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Ideal Spring Problem

  1. Dec 11, 2006 #1
    A mountain climber of mass 60 kg slips and falls a distance of 4 m, at which tmie he reaches the end of his elastic safety rope. The rope then stretches an additional 2 m before the climber comes to rest. What is the spring constant of the rope, assuming it obeys Hooke's law?

    a) 1.76*10^3
    b) 3.52*10^3
    c) 5.28*10^3
    d) 7.04*10^3

    I would think that F = kx = mg, where k(2) = (60)9.8 and k = 294 N/m, but that's not one of the choices. Alternatively, I thought that kx = (1/2)mv(f)^2 where v(f)^2 = 2gx and v(f) = 8.85 m/s, but that gives k = 1.17*10^3, which isn't a choice either.

    Help please! It would be great if I could get help on this by tomorrow morning.
  2. jcsd
  3. Dec 11, 2006 #2


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    Energy conservation is the right idea. The climber falls from rest and in the process loses gravitational potential energy (GPE). Initially, the GPE is converted to kinetic energy, but then the kinetic energy is converted into elastic energy in the rope. How much kinetic energy is there when the climber stops falling? What is the total change in GPE at that point? What is the elastic energy at that point?
  4. Dec 11, 2006 #3
    When the climber reaches the end of the rope his kinetic energy is (1/2)mv(f)^2 = 2350 J. The change in GPE is mg(h(o) - h(f)) = -2352 J.

    But I'm still not sure what the elastic potential energy is because I don't know k in (1/2)kx^2. Is (1/2)kx^2 = (1/2)mv(f)^2 + mg(h(o) - h(f)), or something along those lines? I'm still not getting the right value for k.
  5. Dec 11, 2006 #4


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    You are correct to use [tex]\frac{1}{2}kx^2[/tex] for the elastic energy but you do not need to involve kinetic energy just the difference in potential which is.

    [tex] P.E. = 6mg [/tex]

    now all that potential must be turned into elastic energy.
  6. Dec 11, 2006 #5
    Aha, got it. For some reason I didn't want to use those last two meters of PE before. Thank you!!!
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