SM PDF Chapter6

8/13/2019 SM PDF Chapter6

1/22

143

Energy and EnergyTransfer

CHAPTER OUTLINE

6.1 Systems andEnvironments

6.2 Work Done by a ConstantForce

6.3 The Scalar Product ofTwo Vectors

6.4 Work Done by a Varying

Force6.5 Kinetic Energy and the

Work-Kinetic EnergyTheorem

6.6 The Nonisolated System6.7 Situations Involving

Kinetic Friction6.8 Power6.9 Context

ConnectionHorsepowerRatings of Automobiles

ANSWERS TO QUESTIONS

Q6.1 The force is perpendicular to every increment of displacement.

Therefore,r rF r = 0.

Q6.2 Yes. Force times distance over which the toe is in contact with theball. No, he is no longer applying a force. Yes, both air friction andgravity do work.

Q6.3 Force of tension on a ball rotating on the end of a string. Normalforce and gravitational force on an object at rest or moving across alevel floor.

Q6.4 Work is only done in accelerating the ball from rest. The work is done over the effective length of thepitchers armthe distance his hand moves through windup and until release.

Q6.5 (a) Tension (b) Air resistance

(c) Positive in increasing velocity on the downswing.Negative in decreasing velocity on the upswing.

Q6.6 No. The vectors might be in the third and fourth quadrants, but if the angle between them is less

than 90their dot product is positive.

Q6.7 The scalar product of two vectors is positive if the angle between them is between 0 and 90. Thescalar product is negative when 90 180 < < .

Q6.8 =k k2 . To stretch the smaller piece one meter, each coil would have to stretch twice as much as one

coil in the original long spring, since there would be half as many coils. Assuming that the spring isideal, twice the stretch requires twice the force.

Q6.9 Kinetic energy is always positive. Mass and squared speed are both positive. A moving object canalways do positive work in striking another object and causing it to move along the same directionof motion.

Q6.10 The longer barrel will have the higher muzzle speed. Since the accelerating force acts over a longerdistance, the change in kinetic energy will be larger.


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144 Energy and Energy Transfer

Q6.11 Kinetic energy is proportional to mass. The first bullet has twice as much kinetic energy.

Q6.12 No violation. Choose the book as the system. You did work and the earth did work on the book. Theaverage force you exerted just counterbalanced the weight of the book. The total work on the book iszero, and is equal to its overall change in kinetic energy.

Q6.13 (a) Kinetic energy is proportional to squared speed. Doubling the speed makes an objectskinetic energy four times larger.

(b) If the total work on an object is zero in some process, its speed must be the same at the finalpoint as it was at the initial point.

Q6.14 The larger engine is unnecessary. Consider a 30-minute commute. If you travel the same speed ineach car, it will take the same amount of time, expending the same amount of energy. The extrapower available from the larger engine isnt used.

Q6.15 If the instantaneous power output by some agent changes continuously, its average power in aprocess must be equal to its instantaneous power at least one instant. If its power output is constant,its instantaneous power is always equal to its average power.

Q6.16 The rock increases in speed. The farther it has fallen, the more force it might exert on the sand at thebottom; but it might instead make a deeper crater with an equal-size average force. The farther itfalls, the more work it will do in stopping. Its kinetic energy is increasing due to the work that thegravitational force does on it.

Q6.17 The normal force does no work because the angle between the normal force and the direction ofmotion is usually 90. Static friction usually does no work because there is no distance throughwhich the force is applied.

Q6.18 An argument for: As a glider moves along an airtrack, the only force that the track applies on theglider is the normal force. Since the angle between the direction of motion and the normal force is90, the work done must be zero, even if the track is not level.

Against: An airtrack has bumpers. When a glider bounces from the bumper at the end of theairtrack, it loses a bit of energy, as evidenced by a decreased speed. The airtrack does negative work.

SOLUTIONS TO PROBLEMS

Section 6.1 Systems and Environments

No problems in this section

Section 6.2 Work Done by a Constant Force

P6.1 (a) W F r= = = cos . . cos . . 16 0 2 20 25 0 31 9N m Ja fa f

(b), (c) The normal force and the weight are both at 90 to the displacement in any time interval.

Both do 0 work.

