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POTENTIAL AND KINETIC ENERGY
HOMEWORK SOLUTIONS
January 17, 2017
PROBLEM 1
A car travels at constant speed up a hill from A to B, as shown in the diagram below.
B
v
A
As the car travels from A to B, its gravitational potential energy
a. Remains the same and its kinetic energy remains the same
b. Remains the same and its kinetic energy decreases
c. Increases and its kinetic energy decreases
d. Increases and its kinetic energy remains the same
PROBLEM 1
A car travels at constant speed up a hill from A to B, as shown in the diagram below.
B
v
A
As the car travels from A to B, its gravitational potential energy
a. Remains the same and its kinetic energy remains the same
b. Remains the same and its kinetic energy decreases
c. Increases and its kinetic energy decreases
d. Increases and its kinetic energy remains the same
The correct answer is “d”.
The increase in the elevation of the car as it moves from A to B, increases the gravitational
potential energy (the “h” in PE = mgh increases).
Further, the problem states the cars speed remains the same. The formula for kinetic
energy is KE = ½ mv2. Because the speed of the car remains constant, the kinetic energy
remains constant.
PROBLEM 2
The diagram below shows a 1.5 kilogram kitten jumping from the top of a 1.8 meter
high refrigerator to a 0.9 meter high counter.
180 cm
90 cm
Compared to the kitten’s gravitational potential energy on top of the refrigerator, to
the kitten’s gravitational potential energy on top of the counter is:
a. Half as great
b. Twice as great
c. One-forth as great
d. Four times as great
PROBLEM 2
The diagram below shows a 1.5 kilogram kitten jumping from the top of a 1.8 meter
high refrigerator to a 0.9 meter high counter.
180 cm
90 cm
Compared to the kitten’s gravitational potential energy on top of the refrigerator, the
kitten’s gravitational potential energy on top of the counter is:
The correct answer is “a”.
a. Half as great
Gravitational potential energy is represented by PE = mgh.
b. Twice as great
Prior to the jump, the kitten was at an elevation of 180 cm
c. One-forth as great
relative to the floor. After the jump, the kitten was 90 cm
d. Four times as great
relative to the floor, losing half it’s height, and therefore half
it’s potential energy
B)
Potential
Energy
Potential
Energy
Height
C)
Height
Potential
Energy
Potential
Energy
PROBLEM 3
Which graph best represents the relationship between the gravitational potential
energy of an object near the surface of the earth and it’s height above the surface?
(Analyze as if you did not leave the Earth’s atmosphere.)
A)
B)
Height
Height
Height
Potential
Energy
B)
C)
Height
Potential
Energy
B)
Height
Potential
Energy
A)
Potential
Energy
PROBLEM 3
Which graph best represents the relationship between the gravitational potential energy of a
(Analyze as if you did not leave the Earth’s atmosphere.)
Height
“B” is the correct answer.
The equation for the gravitational
potential energy of an object near the
surface of the Earth is:
PE = mgh
As both “m” and “g” are constants near
the surface of the Earth, and “h” is to the
first power, PE v.s. h is a linear function.
Note: At significantly higher elevations,
“g” changes as a function of distance
from the center of the earth. As such, PE
would not be a linear function.
PROBLEM 4
While riding a chairlift, a 70-kilogram skier is raised a vertical distance of 500 meters.
What is the total change in the skier’s gravitational potential energy?
.
S
PROBLEM 4
While riding a chairlift, a 70-kilogram skier is raised a vertical distance of 500 meters.
What is the total change in the skier’s gravitational potential energy?
For situations occurring near the surface of the Earth, the change in gravitational
energy is calculated using the following equation:
PE = mg[h2 –h1]
Where h2 is the final elevation and h1 is the initial elevation.
S
In this problem we are asking for the change in gravitational potential energy
relative
to the initial elevation (h1), so we may conveniently set the initial elevation as zero.
With this simplification, the final elevation (h2) becomes 500 m.
Substituting values:
PE = [70 kg][9.81 m/s2][500 m – 0 m] = 3.43 x 104 Joules
.
PROBLEM 5
What is the gravitational potential energy with respect to the surface of the water of
a 65-kilogram diver located 5.00 m above the water?
.
