Confused on potential energy losses and regain.

In summary: Both are correct ways of solving the problem, it just depends on which way you are thinking about it.
  • #1
Lucyc2008
2
0

Homework Statement


A 3.20kg rubber ball drops from a height of 4.80m to the ground a bounces back to a height of 2.70m
a) how much potential energy does the ball lose on the trip down?
b) how much energy does the ball regain on the trip back up?
c) what is the net loss of potential energy?

Homework Equations


Ep = mgh

The Attempt at a Solution


I got the question "correct" a) (3.20) x (9.80) x (4.80) = 151J b) (3.20) x (9.80) x (2.70) = 84.7J C) 151 - 84.7 = 66.3J but when studying for my test I found a completely different way of solving it by subtracting Ep2 (84.7) from Ep1 (151) so answer a) would be Ep lost = 67 instead. What is the correct way of solving this equation?. Attached is the second way of solving. The first way makes more sense in my mind because there is no height as the ball hits the ground so Ep = 0.
 
Last edited:
Physics news on Phys.org
  • #2
Lucyc2008 said:

Homework Statement


A 3.20kg rubber ball drops from a height of 4.80m to the ground a bounces back to a height of 2.70m
a) how much potential energy does the ball lose on the trip down?
b) how much energy does the ball regain on the trip back up?
c) what is the net loss of potential energy?

Homework Equations


Ep = mgh

The Attempt at a Solution


I got the question "correct" a) (3.20) x (9.80) x (4.80) = 151J b) (3.20) x (9.80) x (2.70) = 84.7J C) 151 - 84.7 = 66.3J but when studying for my test I found a completely different way of solving it by subtracting Ep2 (84.7) from Ep1 (151) so answer a) would be Ep lost = 67 instead. What is the correct way of solving this equation? The second way subtracting Ep2 from Ep1 makes more sense in my mind. Attached is the second way of solving.

In part (a) you are asked to find the loss of potential energy from the point of release to the point of contact with the floor. The answer to that is 151 J. The 67 J (or 66.3 J) is the answer to part (c), the net loss of potential energy which is also the loss of potential energy from 4.80 m to 2.70 m.
 

1. What is potential energy?

Potential energy is the energy that an object possesses due to its position or state. It is often associated with the forces acting on the object, such as gravity or electric forces.

2. How is potential energy lost?

Potential energy can be lost through various processes, such as friction, heat transfer, or collisions. When energy is transferred from an object to its surroundings, the object's potential energy decreases.

3. How can potential energy be regained?

Potential energy can be regained through external forces acting on the object. For example, if a ball at rest on a hill is pushed, it gains potential energy as it moves up the hill due to the force of gravity.

4. What factors affect potential energy losses and regain?

The factors that affect potential energy losses and regain include the mass of the object, the height or position of the object, and the external forces acting on the object. The type of potential energy, such as gravitational or elastic potential energy, also plays a role.

5. Why is understanding potential energy important?

Potential energy is a fundamental concept in physics and is important for understanding the behavior of objects and systems. It is also crucial in many real-life applications, such as designing structures and predicting the motion of objects.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
935
  • Introductory Physics Homework Help
Replies
9
Views
4K
  • Introductory Physics Homework Help
Replies
9
Views
3K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
4K
  • Introductory Physics Homework Help
Replies
5
Views
1K
Back
Top