(d) W = + + =31 9 0 0 31 9. .J J


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5/22


6/22


P6.12 Compare an initial picture of the rolling car with afinal picture with both springs compressed

K W Ki f+ = . Work by both springs changes the

cars kinetic energy

K k x x k x x K

mv

v

v

i i f i f f

i

i

i

+ + =

+

+ =

=

= =

1

2

1

21

20

1

21600 0 500

01

23400 0 200 0

1

26 000 200

2 268

6 0000 299

1 1

2

1

2

2 2

2

2

2

2 2

2

2

d i d ib ga f

b ga f

b g

a f

N m m

N m m

kg J 68.0 J 0

J

kg m s

.

.

.

F Na f

FIG. P6.12

P6.13 k F

y

Mg

y= = =

=

4 00 9 80

2 50 101 57 10

2

3. .

..

a fa fNm

N m

(a) For 1.50 kg mass y mg

k= =

=

1 50 9 80

1 57 100 938

3

. .

..

a fa fcm

(b) Work =1

22ky

Work = =1

21 57 10 4 00 10 1 253 2

2. . .N m m Je je j

P6.14 W d x y dxi

f

= = +

z z

r rF r i j i4 3

0

5$ $ $e jN

m

4 0 42

50 00

5 2

0

5

N m N m Jm m

b g b gxdx x+ = =z .

P6.15 (a) Spring constant is given by F kx=

k F

x= = =

230

0 400575

N

m N m

a fa f.

(b) Work = = =F xavg N m J1

2

230 0 400 46 0a fa f. .


7/22

Chapter 6 149

P6.16 (a) W d

W x x dx

W x x x

W

i

f

=

= +

= +

= + =

z

z

r rF r

15 000 10 000 25 000 0

15 000 10 0002

25 0003

9 00 1 80 1 80 9 00

2

0

0 600

2 3

0

0 600

N N m N m

kJ kJ kJ kJ

2m

m

e j cos

. . . .

.

.

(b) Similarly,

W

W

= +

=

15 0 1 0010 0 1 00

2

25 0 1 00

3

11 7

2 3

. .. . . .

.

kN m kN m m kN m m

kJ , larger by 29.6%

2

a fa f b ga f e ja f

*P6.17 4 00

1

2 0 100

2

. .J m=

ka f =k 800 N m and to stretch the spring to 0.200 m requires

W = =1

2800 0 200 4 00 12 0

2a fa f. . .J J

*P6.18 If the weight of the first tray stretches all four springs by a distance equal to the thickness of the tray,then the proportionality expressed by Hookes law guarantees that each additional tray will havethe same effect, so that the top surface of the top tray will always have the same elevation above the

floor. The weight of a tray is 0 580 9 8 5 68. . .kg m s N2e j = . The force1

45 68 1 42. .N Na f = should

stretch one spring by 0.450 cm, so its spring constant is k Fxs= = =1 42 316. N

0.004 5 m N m . We did not

need to know the length or width of the tray.

P6.19 (a) The radius to the object makes angle with the

horizontal, so its weight makes angle with thenegative side of the x-axis, when we take the xaxisin the direction of motion tangent to the cylinder.

F ma

F mg

F mg

x x =

=

=

cos

cos

0

(b) W di

f

= zr rF r

r r

r

FIG. P6.19

continued on next page


8/22


We use radian measure to express the next bit of displacement as dr Rd= in terms of thenext bit of angle moved through:

W mg Rd mgR

W mgR mgR

= =

= =

z cos sin

0

2

0

2

1 0a f

*P6.20 The same force makes both light springs stretch.

(a) The hanging mass moves down by

x x x mg

k

mg

kmg

k k= + = + = +

FHG

IKJ1 2 1 2 1 2

1 1

(b) We define the effective spring constant as

k Fx

mgmg k k k k

= =+

= +FHG IKJ

1 11 1

1 2 1 2

1

b g

Section 6.5 Kinetic Energy and the Work-Kinetic Energy Theorem

Section 6.6 The Nonisolated System

P6.21 (a) KA = =1

20 600 2 00 1 20

2. . .kg m s Jb gb g

(b) 12

2mv KB B= : v K

mB

B= = =

2 2 7 500 600

5 00a fa f..

. m s

(c) W K K K m v vB A B A = = = = =1

27 50 1 20 6 302 2e j . . .J J J

P6.22 (a) K mv= = =1

2

1

20 300 15 0 33 82


(b) K= = = =1

20 300 30 0

1

20 300 15 0 4 4 33 8 135

2 2. . . . .a fa f a fa f a f a f J

P6.23r

v i ji = 6 00 2 00. $ . $e j m s

(a) v v vi ix iy= + =2 2 40 0. m s

K mvi i= = =1

2

1

23 00 40 0 60 02 . . .kg m s J2 2b ge j



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Chapter 6 151

(b)rv i jf = +8 00 4 00.