S
PROBLEM 5
What is the gravitational potential energy with respect to the surface of the water of
a 65-kilogram diver located 5.00 m above the water?
For situations occurring near the surface of the Earth, the change in gravitational
energy is calculated using the following equation:
PE = mg[h2 –h1]
Where h2 is the final elevation and h1 is the initial elevation.
In this problem we are asking for the change in gravitational potential energy relative
to the surface of the water, so we may conveniently set the initial elevation
(h1), as
S
zero. With this simplification, the final elevation (h2) becomes 5.00 m.
Substituting values:
PE = [65 kg][9.81 m/s2][5.00 m – 0.00 m] = 3188 Joules
Note: The product of the mass time acceleration of gravity has units of kg-m/s2,
which is equal to a Newton. The product of a Newton and a meter has units of
Newton-meter, which is equal to a Joule.
.
PROBLEM 6
An object weighing 25 Newtons is lifted from the ground to a height of 2.2 m. What is
the increase in the object’s gravitational potential energy?
.
S
PROBLEM 6
An object weighing 25 Newtons is lifted from the ground to a height of 2.2 m. What is
the increase in the object’s gravitational potential energy?
For situations occurring near the surface of the Earth, the change in gravitational
energy is calculated using the following equation:
PE = mg[h2 –h1]
Where h2 is the final elevation and h1 is the initial elevation.
S
In this problem we are asking for the change in gravitational potential energy
relative
to the ground, so we may conveniently set the initial elevation (h1), as zero. With this
simplification, the final elevation (h2) becomes 2.2 m.
Also, in this problem we are given the objects weight (which is equal to mg).
Substituting values:
PE = [25 N][2.2 m – 0.0 m] = 55 Joules
.
PROBLEM 7
A cart weighing 25 Newtons is pushed 25 meters on a level surface by a force of 10
Newtons. What is the increase in the car’s gravitational potential energy?
.
S
PROBLEM 7
A cart weighing 25 Newtons is pushed 25 meters on a level surface by a force of 10
Newtons. What is the increase in the car’s gravitational potential energy?
For situations occurring near the surface of the Earth, the change in gravitational
energy is calculated using the following equation:
PE = mg[h2 –h1]
In this problem, the carts initial and final elevations are equal. Therefore h2 minus h1
is equal to zero. As such, there is no change in gravitational potential energy.
S
.
PROBLEM 8
A mass resting on a shelf 5.0 meters above the floor has a gravitational potential
energy of 490 Joules with respect to the floor. The mass is moved to a shelf 2.0
meters above the floor. What is the new gravitational potential energy.
Gravitational energy is calculated using the following equation:
PE = mg[h2 –h1]
In this problem, the carts initial and final elevations are equal. Therefore h2 minus h1
is equal to zero. As such, there is no change in gravitational potential energy.
S
.
PROBLEM 8
A mass resting on a shelf 5.0 meters above the floor has a gravatstional potential
energy of 490 Joules with respect to the floor. The mass is moved to a shelf 2.0
meters above the floor. What is the new gravitational potential energy.
Gravitational energy is calculated using the following equation:
PE = mg[h2 –h1]
In this problem, we are given the potential energy the object possess as well as the
relative height (5.0 meters above the floor). As such we can find the mass
S of the
object. Inserting values into the equation,
490 J = [m][9.81 m/s2][5.0 m - 0.0 m]
Solving for m:
m = 9.99 kg
.
PROBLEM 8
A mass resting on a shelf 5.0 meters above the floor has a gravitational potential
energy of 490 Joules with respect to the floor. The mass is moved to a shelf 2.0
meters above the floor. What is the new gravitational potential energy.
Knowing the mass of the object, we can use the potential energy equation to find the
gravitational potential energy at the new elevation (2.0 m). Inserting values:
PE = [9.99 kg][9.81 m/s2][2.0 m - 0.0 m]
PE = 196 J
S
.
PROBLEM 8
A mass resting on a shelf 5.0 meters above the floor has a gravitational potential
energy of 490 Joules with respect to the floor. The mass is moved to a shelf 2.0
meters above the floor. What is the new gravitational potential energy.
