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Max Acceleration of a Bungee Jumper

  1. Oct 20, 2009 #1
    1. The problem statement, all variables and given/known data

    A bungee jumper of mass m kg jumps from 20 meters. The bungie cord's spring constant is k N/m. Let y = 0 be the point where the bungee cord begins to become taught, and let y = 20 be the jump point.

    It is known that the point of maximum acceleration is y = -17.

    Question: What is the value of the maximum acceleration?

    Hint: The acceleration is not equal to gravity.

    2. The attempt at a solution

    F = ky - mg = ma
    a = (ky - mg)/m [FAILS!]
     
    Last edited: Oct 20, 2009
  2. jcsd
  3. Oct 20, 2009 #2
    Thanks, but that didn't work.
     
  4. Oct 20, 2009 #3
    The problem is a bit unclear about the altitudes used.
    It says a jump from 20 m, but then they say that the cord only starts to become taught
    at 20 m below the jump point, and the point of maximum acceleration (hence maximum length of the cord) is 17m below that. I hope the jump is from an altitude of at least 37m.

    The method of you and anti-meson is correct. a = F/m = (kx - mg)/m, with
    F = net force on person
    k = spring constant = 100N/m
    x = distance cord is stretched = 17m.
    m = mass of person = 60 kg.
    g = 9.81 m/s^2

    only the distance should be 17m and not 37m

    I don't see how you got 9.81, -9.81, 6, or -6 as none of them is the answer comes
    from the equation with either 17 or 37 m.
     
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