$ . $

vf f f2 64 0 16 0 80 0= = + =

r rv v . . . m s2 2

K K K m v vf i f i= = = =1

2

3 00

280 0 60 0 60 02 2e j a f

.. . . J

P6.24 (a) K K K mv W f i f= = = =1

202 (area under curve from x = 0 to x= 5 00. m)

vm

f = = =2 2 7 50

4 001 94

area J

kg m s

a f a f..

.

(b) K K K mv W f i f= = = =1


vm

f = = =2 2 22 5

4 003 35

area J

kg m s

a f a f..

.

(c) K K K mv W f i f= = = =1


vm

f = = =2 2 30 0

4 003 87

area J

kg m s

a f a f..

.

P6.25 Consider the work done on the pile driver from the time it starts from rest until it comes to rest atthe end of the fall. Let d = 5.00 m represent the distance over which the driver falls freely, andh = 0 12. m the distance it moves the piling.

W K = : W W mv mvf igravity beam+ = 1

2

1

22 2

so mg h d F db ga f d ia f+ + = cos cos0 180 0 0 .

Thus, Fmg h d

d=

+= =

b ga f b ge ja f2 100 9 80 5 120 120

8 78 105kg m s m

mN

2. .

.. . The force on the pile

driver is upward .

P6.26 (a) K W K mvi f f+ = =1

22

01

215 0 10 780 4 563

2+ = =

W . .kg m s kJe jb g

(b) F Wr

= =

=

cos.

cos.

4 56 10

06 343 J

0.720 mkN

a f

(c) av v

x

f i

f

=

=

=

2 2 2

2

780 0

2 0 720422

m s

m km s2

b ga f.

(d) F ma = = =15 10 422 10 6 343 3kg m s kN2e je j .


10/22


P6.27 W K = = 0 : mg dl kx dxL d

sin .35 0 0 =z z 0 0

mg L kdsin .35 01

22

=a f

d mg L

k=

2 35 0sin . a f

d =

=

2 12 0 9 80 35 0 3 00

3 00 100 116

4

( . )( . )(sin . )( . )

..

kg m s m

N mm

2

P6.28 (a) vf = = 0 096 3 10 2 88 108 7. .m s m se j

K mvf f= = = 1

2

1

29 11 10 2 88 10 3 78 102 31 7

2 16. . .kg m s Je je j

(b) K W Ki f+ = : 0 + =F r Kf cos

F 0 028 0 3 78 10 16. cos .m Ja f =

F = 1 35 10 14. N

(c) F ma = ; aF

m= =

=

+1 35 10

9 11 101 48 10

14

3116.

..

N

kg m s2

(d) v v a txf xi x= + 2 88 10 0 1 48 107 16. . = + m s m s2e jt

t= 1 94 10 9. s

Check: x x v v tf i xi xf= + +12d i

0 028 01

20 2 88 107. .m m s= + + e jt

t= 1 94 10 9. s

Section 6.7 Situations Involving Kinetic Friction

P6.29 F may y = : n =392 0N

nf nk k

=

= = =392

0 300 392 118N

N N .a fa f

(a) W F rF = = = cos . cos 130 5 00 0 650a fa f J

(b) E f dkint J= = =118 5 00 588a fa f.

r

r

r

r

FIG. P6.29continued on next page


11/22

Chapter 6 153

(c) W n rn = = = cos . cos 392 5 00 90 0a fa f

(d) W mg rg = = = cos . cos 392 5 00 90 0a fa f a f

(e) K K K W Ef i= = other int

1

20 650 588 0 0 62 02mvf = + + =J J J.

(f) vK

mf

f= = =

2 2 62 0

40 01 76

.

..

J

kg m s

a f

P6.30 (a) W kx kxs i f= = =1

2

1

2

1

2500 5 00 10 0 0 6252 2 2

2a fe j. . J

W mv mv mvs f i f = = 1

2

1

2

1

202 2 2

so vW

mf = = =

2 2 0 625

2 000 791c h a f.

..m s m s

(b)1

2

1

22 2mv f d W mvi k s f + =

0 0 350 2 00 9 80 0 050 0 0 6251

2

0 2821

22 00

2 0 282

2 00

0 531

2

2

+ =

=

= =

. . . . .