Although the previous method works, it was not really necessary to find the mass of
the object. Regardless of the mass of the object, when you change the elevation of an
object from 5.0 m to 2.0 m, the gravitational potential energy is reduced by a factor
of 2/5th (the ratio of 2.0 m/5.0m), as gradational potential energy is directly
S
proportional to the height of the object.
We can calculate the new PE as follows:
PE2 = PE1 x [h2/h1]
PE2 = 490 J x [2 m/5 m]
PE2 = 196 J
.
PROBLEM 9
Base your answer on the diagram below which shows a 20-Newton force pulling an
object up hill at a constant rate of 2 meters per second.
a. What is the work done by the force
in pulling the object from A to B?
b. What is the work done against
gravity in moving the object from
point A to point B?
S
c. What is the gravitational potential
energy of the box at the top of the
A
incline?
.
10 m
PROBLEM 9
Base your answer on the diagram below which shows a 20-Newton force pulling an
object up hill at a constant rate of 2 meters per second.
a. What is the work done by the force
in pulling the object from A to B?
The work done moving the object
from point A to point be can be
determined using:
S
W = F x d x cos Φ
As the force is aligned with the
A
direction of travel, cosine Φ
equals 1.
Inserting values:
.
W = [20 N][30 m][1.0]
W = 600 Joules
10 m
PROBLEM 9
Base your answer on the diagram below which shows a 20-Newton force pulling an
object up hill at a constant rate of 2 meters per second.
b. What is the work done against gravity
in moving the object from A to B?
The work done against gravity is the
equal to the change in potential
energy of the object.
S
This can be calculated using
PE = mg [h2 –h1]
A
Inserting values:
PE = [5 kg][9.81 m/s2][10 m -0.0 m]
PE = 491 J
.
10 m
PROBLEM 9
Base your answer on the diagram below which shows a 20-Newton force pulling an
object up hill at a constant rate of 2 meters per second.
b. What is the work done against gravity
in moving the object from A to B?
mg
Alternately, the work against gravity
could be calculated using
Φ
W = F x d x cos Φ
Φ
S
10 m
The force which must be applied
against gravity is mg.
A
Recognizing that Φ is also the interior
angle near point B (when two parallel lines are cut by a transversal, their opposite
interior angles are equal), cosine Φ becomes 10 m/30 m.
Inserting values:
.
W = [5 kg x 9.81 m/s2][30 m][10 m/30 m] = 491 J
PROBLEM 9
Base your answer on the diagram below which shows a 20-Newton force pulling an
object up hill at a constant rate of 2 meters per second.
c. What is the gravitational potential energy of the box at the top of the incline?
mg
The work done against gravity
increases the gravitational potential
energy of the object by that
Φ
amount. As such, the increase
Φ
S
in the gravitational potential
10 m
energy
A
.
PROBLEM 10
During an emergency stop, a 2.5 x 103 kilogram car lost a total of 1.5 x 104 Joules.
What was the speed of the car at the moment the brakes were applied?
.
S
PROBLEM 10
During an emergency stop, a 2.5 x 103 kilogram car lost a total of 1.5 x 104 Joules.
What was the speed of the car at the moment the brakes were applied?
The equation for kinetic energy is:
KE = ½ mv2
In this problem the kinetic energy and mass of the car have been given. Inserting the
mass and energy into the equation:
S
1.5 x 104 = ½[2.5 x 103][v2]
Solving for v2:
.
v2 = [1.5 x 104][2]/[2.5 x 103]
v = 3.64 m/s
PROBLEM 11
Three people of equal mass climb a
mountain using paths A, B and C shown
in the diagram to the right. Along which
path does a person gain the greatest
amount of potential energy from start to
finish? Explain your answer.
FINISH
.
A
START
B
S
C
PROBLEM 11
Three people of equal mass climb a
mountain using paths A, B and C shown
in the diagram to the right. Along which
path does a person gain the greatest
amount of potential energy from start to
finish? Explain your answer.
The change in potential energy of an
object depends on the mass of the
object and the change in elevation
(h2 – h1).
Because all the climbers started at the
same elevation and ended at the same
elevation, the change in elevation of all
climbers is identical. Further
because the climbers are of the same
START
mass, the change in potential energy
of each climber is identical.
FINISH
.
A
B
S
C