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Bungee jumping problem

  1. Oct 31, 2015 #1
    1. The problem statement, all variables and given/known data
    A daredevil plans to bungee jump from a balloon 65.0m above the ground. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at point 10.0m above the ground. Model his body as a particle and the cord as having negligible mass and obeying Hooke❝s law. In a preliminary test he finds that when hanging at rest from a 5.00-m length of the cord, his body weight stretches it by 1.50 m. he will drop from rest at point where the top end of a longer section of the cord is attached to the stationary balloon. (a)What length of cord should he use? (b) What maximum acceleration will he experience?


    2. Relevant equations
    Potential energy = mgh
    Spring potential energy = kx^2/2
    Spring force = kx


    3. The attempt at a solution
    I tried using conservation of energy to find the spring constant k. I said the stretched length of the short cord was h1 and the distance it stretched was x1:

    mgh1 = kx1^2/ 2 => k = (2mgh1) / x1^2

    I did the same with the longer cord:

    mgh2 = kx2^2 / 2 => x2^2 = 2mg(h2) / k

    Then I replaced k with what I found:

    x2^2 = 2mg(h2) / (2mg(h1)/x1^2) => x2^2 = 2mg(h2) x1^2 / 2mg(h1) = (h2) x1^2 / h1

    Is what I did right or not at all? I'm very confused about this problem, thanks for your help!
     
  2. jcsd
  3. Oct 31, 2015 #2

    Student100

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    For the 5 meter cord what is the force given by the problem?
     
  4. Oct 31, 2015 #3

    haruspex

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    The information regarding the 5m cord does not involve any transformation of energy. The mass does not fall distance h1, or any distance. This is just hanging at equilibrium.
    Be careful here. A spring constant is a property of a specific spring, not of the material of which it is made. A spring of twice the length has a different constant.
     
  5. Nov 29, 2015 #4
    Hello, thanks a lot for your help :-)
    So if I understand well when the jumper is hanging at equilibrium I should use Newton to find k? Sum of forces = kx-mg => k= mg/x
    But I'm still lost about the other part of the problem, because now there are 2 things I don't know (the spring constant for the longer cord and its length, I only know the stretched length). How can I find the spring constant for a rope twice the length for example? Is is k/2? How can I find the ratio between the 2 cords when the lengths I know are completely different (one is at equilibrium (5+1.5m) and the other not (55m))? Please help, it's been several weeks, and I still don't understand anything :(
     
  6. Nov 29, 2015 #5

    haruspex

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    Suppose two identical ropes of spring constant k are tied end to end. A tension T is applied to them. How far does each stretch?
     
  7. Nov 29, 2015 #6
    Do they each stretch a distance x? So the double spring stretches 2x, and F=2kx, and the k of the 2 ropes is half the k of one shorter rope?
     
  8. Nov 29, 2015 #7

    haruspex

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    Yes.
     
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