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Work and Spring Stiffness Constant

  1. Oct 26, 2008 #1
    1. The problem statement, all variables and given/known data

    A 65 kg trampoline artist jumps vertically upward from the top of a platform with a speed of 5.0 m/s. (Ignore small changes in gravitational potential energy.)

    a) How fast is he going as he lands on the trampoline, 3.0 m below?
    ANS: 9.2m/s
    NOTE: I understand how to get this, I just need help on part B.

    (b) If the trampoline behaves like a spring with spring stiffness constant 6.2e4 N/m, how far does he depress it?
    ANS:

    2. Relevant equations

    x = Fs/-k
    k = 6.2e4 N/m

    F = ma
    m = 65 kg

    a = {(v_2)^2 - (v_1)^s}/{2d}

    3. The attempt at a solution

    x = {((v_2)^2 - (v_1)^s)*65}/{2*d*6.2e4}

    My value of d = x = the depression of the trampoline, so,

    x = {((v_2)^2 - (v_1)^s)*65}/{2*(x)*6.2e4}.

    v_1 = 9.2 m/s
    v_2 = 0 m/s

    2x = .0443
    x = 0.22 , which is wrong :-(.
     
    Last edited: Oct 26, 2008
  2. jcsd
  3. Oct 26, 2008 #2
    Does a spring always produce the same force no matter what its displacement is?

    (And why do you think the question asks you to ignore changes in potential energy? that may be a clue...)
     
  4. Oct 27, 2008 #3
    Yes it does produce the same force no matter what displacement is. Where did I use potential energy? I still don't understand how my equation is wrong...
     
  5. Oct 28, 2008 #4
    Imagine you have a great big spring, and you're trying to extend it. Or a piece of rubber and you're trying to stretch it. Imagine it pulling back at you as you pull it, and that it's strong enough not to deform.

    You're saying that the spring / rubber will always pull you with the same force, no matter how far you've extended it. Does that sound right?

    (Have you heard of Hooke's Law? If not, then look it up in your textbook - it will help.)

    The fact that the question mentions "small changes in potential energy" should make you wonder if you can use energy conservation to do the question. If you can use energy conservation, this is often easier than using forces.
     
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