PHYSICS HOMEWORK #41 ENERGY CONSERVATION
WORK & ENERGY
ANSWER KEY
WORK & ENERGY
1. A force of 25.0 Newtons is applied so as to move a 5.0 kg mass a distance of 20.0 meters.
How much work was done?
Ans. W = F × d = 25 N × 20.0 m = 50 J
2. A force of 120 N is applied to the front of a sled at an angle of 28.0 above the horizontal so
as to pull the sled a distance of 165 meters. How much work was done by the applied force?
Ans. W = F •d cos = 120 N • 165 m cos 28.0 = 17,482.36 J
3. A sled, which has a mass of 45.0 kg., is sitting on a horizontal surface. A force of 120 N is
applied to a rope attached to the front of the sled such that the angle between the front of the
sled and the horizontal is 35.0º. As a result of the application of this force the sled is pulled a
distance of 500 meters at a relatively constant speed. How much work was done to this sled
by the applied force?
Ans. W = F• d cos = 120 N • 500 m • cos 35.0 = 49,149.12 N
4. A rubber stopper, which has a mass of 38.0 grams, is being swung in a horizontal circle which
has a radius of R = 1.35 meters. The rubber stopper is measured to complete 10 revolutions in
8.25 seconds.
a. What is the speed of the rubber stopper?
Ans. v =
v =
•
v =
•
v = 10.29 m/s
b. How much force must be applied to the string in order to keep this stopper moving in this
circular path at a constant speed?
Ans. The Centripetal Force, F
c
=
•
=
•
= 29.76 N
c. How far will the stopper move during a period of 25.0 seconds?
Ans. d = vt = 10.29 m/s25 s = 257.25 m
d. How much work is done on the stopper by the force applied by the string during 25.0
seconds?
Ans. 0 J. The force is towards the center and the distance is around the perimeter of the
circle. They are at right angles to each other. For work to be done, the force has to be in
the same direction as the displacement.
5. How much work would be required to lift a 12.0 kg mass up onto a table 1.15 meters high?
Ans. W = F• d = mg d = 12 kg 9.8 m/s
2
1.15 m = 135.24 J
6. A barge is being pulled along a canal by two cables being pulled
as shown to the right. The tension in each cable is T =14,000 N
and each cable is being pulled at an angle = 18.0 relative to the
direction of motion as shown. How much work will be done in
pulling this barge a distance of 3.0 kilometers?
Ans. The component of the force along the direction of motion is F
x
= T cos
F
x
= 14,000 N cos 18 = 13,314.79 N
Since there are two such cables, the total force along the x-direction is
F
x
= 13,314.79 N 2 = 26,629.58 N
W = F d = 26,629.58 N 3000 m = 79,888,747.37 J = 79 MegaJoules
Kinetic Energy KE = ½ mv
2
7. A car, which has a mass of 1250 kg is moving with a velocity of 26.0 m/sec. What is the
kinetic energy of this car?
Ans. TKE = ½mv
2
= ½(1250 kg)(26 m/s)
2
= 422,500 J
8. What will be the kinetic energy of a bullet, which has a mass of 22.0 grams, moving with a
velocity of 650 m/sec.?
Ans. TKE = ½mv
2
= ½(0.022 kg)(650 m/s)
2
= 4647.5 J
9. How fast must a 4.40 kg bowling ball move in order to have a kinetic energy of 185 Joules?
Ans. TKE = ½mv
2
185 J = ½(4.40 kg)v
2
185 = 2.20v
2
84.09 = v
2
84.09 = v
2
9.17 m/s = v
10. A ball, which has a mass of 2.40 kg., is dropped from the top of a building 96.0 meters tall.
a. How long will it take for this ball to reach the ground?
Ans. y = ½gt
2
t =
t =
= 4.42 sec
b. What will be the velocity of the ball just as it reaches the ground?
Ans. v = gt =
4.42 sec = 43.38 m/s
c. What will be the kinetic energy of the ball just as it reaches the ground?
Ans. TKE = ½mv
2
= ½(2.40 kg)(43.38 m/s)
2
= 2258.19 J
d. How much work would be needed to lift this ball back up to the top of the building at a
constant speed?