. .

.

.

.

a fa fa fb g

b g

a f

J J

J kg

m s m s

mv

v

v

f

f

f

FIG. P6.30

P6.31 (a) W mgg = +l cos .90 0 a f

Wg = = 10 0 9 80 5 00 110 1682. . . coskg m s m Jb gd ia f

(b) f n mgk k k= = cos

E mg

E

k kint

int m J

= =

= =

l l cos

. . . . cos .5 00 0 400 10 0 9 80 20 0 184a fa fa fa f

(c) W FF = = =l 100 5 00 500a fa f. J

(d) K W E W W EF g= = + = other int int J148

(e) K mv mvf i= 1

2

1

22 2

v K

mvf i= + = + =

2 2 148

10 01 50 5 652

2a f a f a f.

. . m s

r

r

r

FIG. P6.31


12/22


P6.32 F may y = : n + =70 0 20 0 147 0. sin .N Na f

n= 123 N

f nk k= = = 0 300 123 36 9. .N Na f

(a) W F r= = = cos . . cos . 70 0 5 00 20 0 329N m Ja fa f

(b) W F r= = = cos . cos . 123 5 00 90 0 0N m Ja fa f

(c) W F r= = = cos . cos . 147 5 00 90 0 0N ma fa f

r

r

r

FIG. P6.32

(d) E f dkint N m J= = =36 9 5 00 185. .a fa f

(e) K K K W Ef i= = = = + int J J J329 185 144

P6.33 vi = 2 00. m s k = 0 100.

K f d W K i k f + =other :1

202mv f di k =

1

22mv mgdi k= d

v

gi

k

= = =

2 2

2

2 00

2 0 100 9 802 04

.

. ..

m sm

b ga fa f

Section 6.8 Power

P6.34 (a) The distance moved upward in the first 3.00 s is

y vt= = +L

NM O

QP =

0 1 75

23 00 2 63

.. .

m ss ma f .

The motor and the earths gravity do work on the elevator car:

1

2180

1

21

2650 1 75 0 650 2 63 1 77 10

2 2

2 4

mv W mg y mv

W g

i f+ + =

= + =

motor

motor kg m s kg m J

cos

. . .b gb g b g a f

Also, W t= P so P = =

= =

W

t

1 77 105 91 10 7 92

43.

. .J

3.00 s W hp.

(b) When moving upward at constant speed v= 1 75. m sb g the applied force equals theweight kg m s N2= = 650 9 80 6 37 103b ge j. . . Therefore,

P = = = =Fv 6 37 10 1 75 1 11 10 14 93 4. . . .N m s W hpe jb g .


13/22

Chapter 6 155

P6.35 Power=W

t P = = =

mgh

t

700 10 0

8 00875

N m

sW

a fa f..

P6.36 (a) W K = , but K= 0 because he moves at constant speed. The skier rises a vertical

distance of 60 0 30 0 30 0. sin . .m ma f = . Thus,

W Wgin2kg m s m J kJ= = = =70 0 9 8 30 0 2 06 10 20 64. . . . .b ge ja f .

(b) The time to travel 60.0 m at a constant speed of 2.00 m/s is 30.0 s. Thus,

Pinput

J

30.0 sW hp= =

= =

W

t

2 06 10686 0 919

4.. .

*P6.37 energy power time=

For the 28.0 W bulb:

Energy used = = 28 0 1 00 10 2804. .W h kilowatt hrsa fe j total cost = + =$17. $0. $39.00 280 080 4kWh kWha fb g

For the 100 W bulb:

Energy used = = 100 1 00 10 1 00 104 3W h kilowatt hrsa fe j. .

# bulb used =

=1 00 10

13 34.

.h

750 h bulb

total cost = + =13 3 420 1 00 10 080 63. $0. . $0. $85.b g e jb gkWh kWh

Savings with energy-efficient bulb = =$85. $39. $46.60 40 2

P6.38 (a) Burning 1 lb of fat releases energy 19 4186

1 71 107lb454 g

1 lb

kcal

1 g

J

1 kcalJ

FHG

IKJFHG

IKJFHG

IKJ = . .

The mechanical energy output is 1 71 10 0 207. . cos =Je ja f nF r .

Then 3 42 10 06. cos = J nmg y

3 42 10 50 9 8 80 0 150

3 42 10 5 88 10

6

6 3

. . .

