Download Energy 1 Answers

Energy
Problem Set 1
Kinetic Energy and Work
1.
Name: _______________________________
If a man lifts a 20.0-kg bucket from a well and does 6.00 kJ of work, how deep is the well? Assume that
the speed of the bucket remains constant as it is lifted. (5. 2)
Since the bucket is moving upward at a constant velocity, the upward force of the man on the bucket must be equal in
magnitude to the weight of the bucket.
The weight of the bucket is mg.
The formula for work is Work  Fd cos .
Rearranging this we get
d
Work
6000 J

 30.6m
F cos  (20.0kg )(9.81 m s 2 )
2.
A shopper in a supermarket pushes a cart with a force of 35 N directed at an angle of 25° downward
from the horizontal. Find the work done by the shopper as she moves down a 50-m length of aisle. (5. 4)
N
F
W   F cos  s  35 N  cos25  50 m   1.6  10 J 1.6 kJ
3
W
3.
Starting from rest, a 5.00-kg block slides 2.50 m down a rough 30.0° incline. The coefficient of kinetic
friction between the block and the incline is μk = 0.436. Determine (a) the work done by the force of gravity, (b)
the work done by the friction force between block and incline, and (c) the work done by the normal force. (5.
5)


(a) The force of gravity is given by m g   5.00 kg 9.80 m s2  49.0 N and is directed downwards. The
angle between the force of gravity and the direction of motion is  = 90.0° - 30.0° = 60.0°, and so the
work done by gravity is given as
W g   F cos  s  49.0 N  cos60.0  2.50 m   61.3 J
(b) The normal force exerted on the block by the incline is n  m g cos30.0 , so the friction force is
fk  kn   0.436 49.0 N  cos30.0  18.5 N
This force is directed opposite to the displacement (that is  = 180°), and the work it does is
W f   fk cos  s 18.5 N  cos180  2.50 m    46.3 J
(c) Since the normal force is perpendicular to the displacement;   90, cos  0 , and the work done
by the normal force is zero .
4.
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. (a) What is the tension in the rope? (b)
How much work is done by the rope on the sledge? (c) What is the mechanical energy lost due to friction? (5.
7)
Fy  F sin  n  m g  0
(a)
n  m g  F sin 
Fx  F cos  kn  0
n

F cos
k
m g  F sin 
F cos
k
 0.500 18.0 kg  9.80 m s 
km g
=
= 79.4 N
k sin  cos  0.500 sin 20.0  cos20.0
2
F=
(b) W F   F cos  s  79.4 N  cos20.0  20.0 m   1.49  103 J 1.49 kJ
fk  F cos  74.6 N
(c)
W
f
=  fk cos  s= 74.6 N  cos180  20.0 m   1.49  103 J  1.49 kJ
Section 5.2
5.
Kinetic Energy and the Work–Energy Theorem
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? (5. 10)


2
1
1
2.45  103 kg v2  7.00 kg 3.00 m s , giving v  160 m s
2
2
6.
A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.50 m/s. The pulling force
is 100 N parallel to the incline, which makes an angle of 20.0° with the horizontal. The coefficient of kinetic
friction is 0.400, and the crate is pulled 5.00 m. (a) How much work is done by gravity? (b) How much
mechanical energy is lost due to friction? (c) How much work is done by the 100-N force? (d) What is the
change in kinetic energy of the crate? (e) What is the speed of the crate after being pulled 5.00 m? (5. 12)
(a)
W g   m g cos  s  m g cos 90.0     s
W g   10.0 kg   9.80 m s2  cos110  5.00 m

  168 J
(b) W f   fk cos  s  kn cos180 s
 0.400 98.0 N cos20.0  cos180  5.00 m

  184 J
(c)
W F   F cos  s 100 N  cos0  5.00 m   500 J
(d) KE  W net  W g  W f  W F  148 J
(e)
KE 
v
1 2 1 2
m v  m vi
2
2
2 KE
 vi2 
m
2148 J
2
 1.50 m s  5.64 m s
10.0 kg
7.
A 70-kg base runner begins his slide into second base when he is moving at a speed of 4.0 m/s. The
coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches
the base. (a) How much mechanical energy is lost due to friction acting on the runner? (b) How far does he
slide? (5. 13)
(a) We use the work-energy theorem to find the work.
W  KE 
2
1 2 1 2
1
m vf  m vi  0  70 kg 4.0 m s   5.6  102 J
2
2
2
(b) W   F cos  s  fk cos180 s   kn s   k m g s,
so


5.6  102 J
W
s 

 1.2 m
k m g
 0.70 70 kg 9.80 m s2


8.
A 2.0-g bullet leaves the barrel of a gun at a speed of 300 m/s. (a) Find its kinetic energy. (b) Find the
average force exerted by the expanding gases on the bullet as it moves the length of the 50-cm-long barrel. (5.
15)
(a) The final kinetic energy of the bullet is
KE f 


2
1 2 1
m v  2.0  103 kg  300 m s  90 J
2
2
(b) We know that W  KE , and also W   Fav cos  s.
Thus, Fav 
KE
90 J 0

 1.8  102 N
scos  0.50 m  cos0
9.
A 0.60-kg particle has a speed of 2.0 m/s at point A and a kinetic energy of 7.5 J at point B. What is (a)
its kinetic energy at A? (b) its speed at point B? (c) the total work done on the particle as it moves from A to B?
(5. 16)
(a)
KEA 
(b) KEB 
2
1 2 1
m vA   0.60 kg 2.0 m s  1.2 J
2
2
1 2
m vB , so vB 
2
2 KEB 
m

2 7.5 J
 5.0 m s
0.60 kg
(c) W net  KE  KEB  KEA   7.5  1.2 J 6.3 J
10. A 2 000-kg car moves down a level highway under the actions of two forces: a 1 000-N forward force
exerted on the drive wheels by the road and a 950-N resistive force. Use the work–energy theorem to find the
speed of the car after it has moved a distance of 20 m, assuming that it starts from rest. (5. 17)
W net   Froad cos1  s  Fresist cos2  s 1000 N  cos0 s  950 N  cos180 s
W net   1000 N  950 N
  20 m   1.0 103 J
Also, W net  KE f  KEi 
v
2W net

m
1 2
m v  0 , so
2


2 1.0  103 J
2000 kg
1.0 m s