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Bungee Jump: Conservation of Energy

  1. May 19, 2014 #1
    1. The problem statement, all variables and given/known data

    A bungee jumper of mass m drops o ff a bridge and falls vertically downwards. The bungee cord is elastic with natural length L and stiff ness k. Deduce that at the lowest point of the fall, the cord is stretched by an amount:

    x=mg/k(1+√(1+(2kl/mg))

    2. Relevant equations

    F=-kx
    PE=mgh=0.5kx2

    Where h=L+x

    3. The attempt at a solution

    The total energy before the jump is equal to the total energy after the jump. Since at the bottom there is no kinetic energy I said the Potential Energy before is equal to the Elastic Energy at the bottom.

    So:

    mg(L+x)=0.5kx2
    x2=(2mgx +2mgl)/k
    x=√((2mgx +2mgl)/k)

    I'm having problems here trying to get in the asked form. Have I forgotten something n the conservation of energy?
     
  2. jcsd
  3. May 19, 2014 #2
    There's an x on the right hand side of the equals sign in your final answer.
    The line above that is a quadratic equation.
     
  4. May 19, 2014 #3
    So does that mean I have to solve it quadratically?
     
  5. May 19, 2014 #4
    I suppose you know how to solve a quadratic equation. ax^2 + bx + c = 0 ?
     
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