Elastic Potential Energy and spring

In summary, the conversation discusses how to balance the forces on a vertical spring attached to a ceiling and a cabbage attached to the other end. The loss of gravitational potential energy of the cabbage-Earth system is equal to twice the gain in the spring's potential energy. The conversation also explores why these two values may not be equal, considering the effects of slowly lowering the cabbage and the presence of air resistance.
  • #1
Parth Dave
299
0
Fasten one end of a vertical spring to the ceiling, attach a cabbage (or any other mass) to the other end, and then slowly lower this cabbage until the upward force on it due to the spring balances the gravitational force on it.

1. Show that the loss of gravitational potential energy of the cabbage-Earth system equals twice the gain in the springs potential energy.

2. Why are these two values not equal? (Hint: the slowly lowering in now really the issue here. This would occur if the cabbage had been let go from a height and ultimately came to rest supported by the spring.)
 
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  • #2
First i know that the forces must balance. That is, force of the spring must be force of gravity. Thus mg = kx. I can than figure out how much the spring stretches.

I know elastic potential is 0.5kx^2. So what i tried doing was showing that, using the value of x i obtained create an expression for elastic potential energy. Than i tried to show that it was equal to the difference in gravitational potential energy. This ultimately led me nowhere.

Any suggestions?
 
  • #3
Cabbage, eh...they don't give you a spring constant?
 
  • #4
Unfortunately they don't.
 
  • #5
Ok (1) was very easy, i just misread the question.

the forces must be equal, thus kx = mg. or k = mg/x

E = 0.5kx^2 or 0.5mgx and gravitational is mgx, therefore true.
 
  • #6
Is the reason this happens because when you release it, and let it drop itself, it oscillates and than slows down its speed of oscillation. Thus, there must be a force causing that, Drag/air resistance. So your hand essentially does what the air does also, absorbs potential energy. Is that correct?
 
  • #7
Parth Dave said:
Is the reason this happens because when you release it, and let it drop itself, it oscillates and than slows down its speed of oscillation. Thus, there must be a force causing that, Drag/air resistance. So your hand essentially does what the air does also, absorbs potential energy. Is that correct?
Looks right to me. If there were not air resistance or no "hand", the object (I refuse to say "cabbage"!) would continue to oscillate. It would always have kinetic energy as it passed the "neutral" point.
 

1. What is elastic potential energy?

Elastic potential energy is the energy stored in an elastic material, such as a spring, when it is stretched or compressed. This energy is stored in the form of potential energy, which can be released and converted into kinetic energy when the material returns to its original shape.

2. How is elastic potential energy related to a spring?

A spring is a classic example of an elastic material. When a force is applied to a spring, it stretches or compresses, storing potential energy in the process. The amount of potential energy stored in a spring is directly proportional to the amount of force applied and the distance the spring is stretched or compressed.

3. What is Hooke's Law and how does it relate to elastic potential energy?

Hooke's Law states that the force applied to an elastic material is directly proportional to the amount of stretch or compression of that material. This law applies to springs and is used to calculate the amount of elastic potential energy stored in a spring.

4. How is elastic potential energy used in everyday life?

Elastic potential energy is used in many everyday objects, such as trampolines, bungee cords, and wind-up toys. It is also used in more complex systems, such as shock absorbers in cars and suspension systems in buildings.

5. How can elastic potential energy be calculated?

The formula for calculating elastic potential energy is E = 1/2kx², where E is the potential energy, k is the spring constant (a measure of the stiffness of the spring), and x is the distance the spring is stretched or compressed. This formula assumes that the spring is being stretched or compressed within its elastic limit, meaning that it will return to its original shape when the force is removed.

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