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Homework Help: Bungee Diver Spring Constant

  1. Nov 23, 2005 #1
    From a test I had:

    A 700N bungee diver is released from a 40m bridge attached to a bungee cord which is 25m when unstretched. The cord obeys Hooke's Law. Find the spring constant of the cord if the diver is to stop 4m above the surface.

    What I did was say that gravity pulls this diver down 36m, and W=Fd=700*36. Now, when the cord actually gets him stopped, all that energy is potential spring energy, and with the distance the cord is stretched from equilibrium being 11m, 1/2kx^2 = mgh is easy to find. I ended up with 416N/m. However, the only force acting on the diver is gravity from 0 to 25m. From 25 to 36m, though, the cord is acting. But, isn't the cord only acting to decrease the Ke gained during the 25m fall, and to also decrease the GPE the diver has as he continues to fall from 25 to 36m?? Or did I do this right and all the energy that the cord has at the bottom of the fall is solely due to gravity, thus, mgh?
  2. jcsd
  3. Nov 24, 2005 #2


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    These are the proper guidelines to your analysis:

    Consider some reference point in the y-axis, let the bridge at 40meters above the river equal to potential zero. Going down you will get negative gravitational potential.

    Quickly find the total energy before he jumps (hint: potential only). Let this be time 0. Then start your real analysis when the diver reaches 25 meters below the bridge (what is his speed then?). Let this be time 1, then:

    KE0 + PEgrav0 = KE1 + PEgrav1 + PErope1 = KE2 + PEgrav2 + PErope2

    At time 2, is when the diver reaches the bottom (4 meters above surface), KE2 = 0. What are the values of the other terms in the conservation equation?
  4. Nov 25, 2005 #3
    So, since total energy is conserved, would that just simplify to

    PEgrav0 = PErope2 ????

    I believe it would so long as it's not actually:

    PEgrav0 + Wrope = PErope2

    So....does the work do any rope that would make the potential spring energy of the rope GREATER than the potential energy of the diver 40m high on the bridge (though, since he will stop 4m above, it's 36m worth of PEgrav). If it doesn't, then it doesn't actually matter what is happening at 25m. Now, if the rope did work on the diver, then, it would be 1/2kx^2...and I would endup with mgh + 1/2kx^2 = 1/2kx^2 which means mgh=0 which isn't true.

    If Wrope doesn't exist, or perhaps, equals 0, then 1/2kx^2 = mgh = 700*36. Hence, kx^2=50,400. x=36-25=11. Therefore, 50,400/11^2 = k = 416.5 N/m.

    Ok, so I've done this problem several times and keep getting 416N/m. Since it's from a test and not HW, I don't know the answer. From the problem, is my work right, and thus the answer?
  5. Nov 25, 2005 #4


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    All these Forces are conservative (with a PE function).
    Skipping over the intermediate points (at 25m, 26m, 27m...)
    looks like mgh_top = ½ k x^2 to me. k = 416 N/m . Good job.
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