = − = −( ) = − − =∆ 1

2

1

2

1

2

1

201

202

12

02 2 2 2( )[( ) ( ) ]0.020 kg 30 m/s 30 m/s J

Assess: Negative work is done in slowing down the particle to rest, and an equal amount of positive work is donein bringing the particle to the original speed but in the opposite direction.

11.8. Model: Work done by a force rF on a particle is defined as

r r rW F r= ⋅ ∆ , where ∆

rr is the particle’s

displacement.Visualize:

Solve: (a) The work done by gravity is

W w r mgj jg2 N m 2.0 kg 9.8 m/s 1.50 m J 29.4 J= ⋅ = − ⋅ − = − = −

r r∆ ( ˆ) ( . . ) ˆ ( )( )( )2 25 0 75

(b) The work done by hand is W F rH hand on book= ⋅r r

∆ . As long as the book does not accelerate,r rF F mgj mgj

W mgj j

hand on book earth on book

H2 m 2.0 kg 9.8 m/s 1.50 m = 29.4 J

= − = − − =

⇒ = ⋅ − =

( ˆ) ˆ

( ˆ) ( . . ) ˆ ( )( )( )2 25 0 75

11.9. Model: Model the piano as a particle and use W F r= ⋅r r

∆ , where W is the work done by the force rF

through the displacement ∆rr .

Visualize:

Solve: For the force rw:

W F r w r w r= ⋅ = ⋅ = ° = =r r r r

∆ ∆ ∆( )( )cos ( )( . )( ) ,0 2500 5 0 1 12 500 N m J

For the tension rT1:

W T r T r= ⋅ = ° = − = −r r1 1 150 1830 5 0 0 8660 7920∆ ∆( )( )cos( ) ( )( . )( . ) N N J

For the tension rT2 :

W T r T r= ⋅ = ° = − = −r r2 2 135 1295 4580∆ ∆( )( )cos( ) ( )( )( ) N 5.0 m 0.7071 J

Assess: Note that the displacement ∆rr in all the above cases is directed downwards along − ˆ.j

11.10. Model: Model the crate as a particle and use W F r= ⋅r r

∆ , where W is the work done by a force rF on a

particle and ∆rr is the particle’s displacement.

Visualize:

Solve: For the force rfk :

W f r f r= ⋅ = ° = − = −r rk k 500 N) 3.0 m) ) J∆ ∆( )cos( ) ( ( (180 1 1500

For the tension rT1:

W T r T r= ⋅ = °r r1 1 20 0 9397∆ ∆( )( )cos ( ( ( .= 326 N) 3.0 m) ) = 919 J

For the tension rT 2:

W T r T r= ⋅ = ° = =r r2 2 30 223 579∆ ∆( )( )cos ( )( ) ) N 3.0 m (0.866 J

Assess: Negative work done by the force of kinetic friction ( )rfk means that 1500 J of energy has been transferred

out of the crate.

11.11. Model: Model the 2.0 kg object as a particle, and use the work-kinetic energy theorem.Visualize: Please refer to Figure Ex11.11. For each of the five intervals the velocity-versus-time graph gives theinitial and final velocities. The mass of the object is 2.0 kg.Solve: According to the work-kinetic energy theorem:

Interval AB: m/s m/s 2.0 kg 2 m/s 0 m/s J

Interval BC: m/s m/s 2.0 kg 0 m/s 2 m/s

f f i

i f

i f

W K mv mv m v v

v v W

v v W

= = = = −( )= = ⇒ = − = +

= = ⇒ = − =

∆ 1

2

1

2

1

2

0 21

24 0

2 01

2

2 2 2 2

2 2

2 2

i

, ( )[( ) ( ) ] .

, ( )[( ) ( ) ] −−

= = − ⇒ = − − = +

= − = − ⇒ = − − − =

= − =

4 0

0 21

22 0 4 0

2 21

22 0

2

2 2

2 2

.

, ( )[( ) ( ) ] .

, ( )[( ) ( ) ]

,

J

Interval CD: m/s m/s 2.0 kg m/s m/s J

Interval DE: m/s m/s 2.0 kg 2 m/s m/s J

Interval EF: m/s

i f

i f

i f

v v W

v v W

v v −− ⇒ = − − − = −11

22 3 02 2m/s 2.0 kg 1 m/s m/s JW ( )[( ) ( ) ] .

Assess: The work done is the same in intervals AB and CD. It is not whether v is positive or negative that countsbecause K v∝ 2 . What is important is the magnitude of v and how v changes.

11.12. Model: Use the definition of work.Visualize: Please refer to Figure Ex11.12.Solve: Work is defined as the area under the force-versus-position graph:

W F dss

s

s

= =∫i

f

area under the force curve

Interval 0 – 1 m: 4.0 N 1.0 m 0.0 m J

Interval 1 – 2 m: 4.0 N 0.5 m 4.0 N 0.5 m J

Interval 2 – 3 m: 4.0 N m J

W

W

W

= − == + − =

= − = −

( )( ) .

( )( ) ( )( )

( )( ) .

4 0

0

1

21 2 0

11.13. Model: Use the work-kinetic energy theorem to find velocities.Visualize: Please refer to Figure Ex11.13.Solve: The work-kinetic energy theorem is

∆K mv mv W F dx x x

mv mv F dx x

x

x

x

= − = = =

⇒ − = − = =

∫

∫

1

2

1

2

1

2

1

20 500 2 0

1

21 0

2 2

2 2 2

f i i f

x

f f

0 m

x

area under the force curve from to

kg m/s J area from 0 to

i

f

f

( . )( . ) .

At = 1 m: 0.500 kg J J m/s

At m: 0.500 kg J J 9.17 m/s

At m: 0.500 kg J J m/s

f f

f f

f f

x v v

x v v

x v v

1

21 0 12 5 7 35

21

21 0 20

31

21 0 22 5 9 70

2

2

2

( ) . . .

( ) .

( ) . . .

− = ⇒ =

= − = ⇒ =

= − = ⇒ =

11.14. Model: Use the work-kinetic energy theorem.Visualize: Please refer to Figure Ex11.14.Solve: The work-kinetc energy theorem is

∆K mv mv W F dx x x

mv mv F dx x

x

x

x

= − = = =

⇒ − = − = =

∫

∫

1

2

1

2

1

2

1

22 4 0

1

216

2 2

2 2 2

f i i f

x

f f

0 m

x

area under the force curve from to

kg m/s J area from 0 to

i

f

f

( )( . )

At m kg J N m J m/s

At m kg J J m/s

f f

f f

x v v

x v v

= − = − = − ⇒ =

= − = ⇒ =

21

22 0 16

1

210 2 10 2 45

41

22 0 16 0 4 0

2

2

: ( . ) ( )( ) .

: ( . ) .

11.15. Model: Use the work-kinetic energy theorem.Visualize: Please refer to Figure Ex11.15.Solve: The work-kinetic energy theorem is

∆K W F dx F x x x

mv mv F

x

x

x

x= = =

− =

∫i

f

area of the -versus- graph between and

2 m

i f

f i

1

2

1

2

1

22 2 ( )( )max

Using m = 0.500 kg, vf = 6.0 m/s, and vi = 2.0 m/s, the above equation yields Fmax = 8.0 N.Assess: Problems in which the force is not a constant can not be solved using constant-acceleration kinematicequations.

11.16. Model: Use the definition F dU dss = − / .Visualize: Please refer to Figure Ex11.16.Solve: Fx is the negative of the slope of the potential energy graph at position x. Between x = 0 cm and x = 10 cmthe slope is

slope = ( ) / ( ) ( ) / ( )U U x xf i f i 0 J 10 J 0.1 m 0.0 m N− − = − − = −100

Thus, Fx = 100 N at x = 5 cm. The slope between x = 10 cm and x = 20 cm is zero, so Fx = 0 N at x = 15 cm.Between 20 cm and 40 cm,

slope = ( ) / ( )10 J 0 J 0.40 m 0.20 m N− − = 50

At x = 25 cm and x = 35 cm, therefore, Fx = −50 N.

11.17. Model: Use the definition F dU dss = − .Visualize: Please refer to Figure Ex11.17.Solve: Fx is the negative of the slope of the potential energy graph at position x.

FdU

dxx = −

Between x = 0 m and x = 2 m, the slope is

slope = ( ) / ( ) ( ) / ( )U U x xf i f i 60 J 0 J 2 m 0 m N− − = − − = 30

and between x = 2 m and x = 5 m, the slope is

slope = ( ) / ( ) ( ) / ( )U U x xf i f i 0 J 60 J 5 m 2 m J− − = − − = −20

Thus, Fx = –30 N at x = 1 m and Fx = 20 N at x = 3 m.

