Elastic Potential Energy and spring

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Homework Help Overview

The discussion revolves around the concepts of elastic potential energy and gravitational potential energy in the context of a spring-mass system. The original poster presents a scenario involving a mass (cabbage) attached to a vertical spring, exploring the relationship between the energies involved when the mass is lowered to a position where the forces balance.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the balance of forces between the spring and gravity, with some attempting to derive expressions for the energies involved. Questions arise regarding the lack of a spring constant and the implications of lowering the mass slowly versus letting it drop.

Discussion Status

Some participants have clarified their understanding of the forces and energy equations, while others are exploring the effects of oscillation and air resistance on energy absorption. There is an ongoing examination of the differences in energy values under different conditions, but no consensus has been reached.

Contextual Notes

Participants note the absence of a spring constant as a constraint in their calculations. The discussion also touches on the implications of how the mass is lowered, questioning the assumptions made about energy transfer in different scenarios.

Parth Dave
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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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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?
 
Cabbage, eh...they don't give you a spring constant?
 
Unfortunately they don't.
 
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.
 
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?
 
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.
 

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