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11-1

ELASTICITY AND PERIODIC MOTION 11Answers to Multiple-Choice Problems

1. A 2. B 3. B 4. A 5. C 6. C 7. A, C, D 8. B 9. C 10. C 11. B 12. C 13. B 14. D 15. C

Solutions to Problems

11.1. Set Up: with The force applied to the end of the wire is

and

Solve: (a)

(b)

(c)

Reflect: Our result for Y is about the same size as the values in Table 11.1. In SI units the stress is very large. The

strain is dimensionless and small, so Y is very large and has the same units (Pa) as stress.

11.2. Set Up:

Solve:

11.3. Set Up: The cross-sectional area of the post is The force applied to

the end of the post is The Young’s modulus of steel is

Solve: (a)

(b) The minus sign indicates that the length decreases.(c)

11.4. SetUp: For steel, The force applied to the end of the lower wire is

The force applied to the end of the upper wire is Tensile

strain 5Dl

l0

5F '

AY .

15.0 kg 1 10.0 kg 2 19.80 m / s2 2 5 147.0 N.98.0 N.

1 10.0 kg 2 19.80 m / s2 2 5Y 5 2.0 3 1011 Pa.

D l 5 l0 1 strain 2 5 12.50 m 2 128.0 3 1026 2 5 22.0 3 1025 m

strain 5

stress

Y 5 2

1.60 3 106 Pa

2.0 3 1011 Pa 5 28.0 3 10

26

.

stress 5F '

A5

7.84 3 104 N

0.0491 m25 1.60 3 106 Pa

Y 5 2.0 3 1011 Pa.

F ' 5 18000 kg 2 19.80 m / s

2 2 5 7.84 3 104 N.

A 5 pr 2 5 p 10.125 m 2 2 5 0.0491 m2.

Y 5l0 F '

A D l5

14.00 m 2 15000 N 210.50 3 1024 m2 2 10.20 3 1022 m 2 5 2.0 3 1011 Pa

A 5 0.50 cm2 5 0.50 3 1024 m2

Y 5stress

strain5

1.03 3 109 Pa

1.47 3 10235 7.01 3 1011 Pa

strain 5 D l / l0 51.10 3 1023 m

0.750 m5 1.47 3 1023

stress 5245 N

p 12.75 3 1024 m 2 25 1.03 3 109 Pa

Y 5 stress / strainstrain 5 D l / l0 .stress 5 F T / Amg 5 125.0 kg 2 19.80 m / s2 2 5 245 N.

F Tr 5 2.75 3 1024 m. A 5 pr 2,



Solve: (a) lower wire:

upper wire:

(b) lower wire:

upper wire:

11.5. Set Up:

Solve: relaxed:

maximum tension:

11.6. Set Up: The force applied to the end of the rope is theweight of the climber:

Solve:

11.7. Set Up: For steel,

Solve: gives

Reflect: Thethinner thewire themore it stretchesforthe same applied force.The lengthof this wire changes by 0.12%

11.8. Set Up: A 5.0% elongation means For a spring,

Solve: (a)

(b) so

The change in length is gives

(c) and

11.9. Set Up: so and From Problem 11.8, for the natural

Achilles tendon.

Solve: (a) so

so and the diameter is 2.2 mm.

(b) The natural tendon has and diameter 10.0 mm. The artificial tendon’s diame-

ter is much smaller.

Reflect: The artificial tendon has a larger Y and therefore a smaller diameter.

r 5 " 178.1 mm2 2 / p 5 4.99 mm

r 5 " A / p 5 1.1 mm A 5 pr 2

A 5kl0

Y 5 14.6 3 105 N / m 2 1 0.25 m 2

30 3 109 Pa5 3.8 3 1026 m2k 5

YA

l0

A 5 pr 2

k 5 4.6 3 105 N / mk 5YA

l0

.F T 5 1YA

l02 D lY 5

F T / A

Dl / l0

x 5F T

k 5

9555 N

4.6 3 105 N / m5 2.08 cmF 5 13mg 5 13 175 kg 2 19.80 m / s

2 2 5 9555 N

k 5F T

x5

5.76 3 103 N

1.25 3 1022 m5 4.6 3 105 N / m

F T 5 kx x 5 D l 5 10.050 2 125 cm 2 5 1.25 cm.

F T 5 stress 3 A 5 1 7.37 3 107 Pa 2 178.1 3 1026 m2 2 5 5.76 3 103 Nstress 5 F T / Astress 5 Y 3 strain 5 11474 3 106 Pa 2 10.050 2 5 7.4 3 107 Pa

F T 5 kx.D l / l0 5 0.050.Y 5stress

strain .

d 5 Å 4 A

p5 Å

4 11.6 3 1026 m2 2p

5 1.43 mm

A 5l0 F

'

Y Dl5

12.00 m 2 1400.0 N 212.0 3 1011 Pa 2 10.25 3 1022 m 2 5 1.6 3 1026 m2Y 5

l0 F '

A D l

A 5 pd 2 / 4.Y 5 2.0 3 1011 Pa.

Y 5l0 F

'

A D l5

145.0 m 2 1637 N 213.85 3 1025 m2 2 11.10 m 2 5 6.77 3 108 Pa

F ' 5 165.0 kg 2 1 9.80 m / s2

2 5 637 N. A 5 pr

2

5 p

13.5 3 10

23

m

22

5 3.85 3 10

25

m

2

.