11.18. Model: Use the negative derivative of the potential energy to determine the force acting on a particle.Solve: (a)

The graph of the potential energy U y= 4 3 is shown.(b) The y-component of the force is

FdU

dy

d

dyy yy = − = − = −( )4 123 2

At y = 0 m, Fy = 0 N; at y = 1 m, Fy = – 12 N; and at y = 2 m, Fy = − 48 N.

11.19. Model: Use the negative derivative of the potential energy to determine the force acting on a particle.Solve: (a)

The graph of the potential energy U x= 10 / is shown.(b) The x-component of the force is

FdU

dx

d

dx x x

Fx

Fx

Fx

x

xx

xx

xx

= − = −

=

= = = = = ===

==

==

10 10

102 5

100 40

100 156

2

2 22

5 25

8 28

N N Nmm

mm

mm

. . .

11.20. Model: Assume the carbon-carbon bond acts like an ideal spring that obeys Hooke’s law.Visualize:

The quantity ( )x x− e is the stretching relative to the spring’s equilibrium length. In the present case, bondstretching is analogous to spring stretching.Solve: (a) The kinetic energy of the carbon atom is

K mv= = × = ×− −1

2

1

22 00 10 2 5 102 26 2 21( . )( ) .kg 500 m/s J

(b) The energy of the spring is given by

U k x x K

kK

x x

s e

e

J

0.050 10 m

N

m

= − =

⇒ =−

= ××

=−

−

1

2

2 2 2 50 102 0

2

2

21

9 2

( )

( )

( . )

( ).

11.21. Visualize: One mole of helium atoms in the gas phase contains NA = ×6 02 1023. atoms.Solve: If each atom moves with the same speed v, the microscopic total kinetic energy will be

K N mv vK

mNmicro Amicro

A

J3700 J

kg m/s=

= ⇒ = =

× ×=−

1

23700

2 2

6 68 10 6 02 1013602

27 23

( )

( . )( . )

11.22. Solve: (a) The car has an initial kinetic energy Ki. That energy is transformed into thermal energy of thecar and the road surface. The gravitational potential energy does not change and no work is done by external forces,so during the skid Ki transforms entirely into thermal energy Eth. This energy transfer and transformation is shownon the energy bar chart. Note that

K mvi 1500 kg 20 m/s J= = = ×1

2

1

23 0 102 2 5( )( ) .

(b) The change in the thermal energy of the car and the road surface is 3 0 105. × J .

11.23. Visualize:

Solve: (a) K K mv U U mgy

W K K mv U U mgy

i i g2

ext f f g

J 20 kg 9.8 m/s 3 m J

J 20 kg 2.0 m/s J J

= = = = = = =

= = = = = = = =

0 02

0 0

1 12 2

1 1

1

20 588

01

2

1

240 0

( )( )( )

( )( )

At the top of the slide, the child has gravitational potential energy of 588 J. This energy is transformed into thermalenergy of the child’s pants and the slide and the kinetic energy of the child. This energy transfer and transformationis shown on the energy bar chart.(b)

The change in the thermal energy of the slide and of the child’s pants is 588 J – 40 J = 548 J.

11.24. Visualize: The system loses 1000 J of potential energy. In the process of losing this energy, it does500 J of work on the environment, which means Wext = –500 J. Since the thermal energy increases 100 J, we have∆Eth = 100 J. This is shown in the energy bar chart.

11.25. Visualize:

Note that the conservation of energy equationK U W K U Ei i ext f f th+ + = + + ∆

requires that Wext be equal to +400 J.

11.26. Solve: Please refer to Figure Ex11.26. The energy conservation equation yields

K U W K U E Wi i ext f f th ext ext J J J J J W J+ + = + + ⇒ + + = + + ⇒ = −∆ 4 1 1 2 1 1

Thus, the work done to the environment is –1 J. In other words, 1 J of energy is transferred from the system into theenvironment. This is shown in the energy bar chart.

11.27. Visualize: The tension of 20 N in the cable is an external force that does work on the block Wext =(20 N)(2.0 m) = 40 J, increasing the gravitational potential energy of the block. We placed the origin of our coordinatesystem on the initial resting position of the block, so we have Ui = 0 J and U mgyf f

21.02 kg 9.8 m/s 2.0 m= = =( )( )( )20 J. Also, Ki = 0 J, and ∆Eth = 0 J. The energy bar chart shows the energy transfers and transformations.

Solve: The conservation of energy equation is

K U W K U E mv

v

i i ext f f th f2

f

J 0 J J J J

J kg m/s

+ + = + + ⇒ + + = + +

⇒ = =

∆ 0 401

220 0

20 2 1 02 6 26( )( ) / . .

11.28. Model: Model the elevator as a particle, and apply the conservation of energy equation.Solve: The tension in the cable does work on the elevator to lift it. Because the cable is pulled by the motor, wesay that the motor does the work of lifting the elevator.(a) The energy conservation equation is K U W K U Ei i ext f f th+ + = + + ∆ . Using Ki = 0 J, Kf = 0 J, and ∆Eth = 0 J gives

W U U mg y yext f i f i21000 kg 9.8 m/s 100 m J= − = − = = ×( ) ( ) ( )( )( ) .9 8 105

(b) The power required to give the elevator this much energy in a time of 50 s is

PW

t= = × = ×ext

59.8 10 J

50 s W

∆1 96 104.

Assess: Since 1 horsepower (hp) is 746 W, the power of the motor is 26 hp. This is a reasonable amount of powerto lift a mass of 1000 kg to a height of 100 m in 50 s.

11.29. Model: Model the steel block as a particle subject to the force of kinetic friction and use the energyconservation equation.Visualize:

Solve: (a) The work done on the block is W F rnet net= ⋅r r

∆ where ∆rr is the displacement. We will find the

displacement using kinematic equations and the force using Newton’s second law of motion. The displacement inthe x-direction is

∆x x x v t t a t tx x= = + − + − = − +1 0 0 1 0 1 021

20 3 0 0 0( ) ( ) ( ( . ) m + 1.0 m/s) s s m = 3.0 m

Thus ∆rr i= 3 0. ˆ m.

The equations for Newton’s second law along the x and y components are

F n w n w mg

F F f F f n

W F r F x

y

x

net2

net k k k

net net

N kg m/s N

N N N

N m J

( ) = − = ⇒ = = = =

( ) = − = ⇒ = = = =

⇒ = ⋅ = ° = =

0 10 9 8 98 0

0 0 6 98 0 58 8

0 58 8 3 0 1 176 4

( )( . ) .

( . )( . ) .

cos ( . )( . )( ) .

r r

r r

µ

∆ ∆

(b) The power required to do this much work in 3.0 s is

PW

t= = =176 4

58 8.

. J

3.0 s W

11.30. Solve: The power of the solar collector is the solar energy collected divided by time. The intensity of thesolar energy striking the earth is the power divided by area. We have

PE

t= = × = =

⇒ =

∆∆

150 1041 667

41 66741 7

6 J

3600 s W and intensity 1000

W

m

Area of solar collector = W

1000W / m m

2

22

,

,.

11.31. Solve: The night light consumes more energy than the hair dryer. The calculations are

1 2 10 60

24 60 60 105

. kW 10 min = 1.2 10 J = 7.2 10 J

10 W 24 hours = 10 J = 8.64 J

3 5× × × × ×

× × × × ×

11.32. Solve: (a) A kilowatt hour is a kilowatt multiplied by 3600 seconds. It has the dimensions of energy.(b) One kilowatt hour is energy

1 0 360 3 6 106 kwh 1 00 J /s) 0 s) J= = ×( ( .

Thus

5003 6 10

1 8 106

9 kwh 500 kwh) J

1 kwh J= ×

= ×(.

.

11.33. Model: Model the sprinter as a particle, and use the constant-acceleration kinematic equations and thedefinition of power in terms of velocity.Visualize:

Solve: (a) We can find the acceleration from the kinematic equations and the horizontal force from Newton’ssecond law. We have

x x v t t a t t a a

F ma

x x x x

x x

= + − + − ⇒ = + + − ⇒ =

⇒ = = =

0 0 1 0 1 02 21

250 0 0

1

27 0 0

102

( ) ( ) ( . )

( )( )

m m m s s 2.04 m/s

50 kg 2.04 m/s N

2

2

(b) We obtain the sprinter’s power output by using P F v= ⋅r r

, where rv is the sprinter’s velocity. At t = 2 s the power is

P F v a t tx x x= + − = + − =( )[ ( )] ( )[ ( . )( . )]0 0 102 0 2 04 2 0 0 416 N m/s m/s s s W2

The power at t = 4 s is 832 W, and at t = 6 s the power is 1248 W.

