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Spring and falling body problem

  1. Sep 29, 2008 #1
    1. Neglect air resistance. A suspention bridge spand the colorado river at a height of 321 meters. Consider a 75kg bungee jumper, and a 200m bungee cord which obeys hooke's law (once it reaches its unstreched length).

    a) calculate the spring constant k, so that the jumper halts just above the surface
    b) What is the magnitude and direction of the jumper’s acceleration at the bottom of the jump? Express in G's.
    c) Find the height above the river where the jumper reaches maximum speed.
    d) Find the maximum speed. Terminal velocity for a falling human is about 120 mph.
    Did the jumper reach terminal velocity? If so, we should take into account air resistance but that’s another problem).

    2. F=-kx, x=.5at2 (x initial is 0, v initial is 0), x=xi+vit+.5at2 , V=at

    3. If I write out less of an answer, it is because I think I have it correct. If I do not, I will certainly write everything out in more detail.

    a) I set the force of gravity equal to the force of the spring for position 321 meters, but since the spring doesn't do anything until 200m are passed, I actually used 121 meters as the x point for the spring. Just a little algebra got me a spring constant of about 6 kg/s2, which I believe is correct, until I use it later.

    Here is what I did:
    this give me k= -6.07kg/s2

    Is there anything wrong with this?

    b) At the bottom of the jump, it seems to me that the forces are canceled (the force of the spring versus the force of gravity). Which would mean Acceleration is 0. This seems intuitively correct, but my intuition is not allways correct.

    c) I did all sorts of things with this problem, which I wish I could scan. Basically, I haven't really got anywhere, but here are some key points of my attempt:
    The jumper has a=-9.8m/s2 until the point 200 meters. after this point, I think the accel is: -9.8 + (6/76kg)*(x-200). I calculated the acceleration added by the spring by setting -kx = ma, and solving for a, where I used 6kg//s2 for k. I tried to use the position formula: x=xi+vit+.5at2, where xi was equal to 200m, vi was equal to 62.6m/s (which I calculated was the velocity of the jumper at position equals 200m), and a=-9.8 + (6/76kg)*(x-200). Then i basically got stuck. I cant figure out the time or height of the jumper at his max speed. Presumably this would be when a=0. Perhaps a=o when I solve for my equation for a, but this gives me x=77.5 meters from the start, which makes no sense because no force is acting against the acceleration of gravity until atleast 200 meters.

    This probably is a result of my spring coefficient being wrong.

    d) Once I find out where the max speed occurs, I will be able to easily calculate this I think.

  2. jcsd
  3. Sep 29, 2008 #2


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    Homework Helper

    Hi professor,

    I don't believe this is correct. The forces are not cancelled at the point right above the water. Since the jumper's speed is zero there, if all of the forces cancelled then he would stay there indefinitely; but in fact we know that the cord will pull him back upwards (so at that point the spring force must be greater than the gravitational force).

    I would suggest you think in terms of energy for this part. What does that give?
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