Y 510.200 m 2 1500 N 2

150.0 3 1024 m2 2 13.0 3 1022 m 2 5 6.67 3 105 Pa

Y 510.200 m 2 125.0 N 2

1 50.0 3 1024 m2 2 13.0 3 1022 m 2 5 3.33 3 104 Pa

Y 5l0 F

'

A D l A 5 50.0 cm2 5 50.0 3 1024 m2.

D l 5 l0 1 strain 2 5 10.500 m 2 11.2 3 1023 2 5 6.0 3 1024 m

Dl 5 l0

1strain

25

10.500 m

2 18.2 3 1024

25 4.1 3 1024 m

tensile strain 5147.0 N16.0 3 1027 m2 2 12.0 3 1011 Pa 2 5 1.2 3 1023

tensile strain 598.0 N16.0 3 1027 m2 2 12.0 3 1011 Pa 2 5 8.2 3 1024.

11-2 Chapter 11



11.10. Set Up:

Solve: (a) and

(b) and gives

11.11. Set Up:

Solve: (a)

(b) so

Reflect: Young’s modulus for cartilage is much smaller than typical values for metals and the fractional change in

length is larger.

11.12. Set Up: can be computed with both and expressed in

Solve:

11.13. Set Up: At the surface the pressure is so

At the surface of water has mass

Solve: (a) gives

(b) At this depth of seawater has volume The density is

11.14. Set Up: where d is the depth below the surface.

Solve: (a)

One cubic centimeter of her blood decreases in volume by

(b) gives

The depth would be The ocean is not this

deep; the greatest depth in the ocean is an order of magnitude less than this, about 11 km.

11.15. Set Up: The surface pressure on earth is about 1 atm.

Solve:

The volume would decrease 9.1 mL.

Reflect: The percent change in the volume is A nearly 100-fold increase in pressure has a small

effect on the volume because the bulk modulus is very large.

DV

V 05 0.18%.

DV 5 2V 0 D p B

5 2

15.00 L

2

191

2 11.01 3 10

5

Pa

25.0 3 109 Pa5 29.1 mL

DV

V 05 2

D p

B

d 5D p

1.0 3 104 Pa / m5

1.1 3 109 Pa

1.0 3 104 Pa / m5 1.1 3 105 m 5 110 km.

D p 5 BA12 B 5 1.1 3 109 PaDV

V 05 2

1

2

1.5 3 1024 cm3.

DV 5 2

D p

B V 0 5 2 11.0 3 104 Pa / m 2 133 m 22.2 3 109 Pa

11.0 cm3 2 5 21.5 3 1024 cm3

D p 5 11.0 3 104 Pa / m 2d ,DV

V 0

5 2

D p

B .

1.03 3 103 kg

0.9473 m35 1.09 3 103 kg / m

3.

V 0 1 DV 5 0.9473 m3.1.03 3 103 kg

DV 5 2

1D p 2V 0

B5 2

11.16 3 108 Pa 2 11.00 m3 22.2 3 109 Pa

5 20.0527 m3 B 5 2

1D p 2V 0

DV

1.03 3 103 kg.1.00 m3Density 5 m / V .

V 0 5 1.00 m3.D p 5 1.16 3 108 Pa.1.0 3 105 Pa,

B 5 2 D pDV / V 0

5 2

1D p

2V 0

DV 5 2

13.6 3 10

6

Pa

2 1600 cm

3

220.45 cm35 4.8 3 109 Pa

cm3.V 0DV DV / V 0

Dl

l0

5 12 124% 2 5 12%F T 5 4mg

Dl

l0

5F T

AY 5

5880 N110 3 1024 m2 2 124 3 106 Pa 2 5 0.24 5 24%F T 5 8mg 5 5880 N.

5F T / A

Dl / l0

l0 5 8.6 cm.l 5 12 cm0.40 5 l 2 l0

l0

F ' 5 0.385 N.196 3 106 Pa 5

F '

p 125 3 1026 m 2 2

strain 5D l

l0

.stress 5F '

A ,

Elasticity and Periodic Motion 11-3



11.16. Set Up:

Solve: (a)

(b) The depth for a pressure increase of is 1.5 km. Unprotected dives do not approach this depth sobone compression is not a concern.

11.17. Set Up: For steel the shear modulus is

Solve: A is the area of each edge of the plate:

11.18. Set Up: is the component of the force tangent to the surface, so

1360 N. must be in radians,

Solve:

11.19. Set Up: is in radians. with

Solve:

Reflect: The shear modulus of cartilage is much less than the values for metals given in Table 11.1.

11.20. Set Up: The tensile force equals the weight w of the object suspended from the wire. The maximum

weight that can be supported in one that makes the stress equal to the breaking stress.

Solve: (a) so

(b) so

(c) so

11.21. Set Up: The free-body diagram for the elevator is given in

Figure 11.21. is the tension in the cable.

Figure 11.21

mg

a

x

F '

y

F '

F '

A5 1

3 12.40 3 108 Pa 2 5 0.80 3 108 Pa.

w 5 111.0 3 108 Pa 2 10.040 3 1024 m2 2 5 4.4 3 103 NF ' / A 5 11.0 3 108 Pa

D l 5 l0 F '

YA5

15.00 m

2 11.44 3 10

3

N

212.0 3 1011 Pa 2 10.040 3 1024 m2 25 9.0 3 1023 m 5 9.0 mm.Y 5 l0 F '

A D l

w 5 13.60 3 108 Pa 2 1 0.040 3 1024 m2 2 5 1.4 3 103 N.F '

A5 3.60 3 108 Pa

F '

f 5F i

AS5

8mg sin 12°

110 3 1024 m2 2 112 3 106 Pa 2 5 0.1494 rad 5 8.6°

p rad 5 180°m 5 110 kg.F 5 8mg,fF i 5 F sin 12°.S 5F i

Af .