11.34. Model: Use the definition of work for a constant force rF, W F s= ⋅

r r∆ , where ∆

rs is the displacement.

Visualize: Please refer to Figure P11.34. The force rF i j= +( ˆ ˆ)6 8 N on the particle is a constant.

Solve: (a) W W W F s F s

i j i i j j

ABD AB BD AB BD

N m N m J J 50 J

= + = ⋅ + ⋅

= + ⋅ + + ⋅ = + =

r r r r( ) ( )

( ˆ ˆ ) ( ˆ) ( ˆ ˆ) ( ˆ)

∆ ∆

6 8 3 6 8 4 18 32

(b) W W W F s F s

i j j i j j

ACD AC CD AC CD

N m N m J J 50 J

= + = ⋅ + ⋅

= + ⋅ + + ⋅ = + =

r r r r( ) ( )

( ˆ ˆ ) ( ˆ) ( ˆ ˆ) ( ˆ)

∆ ∆

6 8 4 6 8 3 32 18

(c) W F s i j i jAD AD N m J J 50 J = ⋅ = + ⋅ + = + =r r

( ) ( ˆ ˆ ) ( ˆ ˆ)∆ 6 8 3 4 18 32The force is conservative because the work done is independent of the path.

11.35. Model: The force is conservative, so it has a potential energy.Visualize: Please refer to Figure P11.35 for the graph of the force.Solve: The definition of potential energy is ∆U = –W(i → f). In addition, work is the area under the force-versus-displacement graph. Thus ∆U = Uf – Ui = –(area under the force curve). Since Ui = 0 at x = 0 m, the potentialenergy at position x is U(x) = –(area under the force curve from 0 to x). From 0 m to 3 m, the area increases linearlyfrom 0 N m to –60 N m, so U increases from 0 J to 60 J. At x = 4 m, the area is –70 J. Thus U = 70 J at x = 4 m, andU doesn’t change after that since the force is then zero. Between 3 m and 4 m, where F changes linearly, U musthave a quadratic dependence on x (i.e., the potential energy curve is a parabola). This information is shown on thepotential energy graph below.

(b) Mechanical energy is E = K + U. From the graph, U = 20 J at x = 1 m.The kinetic energy is K = 1

2 mv2 = 12 (0.10 kg) (25 m/s)

2 = 31.25 J. Thus E = 51.25 J.(c) The total energy line at 51.25 J is shown on the graph above.(d) The turning point occurs where the total energy line crosses the potential energy curve. We can see from thegraph that this is at approximately 2.5 m. For a more accurate value, the potential energy function is U = 20x J. TheTE line crosses at the point where 20x = 51.25, which is x = 2.56 m.

11.36. Model: Use the relationship between force and potential energy and the work-kinetic energy theorem.Visualize:

Please refer to Figure P11.36. We will find the slope in the following x regions: 0 cm < x < 1 cm, 1 < x < 3 cm,3 < x < 5 cm, 5 < x < 7 cm, and 7 < x < 8 cm.Solve: (a) Fx is the negative slope of the U-versus-x graph, for example, for 0 m < x < 2 m

J

0.01 m N 400 N

dU

dxFx= − = − ⇒ = +4

400

Calculating the values of Fx in this way, we can draw the force-versus-position graph as shown.(b) Since W F dxx

xx= ∫ =

i

f area of the Fx-versus-x graph between xi and xf, the work done by the force as the particle

moves from xi = 2 cm to xf = 6 cm is –2 J.(c) The conservation of energy equation is Kf + Uf = Ki + Ui. We can see from the graph that Ui = 0 J and Uf = 2 J inmoving from x = 2 cm to x = 6 cm. The final speed is vf = 10 m/s, so

2 0 010 0 0 010 22 412 J kg)(10.0 m/s) J kg) m/s2 1

2 i2

i+ = + ⇒ =( . ( . .v v

11.37. Model: Use the relationship between a conservative force and potential energy.Visualize: Please refer to Figure P11.37. We will obtain U as a function of x and Fx as a function of x by usingthe calculus techniques of integration and differentiation.Solve: (a) For the interval 0 m < x < 0.5 m, Fx = (4x) N, where x is in meters. This means

dU

dxF x U x C xx= − = − ⇒ = − + = −4 2 22

12

where we have used U = 0 J at x = 0 m to obtain C1 = 0. For the interval 0.5 m < x < 1 m, F xx = − +( )4 4 N.Likewise,

dU

dxx U x x C= − ⇒ = − +4 4 2 42

2

Since U should be continuous at the junction, we have the continuity condition

( ) ( ) . .. .− = − + ⇒ − = − + ⇒ == =2 2 4 0 5 0 5 2 120 5

22 0 5 2 2x x x C C Cx x m m

U remains constant for x ≥ 1 m.(b) For the interval 0 m < x < 0.5 m, U = +4x, and for the interval 0.5 m < x < 1.0 m, U = −4x + 4, where x is inmeters. The derivatives give F Fx x= − = +4 N and 4 N, respectively. The slope is zero for x ≥ 1 m.

11.38. Model: Use adv

dtxx= , x v dtx= ∫ , K mvx= 1

22 , and F max= .

Visualize: Please refer to Figure P11.38. We know ax = slope of the vx-versus-t graph and x = area under thevx-versus-x graph between 0 and x.Solve: Using the above definitions and methodology, we can generate the following table:

t(s) ax (m/s2) x(m) K(J) F(N)

00.51.0

101010

01.255

06.25

25

555

1.52.02.5

10+10 or −10

−10

11.252028.75

56.25100

56.25

5−5 or +5

−5

3.03.54.0

−10 or 000

354045

252525

−5 or 000

(e) Let J1 be the impulse from t = 0 s to t = 2 s and J2 be the impulse from t = 2 s to t = 4s. We have

J F dt J F dtx x1 210 5= = = ⋅ = = − = − ⋅∫ ∫( )( ) ( )( )5 N 2 s N s and 5 N 1 s N s0 s

2 s

0 s

4 s

(f) J p mv mv v v J m

t v

t vx

x

= = − ⇒ = += = + ⋅ = + == = + − ⋅ = − =

∆ f i f i

At s, m/s N s 0.5 kg m/s m/s m/s

At s, m/s 5 N s 0.5 kg m/s m/s m/s

/

( ) /

( ) /

2 0 10 0 20 20

4 20 20 10 10The vx-versus-t graph also gives vx = 20 m/s at t = 2 s and vx = 10 m/s at t = 4 s.(g)

(h) From s to s, 5 N 20 m J

From s to s, N 15 m J

t t W F dx

t t W F dx

x

x

= = = = =

= = = = − = −

∫∫

0 2 100

2 4 5 75

( )( )

( )( )

(i) At t = 0 s, vx = 0 m/s so the work-kinetic energy theorem for calculating vx at t = 2 s is

W K mv mv v vx x= = − ⇒ = − ⇒ =∆ 1

2

1

2100

1

2

1

20 5 0 202 2 2 2

f i J 0.5 kg kg m/s m/s( ) ( . )( )

To calculate vx at t = 4 s, we use vx at t = 2 s as the initial velocity:

− = − ⇒ =751

20 5

1

20 5 20 102 2 J kg kg m/s m/s( . ) ( . )( )v vx x

Both of these values agree with the values on the velocity graph.

11.39. Model: Model the elevator as a particle.Visualize:

Solve: (a) The work done by gravity on the elevator is

W U mgy mgy mg y yg g21000 kg 9.8 m/s 10 m J= − = − = − − = − = − ×∆ 0 1 1 0

49 8 10( ) ( )( )( ) .

(b) The work done by the tension in the cable on the elevator is

W T y T y y TT = ° = − =( )cos ( ) ( )∆ 0 1 0 10 m

To find T we write Newton’s second law for the elevator:

F T w ma T w ma m g a

W

y y y y

T

= − = ⇒ = + = + = +

= × ⇒ = × = ×∑ ( ) ( )( . )

. ( . )( ) .

1000 kg 9.8 m/s m/s

N N 10 m J

2 21 0

1 08 10 1 08 10 1 08 104 5 5

(c) The work-kinetic energy theorem is

W W W K K K K mv K W W mv

K

T Tnet g f i f f g

f J J 1000 kg 0 m/s J

= + = = − = − ⇒ = + +

⇒ = − × + × + = ×

∆ 1

2

1

2

9 8 10 1 08 101

21 0 10

02

02

4 5 2 4( . ) ( . ) ( )( ) .