. .

=

=

J kg m s steps m

J J

2n

n

b ge jb ga f

e j

where the number of times she must climb the steps is n=

=

3 42 10

5 88 10582

6

3

.

.

J

J.

This method is impractical compared to limiting food intake.



14/22


(b) Her mechanical power output is

P = =

= = F

HG I

KJ =W

t

5 88 1090 5 90 5

10121

3.. . .

J

65 sW W

hp

746 Whp .

P6.39 (a) The fuel economy for walking is1 3 1 1 30 10

423

8h

220 kcal

mi

h

kcal

4 186 J

J

1 gal mi galFHG IKJFHG

IKJ

FHG

IKJ =

..

(b) For bicycling1 10 1 30 10

7768h

400 kcal

mi

h

1 kcal

4 186 J

J

1 gal mi gal

FHG

IKJFHG

IKJ

FHG

IKJ

=.

.

Section 6.9 Context ConnectionHorsepower Ratings of Automobiles

P6.40 A 1 300-kg car speeds up from rest to 55.0 mi/h =24.6 m/s in 15.0 s. The output work of the engine isequal to its final kinetic energy,

1

21 300 24 6 390

2kg m s kJb gb g. =

with power P =390 000

104J

15.0 sW~ around 30 horsepower.

P6.41 Pa af v= : fv

aa

= =

=P 2 24 10

27 0830

4.

.N

Additional Problems

P6.42 At start,

r

v i j= +

40 0 30 0 40 0 30 0. cos . $

. sin . $

m s m sb g b g At apex,

rv i j i= + =40 0 30 0 0 34 6. cos . $ $ . $ m s m sb g b g

And K mv= = =1

2

1

20 150 34 6 90 02


*P6.43 Concentration of Energy output = F

HG I

KJ =0 600 60 01

24 0. . .J kg step kgstep

1.50 m J mb gb g

F

Fv

v

v

= =

=

=

=

24 0 1 24 0

70 0 24 0

2 92

. .

. .

.

J m N m J N

W N

m s

b gb g

a f

P


15/22

Chapter 6 157

*P6.44 As it moves at constant speed, the bicycle is in equilibrium. The forward frictional force is equal in

magnitude to the air resistance, which we write as av2 where ais a proportionality constant. Theexercising woman exerts the friction force on the ground; by Newtons third law, it is this same size

again. The womans power output is P = = =Fv av ch3 , where cis another constant and his her heart

rate. We are given a c22 903

km h beats minb g b g= . For her minimum heart rate we have

av cmin3 136= beats minb g . By division vmin22

13690

3

km hFHG IKJ = . vmin .= FHG IKJ =

13690

22 25 21 3

km h km hb g .

Similarly, vmax .=FHG

IKJ =

166

9022 27 0

1 3

km h km hb g .

P6.45 (a) x t t= + 2 00 3.

Therefore,

v dx

dtt

K mv t t t

= = +

= = + = + +

1 6 00

1

2

1

2 4 00 1 6 00 2 00 24 0 72 0

2

2 2 2 2 4

.

. . . . .a fe j e jJ

(b) a dv

dtt= = 12 0.a f m s2

F ma t t= = =4 00 12 0 48 0. . .a f a fN

(c) P = = + = +Fv t t t t48 0 1 6 00 48 0 2882 3. . .a fe j e jW

(d) W dt t t dt= = + =z zP0

2 003

0

2 00

48 0 288 1 250. .

.e j J

*P6.46 The work done by the applied force is

W F dx k x k x dx

k x dx k x dx k x

k x

k x

k x

i

f x

x x x x

= = +

= + = +

= +

z z

z z

applied 1 22

0

1

0

22

0

1

2

0

2

3

0

1

2

2

3

2 3

2 3

e jmax

max max max max

max max


16/22


P6.47 (a) The work done by the traveler is mgh Ns where Nis the number of steps he climbs during

the ride.

N=(time on escalator)(n)

where time on escalatorvertical velocity of person

a f = h

and vertical velocity of person = +v nhs

Then, N nh

v nhs=

+

and the work done by the person becomes W mgnhh

v nhs

sperson =

+

(b) The work done by the escalator is

W mgvte = = =power time force exerted speed timeb ga f a fb ga f

where t hv nhs

=+

as above.

Thus, W mgvh

v nhe

s

=+

.

As a check, the total work done on the persons body must add up to mgh, the work anelevator would do in lifting him.

