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Released Spring

  1. Feb 23, 2008 #1
    [SOLVED] Released Spring

    1. The problem statement, all variables and given/known data

    A piece of cheese with a mass of 1.35kg is placed on a vertical spring of negligible mass and a force constant k = 1600N/m that is compressed by a distance of 12.3cm .

    When the spring is released, how high does the cheese rise from this initial position? (The cheese and the spring are not attached.)

    2. Relevant equations

    U1 = Ugrav1 + Uel1

    U2 = Ugrav2

    v1 = 0

    v2 = 0

    K1 = 0

    K2 = 0

    K1 + U1 = K2 + U2

    --> Ugrav1 + Uel1 = Ugrav2

    y1 = -.123m

    3. The attempt at a solution

    (1.35kg)(9.81m/s^2)(-.123m) + 1/2k(-.123m)^2 = (1.35kg)(9.81m/s^2)h

    h = ((1.35kg)(9.81m/s^2)(-.123m) + 1/2(1600N/m)(-.123m)^2))/((1.35kg)(9.81m/s^s))

    h = .80m

    it's incorrect and I'm not sure what I'm doing wrong, at the bottom position where the spring is compressed the initial position should be less than 0 correct?

    there's no velocity at the inital position and at the highest point there is no velocity or potential energy, only gravity should be acting on it correct? any help is greatly appreciated.
  2. jcsd
  3. Feb 23, 2008 #2

    Doc Al

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    Staff: Mentor

    The phrase "this initial position" is ambiguous. I would assume they mean from the lowest position of the compressed spring, not the unstretched position.
  4. Feb 23, 2008 #3
    wouldn't the lowest position be the distance it was compressed?

    what I'm doing is I'm taking the positions to be (-.123m, h), h being the variable I'm trying to find.
  5. Feb 23, 2008 #4
    bump, anyone? I don't see how the initial position can be anything else.
  6. Feb 23, 2008 #5

    Doc Al

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    You have defined your height (h) from the unstretched position, not the lowest position.
  7. Feb 23, 2008 #6
    but wouldn't the unstretched position be when the cheese is not on the spring? I'm trying to grasp how the cheese's mass compressing it a distance of .123m isn't the lowest position? would it then be .246m?

    do I have to find integrals first to find the lowest position and then from that position to the height using the spring force?

    edit: sorry if I'm being difficult, just not getting the positions very well here I guess, I do have the right formulas I think right?
    Last edited: Feb 23, 2008
  8. Feb 23, 2008 #7

    Doc Al

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    It is the lowest position. But you are measuring heights from the unstretched position, not the lowest position: You call the initial height = -.123 m (not zero).

    Give your answer as a distance from the lowest position.
  9. Feb 23, 2008 #8
    but that's what I did; I used y1 = -.123m and plugged it into the initial gravitational and elastic energies, I didn't use 0.
  10. Feb 23, 2008 #9

    Doc Al

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    Answer this question: If something starts at y = - 5m and rises to the point y = 10m, how far does it move relative to the starting point?
  11. Feb 23, 2008 #10
    15 meters
  12. Feb 23, 2008 #11

    Doc Al

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    Good. Now apply that same reasoning to your problem, measuring from the starting point (the lowest point).
  13. Feb 23, 2008 #12
    hmmmm, okay, how about this:

    the original answer I got, .80m

    then add to that the distance from y0 to h, and solve for h.

    though y0 = 0 so I'm not sure what kind of equation I'm supposed to use since gravitational potentional energy will go to zero there.
  14. Feb 23, 2008 #13
    would it be the answer I received, .80, and then add that to .123m?
  15. Feb 23, 2008 #14

    Doc Al

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    In your original conservation of energy equation, there are gravitional PE terms on both sides which involve heights. Those heights must be measured from the same reference point (otherwise it makes no sense). On the left, you chose to measure height from the unstretched position, thus y1 = -.123 m. Thus you must interpret your answer for h from the same reference point.

    Then you can translate your value for h into what the question asked: Height from the starting point (which we assume is the lowest point).

    You can redo your equation, using y1 = 0. (Nothing wrong with gravitational PE = 0!) Or you can just properly interpret your existing value for h in terms of what the question asked. (No need to do anything over.)
  16. Feb 23, 2008 #15
    I'm sorry but I don't understand what you're saying, or rather I don't get how to write it mathematically, are you saying I should be use.

    mg(-.123) + 1/2k(-.123)^2 = mg(-.123)?

    or mgh + 1/2kh^2 = mg(-.123)?

    also using y1 = 0, how would I get an answer than, since:

    mg0 + 1/2k0 = mgh would just give me 0?

    and how is -.123m the 'unstretched' position?

    sorry again but I'm just not understanding what you're saying.
  17. Feb 23, 2008 #16
    okay, I just go the answer, there's no potential energy at positon 0 or -.123m, but there is elastic energy at position -.123m, and there's gravitational energy at positon h, so what I did was:

    1/2ky^2 = mgh

    h = 1/2ky^2/mg

    h = (1/2(1600)(-.123m)^2))/(1.35*9.81) = .91m, it was the exact answer,

    it also turns out that my initial answer of .79 + .123m would have also given me .91m, thanks Doc Al for the help (and putting up with my nonunderstanding :redface: )
    Last edited: Feb 23, 2008
  18. Feb 23, 2008 #17

    Doc Al

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    I knew you'd figure it out, sooner or later. :wink:
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