(d) K mv v vff f f J 1000 kg m/s= ⇒ × = ⇒ =1

21 0 10

1

24 472 4 2. ( ) .

11.40. Model: Model the rock as a particle, and apply the work-kinetic energy theorem.Visualize:

Solve: (a) The work done by Bob on the rock is

W K mv mv mvBob 0.50 kg 30 m / s 225 J= = − = = =∆ 1

2

1

2

1

2

1

212

02

12 2( )( )

(b) For a constant force, W F x F W xBob Bob Bob Bob N= ⇒ = =∆ ∆/ 225 .(c) Bob’s power output is P F vBob Bob rock= and will be a maximum when the rock has maximum speed. This is justas he releases the rock with vrock = v1 = 30 m/s. Thus, P F vmax = = =Bob W 6.75 kW.1 6750

11.41. Model: Model the crate as a particle, and use the work-kinetic energy theorem.Visualize:

Solve: (a) The work-kinetic energy theorem is ∆K mv mv mv W= − = =12 1

2 12 0

2 12 1

2total . Three forces act on the box,

so Wtotal = Wgrav + Wn + Wpush. The normal force is perpendicular to the motion, so Wn = 0 J. The other two forces dothe following amount of work:

W F r F x W W r W x mg xxpush push push grav J J= ⋅ = ° = = ⋅ = = − ° = −r r r r

∆ ∆ ∆ ∆ ∆cos . ( sin ) .20 137 4 20 98 0

Thus, Wtotal = 39.4 J, leading to a speed at the top of the ramp equal to

vW

m1

2 2= = =total 39.4 J

5 kg3.97 m/s

( )

(b) The x-component of Newton’s second law is

a aF

m

F w

m

F mg

mxx= = =

° − °=

° − °=( ) cos sin cos sin

net push push 21.347 m/s20 20 20 20

Constant-acceleration kinematics with x h1 20= ° =/ sin 5.85 m gives the final speed

v v a x x ax v ax12

02

1 0 1 1 12 2 2 2= + − = ⇒ = = =( ) )( )(1.347 m / s 5.85 m 3.97 m/s2

11.42. Model: Model Sam strapped with skis as a particle, and apply the law of conservation of energy.Visualize:

Solve: (a) The conservation of energy equation is

K U E K U W1 1 0 0+ + = + +g th g ext∆

The snow is frictionless, so ∆E th = 0 J. However, the wind is an external force doing work on Sam as he movesdown the hill. Thus,

W W K U K U

mv mgy mv mgy mv mgy mv mgy

v gyW

m

ext wind g g

wind

J J

= = + − +

= +

− +

= +

− + = −

⇒ = +

( ) ( )

( )

1 1 0 0

12

1 02

0 12

0 12

0

1 0

1

2

1

2

1

20 0

1

2

22

We compute the work done by the wind as follows:

W F r F rwind wind wind 200 N 146 m J= ⋅ = ° = ° = −r r

∆ ∆ cos ( )( )cos ,160 160 27 400

where we have used ∆r h= ° =/ sin20 146 m. Now we can compute

v1 2 502= + − =( )( )

( )9.8 m / s m

27,400 J

75 kg15.7 m / s2

(b) We will use a tilted coordinate system, with the x-axis parallel to the slope. Newton’s second law for Sam is

a aF

m

w F

m

mg F

mxx= = = ° − ° = ° − °

= ° − ° =

( ) sin cos sin cos

( )( )sin ( )cos

net wind wind

2275 kg 9.8 m/s 200 N

75 kg0.846 m/s

20 20 20 20

20 20

Now we can use constant-acceleration kinematics as follows:

v v a x x ax v ax12

02

1 0 1 1 12 2 2 2= + − = ⇒ = = =( ) ( )( )0.846 m/s 146 m 15.7 m/s2

Assess: We used a vertical y-axis for energy analysis, rather than a tilted coordinate system, because Ug isdetermined by its vertical position.

11.43. Model: Model Paul and the mat as a particle, assume the mat to be massless, use the model of kineticfriction, and apply the work-kinetic energy theorem.Visualize:

We define the x-axis along the floor and the y-axis perpendicular to the floor.Solve: We need to first determine fk. Newton’s second law in the y-direction is n T w mg n+ ° = = ⇒ =sin30mg T− ° = − ° =sin ( )( ) ( )(sin ) .30 30 83 010 kg 9.8 m/s 30 N N.2 Using n and the model of kinetic friction, f nk k= =µ( . )( ) .0 2 16 6083.0 N N.= Thus,

W f r T r f x T xnet k k

16.60 N 3.0 m 30.0 N 3.0 m J

= ⋅ + ⋅ = ° + °= − + =

r r r r∆ ∆ ∆ ∆( )( )(cos ) ( )( )(cos )

( )( )( ) ( )( )( . ) .

180 30

1 0 8660 28 14

The other forces r rn w and make an angle of 90° with ∆

rr and do zero work. We can now use the work-kinetic

energy theorem to find the final velocity as follows:

W K K K mv v W mnet f i f f f net J J kg m/s= − = − = ⇒ = = =01

22 2 28 14 10 2 372 / ( . )/ .

Assess: A speed of 2.37 m/s or 5.3 mph is reasonable for the present problem.

11.44. Model: Assume an ideal spring that obeys Hooke’s law. Model the box as a particle and use the modelof kinetic friction.Visualize:

Solve: When the horizontal surface is frictionless, conservation of energy means1

2

1

2

1

22 00

212

1 12k x x mv K Kx( ) ( )( ) .− = = ⇒ = − =e 100 N/m 0.20 m 0 m J

That is, the box is launched with 2.0 J of kinetic energy. It will lose 2.0 J of kinetic energy on the rough surface.The work-kinetic energy theorem is

W f r K K

f x x mg x x

x xmg

net k

k k

k2

J 2.0 J J

J

2.0 J 2.0 J

0.15 2.5 kg 9.8 m/s cm

= ⋅ = − = − = −⇒ − ° = − − = −

⇒ − = = =

r r∆ 2 1

2 1 2 1

2 1

0 2 0

180 2 0

54 4

.

( )(cos ) ( ) .

( )( )( ) ( )( )( )

.

µ

µ

Assess: Because the force of friction does negative work on the box, energy is transferred out of the box into theenvironment. In response, the box slows down and comes to rest.

11.45. Model: Model the suitcase as a particle, use the model of kinetic friction, and use the work-kineticenergy theorem.Visualize:

The only force that does work on the suitcase is the force of kinetic friction, rfk . The forces

rw and

rn are

perpendicular to the displacement, and therefore do not do work on the suitcase.Solve: The work-kinetic energy theorem is

W K mv mv f r mv f x x mv

mg x x mvv

g x x

net k k

k k 2

J

1.2 m/s

9.8 m/s 2.0 m 0 m

= = − ⇒ ⋅ = − ⇒ − ° = −

⇒ − − = − ⇒ =−

=−

=

∆ ∆1

2

1

20

1

2180

1

2

1

2 2 20 037

12

02

02

1 0 02

1 0 02 0

2

1 0

2

r r( )( )cos

( )( )

( )

( )( ).µ µ

Assess: Friction does negative work on the suitcase, and thus transfers energy out of the suitcase. In response, thesuitcase slows down and comes to rest.

11.46. Model: Identify the truck and the loose gravel as the system. We need the gravel inside the systembecause friction increases the temperature of the truck and the gravel. We will also use the model of kinetic frictionand the conservation of energy equation.Visualize:

We place the origin of our coordinate system at the base of the ramp in such a way that the x-axis is along the rampand the y-axis is vertical so that we can calculate potential energy. The free-body diagram of forces on the truck isshown.Solve: The conservation of energy equation is K U E K U W1 1 0 0+ + = + +g th g ext∆ . In the present case, Wext = 0 J,

v1x = 0 m/s, Ug0 = 0 J, v0x = 35 m/s. The thermal energy created by friction is

∆E f x x n x x mg x x

x x x x

th k k k

215,000 kg 9.8 m/s 58,478 J /m

= − = − = ° −

= ° − = −

( )( ) ( )( ) ( cos . )( )

( . )( )( )(cos . )( ) ( )( )

1 0 1 0 1 0

1 0 1 0

6 0

0 40 6 0

µ µ

Thus, the energy conservation equation simplifies to

01

20 0

6 01

2124 4

1 1 0 02

1 0 1 02

1 0

J 58,478 J /m J J

15,000 kg 9.8 m/s 58,478 J /m 15000 kg 35 m/s

m

2

+ + − = + +

− ° + − =

⇒ − =

mgy x x mv

x x x x

x x

x( )( )

( )( )[( )sin . ] ( )( ) ( )( )

( ) .

