Chapter 6

PHYS 101: Solutions to Chapter 6 Home Work 4.

REASONING a. The work done by the gravitational force is given by Equation 6.1 as W = (F cos θ) s. The gravitational force points downward, opposite to the upward vertical displacement of 4.60 m. Therefore, the angle θ is 180º. b. The work done by the escalator is done by the upward normal force that the escalator exerts on the man. Since the man is moving at a constant velocity, he is in equilibrium, and the net force acting on him must be zero. This means that the normal force must balance the man’s weight. Thus, the magnitude of the normal force is FN = mg, and the work that the escalator does is also given by Equation 6.1. However, since the normal force and the upward vertical displacement point in the same direction, the angle θ is 0º. SOLUTION a. According to Equation 6.1, the work done by the gravitational force is

W   F cos  s   mg cos  s   75.0 kg   9.80 m/s 2  cos180  4.60 m   3.38 103 J b. The work done by the escalator is

W   F cos  s   FN cos  s   75.0 kg   9.80 m/s 2  cos 0  4.60 m   3.38 103 J 7.

REASONING AND SOLUTION a. In both cases, the lift force L is perpendicular to the displacement of the plane, and, therefore, does no work. As shown in the drawing in the text, when the plane is in the dive, there is a component of the weight W that points in the direction of the displacement of the plane. When the plane is climbing, there is a component of the weight that points opposite to the displacement of the plane. Thus, since the thrust T is the same for both cases, the net force in the direction of the displacement is greater for the case where the plane is diving. Since the displacement s is the same in both cases, more net work is done during the dive . b. The work done during the dive is

Wclimb  (T  W cos 115) s .

Wdive  (T  W cos 75) s , while the work done during the climb is Therefore, the difference between the net work done during the dive and the climb is

Wdive – Wclimb  (T  W cos 75) s – (T  W cos 115) s  W s (cos 75 – cos 115) 4

3

7

= (5.9 10 N)(1.7 10 m)(cos 75 – cos 115) = 6.8 10 J 12. REASONING AND SOLUTION The net work done on the car is WT = WF + Wf + Wg + WN WT = Fs cos 0.0° + f s cos 180° – mgs sin 5.0° + FNs cos 90° Rearranging this result gives

F

WT s

 f  mg sin 5.0°





150  103 J =  524 N + 1200 kg  9.80 m/s 2 sin 5.0°  2.07  103 N 290 m

17.

REASONING The work done by the catapult Wcatapult is one contribution to the work done by the net external force that changes the kinetic energy of the plane. The other contribution is the work done by the thrust force of the plane’s engines Wthrust. According to the work-energy theorem (Equation 6.3), the work done by the net external force Wcatapult + Wthrust is equal to the change in the kinetic 7

energy. The change in the kinetic energy is the given kinetic energy of 4.5 × 10 J at lift-off minus the initial kinetic energy, which is zero since the plane starts at rest. The work done by the thrust force can be determined from Equation 6.1 [W = (F cos θ) s], since 5 the magnitude F of the thrust is 2.3 × 10 N and the magnitude s of the displacement is 87 m. We note that the angle θ between the thrust and the displacement is 0º, because they have the same direction. In summary, we will calculate Wcatapult from Wcatapult + Wthrust = KEf  KE0. SOLUTION According to the work-energy theorem, we have Wcatapult + Wthrust = KEf  KE0 Using Equation 6.1 and noting that KE0 = 0 J, we can write the work energy theorem as follows:

Wcatapult   F cos  s  KE f Work done by thrust

Solving for Wcatapult gives

Wcatapult  KE f   F cos  s Work done by thrust

 4.5 107 J   2.3 105 N  cos 0 87 m   2.5 107 J 19.

SSM REASONING The work done to launch either object can be found from Equation 6.3, the work-energy theorem,

1

1

W  KEf  KE0  mvf2  mv02 . 2 2 SOLUTION a. The work required to launch the hammer is





W  12 mvf2  12 mv02  12 m vf2  v02  12 (7.3 kg)  (29 m/s)2   0 m/s    3.1103 J 

2





b. Similarly, the work required to launch the bullet is





W  12 m vf2  v02  12 (0.0026 kg)  (410 m/s)2   0 m/s    2.2  102 J 27.



