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Elastic and gravitational potential energy

  1. Nov 15, 2016 #1
    • Moved from a technical forum, so homework template missing
    X has a mass of 55 kg and hangs from a rope. As a result of this the rope stretches by 0.6 m. Calculate the energy stored in the rope as a result of stretching.
    Solution:
    F = kx and E = 0.5x^2. Using this fetches the answer as 161.5
    However, the change in potential energy (calculated using mgh ) of X is exactly twice of this energy stored. Is there a relationship between the two?. What is the explanation
     
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  3. Nov 15, 2016 #2

    QuantumQuest

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    What sort of things are taking place as the rope stretches and potential energy of the mass gets lower?
     
  4. Nov 15, 2016 #3
    That is the question I am trying to understand
     
  5. Nov 15, 2016 #4

    BvU

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    Hello Pratik, :welcome:

    There is a relationship: Gravity does work to the tune of 323 J when the weight is lowered. One half ends up in the spring, the other half is delivered to the device (a hand, or whatever) that lowers X from its initial position to 0.6 m lower.
     
  6. Nov 15, 2016 #5
    But isn't the whole body being lowered and not just the hand ? The hand is part of the body so the energy loss has been accounted for already
     
  7. Nov 15, 2016 #6
    And one half ends up in the rope, the other in the hand/device, does not sound quite convincing
     
  8. Nov 15, 2016 #7

    haruspex

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    That's not the hand BvU meant.
    What would happen if you started with an unstretched rope, tied a mass to the end while keeping the rope taut but unstretched, then let go?
     
  9. Nov 15, 2016 #8
    It is not clear what you are trying to say. I will refine the question, "when a point mass is attached to a rope, the rope will stretch, why is the loss in gravitational potential energy of the mass NOT equal to the gain in elastic energy of rope?"
     
  10. Nov 15, 2016 #9

    haruspex

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    To make it clearer, consider a light spring. It hangs, unstretched. You carefully tie a mass to the end, not allowing the spring to become stretched as you do so. Now you let go. What do you expect to happen over the next little while?
     
  11. Nov 15, 2016 #10
    The spring would stretch and the mass would fall, however, I don't see how this answers my question
     
  12. Nov 15, 2016 #11

    haruspex

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    Is that it? What happens after the mass stops falling?
     
  13. Nov 15, 2016 #12
    Can you put a simple explanation or answers instead of counter-questions
     
  14. Nov 15, 2016 #13
    Hi Pratik ,

    You have posed a nice conceptual doubt . Please do not lose patience . @haruspex is guiding you nicely . Give some thought to his questions and answer accordingly . You will surely benefit from this discussion .
     
  15. Nov 15, 2016 #14
    Hi,
    Thank you for the response. I really fail to see how this will get me to the answer. I am a Physics teacher myself and am aware of the necessity of patience while understanding things. However, there are times when a simple/clear explanation is all a student wants instead of building it up painfully slowly with seemingly obscure questions. Instead of patronizing me, if you could elaborate on the point Haruspex is trying to get to, it would benefit all of us.
     
  16. Nov 15, 2016 #15

    haruspex

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    Learning to ask the right questions is crucial. Please try to answer mine. It is not difficult, and it will lead you to the answer you seek.
     
  17. Nov 15, 2016 #16
    Alright. The mass will drop initially, and then it will bounce back, the oscillations will continue for a while before they dampen and the mass settles down at a particular height. At this point, the mass has lost some energy (because of the change in height) and the the spring has gained some energy (because of the extension)
     
  18. Nov 15, 2016 #17

    haruspex

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    Ok, good.
    Now, if it were perfectly elastic, instead of converging to the equilibrium position, it would oscillate forever. What are the relationships between:
    • The position it was initially released from,
    • The equilibrium position, and
    • The lowest point of the oscillation
    ?
    Then think about the energy differences, and what that tells you about the fraction of energy that is eventually lost in the imperfectly elastic case.
     
  19. Nov 15, 2016 #18
    the equilibrium position should be in between the lowest point and the point it was initially released from. So are you saying that, the energy difference is because there is some loss of energy while the mass oscillates ?
     
  20. Nov 16, 2016 #19
    OK.

    No.

    Pratik ,

    There are two ways in which the mass could come down from initial unstretched position of spring.

    1 . The support holding the mass is suddenly removed and the mass is allowed to fall such that the forces acting on it are spring force and force due to gravity .

    2. The mass is very slowly allowed to come down such that it is in equilibrium at all times .The forces acting on the mass are spring force , force due to gravity and some external supporting force ( may be your hand ) .

    Which of the two cases are we talking about in this question :smile: ?
     
  21. Nov 16, 2016 #20

    haruspex

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    Just a moment...

    I am not entirely sure what Pratik's main question is, whether it is where the energy goes, or why it is exactly half. The way I read the OP I thought it was mainly the second, so I have been concentrating on that.

    For where the energy goes, there are two main possibilities. As you suggest, it might be all lost in how the mass is released, but if not, it will be energy lost during the oscillations. The question will then be, why is energy lost in oscillating.
     
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