PHYS-1401: College Physics-I
CRN 55178
HOMEWORK PROBLEMS
Khalid Bukhari
HW-5
Chapter 5: ENERGY
PART-A: Hand in your answers in class on scantron on Monday 04 October-2010
1. A worker pushes a wheelbarrow 5.0 m along a level surface, exerting a constant horizontal force of
50.0 N. If a frictional force of 43 N acts on the wheelbarrow in a direction opposite to that of the worker,
what net work is done on the wheelbarrow? (a) 250 J (b) 215 J (c) 35 J (d) 15 J (e) 45 J
Wnet  Wapplied  W friction   F cos 0  x   f cos 180  x
force
  F  f  x   50.0 N  43 N   5.0 m   35 J
2. A skier leaves a ski jump at 15.0 m/s at some angle θ. At what speed is he traveling at his maximum
height of 4.50 m above the level of the end of the ski jump? (Neglect air friction.) (a) 11.7 m/s (b) 16.3
m/s (c) 12.2 m/s (d) 8.55 m/s (e) 17.4 m/s
In the absence of any air resistance, the work done by nonconservative forces is zero. The work–energy theorem then states
that KE f  PE f  KEi  PEi ,which becomes
1
1
m v2f  m gy f 
m vi2  m gyi
2
2
or
vf 

vi2  2 g yi  y f

Choosing the initial point to be where the skier leaves end of the jump and the final point where he reaches
maximum height, this yields
15.0
vf 

m s   2 9.80 m s2
2
   4.50 m 
 11.7 m s
making (a) the correct answer.
3. A 40.0-N crate starting at rest slides down a rough 6.00-m-long ramp, inclined at 30.0° with the
horizontal. The magnitude of the force of friction between the crate and the ramp is 6.0 N. What is the
speed of the crate at the bottom of the incline? (a) 1.60 m/s (b) 3.32 m/s (c) 4.5 m/s (d) 6.42 m/s (e) 7.75
m/s
The mass of the crate is


m  w g   40.0 N  9.80 m s2  4.08 kg
and we may write the work–energy theorem as
Wnet  Wfric  Wgrav  KE f  KEi
Since the crate starts from rest, KEi 
1
2
m vi2  0,
and we are left with
KE f  Wfric  Wgrav 
 fk cos 180 
 x   w cos 60.0  x
so
KE f   6.00 N   6.00 m    40.0 N  cos 60.0  6.00 m   36.0 J  120 J  84.0 J
and
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PHYS-1401: College Physics-I
vf 
2 KE f
m
CRN 55178

2  84.0 J 
4.08 kg
Khalid Bukhari
HW-5
 6.42 m s
making choice (d) the correct response.
4. What average mechanical power must be delivered by the muscles of a 70.0-kg mountain climber
who climbs a summit of height 325 m in 95.0 min? Note: Due to inefficiencies in converting chemical
energy to mechanical energy, the amount calculated here is only a fraction of the power that must be
produced by the climber’s body. See Chapter 12. (a) 39.1 W (b) 54.6 W (c) 25.5 W (d) 67.0 W (e) 88.4
W
We assume the climber has negligible speed at both the beginning and the end of the climb. Then KE f  KEi  0 , and the
work done by the muscles is





Wnc  0  PE f  PEi  mg y f  yi   70.0 kg  9.80 m s2
  325 m   2.23  105 J
The average power delivered is
P 
Wnc
2.23  105 J

 39.1 W
t
95.0 min  60 s 1 min 
and the correct answer is choice (a).
5. The work required to accelerate an object on a frictionless surface from a speed v to a speed 2v is (a)
equal to the work required to accelerate the object from v = 0 to v, (b) twice the work required to
accelerate the object from v = 0 to v, (c) three times the work required to accelerate the object from v = 0
to v, (d) four times the work required to accelerate the object from 2v to 3v, or (e) not known without
knowledge of the acceleration.
The net work needed to accelerate the object from  = 0 to  is
W1  KE f  KEi 
1
2
m v2 
1
2
m  0 
2
1
2
m v2
The work required to accelerate the object from speed  to speed 2 is
W2  KE f  KEi 
1
2
m  2v  
2
1
2
m v2 
1
2