2





REASONING AND SOLUTION The net work done on the plane can be found from the work-energy theorem: 2 2 2 2 W  12 mvf  12 mv0  12 m vf  v0





The tension in the guideline provides the centripetal force required to keep the plane moving in a circular path. Since the tension in the guideline becomes four times greater,

Tf = 4 T0 or

mvf2 rf

PHYS 101: Solutions to Chapter 6 Home Work

 4

mv02 r0

pg. 2/7

2 Solving for vf gives 2

2 vf



4v0 rf r0

Substitution of this expression into the work-energy theorem gives

W

 4v 2 r

1 m 2

0 f

 r0



 v02  



1 mv 2 0 2

    rf 4  r   0

    1    

Therefore,

 1 W = (0.90 kg)(22 m/s) 2  4 2  30.

  14 m  2  1  = 5.4 ×10 J   16 m   

REASONING At a height h above the ground, the gravitational potential energy of each clown is given by PE  mgh (Equation 6.5). At a height hJ, Juggle’s potential energy is, therefore, PEJ = mJghJ, where mJ is Juggles’ mass. Similarly, Bangles’ potential energy can be expressed in terms of his mass mB and height hB: PEB = mBghB. We will use the fact that these two amounts of potential energy are equal (PEJ = PEB) to find Juggles’ mass mJ.

SOLUTION Setting the gravitational potential energies of both clowns equal, we solve for Juggles’ mass and obtain

mJ g hJ  mB g hB PE

33.

J

PE

mJ 

or

mBhB  86 kg  2.5 m    65 kg hJ  3.3 m 

B

REASONING The work done by the weight of the basketball is given by Equation 6.1 as

W   F cos   s , where F = mg is the magnitude of the weight, is the angle between the s

weight and the displacement, and s is the magnitude of the displacement. The drawing shows that the weight and displacement are parallel, so that  = 0. The potential energy of the basketball is given by Equation 6.5 as PE = mgh, where h is the height of the ball above the ground.

mg SOLUTION a. The work done by the weight of the basketball is

W   F cos   s = mg (cos 0)(h0  hf) = (0.60 kg)(9.80 m/s2)(6.1 m  1.5 m) = 27 J b. The potential energy of the ball, relative to the ground, when it is released is 2

PE0 = mgh0 = (0.60 kg)(9.80 m/s )(6.1 m) = 36 J

(6.5)

c. The potential energy of the ball, relative to the ground, when it is caught is 2

PEf = mghf = (0.60 kg)(9.80 m/s )(1.5 m) = 8.8 J


(6.5)

pg. 3/7

d.

The change in the ball’s gravitational potential energy is

PE = PEf  PE0 = 8.8 J – 36 J = 27 J We see that the change in the gravitational potential energy is equal to –27 J =

W

, where W is the work done by the weight

of the ball (see part a). 36.

REASONING The girl’s gravitational potential energy the expense of her kinetic energy

KE  12 mv 2

PE  mgh

(Equation 6.5) increases as she rises. This increase comes at

(Equation 6.2), which decreases as she rises. Because air resistance is negligible,

all of the kinetic energy she loses is transformed into potential energy, so that her total mechanical energy constant. We will use this principle to determine the distance she rises during this interval.

E  KE  PE

remains

SOLUTION The conservation principle gives

KEf  PEf  KE0  PE0 Ef

PEf  PE0  KE0  KEf

or

mg  hf  h0   KE0  KEf

or

E0

Solving this expression for hf  h0 , which is the distance she rises, we obtain

hf  h0 

45.

KE0  KEf mg



440 J  210 J

 35 kg   9.80 m/s2 

 0.67 m

SSM REASONING Friction and air resistance are being ignored. The normal force from the slide is perpendicular to the motion, so it does no work. Thus, no net work is done by nonconservative forces, and the principle of conservation of mechanical energy applies. SOLUTION have

Applying the principle of conservation of mechanical energy to the swimmer at the top and the bottom of the slide, we

1 mv2f  mghf 2 Ef

1 2

 mv20  mgh0

If we let h be the height of the bottom of the slide above the water,

v0  0

E0

hf  h , and h0  H .

Since the swimmer starts from rest,

m/s, and the above expression becomes

1 2 v 2 f

 gh  gH

Solving for H, we obtain

vf2 H  h 2g Before we can calculate H, we must find vf and h. Since the velocity in the horizontal direction is constant,

vf 

x 5.00 m  10.0 m/s t 0.500 s

The vertical displacement of the swimmer after leaving the slide is, from Equation 3.5b (with down being negative),

y  2 ayt 2  2 (–9.80 m/s2 )(0.500 s)2  1.23 m 1


1

pg. 4/7

Therefore, h = 1.23 m. Using these values of vf and h in the above expression for H, we find

vf2 (10.0 m/s )2 H  h  1.23 m   6.33 m 2g 2(9.80 m/s2 ) 56.