Ans. 2258.19 J. Energy is conserved. The amount of kinetic energy is the same as the amount
of work to reset the ball to its original height.
11. A cart, which has a mass of m = 2.50 kg., is sitting at the
top of an inclined plane which is 3.30 meters long and
which meets the horizontal at an angle of = 18.5.
a. How long will it take for this cart to reach the bottom of the inclined plane?
Ans. The acceleration of a cart down an inclined plane is given by a = g sin , where g is
gravity and is the angle of the inclined plane.
a =
sin 18.5 = 3.11 m/s
2
The distance an object moves down an inclined plane when it is accelerating is given
by x = ½at
2
Solving for t: t =
t =
= 1.46 sec
b. What will be the velocity of the cart when it reaches the bottom of the incline?
Ans. The velocity of an object at the bottom of an inclined plane is
v = at = 3.11 m/s
2
1.46 s = 4.53 m/s
c. What will be the kinetic energy of the cart when it reaches the bottom of the incline?
Ans. TKE = ½mv
2
= ½(2.50 kg)(4.53 m/s)
2
= 25.66 J
Gravitational Potential Energy GPE = mgh
12. A 5.0 kg mass is initially sitting on the floor when it is lifted onto a table 1.15 meters high at
a constant speed.
a. How much work will be done in lifting this mass onto the table?
Ans. W = F d = mg d = 5.0 kg
1.15 m = 56.35 J
b. What will be the gravitational potential energy of this mass, relative to the floor, once it is
placed on the table?
Ans. 56.35 J. By Conservation of Energy, the Work had to be converted into another
form of energy. In this case, it is gravitational potential energy.
c. What was the initial gravitational potential energy, relative to the floor, of this mass while
sitting on the floor?
Ans. 0 J. If the mass had no height, it had no GPE.
13. A crate, which has a mass of 48.0 kg., is sitting at rest at the bottom
of a frictionless inclined plane which is L = 2.85 meters long and
which meets the horizontal at an angle of = 31.5. A force F is
applied so as to push the crate up this incline at a constant speed.
a. What is the magnitude of the force F required to push the crate to the top of the incline at a
constant speed?
Ans. To get the crate to go up the inclined plane, you would have to apply a
force equal and opposite to F
x
, the component of the crate’s weight
along the inclined plane.
F
x
= F
w
sin = mgsin = 48 kg
sin 31.5 = 245.78 N
18.5
31.5
F
w
F
x
F
y
b. How much work will be done in pushing the crate to the top of the incline?
Ans. W = F d = 245.78 N 2.85 m = 700.48 J
c. What is the height of this incline?
Ans. We have the hypotenuse and the angle and we are
trying to find the opposite side.
We should use
hyp sin = opp. 2.85 m sin (31.5) = opp.
1.48 m = opp.
d. What will be the GPE of this crate when it reaches the top of the incline?
Ans. GPE = 700.48 J. By Conservation of energy, Ans. b. is the same as Ans. d. But to
verify, we can find the GPE using our formula.
GPE = mgh = 48.0 kg
1.48 m = 700.48 J
14. Suppose that you have a mass of 62.0 kg and that you walk to the top of a stairway which is
h = 12.0 meters high and L= 15.0 meters deep.
a. How much work will you have to do in walking to the top of the stairway?
Ans. To get to the top of the stairs,
W = F d=mgh = 62.0 kg
12 m =7291.2 J
b. What will be your GPE when you reach the top of the stairway?
Ans. 7291.2 J. By Conservation of Energy, the work you did going up the stairs will give
you that much GPE.
Elastic Potential Energy EPE = ½ k(x)
2
and [F = kx]
15. A force of F = 35.0 N is applied to a spring and as a result the spring stretches a distance of
x = 12.0 cm.
a. What is the spring constant for this spring?
Ans. The formula for the spring constant is k =
k =
= 291.67 N/m
b. How much energy will be stored in this spring?
Ans. The energy of a spring is elastic potential energy.
EPE = ½kx
2
= ½(291.67 N/m)(0.12 m)
2
= 2.1 J
16. A spring, which has a spring constant of k = 120 N/m, is being stretched a distance of
x = 15.0 cm by a force F.
a. How much force F is being applied to this spring?