Assess: A length of 124.4 m at a slope of 6º seems reasonable.

11.47. Model: We will use the spring, the package, and the ramp as the system. We will model the package asa particle.Visualize:

We place the origin of our coordinate system on the end of the spring when it is compressed and is in contact withthe package to be shot.Model: (a) The energy conservation equation is

K U U E K U U W

mv mgy k x x E mv mgy k x W

1 1 1 0 0 0

12

12

02

021

2

1

2

1

2

1

2

+ + + = + + +

+ + − + = + + +

g s th g s ext

e e th ext

∆

∆ ∆( ) ( )

Using y1 = 1 m, ∆Eth = 0 J (note the frictionless ramp), v0 = 0 m/s, y0 = 0 m, ∆x = 30 cm, and Wext = 0 J, we get

1

20 0 0 0

1

20

1

2

1

21 70

12 2

12 2

1

mv mg k

v

v

+ + + = + + +

+ =

⇒ =

( ) ( )

( ) ( )( )( ) ( )( )

.

1 m J J J J 0.30 m J

2.0 kg 2.0 kg 9.8 m/s 1 m 500 N/m 0.30 m

m/s

2

(b) How high can the package go after crossing the sticky spot? If the package can reach y1 ≥ 1.0 m before stopping(v1 = 0), then it makes it. But if y1 < 1.0 m when v1 = 0, it does not. The friction of the sticky spot generates thermalenergy

∆ = − = − ⋅ ∆ = − − ∆ =E W f r mg xth diss k k2 kg)(9.8 m / s m) = 2.94 J

r r( ) ( . )( . )( .µ 0 30 2 0 0 50

The energy conservation equation is now12 1

21

12

2mv mgy E k x+ + ∆ = ∆th ( )

If we set v1 = 0 m/s to find the highest point the package can reach, we get

y k x E mg112

2 12 500 2 94 2 0 0 998= ∆ − ∆( ) = −( ) =( ) / ( . / ( . ) .th

2 2 N / m)(0.30 m) J kg)(9.8 m / s m

The package does not make it. It just barely misses.

11.48. Model: Model the two blocks as particles. The two blocks make our system.Visualize:

We place the origin of our coordinate system at the location of the 3.0 kg block.Solve: (a) The conservation of energy equation is K U E K U Wf gf th i gi ext+ + = + +∆ . Using ∆Eth = 0 J and Wext = 0 J

we get

1

2

1

2

1

2

1

22 22

3 32

2 2 22

3 32

2m v m v m g y m v m v m g y( ) ( ) ( ) ( ) ( ) ( )f f f i i i+ + = + +

Noting that ( ) ( )v v vf f f2 3= = and ( ) ( )v vi i m / s,2 3 0= = this becomes

1

2

2 23 43

2 32

2

2

2 3

( ) ( )

( ) ( )( )( )

( ).

m m v m g y y

vm g y y

m m

+ = − −

= −+

=+

=

f f i

fi f

22.0 kg 9.8 m/s 1.50 m

2.0 kg 3.0 kg m/s

(b) We will use the same energy conservation equation. However, this time

∆ ∆ ∆ ∆E f x n x m g xth k k k2 kg m/s m J= = = = =( )( ) ( )( ) ( )( ) ( . )( . )( . )( . ) .µ µ 3 0 15 3 0 9 8 1 50 6 615

The energy conservation equation is now

1

2

1

26 615

1

2

1

202

23

22 2 2

23 1 3

22m v m v m gy m v m v m gyf f f i i J J+ + + = + + +. ( ) ( )

1

26 615

26 615

26 615 3 02

2 32

22 3

2( ) . ( ) [ ( ) . ]

[( )( )( ) . ] .

m m v m g y y vm m

m g y y+ + = − ⇒ =+

− −

=

− =

f i f f i f

2

J J

5.0 kg2.0 kg 9.8 m/s 1.50 m J m/s

Assess: A reduced speed when friction is present compared to when there is no friction is reasonable.

11.49. Model: Use the particle model, the definition of work W F s= ⋅r r

∆ , and the model of kinetic friction.Visualize: We place the coordinate frame on the incline so that its x-axis is along the incline.

Solve: (a) W T r T xT N m J= ⋅ = ° = ° =r r

∆ ∆( )( )cos ( )( . )cos18 120 5 0 18 571

(b) W w r mg xg2 kg m/s m J= ⋅ = ° = ° = −( ) ( )cos( ) ( )( . )( . )cos

r r∆ ∆ 120 8 9 8 5 0 120 196

(c) W f r f x n xfriction k k k= ⋅ = ° = −r r

∆ ∆ ∆( )( )cos ( )( )180 µTo find n, we write Newton’s second law as follows:

F n w T n w T

n mg T

y = − ° + ° = ⇒ = ° − °

⇒ = ° − ° = ° − ° =∑ cos sin cos sin

cos sin ( )( )cos ( )sin .

30 18 0 30 18

30 18 30 18 30 814

N

8.0 kg 9.8 m/s 120 N N2

Thus, Wfriction 30.814 N 5.0 m J= − = −( . )( )( ) .0 25 38 5 .

(d) W n r n xn = ⋅ = ° =( ) ( )( )cosr r

∆ ∆ 90 0 J

Assess: Any force that acts perpendicular to the displacement does no work.

11.50. Model: Assume the spring is ideal so that Hooke’s law is obeyed, and model the weather rocket as aparticle.Visualize:

The origin of the coordinate system is placed on the free end of the spring. Note that the bottom of the spring isanchored to the ground.Solve: (a) The rocket is initially at rest. The free-body diagram on the rocket helps us write Newton’s second law as

F F w mg k y mg

ymg

k

y∑( ) = ⇒ = = ⇒ =

⇒ = = =

0

20

N

10.2 kg 9.8 m/s

500 N/m cm

sp

2

∆

∆ ( )( )

( )

(b) The thrust does work. Using the energy conservation equation when y y2 0 40− = = cm 0.40 m:

K U U K U U W

W F y y mv mgy k y y mv mgy k y y

v v

2 2 2 1 1 1

2 22

2 22

12

1 12

22

2

1

2

1

2

1

2

1

2

0 2 43

+ + = + + +

= − ⋅ + + − = + + − +

+ + = − + + ⇒ =

g sp g sp ext

ext 1 e e 200 N 0.60 m

5.10 kg 40.0 J 40.0 J 20.0 J 10.0 J 120 J m/s

( ) ( ) ( ) ( )( )

( ) .

If the rocket were not attached to the spring, the energy conservation equation would not involve the spring energyterm Usp2. That is,

K U K U U W

v

2 2 1 1 1

22

2

1

20

1

20 60

+ = + + +

+ = −

+ +

g g sp ext

2 210.2 kg 10.2 kg 9.8 m/s 0.40 m J 10.2 kg 9.8 m/s 0.20 m

500 N/m 0.20 m 200 N m

( ) ( )( )( ) ( )( )( )

( )( ) ( )( . )

⇒ = ⇒ =( ) . .5.1 kg J m/sv v22

270 0 3 70

11.51. Model: Use the particle model for the ice skater, the model of kinetic/static friction, and the work-kineticenergy theorem.Visualize:

Solve: (a) The work-kinetic energy theorem is

∆K mv mv W W= − = +1

2

1

212

02

wind fric

There is no kinetic friction along her direction of motion. Static friction acts to prevent her skates from slippingsideways on the ice, but this force is perpendicular to the motion and does no work: Wfric = 0 J. The angle betweenrFwind and ∆

rr is θ = °135 , so

W F r F ywind wind wind 4 N 100 m J= ⋅ = ° = ° = −r r

∆ ∆ cos ( )( )cos .135 135 282 8

Thus, her final speed is

v vW

m1 02 2= + =wind 2.16 m/s

(b) If the skates don’t slip, she has no acceleration in the x-direction and so (Fnet)x = 0 N. That is:

f F f Fs wind s wind N 2.83 N− ° = ⇒ = ° =cos cos45 0 45

Now there is an upper limit to the static friction: f f mgs s s≤ =( ) .max µ To not slip requires

µss

2

2.83 N

50 kg 9.8 m/s 0.0058≥ = =f

mg ( )( )

Thus, the minimum value of µs is 0.0058.Assess: The work done by the wind on the ice skater is negative, because the wind slows the skater down.