m 4 v2  v2  3

1
2

mv2  3W1
Thus, the correct choice is (c).
6. Alex and John are loading identical cabinets onto a truck. Alex lifts his cabinet straight up from the
ground to the bed of the truck, whereas John slides his cabinet up a rough ramp to the truck. Which
statement is correct? (a) Alex and John do the same amount of work. (b) Alex does more work than
John. (c) John does more work than Alex. (d) None of these statements is necessarily true because the
force of friction is unknown. (e) None of these statements is necessarily true because the angle of the
incline is unknown.
Assuming that the cabinet has negligible speed during the operation, all of the work Alex does is used increasing the
gravitational potential energy of the cabinet. However, in addition to increasing the gravitational potential energy
of the cabinet by the same amount as Alex did, John must do work overcoming the friction between the cabinet and
ramp. This means that the total work done by John is greater than that done by Alex and the correct answer is (c).
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PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-5
7. Mark and David are loading identical cement blocks onto David’s pickup truck. Mark lifts his block
straight up from the ground to the truck, whereas David slides his block up a ramp on massless,
frictionless rollers. Which statement is true? (a) Mark does more work than David. (b) Mark and David
do the same amount of work. (c) David does more work than Mark. (d) None of these statements is
necessarily true because the angle of the incline is unknown. (e) None of these statements is necessarily
true because the mass of one block is not given.
Since the rollers on the ramp used by David were frictionless, he did not do any work overcoming nonconservative forces as
he slid the block up the ramp. Neglecting any change in kinetic energy of the block (either because the speed was
constant or was essentially zero during the lifting process), the work done by either Mark and David equals the
increase in the gravitational potential energy of the block as it is lifted from the ground to the truck bed. Because
they lift identical blocks through the same vertical distance, they do equal amounts of work and the correct choice
is (b).
8. An athlete jumping vertically on a trampoline leaves the surface with a velocity of 8.5 m/s upward.
What maximum height does she reach? (a) 13 m (b) 2.3 m (c) 3.7 m (d) 0.27 m (e) The answer can’t be
determined because the mass of the athlete isn’t given.
Once the athlete leaves the surface of the trampoline, only a conservative force (her weight) acts on her. Therefore, her total
mechanical energy is constant during her flight, or KE f  PE f  KEi  PEi . Taking the y = 0 at the surface of
the trampoline, PEi  mgyi  0 . Also, her speed when she reaches maximum height is zero, or KE f  0 . This
leaves us with PE f  KEi , or m gymax 
v2
 8.5 m s 
 i 
2g
2 9.80 m s2
1
2
m vi2 , which gives the maximum height as
2
ymax


 3.7 m
making (c) the correct choice.
9. A certain truck has twice the mass of a car. Both are moving at the same speed. If the kinetic energy
of the truck is K, what is the kinetic energy of the car? (a) K/4 (b) K/2 (c) 0.71K (d) K (e) 2K
KEcar 
1
2
mcar v2 
1
2
 mtruck 2  v2

1
2

1
2

mtruck v2  KEtruck 2 , so (b) is the correct answer.
10. If the speed of a particle is doubled, what happens to its kinetic energy? (a) It becomes four times
larger. (b) It becomes two times larger. (c) It becomes √2 times larger. (d) It is unchanged. (e) It
becomes half as large.
11. If the net work done on a particle is zero, which of the following statements must be true? (a) The
velocity is zero. (b) The velocity is decreased. (c) The velocity is unchanged. (d) The speed is
unchanged. (e) More information is needed.
3
PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
The work–energy theorem states that Wnet  KE f  KEi . Thus, if Wnet  0 , then KE f  KEi or

