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Shm 2

  1. Mar 27, 2006 #1
    "A 75kg circus performer jumps from a height of 5.0m onto a trampoline and stretches it a depth of 0.30m. Assume that the trampoline obeys Hooke's law. (a) How far will it stretch if the performer jumps from a height of 8.0m? (b) How far will it be stretched when the performer stands still on it while taking a bow?"
    I'm not sure how to solve this problem, but here's some ideas I have:
    Since Hooke's law is F = -kx,
    -mg = -kx, where x = 0.3
    k = 2450 N/m
    is that right?
    does the spring constant vary when the performer jumps from 5.0m and from 8.0m?
    by only using Hooke's law, I can't seem to determine part (a), because hte equation doesn't use the height...

    I thought about using
    [tex] v_{max} = \sqrt{\frac{k}{m}} (A)[/tex]
    to find the spring constant, but then it seems like the spring constant varies with maximum velocity...
    is it correct to use max velocity = square root of (2gh), where h = 5.0m or 8.0m?

    I'm not sure what to do...
  2. jcsd
  3. Mar 27, 2006 #2


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    does not seem right to me. You are treating the system as if the 0.3 m was the new equilibrium position. It is not.

    It seems that you have to use the work energy theorem: find the speed of the performer just before he touches the trampoline. That's the initial kinetic energy (treat now this as the initial position of the problem). When he reaches the lowest point, he will have no kinetic energy left, he will have lost some gravitational potential energy (mg times 0.3 meter) and energy will have been converted into potential energy stored in the spring (1/2 kx^2). That will allow you to find k.
    Yes..(well, in real life no, but I am sure they had a constant k in mind).

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