S 51360 N

10.0925 m

22

10.0216 rad

25 7.36 3 106 Pa

f 5 1.24° 5 0.0216 rad.f

F i 5 11375 N 2 cos 8.50° 5F iS 5F i

Af .

F i 5 SA 1 shear strain 2 5 10.84 3 1011 Pa 2 11.00 3 1023 m2 2 10.0400 2 5 3.4 3 106 N

A 5 10.100 m 2 10.0100 m 2 5 1.00 3 1023 m2.

S 5shear stress

shear strain5

F i / A

shear strain .S 5 0.84 3 1011 Pa.

1.5 3 107

Pa

D p 5 2 B

DV

V 05 2 115 3 109 Pa 2 120.0010 2 5 1.5 3 107 Pa 5 150 atm

1 atm 5 1.01 3 105 Pa.DV

V 05 2

D p

B .

11-4 Chapter 11



Solve: applied to

the elevator gives and

Reflect: The tension in the cable is about twice the weight of the elevator.

11.22. Set Up: The cross-sectional area of a hollow cylinder with inner radius and outer radius is

The total weight his legs support is his weight and the weight he lifts. Assume each leg supports

half of

Solve: and

He can lift a mass of which is a weight of

11.23. Set Up: The frequency in Hz is the number of cycles per second. The angular frequency is

and has units of radians. The period T and frequency are related by

Solve: (a)

(b)

(c) so ranges from to

so T ranges from to

(d) and

Reflect: Visible light has much higher frequency than either sounds we can hear or ultrasound. Ultrasound is sound

with frequencies higher than what the ear can hear. Large corresponds to small T .

11.24. Set Up: The period is the time for the hand to make one complete revolution. and

Solve: (a)

(b)

11.25. Set Up: The amplitude is the maximum displacement from equilibrium. In one period the object goes from

to and returns.

Solve: (a)

(b) so the period is 1.60 s

(c)

11.26. Set Up: The amplitude A is themaximum displacement from equilibrium.The periodT is the timefor 1 cycle.

and

Solve: (a)

(b)

(c)

(d) v 5 2p f 5 3140 rad / s

f 51

T 5 500 Hz.

T 5 2.0 ms.

A 5 0.40 mm.

v 5 2p f . f 51

T

f 51

T 5 0.625 Hz

0.800 s 5 T / 2 A 5 0.120 m

x 5 2 A x 5 1 A

v 5 2p 12.78 3 1024 Hz 2 5 1.75 3 1023 rad / s f 5 2.78 3 1024 Hz.T 5 1.00 hr 5 3600 s.

v 5 2p 10.0167 Hz 2 5 0.105 rad / s f 5 0.0167 Hz.T 5 1.00 min 5 60.0 s.

v 5 2p f . f 51

T

f

v 5 2p f 5 2p 15.0 3 106 Hz 2 5 3.1 3 107 rad / sT 51

f 5

1

5.0 3 106 Hz5 2.0 3 1027 s

1

4.3 3 1014 Hz5 2.3 3 10215 s

1

7.5 3 1014 Hz5 1.3 3 10215 sT 5

1

f

4.7 3 1015 Hz

2p rad5 7.5 3 1014 Hz

2.7 3 1015 Hz

2p rad5 4.3 3 1014 Hz f f 5

v

2p

v 5 2p f 5 1.26 3 105 rad / s f 51

T 5

1

50.0 3 1026 s5 2.00 3 104 Hz.

v 5 2p f 5 2p 1466 Hz 2 5 2.93 3 103 rad / sT 51

f 5

1

466 Hz5 2.15 3 1023 s.

T 51

f . f

v 5 2p f v f

2.1 3 105 N.2.2 3 104 kg,

mtot 5 2.2 3 104 kg200 3 106 Pa 5

12 W tot

A5

12 mtot g

p 1 30.018 m 4 2 2 30.0125 m 4 2 2W tot .

W totp 1 R22 2 R1

2 2 . A 5 R2 R1

a 5F '

m2 g 5

2.40 3 104 N

1200 kg2 9.80 m / s

2 5 10.2 m / s2

F ' 2 mg 5 ma

gF y 5 ma yF ' 5 A 1 0.80 3 108 Pa 2 5 13.00 3 1024 m2 2 10.80 3 108 Pa 2 5 2.40 3 104 N.




(b)

(c) Or, gives

which checks.

Reflect: The maximum force and maximum acceleration occur when the displacement is maximum and the velocityis zero.

11.32. Set Up: The fly has the same speed as the tip of the tuning fork.

Solve: (a) and

(b)

11.33. Set Up: with and

Solve: says This gives so

Reflect: When the kinetic energy is three times the elastic potential energy.

11.34. Set Up: The total energy is

Solve: (a) says E increases by a factor of The new energy will be

(b) The new amplitude A is

11.35. Set Up: The period is the time for one cycle. A is the maximum value of x.

Solve: (a) From the figure,

(b)

(c)

(d) From the figure,

(e) so

11.36. Set Up: When the fish hangs at rest the upward spring force equals the weight mg of the fish.

Solve: (a) so

(b) Note that T depends only on m and k and is independent of

the distance the fish is pulled down.