It does add up as follows: W W W mgnhh

v nh

mgvh

v nh

mgh nh v

v nhmghe

s

s s

s

s = + =

++

+=

+

+=person

b g

*P6.48 During its whole motion from=

10 0. m to=

3 20. mm, the force of gravity and the force of theplate do work on the ball. It starts and ends at rest

K W K

F y F x

mg F

F

i f

g p

p

p

+ =

+ + =

=

=

=

0 0 180 0

10 003 2 0 003 20 0

5 10

3 2 101 53 10

35

cos cos

. .

..

m m

kg 9.8 m s m

mN upward

2

b g b g

e ja f

P6.49 F max x= : kx ma=

k max

= =

=

(4.70 10 kg)0.800(9.80 m s )

0.500 10 m N m

3 2

27 37.

r

r

r

r

FIG. P6.49


17/22

Chapter 6 159

*P6.50 The spring exerts on each block an outward force of magnitude

F kxs = = =3 85 0 08 0 308. . .N m m Nb ga f .

For the light block on the left, the vertical forces are given by F mgg = = =0 25 9 8 2 45. . .kg m s N2b ge j ,

Fy = 0 , n =2 45 0. N , n= 2 45. N . Similarly for the heavier blockn Fg= = =05 9 8 4 9. . .kg m s N

2b ge j .

(a) For the block on the left, F max x = , =0 308 0 25. .N kgb ga ,a= 1 23. m s2 . For the heavier block, + =0 308 0 5. .N kgb ga ,

a= 0 616. m s2 .

(b) For the block on the left, f nk k= = = 0 1 2 45 0 245. . .N Na f

F ma

a

a

x x =

+ =

=

0 308 0 245 0 25

0 252

. . .

.

m s N kg

m s

2

2

b g

2.45 N

n

Fs

FIG. P6.50

if the force of static friction is not too large. For the block on the right, f nk k= = 0 490. N .

The maximum force of static friction would be larger, so no motion would begin and the

acceleration is zero .

(c) Left block: fk = =0 462 2 45 1 13. . .N Na f . The maximum static friction force would be larger, sothe spring force would produce no motion of this block or of the right-hand block, which

could feel even more friction force. For both a= 0 .

P6.51 (a) P = = + = +FHG

IKJ =

FHG

IKJ

Fv F v at F F

mt

F

mtib g 0

2

(b) P =L

NMM

O

QPP

=20 0

5 003 00 240

2.

..

N

kgs W

a f a f

P6.52 (a) The new length of each spring is x L2 2+ , so its extension is

x L L2 2+ and the force it exerts is k x L L2 2+ FH I

K toward its

fixed end. Theycomponents of the two spring forces add tozero. Their xcomponents add to

rF i i= + FH

IK

+

=

+

FHG

IKJ

2 2 12 22 2 2 2

$ $k x L L x

x Lkx

L

x L. FIG. P6.52



18/22


(b) W F dxxi

f

= z W kx Lx L

dxA

=

+

FHG

IKJz 2 1 2 2

0

W k x dx kL x L x dx

A A

= + +z z 2 20

2 2 1 20

e j W kx

kLx L

A A

= +

+2

2 1 2

2 0 2 21 2 0

e jb g

W kA kL kL A L= + + +0 2 22 2 2 2 W kL kA kL A L= + +2 22 2 2 2

P6.53 (a) W K = : W Ws g+ = 0

1

20 90 60 0

1

21 40 10 0 100 0 200 9 80 60 0 0

4 12

2

3 2

kx mg x

x

x

i + + =

=

=

cos

. . . . sin .

.

a f

e j a f a fa fa fN m

m

(b) W K E = + int : W W Es g+ = int 0

1

2150 60 0

1

21 40 10 0 100 0 200 9 80 60 0 0 200 9 80 0 400 60 0 0

3 35

2

3 2

kx mg x mg x

x x

x

i k+ =

=

=

cos cos

. . . . sin . . . . cos .

.

N m

m

e j a f a fa fa f a fa fa fa f

P6.54 (a)rF i j i j1 25 0 35 0 35 0 20 5 14 3= + = +. cos .

$ sin . $ . $ . $N Na fe j e j

r

F i j i j2 42 0 150 150 36 4 21 0= + = +. cos $

sin $

. $

. $

N Na fe j e j

(b)r r rF F F i j = + = +1 2 15 9 35 3.

