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Tension acceleration

  1. Oct 19, 2007 #1
    1. The problem statement, all variables and given/known data

    You are designing a "bungee jump" apparatus for adults. A bungee jumper falls from a high platform with two elastic cords tied to the ankles. The jumper falls freely for a while, with the cords slack. Then the jumper falls an additional distance with the cords increasingly tense. Assume that you have cords that are 13 m long, and that the cords stretch in the jump an additional 21 m for a jumper whose mass is 80 kg, the heaviest adult you will allow to use your bungee jump (heavier customers would hit the ground).

    (c) Focus on the instant of greatest tension and, starting from a fundamental principle, determine the spring stiffness ks for each of the two cords.
    ks = N/m

    (d) What is the maximum tension that each one of the two cords must support without breaking? (This tells you what kind of cords you need to buy.)
    FT = N

    (e) What is the maximum acceleration |ay| = |dvy/dt| (in "g's") that the jumper experiences? (Note that |dpy/dt| = m|dvy/dt| if v is small compared to c.)
    |ay| = g's (acceleration in m/s2 divided by 9.8 m/s2)

    3. The attempt at a solution

    I'm very lost as far as where to go with this problem. Any direction as to where to start and steps to follow would be great. THANKS.
  2. jcsd
  3. Oct 19, 2007 #2
    Could you try a conservation of energy for part C? Choose someplace to be your reference point where all the energy is potential, then, at the bottom of the fall, all of that energy has been transferred to spring potential energy.
  4. Oct 20, 2007 #3
    im not sure what u mean by this...could you please elaborate a bit about the potential energy concept/equation that u are talking about...thanks
  5. Oct 21, 2007 #4
    Well, the jumper falls from a height of 13 m before the cords begin to tighten. When they do, the jumper falls another 21 m, which is the maximum fall distance, where the tension in the cords is the greatest, and where the jumper momentarily stops at the bottom. You can choose this bottom point to be the reference point where there is no potential energy; it is all in the form of spring potential energy. There is only potential energy when the jumper is at the top of the cliff.

    mgh=(kx^2)/2. You know h, you know m, and you know x.
  6. Oct 21, 2007 #5
    oh ok so this is the calculation i did, lemme know if its correct(i only have one submission left):


    (80)(9.8)(34) = (k)(21^2) / 2

    k = 120.8888 N/m

    *also, any ideas for tension force or acceleration?

    *i tried Ft=kx = 392N and acceleration = change in velocity / change in time = 4.9

    *these values were incorrect though, and im not sure what else to use for it

  7. Oct 21, 2007 #6
    Hmm, now I'm wondering if you need to divide that k by two since you have two bungee cords. Maybe someone else knows?
  8. Oct 21, 2007 #7
    yeah that might make more sense because i submitted 120.8888 N/m and it was incrrect.

    so the correct k = 60.4444 ???
  9. Oct 21, 2007 #8
    Probably. I don't know what sort of assignment this is but if you don't lose points for submitting wrong answers, I would try it.

    If you do lose points, I would wait for a second opinion.
  10. Oct 21, 2007 #9
    ok...well 120 didnt work so 60 is my best guess now

    any help for tension force or acceleration?

    the force isnt simply mass times 9.8 is it?

    or is it stiffness times stretch?
  11. Mar 7, 2008 #10
    does anyone know how to find the tension force or acceleration?
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