11.37. Set Up: From Problem 11.35, and

Solve: (a) with gives

(b) with gives

(c) at and at

Reflect: At the point in the motion where the speed is a maximum the acceleration is zero and at the point where the

magnitude of the acceleration is a maximum the speed is zero.

x 5 6 A0 a x 0 5 amax x 5 00 v x 0 5 vmax

a 5k

m A 5 1 148 N / m

2.40 kg 2 13.0 cm 2 5 185 cm / s2 x 5 6 A2kx 5 ma x

vmax 5

Å k

m A 5

Å 148 N / m

2.40 kg

13.0 cm

25 23.6 cm / s x 5 0

1

2 mv

x

2

1

1

2 kx

2

5

1

2 kA

2

A 5 3.0 cm.m 5 2.40 kgk 5 148 N / m,

T 5 2p Å m

k 5 2p Å

65.0 kg

5.31 3 103 N / m5 0.695 s.

k 5mg

x5

165.0 kg 2 19.80 m / s2 2

0.120 m5 5.31 3 103 N / m.mg 5 kx

0 F x 0 5 kx

k 5 m 12p

T 22

5

12.40 kg

2 12p

0.80 s 22

5 148 N / mT 5 2p

Å m

k

A 5 3.0 cm

v 5 2p f 5 7.85 rad / s

f 51

T 5 1.25 Hz

T 5 0.80 s.

A 5 A0 " 3 .12 kA2 5 3A12 kA0

2B.4 E 0 .22 5 4. A S 2 A

E 5 12 kA2.

x 5 A / 2 x 5 A / " 2 .2A12 kx2B 5 1

2 kA2,2U 5 E .U 5 K

U 5 12 kx2 E 5 1

2 kA2K 1 U 5 E ,

K max 5 12 mvmax

2 5 12 10.0270 3 1023 kg 2 11.48 m / s 2 2 5 2.96 3 1025 J

vmax 5 2p fA 5 2p 1392 Hz 2 10.600 3 1023 m 2 5 1.48 m / sÅ k

m5 2p f

vmax 5 Å k

m A. f 5

1

2p Å

k

m .

118 N,

F max 5 kA 5 1315 N / m 2 10.376 m 2 5F x 5 2kxF max 5 mamax 5 12.00 kg 2 159.2 m / s2 2 5 118 N.

amax 5k

m A 5 1315 N / m

2.00 kg 2 10.376 m 2 5 59.2 m / s2




11.38. Set Up: The circle of reference and the two positions of the object are sketched in Figure 11.38.

Figure 11.38

Solve: and and

11.39. Set Up:

Solve:

11.40. Set Up:

Solve: (a) Consider the chair alone:

(b) The mass of the astronaut is

11.41. Set Up: The period is the time for 1 cycle; after time T the motion repeats. The graph shows

Solve: (a)

(b)

(c)

(d) when so so From the graph in

the problem,

The mass is at when and this occurs at 1.2 s, and 1.8 s.

(e) gives

amax 5kA

m5 12p f 2 2 A 5 14p2 2 10.625 Hz 2 2 10.051 m 2 5 0.79 m / s

2 5 79 cm / s2.

2kx 5 ma x

t 5 0.4 s,v x 5 0, x 5 6 A

A 5 10.20 m / s 2 / 12p 2 10.625 Hz 2 5 0.0.051 m 5 5.1 cm

vmax 5 0.20 m / s.

A 5 vmax / 12p f 2 . f 51

2p Å

k

m A 5 vmax Å

m

k .

12 kA2 5 1

2 mvmax2. x 5 0v x 5 vmax

v 5 2p f 5 3.93 rad / s

f 51

T 5 0.625 Hz

T 5 1.60 s

12 mv x

2 1 12 kx2 5 1

2 kA2vmax 5 20.0 cm / s.

113 kg 2 35.4 kg 5 77.6 kgm 5 k 1 T

2p 22

5 1894 N / m 2 12.23 s

2p 22

5 113 kg.

k 5 12p

T 22

m 5 1 2p

1.25 s 22

135.4 kg 2 5 894 N / m

T 5 2p

Å

m

k

k 5 12p

T 22

5 1 2p

0.150 s 22 10.600 kg 2 5 1.05 3 103 N / m

T 5 2p Å m

k

t 5 T 1 f

2p rad 2 5 10.600 s 2 1 0.253 rad

2p rad 2 5 0.0242 s.

f 5 vt 5 12p radT 2 t f 5 0.253 rad.sin f 5 A / 4 A

5 0.250

y

x

1 2

A

A / 4

f

f

11-8 Chapter 11



The acceleration is maximum when and this occurs at the times given in (d).

(f) so

Reflect: The speed is maximum at when The magnitude of the acceleration is maximum at

where

11.42. Set Up:

Solve: (a)

(b)

(c)

11.43. Set Up:

Solve: (a)

(b) Solving for gives

Reflect: When the mass increases, the frequency of oscillation increases.

11.44. Set Up:

Solve: so

11.45. Set Up: The motion specified is one-half of a cycle so the time is

Solve: so

Reflect: The period of a simple pendulum does not depend on its amplitude, as long as the amplitude is small, and

does not depend on the mass of the bob. The period depends only on the length of the pendulum and the value of g.

11.46. Set Up: The motion specified is one-fourth of a cycle, so

Solve:

so

The answer doesn’t depend on the mass of the bob or the amplitude of the swing.