11.52. Model: Model the ice cube as a particle, the spring as an ideal that obeys Hooke’s law, and the law ofconservation of energy.Visualize:

Solve: (a) The normal force does no work and the slope is frictionless, so mechanical energy is conserved. We’vedrawn two separate axes: a vertical y-axis to measure potential energy and a tilted s-axis to measure distance alongthe slope. Both have the same origin which is at the point where the spring is not compressed. Thus, the two axesare related by y s= sin .θ Also, this choice of origin makes the elastic potential energy simply U k s ss = − =1

2 02( )

12

2ks . Because energy is conserved, we can relate the initial point—with the spring compressed—to the final pointwhere the ice cube is at maximum height. We do not need to find the speed with which it leaves the spring. Wehave

K U U K U U

mv mgy ks mv mgy ks

2 2 2 1 1 1

22

2 02

12

1 121

2

1

2

1

2

1

2

+ + = + +

+ + = + +

g s g s

It is important to note that at the final point, when the ice cube is at y2, the end of the spring is only at s0. The springdoes not stretch to s2, so Us2 is not 1

2 22ks . Three of the terms are zero, leaving

mgy mgy ks y y yks

mg2 1 12

2 1121

2 2= + + ⇒ − = = = = =∆ height gained 0.255 m 25.5 cm

The distance traveled is ∆s = ∆y/sin 30° = 51.0 cm.(b) Using the energy equation and the model for kinetic friction:

K U U E K U U W E f s n s2 2 2 1 1 1+ + + = + + + = =g s th g s ext th k k∆ ∆ ∆ ∆ µ

From the free-body diagram,

( ) cos cosrF n mg n mgynet N= = − ° ⇒ = °0 30 30

Now, having found ∆ ∆E mg sth k= °µ ( cos ) ,30 the energy equation can be written

0 0 30 01

20

1

230 0

2 1 12

2 1 12

J J J Jk

k

+ + + ° = + + +

⇒ − − + ° =

mgy mg s mgy ks

mg y y ks mg s

µ

µ

( cos )

( ) cos

∆

∆

Using ∆ ∆y s= °( )sin ,30 the above equation simplifies to

mg s mg s ks sks

mg∆ ∆ ∆sin cos

(sin cos ). .30 30

1

2 2 30 300 379 37 91

2 12

° + ° = ⇒ =° + °

= =µµk

k

m cm

11.53. Model: Assume an ideal spring, so Hooke’s law is obeyed. Treat the box as a particle and apply theenergy conservation law. Box, spring, and the ground make our system, and we also use the model of kineticfriction.Visualize: We place the origin of the coordinate system on the ground directly below the box’s starting position.

Solve: (a) The energy conservation equation is

K U U E K U U W

mv mgy mv mgy mv mgy

v gy

1 1 1 0 0 0

12

1 02

0 12

0

1 0

1

20 0

1

20 0

1

20 0

2 2 9 8 5 9 90

+ + + = + + +

+ + + = + + + ⇒ + = +

⇒ = = =

g s th g s ext

2

J J J J J J

m/s m m/s

∆

( . )( ) .

(b) The work of friction creates thermal energy. The energy conservation equation for this part of the problem is

K U U E K U U W mv f x x mv

mv n x x mv mv mg x x mv

2 2 2 1 1 1 22

2 1 12

22

2 1 12

22

2 1 12

1

20 0

1

20 0 0

1

2

1

2

1

2

1

2

+ + + = + + + + + + − = + + +

+ − = ⇒ + − =

g s th g s ext k

k k

J J J J J∆ ( )( )

( ) ( )( )µ µ

⇒ = − − = − =v v g x x2 12

2 122 2 0 25 9 39µk

29.90 m/s 9.8 m/s 2.0 m m/s( ) ( ) ( . )( )( ) .

(c) To find how much the spring is compressed, we apply the energy conservation once again:

K U U E K U U W k x x mv3 3 3 2 2 2 3 22

220 0

1

20

1

20 0 0+ + + = + + + + + − + = + + +g s th g s ext J J J J J J∆ ( )

Using v2 = 9.39 m/s, k = 500 N/m and m = 5.0 kg, the above equation yields ( )x x x3 2− = =∆ 93.9 cm.(d) The initial energy = =mgy0 ( )( )( )5.0 kg 9.8 m / s 5.0 m2 = 245 J. The energy transformed to thermal energyduring each passage is

f x x mg x xk k25.0 kg 9.8 m / s 2.0 m J( ) ( ) ( . )( )( )( ) .2 1 2 1 0 25 24 5− = − = =µ

The number of passages is equal to 245 J / J24 5. or 10.

11.54. Model: Assume an ideal spring, so Hooke’s law is obeyed. Treat the physics student as a particle andapply the law of conservation of energy. Our system is comprised of the spring, the student, and the ground. Wealso use the model of kinetic friction.Visualize: We place the origin of the coordinate system on the ground directly below the end of the compressedspring that is in contact with the student.

Solve: (a) The energy conservation equation isK U U E K U U W

mv mgy k x x mv mgy k x x

1 1 1 0 0 0

12

1 12

02

0 1 021

2

1

20

1

2

1

20

+ + + = + + +

+ + − + = + + − +

g s th g s ext

e J J

∆

( ) ( )

Since y1 = y0 = 10 m, x1 = xe, v0 = 0 m/s, k = 80,000 N/m, m = 100 kg, and (x1 − x0) = 0.5 m,

1

2

1

212

1 02

1 1 0mv k x x vk

mx x= − ⇒ = − =( ) ( ) 14.14 m/s

(b) The work of friction creates thermal energy. Applying the conservation of energy equation once again:

K U U E K U U W

mv mgy f s mgy k x x

2 2 2 0 0 0

22

2 0 1 021

20 0

1

20

+ + + = + + +

+ + + = + + − +

g s th g s ext

k J J J

∆

∆ ( )

With v2 = 0 m/s and y2 = (∆s)sin30º, the above equation is simplified to

mg s n s mgy k x x( )sin ( )∆ ∆301

20 1 02° + = + −µk

From the free-body diagram for the physics student, we see that n w= °cos .30 Thus, the conservation of energyequation gives

∆s mg mg mgy k x x( sin cos ) ( )30 301

20 1 02° + ° = + −µk

Using m = 100 kg, k = 80,000 N/m, (x1 − x0) = 0.50 m, y0 = 10 m, and µk = 0.15, we get

∆smgy k x x

mg=

+ −

° + °=

0 1 021

230 30

32 1( )

(sin cos ).

µk

m

Assess: y s2 30 16 05= ° =( )sin .∆ m, which is greater than y0 = 10 m. The higher value is due to the transforma-tion of the spring energy into gravitational potential energy.

11.55. Model: Treat the block as a particle, use the model of kinetic friction, and apply the energy conservationlaw. The block and the incline comprise our system.Visualize: We place the origin of the coordinate system directly below the block’s starting position at the samelevel as the horizontal surface. On the horizontal surface the model of kinetic friction applies.

Solve: (a) For the first incline, the conservation of energy equation gives

K U E K U W mv mgy v gy gh1 1 0 0 12

0 1 0

1

20 0 0 0 2 2+ + = + + + + = + + ⇒ = =g th g ext J J J J∆

(b) The work of friction creates thermal energy. Applying once again the conservation of energy equation, we have

K U E K U W mv mgy f x x mv mgy W3 3 1 1 32

3 2 1 12

1

1

2

1

2+ + = + + + + − = + +g th g ext k ext∆ ( )

Using v3 = 0 m/s, y1 = 0 m, Wext = 0 J, fk = µkmg, v gh1 2= , and (x2 – x1) = L, we get

mgy mgL m gh y h L3 3

1

22+ = ⇒ = −µ µk k( )

Assess: For µk = 0, y3 = h which is predicted by the law of the conservation of energy.

11.56. Model: Assume an ideal spring, so Hooke’s law is obeyed.Visualize:

Solve: For a conservative force the work done on a particle as it moves from an initial to a final position isindependent of the path. We will show that W WA C B A B→ → →= for the spring force. Work done by a spring force F = −kxis given by

W Fdx kx dxx

x

i

f

= = −∫ ∫This means

W kx dxk

x x W kx dxk

x x W kx dxk

x xx

x

x

x

x

x

A B B A A C C A C B B C

A

B

A

C

C

B

, , and → → →= − = − −( ) = − = − −( ) = − = − −( )∫ ∫ ∫2 2 22 2 2 2 2 2

Adding the last two:

W W Wk

x x x x WCA C B A C C B C A B2

A B→ → → → →= + = − − + −( ) =2

2 2 2

11.57. Model: A “sprong” that obeys the force law F q x xx = − −( )e3 , where q is the sprong constant and xe is

the equilibrium position.Visualize: We place the origin of the coordinate system on the free end of the sprong, that is, xe = xf = 0 m.

Solve: (a) The units of q are N/m3.(b) A cubic curve rises more steeply than a parabola. The force increases by a factor of 8 every time x increases bya factor of 2.