HW-5
1
2
m v2f 
1
2
m vi2 , which

leads to the conclusion that the speed is unchanged v f  vi . The velocity of the particle involves both
magnitude (speed) and direction. The work–energy theorem shows that the magnitude or speed is unchanged when
Wnet  0 , but makes no statement about the direction of the velocity. Therefore, choice (d) is correct but choice
(c) is not necessarily true.
12. A block of mass m is dropped from the fourth floor of an office building, subsequently hitting the
sidewalk at speed v. From what floor should the mass be dropped to double that impact speed? (a) the
sixth floor (b) the eighth floor (c) the tenth floor (d) the twelfth floor (e) the sixteenth floor
As the block falls freely, only the conservative gravitational force acts on it. Therefore, mechanical energy is conserved,
or. KE f  PE f  KEi  PEi Assuming that the block is released from rest (KEi  0) , and taking y = 0 at
ground level (PE f  0) , we have that
KE f  PEi
or
1
m v2f  mgyi
2
and
yi 
v2f
2g
Thus, to double the final speed, it is necessary to increase the initial height by a factor of four, or the correct choice
for this question is (e).
13. A car accelerates uniformly from rest. When does the car require the greatest power? (a) when the
car first accelerates from rest (b) just as the car reaches its maximum speed (c) when the car reaches half
its maximum speed (d) The question is misleading because the power required is constant. (e) More
information is needed.
If the car is to have uniform acceleration, a constant net force F must act on it. Since the instantaneous power delivered to the
car is P  F v , we see that maximum power is required just as the car reaches its maximum speed. The correct
answer is (b).
14. A sledge loaded with bricks has a total mass of 18.0 kg and is pulled at constant speed by a rope
inclined at 20.0° above the horizontal. The sledge moves a distance of 20.0 m on a horizontal surface.
The coefficient of kinetic friction between the sledge and surface is 0.500. What is the tension in the
rope?
a)
59.4 N
b)
69.4 N
c)
79.4 N
d)
89.4 N
(a)
Fy  F sin   n  mg  0
n  mg  F sin 
Fx  F cos   k n  0
n 
F cos 
k
4
PHYS-1401: College Physics-I
mg  F sin  

F=
CRN 55178
Khalid Bukhari
HW-5
F cos 
k
 0.500  18.0 kg   9.80 m s2 
k mg
=
= 79.4 N
k sin   cos   0.500  sin 20.0  cos 20.0 
15. A mechanic pushes a 2.50 X 103-kg car from rest to a speed of v, doing 5 000 J of work in the
process. During this time, the car moves 25.0 m. Neglecting friction between car and road, find the
horizontal force exerted on the car.
a)
100 N
b)
150 N
c)
200 N
d)
250 N
(a)
The work–energy theorem, Wnet  KE f  KEi , gives
5000 J 
(b)
1
2 .50  103 kg v2  0 , or
2


v  2.00 m s

W   F cos   s   F cos 0  25.0 m   5000 J , so F  200 N
16. A 7.00-kg bowling ball moves at 3.00 m/s. How fast must a 2.45-g Ping-Pong ball move so that the
two balls have the same kinetic energy?
a)
80 m/s
b)
100 m/s
c)
120 m/s
d)
140 m/s
e)
160 m/s
Requiring that KEping pong  KEbowling with KE 
1
2
mv2 , we have
1
1
2
2 .45  10 3 kg v2   7.00 kg   3.00 m s 
2
2


giving v  160 m s .
17. A child and a sled with a combined mass of 50.0 kg slide down a frictionless slope. If the sled starts
from rest and has a speed of 3.00 m/s at the bottom, what is the height of the hill?
a)
0.459 m
b)
0.487 m
c)
0.919 m
d)
0.989 m
Since no nonconservative forces do work, we use conservation of mechanical energy, with the zero of potential energy
selected at the level of the base of the hill. Then,
yi 
v2f  vi2
2g

 3.00

m s  0
1
2
mv2f  mgy f 
1
2
mvi2  mgyi with y f  0 yields
2
2 9.80 m s2

 0.459 m
5
PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-5
18. A 2.1 X 103-kg car starts from rest at the top of a 5.0-m long driveway that is inclined at 20° with the
horizontal. If an average friction force of 4.0 X 103 N impedes the motion, find the speed of the car at
the bottom of the driveway.
a)
2.8 m/s
b)
3.2 m/s
c)
3.5 m/s
d)
3.8 m/s
5.45
Choose PEg = 0 at the level of the bottom of the driveway.