L 5 g 1 T

2p 22

5 1 9.80 m / s2 2 14.52 s

2p 22

5 5.07 m

T 5 2p Å L

g

T 5 4t 5 4 11.13 s 2 5 4.52 s

t 5 T / 4.

t 5 1.88 sT 5 2p Å 3.50 m

9.80 m / s2

5 3.75 s,

t 5 T / 2.T 5 2p Å L

g .

L 5 g 1 T

2p 22

5 19.80 m / s2 2 13.40 s

2p 22

5 2.87 mT 585.0 s

25.05 3.40 s

T 5 2p Å L

g

mV 5 9.99 femtograms

mv 5 ms 1 S f s

f s1vT 2

2 1 2 5 1 2.10 3 10216 g 2 1 S 2.00 3 1015 Hz

2.87 3 1014 HzT 2

2 1 2 5 9.99 3 10215 g

mv1 f s1v

f s2

2

51

1 1 mv / ms

.

f s1v

f s5 1 1

2p Å

k

ms 1 mv2 12p Å

ms

k 2 5 Å ms

ms 1 mv

51

" 1 1 mv / ms

f s1v 51

2p Å

k

ms 1 mv

f s 51

2p Å

k

ms

,

f 51

2p Å

k

m

m 5k

v25

120 N / m137.7 rad / s 2 25 0.0844 kg

v 5 2p f 5 2p 16.00 Hz 2 5 37.7 rad / s

T 51

f 5

1

6.00 Hz5 0.167 s

T 51

f . f 5

v

2p5

1

2p Å

k

m .

v x 5 0.

x 5 6 A,a x 5 0. x 5 0,

m 5 k 1 T

2p 22

5 175 N / m 2 11.60 s

2p 22

5 4.9 kgT 5 2p Å m

k

x 5 6 A




The mechanical energy lost is

(b) The mechanical energy has been converted to other forms by air resistance and by dissipative forces within the

rope.

Reflect: After a while the rock will have come to rest and then all its initial mechanical energy has been “lost”, has

been converted to other forms.

11.52. Set Up:

Solve: (a) When is a maximum and the tangent to the curve is horizontal the speed of the mass is zero. This

occurs at and

(b) At and so

(c) At and so At

and so The mechanical energy “lost” is

The mechanical energy lost was converted to other forms by nonconservative forces, such as fric-

tion, air resistance and dissipative forces.

11.53. Set Up: As shown in the figure in the solution to Problem 11.51, the height h of the tool above its lowest

point is When

Solve: (a) When is a maximum and the tangent to the curve is horizontal, the angular speed of the mass is zero.

And so is also zero. This occurs at 0.80 s, 1.60 s, 2.40 s, and 3.20 s.

(b) At and

(c) At and

At and

The mechanical energy “lost” is The mechanical energy lost was converted to other forms

by nonconservative forces, such as air resistance and dissipative forces within the wire.

11.54. Set Up: K is maximum when and at that point U is a maximum when

and at that point Problem 11.35 gives that and

Solve:

when

11.55. Set Up: so is the

magnitude of the acceleration when

Solve: (a)

(b)

(c)

Reflect: For a given amplitude, the maximum acceleration and maximum velocity increase when the frequency of

the motion increases and the period decreases.

K max 5 12 mvmax

2 5 12 10.450 kg 2 118.3 m / s 2 2 5 75.4 J

vmax 5 v A 5

1366.5 rad / s

2 10.0500 m

25 18.3 m / s.

F max 5 mamax 5 10.450 kg 2 1 6.72 3 103 m / s2 2 5 3.02 3 103 N

amax 5 v2 A 5 1366.5 rad / s 2 2 10.0500 m 2 5 6.72 3 103 m / s2

vmax 5 Å k

m A 5 v A. x 5 6 A.

amax 5k

m A 5 v2 Ama x 5 2kxv 5 3500 rpm 5 366.5 rad / s. A 5 0.0500 m.

x 5 0.K 5 K max

K max 5 U max 5 12 kA2 5 1

2 1148 N / m 2 10.030 m 2 2 5 0.0666 J

A 5 3.0 cm.k 5 148 N / mU max 5 E .K 5 0

K max 5 E .U 5 0 E 5 K 1 U .

E 0.8 2 E 2.4 5 0.0218 J.

E 2.4 5 mgL 1 1 2 cos u 2 5 11.15 kg 2 1 9.80 m / s2 2 10.635 m 2 11 2 cos 4.0° 2 5 0.0174 J.

v x 5 0t 5 2.4 s,

E 0.8 5 mgL 1 1 2 cos u 2 5 11.15 kg 2 1 9.80 m / s2 2 10.635 m 2 11 2 cos 6.0° 2 5 0.0392 J.

v x 5 0t 5 0.80 s,

E 5 mgh 5 mgL 11 2 cos u 2 5 1 1.15 kg 2 19.80 m / s2 2 10.635 m 2 11 2 cos 7.0° 2 5 0.0533 J

v x 5 0t 5 0,

t 5 0,vv 5 r v, 0u

0E 5 mgh.v x 5 0,h 5 L 11 2 cos u 2 .