(c) Since we have FdU

dxU x F dx qx dx

qxx x

x= − = − = − − =∫ ∫, ( ) ( ) .3

0

4

4(d) Applying the energy conservation equation to the ball and sprong system:

K U K U

mvqx

vq

m

x

f f i i

fi

f

3

J J

40,000 N/m

0.020 kg

m m/s

+ = +

+ = +

⇒ = = ⋅ − =

1

20 0

4

2

0 10

210

24

4 4( )

( )

( . )

11.58. Solve: (a) Because sin (cx) is dimensionless, F0 must have units of force in newtons.(b) The product cx is an angle because we are taking the sine of it. An angle has no real physical units. If x hasunits of m and the product cx is unitless, then c has to have units of m−1.(c) At x0 = 0 m, the force is F0 sin(0) = 0 N. The particle is able to leave x0 = 0 m only because it has an initial velocity.(d) The force is a maximum when sin(cx) = 1. This occurs when cx = π/2, or for xmax = π/2c.(e) The graph is the first quarter of a sine curve.

(f) We can find the velocity vf at xf = xmax from the work-kinetic energy theorem:

∆K mv mv mv mv W v vW

m= − = − = ⇒ = +1

2

1

2

1

2

1

2

22 2 202

02

f i f f

This is a variable force. As the particle moves from xi = 0 m to xf = xmax = π/2c, the work done on it is

W F x dx F cx dxF

ccx

F

c

F

ccc

x

x= = = − = − −

=∫∫ ( ) sin( ) cos( )| cos cos//

00

02

0

20 0

20ππ π

i

f

Thus, the particle’s speed at xf = xmax = π/2c is v v F mcf = +02

02 / .

11.59. Visualize: We place the origin of the coordinate system at the base of the stairs on the first floor.

Solve: (a) We might estimate y y y y2 1 3 24 0 12− ≈ ≈ ≈ −. , m ft thus, y3 – y1 ≈ 8.0 m.(b) We might estimate the time to run up these two flights of stairs to be 20 s.(c) Estimate your mass as m ≈ 70 kg ≈ 150 lb. Your power output while running up the stairs is

work done by you

time

change in potential energy

time time

0 kg 9.8 m/s 8.0 m

20 s270 W 270 W

1 hp

746 W0.35 hp

2

= = −

= ≈ =

≈

mg y y( )

( )( )( )( )

3 1

7

Assess: Your estimate may vary, depending on your mass and how fast you run.

11.60. Model: Model the lawnmower as a particle and use the model of kinetic friction.Visualize:

We placed the origin of our coordinate system on the lawnmower and drew the free-body diagram of forces.Solve: The normal force

rn, which is related to the frictional force, is not equal to

rw. This is due to the presence

of rF. The rolling friction is f n n fr r r r or = =µ µ, . The lawnmower moves at constant velocity, so

r rFnet = 0. The

two components of Newton’s second law are

F n w F ma f mg F f mg F

F F f F mg F

Fmg

y y

x

∑∑

( ) = − − ° = = ⇒ − − ° = ⇒ = + °

( ) = ° − = ⇒ ° − − ° =

⇒ =° − °

=

sin / sin sin

cos cos sin

cos sin

( . )( )( )

( .

37 0 37 0 37

37 0 37 37 0

37 37

0 15

0

N N

N N

12 kg 9.8 m/s

r r r r r

r r r

r2

µ µ µ

µ µ

µµ 79867986 0 15 0 6018

29 4) ( . )( . )

.−

= N

Thus, the power supplied by the gardener in pushing the lawnmower at a constant speed of 1.2 m/s is P =Fv = =( )( ) .24.9 N 1.2 m / s W.29 9

11.61. Solve: Power output during the push-off period is equal to the work done by the cat divided by the timethe cat applied the force. Since the force on the floor by the cat is equal in magnitude to the force on the cat by thefloor, work done by the cat can be found using the work-kinetic energy theorem during the push-offperiod, W W Knet floor= = ∆ . We do not need to explicitly calculate Wcat, since we know that the cat’s kinetic energy istransformed into its potential energy during the leap. That is,

∆U mg y yg25.0 kg 9.8 m / s 0.95 m J= − = =( ) ( )( )( ) .2 1 46 55

Thus, the average power output during the push-off period is

PW

t= = =net 46.55 J

0.20 s W233

11.62. Solve: Using the conversion 746 W = 1 hp, we have a power of 1492 J/s. This means W = Pt =(1492 J/s)(1 hr) = 5.3712 × 106 J is the total work done by the electric motor in one hour. Furthermore,

W W U U mg y y mg

mW

g

motor g gf gi f i

motor2

4

10 m

m

J

m/s m kg = 5.481 10 kg

liter

1 kg liters

= − = − = − =

= = × = × × × =

( ) ( )

( )

.

( . )( ). ,

10

5 3712 10

9 8 105 481 10

154 800

64

11.63. Solve: (a) The change in the potential energy of 1.0 kg of water in falling 25 m is ∆U mghg = − =− = −( )( )( )1.0 kg 9.8 m/s 25 m J.2 245(b) The power required of the dam is

PW

t

WW= = = × ⇒ = ×

1 s Watts J50 10 50 106 6

That is, 50 × 106 J of energy is required per second for the dam. Out of the 245 J of lost potential energy,(245 J)(0.80) = 196 J is converted to electrical energy. Thus, the amount of water needed per second is50 10 255 0006× = J 1 kg /196 J) kg( , .

11.64. Solve: The force required to tow a water skier at a speed v is F Avtow = . Since powerP = Fv, the power required to tow the water skier is P F v Avtow tow= = 2 . We can find the constant A by noting that aspeed of v = 2.5 mph requires a power of 2 hp. Thus,

( ) ( ) .( )

2 hp 2.5 mphhp

mph= ⇒ =A A2

20 32

Now, the power required to tow a water skier at 7.5 mph is

P Avtow

hp

mph7.5 mph hp= = ⋅ =2

220 32 18.

( )( )

Assess: Since P v∝ 2 , a three-fold increase in velocity leads to a nine-fold increase in power.

11.65. Solve: By definition, the maximum power output of a horse is P ≈ 1 hp = 746 W. At maximum speed,

when the horse is running at constant speed, rFnet =

r0 . The propulsion force Fhorse, provided by the horse pushing

against the ground, is balanced by the drag force of air resistance: Fhorse = D. We learned in Chapter 5 that areasonable model for drag is D ≈ 1

4 Av2, where A is the cross section area. Since the power needed for force F topush an object at velocity v is P = Fv, we have

P F v Av v Av

vP

A

≈ = = = ( ) =

≈

=

=

1 746

4 4 74615

14

2 14

3

1 3 1 3

hp W

W)

(0.5 m)(1.8 m) m / s

horse

/ /(

Assess: 15 m/s ≈ 30 mph is a reasonable top speed for a well-trained horse.

11.66. Solve: The net force on a car moving at a steady speed is zero. The motion is opposed both by rollingfriction and by air resistance. Thus the propulsion force provided by the drive wheels must be F mg Avcar r= +µ 1

42 ,

where µr is the rolling friction, m is the mass, A is the cross-section area, and v is the car’s velocity. The powerrequired to move the car at speed v is

P F v mgv Av= = +car rµ 1

43

Since the maximum power output is 200 hp and 75% of the power reaches the drive wheels, P = (200 hp)(0.75) =150 hp. Thus,

( ) ( . )( )( ) ( )( )

. .

150 hp746 W

1 hp1500 kg 9.8 m/s 1.6 m 1.4 m

m/s

2

= +

⇒ + − = ⇒ =

0 021

4

0 56 294 111900 0 55 5

3

3

v v

v v v

The easiest way to solve this equation is through iterations by trial and error.Assess: A speed of 55.5 m/s ≈ 110 mph is very reasonable.

11.67. Model: Use the model of static friction, kinematic equations, and the definition of power.Solve: (a) The rated power of the Porsche is 217 hp = 161,882 W and the weight of the car is (1480 kg)(9.8 m/s2) =14504 N. The weight of the car on the drive wheels is (14504)(2/3) = 9670 N. Because the static friction of the tireson road pushes the car forward,

F f n mg ma

a

max max max

max

( . )( )

.

= = = = =

⇒ = =

s s s

2

9670 N

9670 N

1480 kg m/s

µ µ 1 0

6 53

(b) P Fv vP

F= ⇒ = = =max max .

161882 W

9670 N m/s16 7

(c) Using the kinematic equation, v v a t tmax max min( )= + −0 0 with v0 = 0 m/s and t0 = 0 s, we obtain

tv

aminmax

max

.= = =16.7 m/s

6.53 m/s s2 2 56

Assess: An acceleration time of 2.56 s for the Porsche to reach a speed of ≈35 mph from rest is reasonable.

