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Conservation of Elastic and Gravitational Energy

  1. Feb 26, 2012 #1
    1. The problem statement, all variables and given/known data

    A 70 kg bungee jumper jumps from a bridge. She is tied to a bungee cord whose unstretched length is 13 m , and falls a total of 37 m .

    Calculate the spring stiffness constant of the bungee cord, assuming Hooke's law applies.

    2. Relevant equations

    0.5kx^2 (final) + mgy (final) + 0.5mv^2 (final) = mgy (initial) + 0.5kx^2 (initial) + 0.5mv^2 (initial)


    3. The attempt at a solution

    Her mass won't change throughout the problem but for the sake of keeping everything organized, I'll just share what I plugged in. K is the unknown, but aren't the initial "x" and "y" positions 13 m? That's a little weird to me and that's why I'm thinking I used in the correct equation. I found out I didn't need the 0.5mv^2 at all because kinetic energy is zero at the beginning and end.
     
  2. jcsd
  3. Feb 26, 2012 #2

    gneill

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    Staff: Mentor

    Ask yourself whether the bungee cord is initially slack, tied to bridge at her jumping height, or whether it is tied 13m above her. In one case she'll free-fall for 13 m before the bungee cord starts to act, in the other case it will start immediately to apply a retarding force.
     
  4. Feb 26, 2012 #3

    Andrew Mason

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    Science Advisor
    Homework Helper

    x is the amount of stretch (ie. 37-13) and y is the change in vertical position (37). Essentially the decrease in gravitational potential energy results in an equal increase in the elastic potential energy of the bungee cord. When the cord is 13 m. it has no potential energy so treat x0 (unstretched) as 0.

    AM
     
  5. Feb 27, 2012 #4
    I ended up getting the correct answer. This means the bungee cord was initially slack because she hadn't jumped but the moment she DID jump the bungee cord stretched 24 m because it was 13 m in length.
     
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