1. POTENTIAL AND KINETIC ENERGY HOMEWORK SOLUTIONS January 17, 2017

2. PROBLEM 1
A car travels at constant speed up a hill from A to B, as shown in the diagram below.
As the car travels from A to B, its gravitational potential energy
Remains the same and its kinetic energy remains the same
Remains the same and its kinetic energy decreases
Increases and its kinetic energy decreases
Increases and its kinetic energy remains the same v B A

3. PROBLEM 1
A car travels at constant speed up a hill from A to B, as shown in the diagram below.
As the car travels from A to B, its gravitational potential energy
Remains the same and its kinetic energy remains the same
Remains the same and its kinetic energy decreases
Increases and its kinetic energy decreases
Increases and its kinetic energy remains the same v B A
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.

4. 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.
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:
Half as great
Twice as great
One-forth as great
Four times as great
180 cm
90 cm

5. 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.
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:
Half as great
Twice as great
One-forth as great
Four times as great
180 cm
90 cm
The correct answer is “a”.
Gravitational potential energy is represented by PE = mgh.
Prior to the jump, the kitten was at an elevation of 180 cm relative to the floor. After the jump, the kitten was 90 cm relative to the floor, losing half it’s height, and therefore half it’s potential energy

6. 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.) B)
B) C)
Potential Energy
Potential Energy
Height
Height
Potential Energy
Potential Energy
Height
Height

7. “B” is the correct answer.
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.) B)
B) C)
Height
Potential Energy
Height
Potential Energy
“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.
Height
Potential Energy
Potential Energy B)
Height

8. 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

9. 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.
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 . S

10. 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

11. 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 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. . S

12. 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

13. 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.
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 . S

14. 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

15. 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

16. 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

17. 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 of the object. Inserting values into the equation,
490 J = [m][9.81 m/s2][5.0 m m]
Solving for m:
m = 9.99 kg . S

18. 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 m]
PE = 196 J . S

19. 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 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 . S

20. 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.
What is the work done by the force in pulling the object from A to B?
What is the work done against gravity in moving the object from point A to point B?
What is the gravitational potential energy of the box at the top of the incline?
30 m .
20 N
5 kg S
10 m A

21. 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.
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:
W = F x d x cos Φ
As the force is aligned with the direction of travel, cosine Φ equals 1.
Inserting values:
W = [20 N][30 m][1.0]
W = 600 Joules
30 m .
20 N
5 kg S
10 m A

22. 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.
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.
This can be calculated using PE = mg [h2 –h1]
Inserting values:
PE = [5 kg][9.81 m/s2][10 m -0.0 m]
PE = 491 J
30 m .
20 N
5 kg S
10 m A

23. 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.
What is the work done against gravity in moving the object from A to B?
Alternately, the work against gravity could be calculated using W = F x d x cos Φ
The force which must be applied against gravity is mg.
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 mg
30 m Φ .
20 N
5 kg Φ S
10 m A

24. 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.
What is the gravitational potential energy of the box at the top of the incline?
The work done against gravity increases the gravitational potential energy of the object by that amount. As such, the increase in the gravitational potential energy mg
30 m Φ .
20 N
5 kg Φ S
10 m A

25. 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

26. 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:
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 . S

27. 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 . S
A B C
START

28. 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 mass, the change in potential energy of each climber is identical.
FINISH . S
A B C
START