11.68. (a) A student uses a string to pull her 2.0 kg physics book, starting from rest, across a 2.0-m-long labbench. The coefficient of kinetic friction between the book and the lab bench is 0.15. If the book’s final speed is4.0 m/s, what is the tension in the string?(b)

(c) The tension does external work Wext. This work increases the book’s kinetic energy and also causes an increase∆Eth in the thermal energy of the book and the lab bench. Solving the equation gives T = 10.9 N.

11.69. (a) A 20 kg chicken crate slides down a 2.5-m-high, 40° ramp from the back of a truck to the ground. Thecoefficient of kinetic friction between the crate and the ramp bench is 0.15. How fast are the chickens going at thebottom of the ramp?(b)

(c) v1 = 6.34 m/s.

11.70. (a) If you expend 75 W of power to push a 30 kg sled on a surface where the coefficient of kinetic frictionbetween the sled and the surface is µk = 0.20, what speed will you be able to maintain?(b)

(c) F v vpush230 kg 9.8 m/s N 75 W = 58.9 N)

75 W

58.8 Nm/s= = ⇒ ⇒ = =( . )( )( ) . ( .0 20 58 8 1 28

11.71. (a) A 1500 kg object is being accelerated upward at 1.0 m/s2 by a rope. How much power must the motorsupply at the instant when the velocity is 2.0 m/s?(b)

(c) T

P T

= + == = =

( )( ) ( ,

( ) ( )( ) ,

1500 kg 9.8 m/s kg 1.0 m/s ) N

m/s N 2.0 m/s W = 32.4 kW

2 21500 16 200

2 16200 32 400

11.72. Model: Model the water skier as a particle, apply the law of conservation of mechanical energy, and usethe constant-acceleration kinematic equations.Visualize:

We placed the origin of the coordinate system at the base of the frictionless ramp.Solve: We’ll start by finding the smallest speed v1 at the top of the ramp that allows her to clear the shark tank.From the vertical motion for jumping the shark tank,

y y v t t a t t

t t t t

y y2 1 1 2 1 2 12

2 12

2 1

1

2

0 01

29 8 0 639

= + − + −

⇒ = + + − − ⇒ − =

( ) ( )

( ) ( . )( ) ( ) . m 2.0 m m m/s s2

From the horizontal motion,

x x v t t a t t

x x v v

x x2 1 1 2 1 2 12

1 1 1 1

1

2

5 0 0 7 825

= + − + −

⇒ + = + + ⇒ = =

( ) ( )

( . ) ( ) . m 0.639 s m5.0 m

0.639 s m/s

Having found the v1 that will take the skier to the other side of the tank, we now use the energy equation to find theminimum speed v0. We have

K U K U mv mgy mv mgy

v v g y y

1 1 0 0 12

1 02

0

0 12

1 02

1

2

1

2

2 2 10 0

+ = + ⇒ + = +

= + − = + =

g g

27.825 m/s 9.8 m/s 2.0 m m/s( ) ( ) ( )( ) .

11.73. Model: Assume the spring to be ideal that obeys Hooke’s law, model the block as a particle, and applythe model of kinetic friction.Visualize: We placed the origin of the coordinate system on the free end of the compressed spring which is incontact with the block. Because the horizontal surface at the bottom of the ramp is frictionless, the spring energyappears as kinetic energy of the block until the block begins to climb up the incline.

Solve: Although we could find the speed v1 of the block as it leaves the spring, we don’t need to. We can use energyconservation to relate the initial potential energy of the spring to the energy of the block as it begins projectile motionat point 2. However, friction requires us to calculate the increase in thermal energy. The energy equation is

K U E K U W mv mgy f s k x x2 012 2

22

12 0

2+ + ∆ = + + ⇒ + + ∆ = −g2 th g0 ext k e( )

The distance along the slope is ∆s = y2/sin45°. The friction force is fk = µkn, and we can see from the free-bodydiagram that n = mgcos45°. Thus

vk

mx x gy gy2 0

22 2

1 2

1 2

2 2 45 45

10000 15 2 9 8 2 0 2 0 20 9 8 2 0 45 45 8 091

= − − − ° °

= − − ° °

=

( ) cos / sin

( . ( . )( . ( . )( . )( . cos / sin .

/

/

e k

2 2 2 N / m

0.20 kg m) m / s m) m / s m) m / s

µ

Having found the velocity v2, we can now find (x3 − x2) = d using the kinematic equations of projectile motion:

y y v t t a t t

v t t t t

y y3 2 2 3 2 2 3 22

2 3 2 3 22

1

2

2 0 451

2

= + − + −

= + ° − + − −

( ) ( )

. ( sin )( ) ( )( ) m 2.0 m 9.8 m / s2

⇒ − =( )t t3 2 0 s and 1.168 s

Finally,

x x v t t a t t

d x x v

x x3 2 2 3 2 2 3 22

3 2 2

1

245 0 6 68

= + − + −

= − = ° + =

( ) ( )

( ) ( cos )( ) .1.168 s m m

11.74. Model: Assume an ideal spring that obeys Hooke’s law, the particle model for the ball, and the modelof kinetic friction. The ball, the spring, and the barrel comprise our system.Visualize:

We placed the origin of the coordinate system on the free end of the spring which is in contact with the ball.

Solve: Force rF does external work Wext. This work compresses the spring, gives the ball kinetic energy, and is

partially dissipated to thermal energy by friction.(a) The energy equation for our system is

K U E K U W

mv k x x f x x mv k x x F x x

mg F

1 1 0 0

12

1 02

1 0 02 2

1 0

2 2

1

2

1

2

1

2

1

21

2

1

20 0

+ + = + +

+ − + − = + − + −

+ + = + +

s th s ext

k e e

k1.0 kg 2.0 m/s 3000 N/m 0.3 m 0.30 m J J 0.30 m

∆

( ) ( )( ) ( ) ( )

( )( ) ( )( ) ( )( ) ( )µ

The friction force is fk = µkmg. With µk = 0.30 and m = 1.0 kg, we can solve this equation to obtain F = 460 N.(b) Using the energy equation for our system once again, we have

K U E K U W mv k x x f x x mv k x x

mv mg x x k x x

v

2 2 1 1 22 2

2 1 12

1 02

22

2 1 1 02

22

1

2

1

2

1

2

1

20

1

20

1

21

20 30

+ + = + + + − + − = + − +

+ − = −

+

s th s ext e e k

k

J

J +

1.0 kg 1.0 kg

∆ ( ) ( )( ) ( )

( ) ( )

( ) ( . )(

µ

)()( )( ) ( )( . )

( . ) . .

9.8 m/s 1.5 m 3000 N/m m

kg J = 135 J m/s

2 =

+ ⇒ =

1

20 30

0 5 4 41 16 2

2

22

2v v

11.75. Solve: (a)

The graph is a hyperbola.(b) The separation for zero potential energy is r = ∞, since

UGm m

rr= − → → ∞1 2 0 J as

This makes sense because two masses don’t interact at all if they are infinitely far apart.(c) Due to the absence of nonconservative forces in our system of two particles, the mechanical energy isconserved.

The equations of energy and momentum conservation are

K U K U m v m vGm m

rm v m v

Gm m

r

m v m v Gm mr r

f gf i gi f ff

i ii

f ff i

+ = + + + −

= + + −

+ = −

1

2

1

2

1

2

1

2

1

2

1

2

1 1

1 12

2 22 1 2

1 12

2 22 1 2

1 12

2 22

1 2

p p m v m v vm

mvf i f f f f kg m / s= ⇒ + = ⇒ = −1 1 2 2 2

1

210

Substituting this expression for v2f into the energy equation, we get

1

2

1

2

1 1 2

1

1 11 1

22

1

21

2

1 2 12 2

1 2

m v mm

mv Gm m

r rv

Gm

m m r rf ff i

ff i

+

= −

⇒ =+

−

( / )

With G r R R r m m= × ⋅ = + = × = × = × = ×− −6 67 10 18 10 1 0 10 8 0 10 2 011 21 2

8 141

302. ( , . . . N m kg) m, m, kg, and f i 1030

kg, the above equation can be simplified to yield

v vm

mv1

52

21

30

305 51 72 10

8 0 10

2 0 101 72 10 6 89 10f f

1f m/s, and

kg

kg m/s m/s= × = − = ×

×

× ×( ) = ×..

.. .

The speed of the heavier star is 172,000 m/s. That of the lighter star is 689,000 m/s.