$ . $e jN

(c)r

r

aF

i j= = +m

3 18 7 07. $ . $e j m s2

(d)r r rv v a i j i jf i t= + = + + +4 00 2 50 3 18 7 07 3 00.

$ . $ . $ . $ .e j e je ja fm s m s s2

rv i jf = +5 54 23 7.

$ . $e j m s

(e)r r r rr r v af i it t= + +

1

22

r

r r

r i j i j

r r i j

f

f

= + + + +

= = +

0 4 00 2 50 3 001

23 18 7 07 3 00

2 30 39 3

2. $ . $ . . $ . $ .

. $ . $

e jb ga f e je ja f

e j

m s s m s s

m

2



19/22

Chapter 6 161

(f) K mvf f= = + =1

2

1

25 00 5 54 23 7 1 482

2 2. . . .kg m s kJ2b ga f a f e j

(g) K mvf i= + 1

22

r rF r

K

K

f

f

= + + +

= + =

1

2 5 00 4 00 2 50 15 9 2 30 35 3 39 3

55 6 1 426 1 48

2 2 2

. . . . . . .

. .

kg m s N m N m

J J kJb ga f a f b g a fa f a fa f

P6.55 K W W K

mv kx kx mg x mv

kx mgx mv

i s g f

i i f f

i i f

+ + =

+ + =

+ + =

1

2

1

2

1

2

1

2

01

20 100

1

2

2 2 2 2

2 2

cos

cos

FIG. P6.55

1

21 20 5 00 0 050 0 0 100 9 80 0 050 0 10 0

1

20 100

0 150 8 51 10 0 050 0

0141

005001 68

2

3 2

. . . . . . sin . .

. . .

.

..

N cm cm m kg m s m kg

J J kg

m s

2b ga fb g b ge jb g b g

b g

=

=

= =

v

v

v

P6.56 (a) F L F LN mm N mm

2.00

4.00

6.00

8.00

10.0

12.0

15.0

32.0

49.0

64.0

79.0

98.0

14.0

16.0

18.0

20.0

22.0

112

126

149

175

190

a f a f a f a f

F Na f

FIG. P6.56

(b) A straight line fits the first eight points, together with the origin. By least-square fitting, its

slope is 0 125 2% 125 2%. N mm N m = . In F kx= , the spring constant is k F

x= , the

same as the slope of the F-versus-xgraph.

(c) F kx= = =125 0 105 13 1N m m Nb ga f. .


20/22


P6.57 If positive Frepresents an outward force, (same as direction as r), then

W d F r F r dr

W F r F r

WF r r F r r F

r r F

r r

W r r r r

W

i

f

r

r

r

r

f i f i

f i f i

f i f i

i

f

i

f

= =

=

=

+

=

=

=

z z

r rF r 2

2

12 6

6 6 6 6

103 10 189 10

103 10 188 10

013 13

07 7

0

13 12

0

7 6

013 12 12

07 6 6

07

6 6 013

12 12

77 6 6 134 12 12

77 6

e j

e j e j

. .

. .

= + =

2 44 10 10 189 10 354 10 596 10 10

2 49 10 112 10 137 10

4 60 134 12 8 120

21 21 21

. . . .

. . .W J J J

P6.58 E K m v vf iint = = 1

2

2 2e j: Eint kg m s J= =1

2

0 400 6 00 8 00 5 602 2 2

. . . .b ga f a fe jb g

(b) E fd mg rkint = = 2a f: 5 60 0 400 9 80 2 1 50. . . .J kg m s m2= kb ge j a f

Thus, k = 0152. .

(c) After Nrevolutions, the object comes to rest and Kf = 0 .

Thus, E K K mvi iint = = + =01

22

or k img N r mv21

22a f = .

This gives Nmv

mg r

i

k

= = =

12

2 12

2

2

8 00

0 152 9 80 2 1 502 28

a fb g

a fe j a f.

. . ..

m s

m s mrev

2.

P6.59 We evaluate375

3 75312 8

23 7dx

x x+z

..

.

by calculating

375 0 100

12 8 3 75 12 8

375 0 100

12 9 3 75 12 9

375 0 100

23 6 3 75 23 60 806

3 3 3

.

. . .

.

. . .

.

. . ..

a fa f a f

a fa f a f

a fa f a f+

++

++

=K

and

375 0 100

12 9 3 75 12 9

375 0 100

13 0 3 75 13 0

375 0 100

23 7 3 75 23 70 791

3 3 3

.

