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Homework Help: Bungee Jumping Question

  1. Apr 8, 2014 #1

    patrickmoloney

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    Gold Member

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

    A bungee jumper of mass m drops off a bridge and falls vertically downwards.
    The bungee cord is elastic with natural length L and stiffness k. Deduce that
    at the lowest point of the fall, the cord is stretched by an amount [tex] x = \frac {mg}{k} \Big( 1 + \sqrt {1+ \frac {2kL}{mg}} \Big) [/tex]



    2. Relevant equations

    [itex] mgh = \frac{1}{2} k x^2 [/itex]



    3. The attempt at a solution

    Energy considerations dictate that the gravitational potential energy of the jumper in the initial state is equal to the elastic potential of the cord in the final state

    [itex] mg ( L + x ) = \frac{1}{2} k x^2 [/itex]

    [itex] mgL + mgx = \frac{1}{2} k x^2 [/itex]

    [itex] 2mgL + 2mgx = k x^2 [/itex]

    [itex] x^2 - \Big( \frac{2mg}{k} \Big )x - \Big( \frac{2mgL}{k} \Big) = 0 [/itex]

    [itex] x = \frac{- \Big(- \frac{2mg}{k} \Big) + \sqrt{ \Big( - \frac{2mg}{k} \Big)^2 - 4 \Big( - \frac{2mgL}{k} \Big )}}{2} [/itex]

    [itex] x = \frac{1}{2} \Big( \frac{2mg}{k} + \sqrt{ \frac{4m^2g^2}{k^2} + \frac{8mgL}{k}} \Big) [/itex]

    [itex] x = \frac{mg}{k} + \frac{1}{2} \Big ( \sqrt{ \frac{4m^2g^2}{k^2} \Big ( 1 + \frac{2kL}{mg} \Big )} \Big ) [/itex]

    [itex] x = \frac{mg}{k} + \frac{1}{2} \Big( \sqrt{\frac{4m^2g^2}{k^2}} \sqrt{1+\frac{2kL}{mg}} \Big ) [/itex]

    [itex] x = \frac{mg}{k} + \frac{1}{2} \Big( \frac{2mg}{k} \Big) \sqrt{1 + \frac{2kL}{mg}} [/itex]

    [itex] x = \frac{mg}{k} + \frac{mg}{k} \sqrt{1 + \frac{2kL}{mg}} [/itex]

    [itex] x = \frac{mg}{k} \Big( 1 + \sqrt{1+ \frac{2kL}{mg}} \Big) [/itex]

    Is this the correct method? There is no solution online to this problem.
     
  2. jcsd
  3. Apr 8, 2014 #2

    CWatters

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    Science Advisor
    Homework Helper
    Gold Member

    Seems reasonable to me.
     
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