1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Bungee jumper

  1. Mar 5, 2008 #1
    1. The problem statement, all variables and given/known data
    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.

    Use g for the magnitude of the acceleration due to gravity.

    How far below the bridge will Kate eventually be hanging, once she stops oscillating and comes finally to rest? Assume that she doesn't touch the water.

    2. Relevant equations
    conservation of energy

    3. The attempt at a solution
    the problem states to use only terms introduced in the problem, so dont use X in kx

    [tex] mgd = \frac{1}{2}k(d-L)^{2}[/tex]

    [tex] \frac{2mgd}{k} = (d-L)^{2}[/tex]

    [tex] \sqrt{\frac{2mgd}{k}} = d - L[/tex]

    [tex] \sqrt{\frac{2mgd}{k}} + L = d[/tex]

    [tex] \sqrt{d} * \sqrt{\frac{2mg}{k}} + L = d[/tex]

    [tex] \sqrt{\frac{2mg}{k}} + \frac{L}{\sqrt{d}} = \frac{d}{\sqrt{d}} = \sqrt{d}[/tex]

    [tex] (\sqrt{\frac{2mg}{k}} + \frac{L}{\sqrt{d}})^{2} = d [/tex]

    im stuck here since i have to solve for d, and i dont even think the result will be similar to the answer

    the answer is [tex] \frac{mg}{k} + L [/tex]

    where did i go wrong?
    Last edited: Mar 5, 2008
  2. jcsd
  3. Mar 5, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    Why do you think those energies should be equal? When she is at rest her weight mg is equal to the force of the spring pulling up, k*(d-L). The energy approach is used if you want to find the value of d where she is closest to hitting the water.
    Last edited: Mar 5, 2008
  4. Mar 5, 2008 #3
    wow that was simple, hm

    well the next part of the problem asks to find the k constant if she jumps and JUST reachs the water,, is the energy approach, then correct?

    BTW how did you know to set the forces equal to each other and not the energys?

    Last edited: Mar 5, 2008
  5. Mar 6, 2008 #4


    User Avatar
    Science Advisor
    Homework Helper

    Because of the phrase "once she stops oscillating and comes finally to rest". That implies that frictional forces have been acting on her and dissipating energy. The energy calc wouldn't account for that lost energy. On the other hand, for the 'just reaches the water' case, it's reasonable to assume that there hasn't been that much energy lost to friction, and since KE is zero you can equate the PE's. Now you tell me why I can't just balance the forces in that case?
    Last edited: Mar 6, 2008
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Bungee jumper
  1. Bungee Jumper (Replies: 14)

  2. Bungee Jumper (Replies: 3)

  3. Bungee jumper problem (Replies: 1)