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Conservation of Elastic and Gravitation Energy - 2

  1. Feb 27, 2012 #1
    1. The problem statement, all variables and given/known data

    A 70 kg bungee jumper jumps from a bridge. She is tied to a bungee cord whose unstretched length is 13 m , and falls a total of 37 m .

    Calculate the spring stiffness constant of the bungee cord, assuming Hooke's law applies.

    88 N/M

    Calculate the maximum acceleration she experiences.

    2. Relevant equations
    Spring Force = -k * x
    F = m * a

    3. The attempt at a solution

    Spring Force = (88 N/M)(24 m)

    I used 24 m because that was the maximum stretch of the cord.
    Spring Force = 2,112 N

    a = F / m
    a = (2112 N) / (70 kg)
    a = 88 m/s^2
    I added 9.8 m/s^2 because she was in free fall.
    a = 97.8 m/s^2

    Could someone explain why this is incorrect?
  2. jcsd
  3. Feb 27, 2012 #2


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    I get 30, not 88.
    Why add 9.8? My thinking is that
    F = ma
    kx - mg = ma
    a = kx/m - g
    Maximum (upward) acceleration is when x is at its maximum but
    9.8 would be subtracted rather than added.
  4. Feb 27, 2012 #3
    I made a typo on the acceleration, oops. I added 9.8 m/s^2 because I thought the jumper was moving in the same direction as gravity as she jumped. Could you explain why you subtracted?
    Last edited: Feb 27, 2012
  5. Feb 27, 2012 #4


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    kx - mg = ma
    The spring force kx is upward. Gravity mg is downward.
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