Ans. Hooke’s law (which is the formula for the spring constant is F = kx)
F = kx = 120 N/m 0.15 m = 18 N
b. How much energy will be stored in this spring?
Ans. The energy of a spring is elastic potential energy.
EPE = ½kx
2
= ½(120 N/m)(0.15 m)
2
= 1.35 J
31.5
? m
15 m
12 m
F
y
17. A spring, which has a spring constant k, is hung from the ceiling as shown
to the right. A mass m = 3.00 kg is added to the end of the spring and is
then slowly lowered until equilibrium is reached. At this point the bottom
of the mass has been lowered a distance of h = 52.0 cm.
a. What is the magnitude of the force being exerted by the spring when
the system reaches equilibrium?
Ans. The spring must be exerting a force equal and opposite to the
weight of the hanging mass for equilibrium to occur.
F
w
= mg = 3 kg
= 29.4 N
b. What is the spring constant of this spring?
Ans. The formula for the spring constant is k =
k =
= 56.54 N/m
c. How much energy is stored in the spring when equilibrium is reached?
Ans. The energy of a spring is elastic potential energy.
EPE = ½kx
2
= ½(56.54 N/m)(0.52 m)
2
= 7.64 J
18. A mass of 5.00 kg is dropped from a height of 2.20 meters above a vertical spring sitting on a
horizontal surface. Upon colliding with the spring the mass compresses the spring x = 30.0
cm before it momentarily comes to halt. [Assume h = 0 at the lowest point!]
a. How much gravitational potential energy was contained in the 5.0 kg mass before it was
dropped?
Ans. GPE = mgh = 5.00 kg
2.20 + 0.30 m = 122.5 J
b. How much energy will be stored in the spring when the mass comes briefly to a halt?
Ans. All the GPE will convert to EPE. 122.5 J
c. What is the spring constant of this spring?
Ans. The formula for the spring constant is k =
k =
k =
= 19.6 N/m
ENERGY CONSERVATION:
19. A cart, which has a mass of 2.30 kg is sitting at the top of
an inclined plane, which is 4.50 meters long and meets the
horizontal at an angle of 14.0º. The car is then allowed to
roll to the bottom of the incline;
a. What was the gravitational energy of the cart before it rolls down the incline?
Ans. GPE = mgh. We know the mass is 2.30 kg and we know gravity is
. We need h.
We have the hypotenuse and the angle and we are trying to find the opposite side, h.
We should use
hyp sin = opp. 4.50 m sin (14) = opp.
1.09 m = opp. = h
GPE = 2.30 kg
4.50 m = 101.43 J
14
h =? m
b. What will be the magnitude of the force that tends to cause the cart to accelerate down the
incline?
Ans. F
x
, the component of the crate’s weight along the inclined plane, is what makes the cart
accelerate down the plane:
F
x
= F
w
sin = mgsin = 2.30 kg
sin 14 = 5.45 N
c. What will be the acceleration of the cart as it moves down the incline?
Ans. The acceleration of a cart down an inclined plane is given by a = g sin , where g is
gravity and is the angle of the inclined plane.
a =
sin 14 = 2.37 m/s
2
d. How much time to it take for the cart to reach the bottom of the incline?
Ans. The distance an object moves down an inclined plane when it is accelerating is given
by x = ½at
2
Solving for t: t =
t =
= 1.95 sec
e. What will be the velocity of the cart as it reaches the bottom of the incline?
Ans. The velocity of an object at the bottom of an inclined plane is
v = at = 2.37 m/s
2
1.95 s = 4.62 m/s
f. What will be the kinetic energy of the cart as it reaches the bottom of the incline?
Ans. TKE = ½mv
2
= ½(2.30 kg)(4.62 m/s)
2
= 24.55 J
g. How much work was done by the gravitational force on the cart as it rolls to the bottom of
the incline?
Ans. W = 24.55 J. The gravitational force, by Conservation of Energy, must have been
equal to the Kinetic Energy gained.
20. A car is sitting at the top of an inclined plane, which is 5.2
meters long and meets the horizontal at an angle of 12.0º.
The cart is then allowed to roll to the bottom of the incline.
What will be the velocity of the cart as it reaches the
bottom of the incline?
Ans. To find the velocity at the bottom, we can use v =
but we need the height. We need h.
