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Elastic Potential Energy

  1. Sep 23, 2004 #1
    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.)
    Last edited: Sep 23, 2004
  2. jcsd
  3. Sep 23, 2004 #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?
  4. Sep 23, 2004 #3
    Cabbage, eh...they don't give you a spring constant?
  5. Sep 23, 2004 #4
    Unfortunately they don't.
  6. Sep 23, 2004 #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.
  7. Sep 23, 2004 #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?
  8. Sep 24, 2004 #7


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    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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