. . .

.

. . .

.

. . ..

a fa f a f

a fa f a f

a fa f a f+

++

++

=K .

The answer must be between these two values. We may find it more precisely by using a value for

x smaller than 0.100. Thus, we find the integral to be 0 799. N m .


21/22

Chapter 6 163

P6.60 P

t W K m v

= = =a f 2

2

The density is = =

m m

xvol.

Substituting this into the first equation and solving for P, since

x

t v= ,

for a constant speed, we get P =Av3

2.

r

FIG. P6.60

Also, since P =Fv, F Av

= 2

2.

Our model predicts the same proportionalities as the empirical equation, and gives D= 1 for thedrag coefficient. Air actually slips around the moving object, instead of accumulating in front of it.For this reason, the drag coefficient is not necessarily unity. It is typically less than one for astreamlined object and can be greater than one if the airflow around the object is complicated.

*P6.61 P =1

22 3D r v

(a) Pa = = 1

21 1 20 1 5 8 2 17 10

2 3 3. . .kg m m m s W3e j a f b g

(b)P

P

b

a

b

a

v

v= =

FHG

IKJ

= =

3

3

3

324

83 27

m s

m s

Pb = = 27 2 17 10 5 86 103 4. .W We j

*P6.62 (a) So long as the spring force is greater than the friction force,the block will be gaining speed. The block slows down whenthe friction force becomes the greater. It has maximum speed

when = =kx f maa k 0 .

=10 10 4 0 03. .N m Ne jxa x= 4 0 10 3. m

(b) By the same logic,

10 10 10 03. .N m N = 0e jxb x= 1 0 10 2. m

0

0

FIG. P6.62

ANSWERS TO EVEN PROBLEMS

P6.2 1 59 103. J

P6.4 (a) 32.8 mJ; (b) 32.8 mJ

P6.6 see the solution

P6.8 16.0

P6.10 (a) see the solution; (b) 12 0. J

P6.12 0 299. m s


22/22


P6.14 50.0 J

P6.16 (a) 9.00 kJ; (b) 11.7 kJ, larger by 29.6%

P6.18 see the solution, 316 N m

P6.20 (a) mgk k

1 1

1 2

+FHG

IKJ

; (b)1 1

1 2

1

k k+

FHG

IKJ

P6.22 (a) 33.8 J; (b) 135 J

P6.24 (a) 1 94. m s ; (b) 3 35. m s ; (c) 3 87. m s

P6.26 (a) 4.56 kJ; (b) 6.34 kN; (c) 422 km s2 ;

(d) 6.34 kN

P6.28 (a) 3 78 10 16. J ; (b) 1 35 10 14. N ;

(c) 1 48 10 16. + m s2 ; (d) 1 94 10 9. s

P6.30 (a) 0 791. m s ; (b) 0 531. m s

P6.32 (a) 329 J; (b) 0; (c) 0; (d) 185 J; (e) 144 J

P6.34 (a) 5 91 103. W ;

(b) it is 53.0% of 1 11 104. W

P6.36 (a) 20.6 kJ; (b) 686 W

P6.38 (a) 582, impractical;(b) 90 5 0 121. .W hp=

P6.40 ~104 W

P6.42 90.0 J

P6.44 25 2. km h , 27 0. km h

P6.46 k x

k x

1

2

2

3

2 3max max

+

P6.48 1 53 105

.

N upward

P6.50 (a) 1 23. m s2 and 0 616. m s2 ;

(b) 0 252. m s2 and 0; (c) 0 and 0

P6.52 (a) see the solution;

(b) 2 22 2 2 2kL kA kL A L+ +

P6.54 (a)rF i j1 20 5 14 3= +.

$ . $e jN ,rF i j2 36 4 21 0= +.

$ . $e jN;

(b) +15 9 35 3. $ . $i je jN ;

(c) +3 18 7 07. $ . $i je j m s2 ;

(d) +5 54 23 7. $ . $i je j m s ;

(e) +2 30 39 3. $ . $i je jm ; (f) 1 48. kJ ;(g) 1 48. kJ

P6.56 (a) see the solution; (b) 125 2%N m ;

(c) 13.1 N

P6.58 (a) 5.60 J; (b) 0.152; (c) 2.28 rev

P6.60 see the solution

P6.62 (a) x= 4 0. mm ; (b) 1.0 cm

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