We have the hypotenuse and the angle and we are trying to find the opposite side, h.
We should use
hyp sin = opp. 5.20 m sin (12) = opp.
1.08 m = opp. = h
v =
=
= 4.6 m/s
12
h =? m
21. A sled, which has a mass of 45.0 kg., is sitting on a
horizontal surface. A force of 120 N is applied to a rope
attached to the front of the sled such that the angle between
the front of the sled and the horizontal is 35.0º. As a result
of the application of this force the sled is pulled a distance
of 500 meters at a relatively constant speed. How much
work was done to this sled by the applied force?
Ans. W = F d = F
x
d = F cos d = 120 N cos 35 500 m = 49,149.12 J
We are only interested in the component of the Force along the horizontal surface. This
would be F
x
(see figure at right).
22. A 25.0 kg crate is sitting at the bottom of an inclined plane. The
inclined plane is 12.0 meters long, meets the horizontal at an angle
of 15.0º and has a coefficient of sliding friction of = 0.55. A
force is applied to the crate so as to slide the crate up the incline at
a constant speed.
a. What will be the magnitude of the frictional force between the crate and the incline?
Ans. The formula for the frictional force on an inclined plane is F
f
= F
N
= F
y
= mg cos
F
f
= 0.55 25.0 kg
= 130.16 N
b. What will be the magnitude of the gravitational force component opposing the motion of
the sled up the incline?
Ans. The component of the weight of the crate along the inclined plane
is just the opposite side of the right triangle formed by F
x
, F
y
,
and F
w
(see figure at right).
F
x
= mg sin = 25.0 kg
= 63.41 N
c. How much work will be done against the gravitational force in moving the crate to the top
of the incline?
Ans. The work required against the gravitational force alone is W = F d = F
x
d
W = 63.41 N (12 m) = 760.93 J
d. What is the magnitude of the force F required to push the sled up the incline at a constant
speed?
Ans. Since the crate is moving up the plane at a constant
speed, the forces up the plane must be equal to the
forces down the plane.
F = F F
p
= F
x
+ F
f
= 63.41 N + 130.16 N
= 193.57 N
The forces down are the force of friction (F
f
) but also the component of the weight of
the crate that is along the inclined plane (F
x
).
= 35
F
x
F
y
15.0
F
w
F
x
F
y
F
w
F
x
F
y
F
f
e. How much work will be done by the applied force in pushing the mass to the top of the
incline?
Ans. The work required against the gravitational force (F
x
) AND the frictional force (F
f
)
is W = F d = 193.57 N (12 m) = 2322.84 J
f. What will be the gravitational potential energy of the crate when it reaches the top of the
incline?
Ans. GPE = mgh. We know the mass is 25 kg and we know gravity is
. We need h.
We have the hypotenuse and the angle and we are trying to find the opposite side, h.
We should use
hyp sin = opp. 12 m sin (15) = opp.
3.11 m = opp. = h
GPE = 25 kg
3.11 m = 760.93 J
g. How much work was done against the frictional force as the crate is pushed to the top of the
incline?
Ans. The work required against the frictional force alone is W = F d = F
f
d
W = 130.16 N (12 m) = 1561.92.93 J
h. How are the work done by the external force, the work done against friction and the work
done against gravity related?
Ans. Work
ext
= W
frict
+ W
grav
23. A box, which has a mass of 14.0 kg , is sliding along a horizontal
surface with a velocity of 18.0 m/sec when it encounters a
frictionless inclined plane which meets the horizontal at an angle
of 28.0. The box slides up the incline until it comes to a halt.
a. What will be the kinetic energy of the crate before it reaches the bottom of the incline?
Ans. TKE = ½mv
2
= ½(14 kg)(18 m/s)
2
= 2268 J
b. What will be the gravitational potential energy of the crate when it finally stops on the
incline?
Ans. By Conservation of energy, the GPE
out
= TKE
in
= 2268 J
c. How far up the incline will the box slide before it stops?
Ans. GPE = mgh h =
h =
= 16.53 m.
This is the height to which the box will rise.
To find how far up, we need to use S.O.H.C.A.H.T.O.A.
and in this case sin =
hyp. =
hyp. =
= 35.21 m
28.0
28.0
h =16.53 m
