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Restoring Forces and Hookes law

  1. Oct 28, 2008 #1
    Kate, a bungee jumper, wants to jump off the edge of a bridge that spans a river below. Kate has a mass m, and the surface of the bridge is a height h above the water. The bungee cord, which has length L when unstretched, will first straighten and then stretch as Kate falls.

    Assume the following:

    * The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant k.
    * Kate doesn't actually jump but simply steps off the edge of the bridge and falls straight downward.
    * Kate's height is negligible compared to the length of the bungee cord. Hence, she can be treated as a point particle.


    and Fsp=kdelta(s)
    Fsp= force of spring
    k=spring constant

    i arrived at

    but its wrong it says:
    At this lowest point, forces are not balanced. If they were, Kate's momentum would carry her farther down, into the water. She actually has zero velocity at this point, much like a thrown ball does at the top of its trajectory.
    But i dont know what that means
  2. jcsd
  3. Oct 28, 2008 #2


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    What is the question? Find k in terms of H,L,m,g?
  4. Oct 28, 2008 #3
    oh, If Kate just touches the surface of the river on her first downward trip (i.e., before the first bounce), what is the spring constant k? Ignore all dissipative forces.
    Express k in terms of L, h, m, and g.
  5. Oct 28, 2008 #4


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    Well think about it then.

    When she jumps she has m*g*h in PE. And when she falls the bungee doesn't begin to retard her until she has fallen L.

    So that means that the work done by the bungee to just stop her at h below the top must be

    m*g*h = 1/2*k*(h-L)2
  6. Oct 28, 2008 #5
    Oh ok Thank you very much.
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