REASONING According to Equation 6.6, the work Wnc done by nonconservative forces, such as kinetic friction and air resistance,

2 is equal to the object’s change in kinetic energy, 1 mvf 2

 12 mv02 Wnc 

plus the change in its potential energy,

mghf  mgh0 :

 12 mvf2  12 mv02    mghf  mgh0 

(6.6)

This relation will be used in both parts (a) and (b) of the problem. SOLUTION a. Since nonconservative forces are absent, the work done by them is zero, so Wnc = 0 J. In this case, we can algebraically rearrange Equation 6.6 to show that the initial total mechanical energy (kinetic energy plus potential energy) E0 at the top is equal to the final total mechanical energy Ef at the bottom: 1 mv 2 f 2

 mghf 

1 mv 2 0 2

Ef

 mgh0

E0

Solving for the final speed vf of the skeleton and rider gives

vf  v02  2 g  h0  hf  The initial speed of the rider is v0 = 0 m/s, and the drop in height is

h0  hf  104 m , so the final speed at the bottom of the run is

2 vf  v02  2 g  h0  hf    0 m/s   2  9.80 m/s 2  104 m   45.1 m/s b. The work Wnc done by the nonconservative forces follows directly from Equation 6.6:





Wnc  12 m vf2  v02  mg  hf  h0  

1 2

86.0 kg   35.8 m/s 2   0 m/s 2   86.0 kg   9.80 m/s 2   104 m  

Note that the difference in heights,

3.25 104 J

hf  h0  104 m , is a negative number because the final height hf is less than the initial

height h0. 58.

REASONING AND SOLUTION According to the work-energy theorem as given in Equation 6.8, we have 1 1 2 2 Wnc  mvf  mghf  mv0  mgh0 2 2

 





The metal piece starts at rest and is at rest just as it barely strikes the bell, so that vf  v0  0 m/s. In addition, hf  h and 2 h0  0 m, while Wnc  0.25 12 M v , where M and v are the mass and speed of the hammer. Thus, the work-energy theorem





becomes

0.25 PHYS 101: Solutions to Chapter 6 Home Work



1 M v2 2

  mgh pg. 5/7

Solving for the speed of the hammer, we find

2mgh 2(0.400 kg)(9.80 m/s2 )(5.00 m) v    4.17 m/s 0.25M 0.25 (9.00 kg) 63.

SSM REASONING AND SOLUTION One

kilowatt houris the amount of work or energy generated when one kilowatt of

power is supplied for a time of one hour. From Equation 6.10a, we know that

W  Pt .

Using the fact that

3

1 kW = 1.0 10 J/s and that 1h = 3600 s, we have 3

3

6

1.0 kWh = (1.0 10 J/s)(1 h)= (1.0 10 J/s)(3600 s)= 3.6 10 J

65.

REASONING The average power is given by Equation 6.10b ( P

in energy  ChangeTime ).

The time is given. Since the road is level, there

is no change in the gravitational potential energy, and the change in energy refers only to the kinetic energy. According to Equation 6.2, the kinetic energy is

1 2

mv 2 . The speed v is given, but the mass m is not. However, we can obtain the mass from the given

weights, since the weight is mg. SOLUTION a. Using Equations 6.10b and 6.2, we find that the average power is 2 2 Change in energy KEf  KE0 12 mvf  12 mv0 P   Time Time Time

Since the car starts from rest, v0 = 0 m/s, and since the weight is W = mg, the mass is m = W/g. Therefore, the average power is

P

1 2

mvf2  12 mv02 Time

Wvf2  2 g  Time 

(1)

Using the given values for the weight, final speed, and the time, we find that

9.0 103 N   20.0 m/s  Wvf2  P   3.3 104 W (44 hp) 2 2 g  Time  2  9.80 m/s   5.6 s  2

b. From Equation (1) it follows that

1.4 104 N   20.0 m/s  Wvf2  P   5.1104 W (68 hp) 2 2 g  Time  2  9.80 m/s   5.6 s  2


pg. 6/7

66.

REASONING The work W done is given by Equation 6.1 as

W   F cos   s , where F is the magnitude of the force that you

exert on the rowing bar, s is the magnitude of the bar’s displacement, and  is the angle between the force and the displacement. Solving this equation for F gives

F The work is also equal to the average power

P

W  cos   s

(6.1)

multiplied by the time t, or

W  Pt

(6.10a)

Substituting Equation (6.10a) into (6.1) gives

F

W Pt   cos   s  cos   s

SOLUTION The angle  is 0, since the direction of the pulling force is the same as the displacement of the rowing bar. Thus, the magnitude of the force exerted on the handle is 

F

67.

 82 W 1.5 s  Pt   1.0 102 N  cos   s  cos 0 1.2 m 

 REASONING The average power developed by the cheetah is equal to the work done by the cheetah divided by the elapsed time (Equation 6.10a). The work, on the other hand, can be related to the change in the cheetah’s kinetic energy by the work-energy theorem, Equation 6.3. SOLUTION a. The average power is

P

W t

(6.10a)

where W is the work done by the cheetah. This work is related to the change in the cheetah’s kinetic energy by Equation 6.3,

W  12 mvf2  12 mv02 , so the average power is 2 2 W 12 mvf  12 mv0 P  t t 1 110 kg 27 m / s 2  1 110 kg 0 m / s 2    2    2  1.0  104 W 4.0 s b. The power, in units of horsepower (hp), is





 1 hp  P  1.0  104 W    13 hp  745.7 W 


pg. 7/7