Then Wnc  KE  PEg
 f cos180°  s
   KE  PE 
g i
f
becomes
1

  mv2f  0    0  mg  s sin 20°  
2

Solving for the final speed gives
vf 
 2gs  sin 20
vf 
2 9.80 m s2
2f s
m
or

  5.0 m  sin 20


2 4.0  103 N  5.0 m 
2.10 
103
kg
 3.8 m s
PART-B: Hand in your solutions to the following questions in class, on Monday 04 October-2010.
Show the detailed calculations. Write your final answers in the box.
23. A 2 300-kg pile driver is used to drive a steel beam into the ground. The pile driver falls 7.50 m
before coming into contact with the top of the beam, and it drives the beam 18.0 cm farther into the
ground as it comes to rest. Using energy considerations, calculate the average force the beam exerts
on the pile driver while the pile driver is brought to rest.
The work the beam does on the pile driver is given by
Wnc   F cos180  x  F  0.180 m 
Here, the factor cos180 is included because the force F exerted on the driver by the beam is directed upward, but
the  x  18.0 cm  0.180 m displacement undergone by the driver while in contact with the beam is directed
downward.
From the work–energy theorem, this work can also be expressed as

 

Wnc  KE f  KEi  PE f  PEi 



1
m v2f  vi2  mg y f  yi
2

Choosing y = 0 at the level where the pile driver first contacts the top of the beam, the driver starts from rest
 vi


 0  at yi  7.50 m and comes to rest again at v f  0 at y f   0.180 m . Therefore, we have
6
PHYS-1401: College Physics-I
 F  0.180 m  
yielding
F 
CRN 55178
Khalid Bukhari
1
m  0  0    2 300 kg  9.80 m s 2
2

HW-5
   0.180 m  7.50 m 
1.73  105 J
 9.62  105 N directed upward
 0.180 m
60. An object of mass 3.00 kg is subject to a force Fx that varies with position as in Figure P5.60. Find
the work done by the force on the object as it moves (a) from x = 0 to x = 5.00 m, (b) from x = 5.00
m to x = 10.0 m, and (c) from x = 10.0 m to x = 15.0 m. (d) If the object has a speed of 0.500 m/s at x
= 0, find its speed at x = 5.00 m and its speed at x = 15.0 m.
The work done by a force equals the area under
the force versus displacement curve.
(a)
For the region 0  x  5.00 m ,
W0 to 5 
(b)
 3.00 N   5.00 m 
2
 7.50 J
For the region 5.00 m  x  10.0 m ,
W5 to 10   3.00 N   5.00 m   15.0 J
(b) For the region 10.0 m  x  15.0 m , W10 to 15 
(c) KE
x xf
 KE
x0
 3.00 N   5.00 m 
2
 7.50 J
 W0 to x f  area under F vs. x curve from x  0 m to x  x f , or
1
1
mv2f  mv02  W0 to x f
2
2
giving
vf 
 2
v02    W0 to x f
 m
For x f  5.00 m :
vf 
 2
v02    W0 to 5 
 m
 0.500

2 
2
m s  
 7.50 J   2.29 m s
 3.00 kg 
For x f  15.0 m :
7
PHYS-1401: College Physics-I
vf 
 2
v02    W0 to 15 
 m
CRN 55178
Khalid Bukhari
HW-5
 2
v02     W0 to 5  W5 to 10  W10 to 15 
 m
or
vf 
 0.500

2 
2
m s  
 7.50 J  15.0 J  7.50 J   4.50 m s
 3.00 kg 
89. Three objects with masses m1 = 5.0 kg, m2 = 10 kg, and m3 = 15 kg, respectively, are attached by
strings over frictionless pulleys as indicated in Figure P5.89. The horizontal surface exerts a force of
friction of 30 N on m2. If the system is released from rest, use energy concepts to find the speed of
m3 after it moves down 4.0 m.
(Kf – Ki) + ( Pf – Pi) = -fd
[ 0.5(m1+m2+m3)Vf2 – 0 ] + [m1g4 +m2x0 –m3g4] = -30x4
….
V = 4.26 m/s
8