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Hooke's law

  1. May 7, 2010 #1
    Hey, i need some help here
    I'm trying to figure out the spring constant for bungee jump

    Say, I'm 735N and the rope will apparently double in length when i jump off it.
    I don't know the length of the rope thought.
    but i do know it does have a spring constant.
    How to find out is the spring constant using the Hooke's Law F=-KX

    Please help :)
  2. jcsd
  3. May 7, 2010 #2
    oh, we are trying to find the N/m ratio
    how could i do that
  4. May 7, 2010 #3
    F= -k/x --> |k| = F/x you cannot find it without some information on its stretching. O I just read that it doubles in length.
  5. May 7, 2010 #4
    The rope doubles in length when 735N is added, so does that mean that
    k would be equal to 735N/the length of the rope.
    k = 735 / L
  6. May 7, 2010 #5
    yes. When it has doubled its length it has displaced L from it equilibrium position.
  7. May 7, 2010 #6
    Thanks homie, so is the
    (the change in F) proportional to (the change in x) = k
  8. May 7, 2010 #7
    If i were using the f = k x equation, and then use
    EPE formula = half k x squared
    Which is also equal to GPE so could i say
    mgh = half k x squared,
    and is the x still L
  9. May 7, 2010 #8
    Yes, this is the point where all the kinetic energy was converted into elastic potential energy.
  10. May 8, 2010 #9
    If this is more that just a thought exercise you should be aware that at the bottom of the jump impulsive forces equal to approximately twice your weight act on the bungee cord.

    There have been fatalities due to cord breakages from this cause.

    Anyone who has had dealings with rope rigging will be familiar with snatch forces.
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