E 4 5 0.30 J. E 1 2 E 4 5 1

2 kx2 5 12 1 225 N / m 2 10.030 m 2 2 5 0.101 J. x 5 3.0 cmv x 5 0

t 5 4.0 s, E 1 5 12 kx2 5 1

2 1225 N / m 2 120.060 m 2 2 5 0.405 J. x 5 26.0 cmv x 5 0t 5 1.0 s,

E 0 5 12 kx2 5 1

2 1 225 N / m 2 10.070 m 2 2 5 0.55 J x 5 7.0 cmv x 5 0t 5 0,

t 5 4.0 s.t 5 3.0 st 5 2.0 s,t 5 1.0 s,t 5 0,

0 x 0 E 5 12 mv x

2 1 12 kx2

E i 2 E f 5 0.543 J.




11.56. Set Up: M is the mass of the empty car and the mass of the loaded car is

Solve: The period of the empty car is The period of the loaded car is

11.57. Set Up: Relative to the accelerating rocket, the downward acceleration of an object that is

dropped inside the rocket is the rock is accelerated downward by gravity and the rocket accelerates

upward toward the rock.

Solve: Use the period when the rocket is at rest to calculate the length of the pendulum.

When the rocket is accelerating, and

Reflect: If the rocket had a downward acceleration, as an elevator might, then the period decreases. If the rocket is in

free-fall, with downward acceleration g, then the pendulum bob doesn’t swing at all and the period is infinite.

11.58. Set Up:Solve: so so

11.59. Set Up: the motion from one position of to the next is half a period. The period depends only on g

and where m and R are the mass and radius of Newtonia. The circumference c is related to R by

Solve: gives

Reflect: The value of G on Newtonia is within 10% of the value on earth. The mass and radius are somewhat larger

than those of earth.

m 5gR2

G5

19.06 m / s2 2 1 8.18 3 106 m 2 2

6.673 3 10211 N # m2 / kg2

5 9.08 3 1024 kg

R 5c

2p5

5.14 3 107 m

2p5 8.18 3 106 m.

g 5

1

2p

T 2

2

L 5

1

2p

2.84 s 2

2 11.85 m 2 5 9.06 m / s2.

T 5 2p Å L

gT 5 2 11.42 s 2 5 2.84 s.

G 5 6.673 3 10211 N # m2 / kg2.

c 5 2p R.g 5 G

m

R2 ,

v 5 0

K 5 34 E .K 1 U 5 E U 5 1

4 A12 kA2B 5 14 E . x 5

A

2

U 5

1

2

kx

2

E 5

1

2

kA

2

5 K 1 U .

a 5 39.2 m / s2 2 9.80 m / s

2 5 29.4 m / s2.

g 1 a 5 L 12p

T 22

5 11.55 m 2 1 2p

1.25 s 22

5 39.2 m / s2T 5 2p Å

L

g 1 a .

L 5 g 1 T

2p 22

5 19.80 m / s2 2 12.50 s

2p 22

5 1.55 m

T 5 2.50 s

a 1 9.80 m / s2;

T 5 2p Å L

g .

T E 5 2p Å 1.56 3 103 kg

6.125 3 104 N / m5 1.00 s

M 5 1 T L

2p 22

k 2 250 kg 5 11.08 s

2p 22 16.125 3 104 N / m 2 2 250 kg 5 1.56 3 103 kg.

k 5 1250 kg

2 19.80 m / s

2

24.00 3 1022 m5 6.125 3 104 N / m

T L 5 2p Å M 1 250 kg

k .T E 5 2p Å

M

k .

M 1 250 kg.T 5 2p Å m

k .

11-12 Chapter 11



11.60. Set Up: The period depends on the force constant k of the spring. At The

acceleration is related to position by Take to be downward. The amplitude of the motion is

Solve: gives

(a)

(b)

The acceleration is downward.

11.61. Set Up: The maximum acceleration of the lower block can’t exceed the maximum acceleration that can be

given to the other block by the friction force.

Solve: For block m, the maximum friction force is gives and

Then treat both blocks together and consider their simple harmonic motion. Set and

solve for A: and

Reflect: If A is larger than this the spring gives the block with mass M a larger acceleration than friction can give the

other block, and the first block accelerates out from underneath the other block.

11.62. Set Up: He first comes to rest after beginning his swing at the end of one-half of a cycle, so

Apply conservation of linear momentum to find the speed and kinetic energy of the system just after Tarzan has

grabbed the chimp. The figure in the solution to Problem 11.51 shows that the height h above the lowest point of the

swing is

Solve: (a) so

(b) The amplitude is

(c) Apply conservation of energy to find Tarzan’s speed just before he grabs the chimp:

Apply conservation of momentum to the inelastic collision between Tarzan and the chimp:

and

Apply conservation of energy to find the maximum angle of swing after the collision:

and

The length doesn’t change so remains 0.125 Hz. doesn’t depend on the mass or on the amplitude

of swing.

f f f 51

2p Å

g

L .

u 5 7.8°1 2 cos u 5 V 2

2gL5

11.70 m / s

22

2 19.80 m / s2 2 115.9 m 2

5 0.00927

12 mtot V 2 5 mtot gL 11 2 cos u 2V 5 1.70 m / s165 kg 2 12.61 m / s 2 5 165 kg 1 35 kg 2V

v 5 " 2gL 11 2 cos u 2 5 " 2 19.80 m / s2 2 115.9 m 2 11 2 cos 12° 2 5 2.61 m / s

mgL 11 2 cos u 2 5 12 mv

2.

U i 5 K f .

12°. f 51

T 5

1

8.0 s5 0.125 Hz.

L 5 g 1 T

2p 22

5 19.80 m / s2 2 1 8.0 s

2p 22

5 15.9 mT 5 2p Å L

g

h 5 L 11 2 cos u 2 .

