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Spring Constant - Bungee Jumping

  1. Jun 15, 2014 #1
    1. The problem statement, all variables and given/known data

    A person with a mass of 65 kg goes bungee jumping. At the lowest point, he is located 30 m below his starting point. If, at the equilibrium point, the bungee cord measures 15m, what is its spring constant?

    2. Relevant equations

    F = k*x

    Ep = (1/2)(k)(x^2)

    3. The attempt at a solution

    At the bottom, all the gravitational potential energy is converted to elastic potential energy.

    mgh = (1/2)(k)(x^2)
    (60 kg)(9.8 m/s^2)(30 m) = (1/2)(k)(30^2)

    k = 39.2 N/m

    I am not sure if this is correct, why did they give the equilibrium information? What does that mean?
     
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  3. Jun 15, 2014 #2

    Orodruin

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    What is x in your expression for the bungee cord's potential energy measured relative to?
     
  4. Jun 15, 2014 #3

    haruspex

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    Just because two relevant equations use the same symbol (x here), doesn't mean they refer to the same quantity. Whenever you quote an equation, you ought to state what each symbol represents for that equation.
     
  5. Jun 15, 2014 #4

    tms

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    First thing, a typo somewhere: in the problem the mass is 65 kg, but you plugged in 60 kg at the end.

    Second, to address your question: in this case equilibrium means that the forces are in balance. That should point you to a slightly easier method of solution.,
     
  6. Jun 16, 2014 #5
    At equilibrium point, the gravitational force is equal to the force in the spring:

    mgh = k x
    (65 kg)(9.8 m/s^2)(15 m) = (k)(15 m)

    k = 637 N/m
     
  7. Jun 16, 2014 #6
    You have used incorrect mass .Mass is 65 kg .While using (1/2)kx2 ,you need to be careful in the sense that 'x' represents the displacement from the unstretched length.

    30m is the distance between the topmost and lowest point ,not between the lowest point and the equilibrium position .

    Wrong.

    How can you equate mgh i.e energy with 'kx' i.e force ?
     
    Last edited: Jun 16, 2014
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