Elastic Potential Energy and spring

AI Thread Summary
When a cabbage is attached to a vertical spring and lowered until the spring's force balances gravitational force, the gravitational potential energy loss equals twice the elastic potential energy gain. This discrepancy arises because, during the slow lowering, energy is dissipated through air resistance and other forces, unlike a free fall scenario where oscillation occurs. The forces must balance, leading to the relationship kx = mg, allowing for the calculation of spring stretch. The elastic potential energy can be expressed as E = 0.5kx^2 or 0.5mgx, while gravitational potential energy is mgx. Ultimately, the presence of damping forces like air resistance alters the energy dynamics, preventing equal values in potential energy changes.
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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