T 5 8.0 s.

A 5m sg 1 M 1 m 2

k .m s g 5 1 k

M 1 m 2 Aamax 5 aamax 5 1 k

M 1 m 2 A.

a 5 m s g.msmg 5 magF x 5 ma x f s 5 m s n 5 m s mg.

a x 5 2

kx

m5 2

111.2 N / m 2 120.050 m 25.00 kg

5 10.112 m / s2.

vmax 5 A Å k

m5 10.100 m 2 Å

11.2 N / m

5.00 kg5 0.150 m / s

k 5 12p

T 22

m 5 1 2p

4.20 s 22 15.00 kg 2 5 11.2 N / m.

T 5 2p

Å mk

A 5 0.100 m.

1 x2kx 5 ma x.

v 5 vmax 5 A Å k

m . x 5 0,




11.63. Set Up: For steel, The elongation is a small fraction of

the length, so use to calculate

Solve: Apply to the mass when it is at the bottom of the circle. The acceleration is

and is upward, so take upward. is the tension in the

wire.

so

Reflect: When the mass hangs at rest at the end of the wire, and the wire is stretched 0.37 mm.

The radial acceleration is large and is much greater than mg.

11.64. Set Up: so the slope of the graph in part (a) depends on Young ’s modulus. is the total

load, 20 N plus the added load.

Solve: (a) The graph is given in Figure 11.64.

Figure 11.64

(b) The slope is

(c) The stress is The total load at the proportional limit is

11.65. Set Up: The height from which he jumps determines his speed at the ground. The acceleration

as he stops depends on the force exerted on his legs by the ground. In considering his motion take downward.

Solve: (a)

(b) As he is stopped by the ground, the net force on him is where is the force exerted on him by

the ground. From part (a), and

gives since the acceleration is upward.a y 5 21.19 3 103 m / s2a 5 1.19 3 103 m / s

2.F net 5 ma8.33 3 104 N.

F 5 8.4 3 104 N 2 170 kg 2 1 9.80 m / s2 2 5F ' 5 2 14.2 3 104 N 2 5 8.4 3 104 N

F 'F net 5 F ' 2 mg,

F ' 5 YA 1D l

l02 5 13.0 3 1024 m2 2 114 3 109 Pa 2 10.010 2 5 4.2 3 104 N

1 y

F '

A5 Y 1Dl

l0 2 .

stress 580 N

p 10.35 3 1023 m 2 25 2.1 3 108 Pa

60 N 1 20 N 5 80 N.F ' / A.

Y 5 1 l0

pr 2 2 12.0 3 104 N / m 2 5 1 3.50 m

p 30.35 3 1023 m 4 2 2 12.0 3 104 N / m 2 5 1.8 3 1011 Pa

60 N13.32 2 3.02 2 3 1022 m5 2.0 3 104 N / m.

3 3.2 3.4 3.6 3.8 4 4.2 4.4

l 1cm2

80757065

605550

45403530252015105

0

F 1N2

F '

F ' 5 1YA

l02 D l

F '

F ' 5 mg 5 147 N

D l 5l0 F '

AY 5

10.50 m 2 11.34 3 103 N 210.010 3 1024 m2 2 12.0 3 1011 Pa 2 5 3.35 3 1023 m 5 3.35 mm.Y 5

l0 F '

A D l

F ' 5 m 1g 1 arad 2 5 115.0 kg 2 19.80 m / s2 1 79.4 m / s

2 2 5 1.34 3 103 N.

F ' 2 mg 5 marad

.

F '1 yarad 5 r v2 5 10.50 m 2 112.6 rad / s 2 2 5 79.4 m / s2

gF y 5 ma y

arad .l0 5 0.50 m

v 5 2.00 rev / s 5 12.6 rad / s.Y 5 2.0 3 1011 Pa.

11-14 Chapter 11



gives His speed at the ground there-

fore is This speed is related to his initial height h above the floor by and

11.66. Set Up: Let t be the duration of the chemical process and let T be the period of the pendulum. Then, the

number of swings N is Let a subscript of E refer to the earth and M to the moon.

Solve: is the same on earth and moon, so and

11.67. Set Up: A force of is applied to each section of the rod. From Table 11.1,

and

Solve: (a) gives

(b) brass: steel:

(c) brass: steel:

11.68. Set Up: The foot is in rotational equilibrium. Let counterclockwise torques be positive.

Solve: (a) The free-body diagram for the foot is given in Figure 11.68, T is the tension in the tendon and A is the

force exerted on the foot by the ankle. the weight of the person.

Figure 11.68

(b) Apply with the pivot at the ankle:

2.72 times his weight.

(c) The foot pulls downward on the tendon with a force of 2000 N.

D l 5 1 F T

YA 2 l0 52000 N11470 3 106 Pa 2 178 3 1026 m2 2 125 cm 2 5 4.4 mm

T 5 1 12.5 cm

4.6 cm 2 175 kg 2 19.80 m / s2 2 5 2000 N,

T 14.6 cm 2 2 n 112.5 cm 2 5 0gt 5 0,

T

n

A

4.6cm 12.5cm

Pivot

n 5

175 kg

2g,

Y 5F T / A

Dl / l0

.

strain 51.00 3 108 Pa

2.0 3 1011 Pa5 5.0 3 1024strain 5

5.00 3 107 Pa

0.91 3 1011 Pa5 5.5 3 1024;strain 5

stress

Y .

