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Elastic Potential = Mass?

  1. Nov 20, 2007 #1
    A climber of mass m is scaling the vertical wall of a mountain. His safety rope is made of nylon that, when stretched, behaves like a sping with a spring constant k. He stops to rest, but then accidentally slips and falls freely for a distance h before the rope runs out of slack. The rope then stretches an amount x as it breaks his fall and momentarily brings him to rest. Derive an expression for the climbers mass, m, in terms of k, h, x, and g (the acceleration due to gravity).

    When i think about this i automatically think to use the equation:

    1/2(k)(y final^2) = (m)(g)(h initial)...
    but I do not think that you can use two distances, it wouldn't make sense. Or is the y = to the height?
     
  2. jcsd
  3. Nov 20, 2007 #2

    rl.bhat

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    The fall of potential = m*g*(h + x )
    The energy stored in the spring = 1/2*k*x^2
     
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