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Bungee physics problem

  1. Nov 11, 2003 #1
    The problem is as follows:

    "Bungee Barbara jumps off a crane platform 30m above the ground. The bungee cord to which she is hooked has an unstretched length of 9.0m We assume the bungee cord behaves like an ideal spring with spring constant k=100 N/m. Barbara has a weight of 600 N. At what height will the net force acting on Barbara be zero?"

    I don't understand why this point is not the lowest point in the fall. I mean, wouldn't the net force be zero when the force of gravity and the tension in the spring are balanced, and wouldn't that be when Barbara momentarily comes to rest at the lowest point?
  2. jcsd
  3. Nov 11, 2003 #2


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    At the lowest point in the fall, the net force on her is pulling her up.

    While she's falling, she builds up momentum that carries her past the equilibrium point.
  4. Nov 11, 2003 #3
    So, how do I find out exactly what the net force is acting on Barbara at the lowest point in the fall?
  5. Nov 11, 2003 #4
    At her lowest point I think her ΣFy = 0, because there is a surface holding her up with the same force she is pulling the earth with.
  6. Nov 11, 2003 #5


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    The net force will be the force from gravity plus the force from the bungee.
  7. Nov 12, 2003 #6


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    No, when the total force is 0, there is no acceleration so she continues downward (remind me not to bungee jump with you! :smile:). That had already been said. The lowest point comes when her velocity is 0, not acceleration.

    If I am reading the problem correctly, Barbara will first "free fall" until she reaches 9 m below the platform. Then the cord will start stretching. For those first 9 m her acceleration is -9.8 m/s2 so her speed after t seconds is -9.8t and her height above the ground will be 30- 4.9 t2. That will be 9 m when 30- 4.9t2= 30-9= 21 or t2= 9/4.9= 1.83 seconds so t= 1.35 seconds At that time her speed will be -9.8(1.35)= -13.3 m/s.

    After that first 9 m the cord will act with force 100(21-x) Newtons (21- x is the stretch of the cord in m when the Barbara's height is x meters). The total force on Barbara will be -600+ 100(21-x)= 1500- 100x Newtons. Her mass is 600/9.8= 61.2 kg so her accleration is
    a= (1500- 100x)/61.2= 24.5- 1.63x.
    We have the differential equation x"= 24.5- 1.63x or x"+ 1.63x= 24.5

    The General solution to that is x(t)= C1Cos(1.28t)+ C2Sin(1.28t)+ 14.8. Taking t= 0 at the point at which the bungee cord starts stretching, x(0)= 21 m and x'(0)= -13.3 m/s.
    x(0)= C1+ 14.8= 21 so C1= 6.2.
    x'(t)= -1.28C1Sin(1.28t)+ 1.28C2Cos(1.28t)
    so x'(0)= 1.28 C2= -13.3 so C2= -10.4.

    That is: x(t)= 6.2 Cos(1.28t)- 10.4 Sin(1.28t)+ 14.8 meters.
    her speed at time t is x'(t)= -7.94 Sin(1.28t)- -13.3 Cos(1.28t).
    Barbara's lowest point will come when her speed is 0: when
    -7.94 Sin(1.28t)- 13.3 Cos(1.28t)= 0 or Tan(1.28t)= -13.3/7.94= -1.68.
    That means 1.28t= 2.1 or t= 1.6 seconds.
    Putting that into x(t), x(1.6)= 6.2Cos(2.1)- 10.4Sin(2.1)+ 14.8= 2.7 meters.
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