F '

A5

1.00 3 104 N

1.00 3 1024 m25 1.00 3 108 Pa

F '

A5

1.00 3 104 N

2.00 3 1024 m25 5.00 3 107 Pa;

L 5 l0, steel 5 l0, brass 1 Y steel

Y brass2 1 Asteel

Abrass2 5 11.40 m 2 1 2.0 3 1011 Pa

0.91 3 1011 Pa 2 11.00 cm2

2.00 cm2 2 5 1.54 m

F 'l0, brass

Y brass Abrass

5F 'l

0, steel

Y steel Asteel

.D lsteel 5 D lbrassD l 5 F 'l0

YA .

Y brass 5 0.91 3 1011 Pa.2.0 3 1011 Pa

Y steel 51.00 3 104 N

N M 5 N E Å gM

gE

5 110.0 swings 2 Å 1.67 m / s

9.80 m / s2

5 4.13 swings

N E

" gE

5 N M

" gM

N E 2p Å L

gE

5 N M2p Å L

gM

.t 5 NT 5 N 2p " L / g

T 5 2p Å L

gt / T .

h 5v

2

2g5

135.7 m / s 2 2

2 19.80 m / s2 2 5 65 m.

12 mv

2 5 mghv 5 35.7 m / s.

v0 y 5 2a yt 5 121.19 3 103 m / s2 2 10.030 s 2 5 35.7 m / s.v y 5 v0 y 1 a yt




11.69. Set Up: When the ride is atrestthe tension in the rodis theweight 1900N ofthe car and occupants.When

the ride is operating, the tension in the rod is obtained by applying to a car and its occupants. The free-

body diagram is shown in Figure 11.69. The car travels in a circle of radius where l is the length of the rod

and is the angle the rod makes with the vertical. For steel,

Figure 11.69

Solve: (a)

(b) gives and

Reflect: gives and As increases increases and becomes

small. Smaller means increases, so the rods move toward the horizontal as increases.

11.70. Set Up: Let be downward. The tension in the wire equals the 980 N weight of the mass.

gives

Solve: (a)

(b)

11.71. Set Up: Density is For water, and air pressure at the surface of

the earth is

Solve: (a) On earth On Venus this same mass m of water has volume

On Venus,

so and The density increases by 0.4%.

(b) The difference is very small and not noticeable.

Reflect: The increase in pressure decreases V and increases density.

rv / rE 5 100.4%.rV 5 1004 kg / m3rE 5 1000 kg / m

3

rV 5m

V 5

m

0.9958V 05 1.004

m

V 05 1.004rE.

V 5 V 0 1 DV 5 V 0 11 2D p

B 2 5 V 0 11 291 11.01 3 105 Pa 2

2.2 3 109 Pa 2 5 0.9958V 0 .

rE 5 m / V 0 .

D p 5 92 p0 2 p0 5 91 p0 . p0 5 1.01 3 105 Pa.

B 5 2.2 3 109 Par 5m

V .DV 5 2V 0

D p

B .

Y 5l0 F

'

A D l5

14.00 m 2 1980 N 210.10 3 1024 m2 2 16.0 3 1023 m 2 5 6.5 3 1010 Pa.

f 51

2p Å

k

m5

1

2p Å

1.63 3 105 N / m

100 kg5 6.4 Hz

k 5 2

F '

x5 2

2980 N

6.0 3 1023 m5 1.63 3 105 N / m.

F x 5 2kxF '1 x

vucos u

cos uF 'vcos u 5 mg / F '.F ' cos u 5 mggF y 5 ma y

Dl 5 1 2.04 3 103 N

1900 N 2 10.18 mm 2 5 0.19 mm

F ' 5 mlv2 5 1 1900 N

9.80 m / s2 2 115.0 m 2 10.838 rad / s 2 2 5 2.04 3 103 N.

F ' sin u 5 mr v2 5 ml sin uv2gF x 5 ma x

D l 5l0 F '

YA5

115.0 m 2 11900 N 212.0 3 1011 Pa 2 18.00 3 1024 m2 2 5 1.78 3 1024 m 5 0.18 mm

y

u

x

F 'cosu

F '

sinu

mg

F '

arad5r v2

v 5 8.00 rev / min 5 0.838 rad / s.Y 5 2.0 3 1011 Pa.u

r 5 l sin u,

g FS

5 m aS

F '

F '

11-16 Chapter 11



11.72. Set Up: Constant speed means and the force he applies to her hair equals his weight.

Solve: (a)

(b) so

11.73. Set Up: The pressure increase is where w is the weight of the bricks and A is the area of the

piston.

Solve:

gives

DV 5 2

1D p

2V 0

B5 2

11.97 3 10

5

Pa

2 1250 L

29.09 3 108 Pa5 20.0542 L

D p 5 2 B

DV

V 0

D p 511420 kg 2 19.80 m / s

2 2p 10.150 m 2 2

5 1.97 3 105 Pa

pr 2w / A,

m 51196 3 106 Pa 2 13.14 3 1024 m2 2

9.80 m / s2

5 6280 kgF T

A5

mg

A5 196 3 106 Pa

D l 5 1 F T AY 2 l0 5 1mg

AY 2 l0 5

160 kg

2 19.80 m / s

2

213.14 3 1024 m2

2 1490 3 106 Pa 2 120 m

25 7.6 cm

A 5 pr 2 5 p 11.0 cm 2 2 5 3.14 cm2 5 3.14 3 1024 m2l0 5 20 m.

Y 5F T / A

Dl / l0

.a 5 0


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