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Vertical spring (trampoline) compresion.

  1. Mar 22, 2014 #1
    1. The problem statement, all variables and given/known data
    A student jumping on a trampoline reaches a maximum height of h = 0.96 m. The student has a mass of m = 58 kg.
    What is the student's speed immediately before she reaches the trampoline after the jump in m/s?
    answer: 4.338
    If, when she lands on the trampoline, she stretches the trampoline down d = 0.75 m, what is the spring constant k in N/m of the trampoline?
    h=0.96m
    m=58 kg
    d= 0.75
    v=4.338
    2. Relevant equations
    KE=(1/2)mv2
    F=(1/2)kx
    SpringPE=(1/2)kx2
    ΔPe=mgh
    W=fd cos(Θ)
    KE=-PE

    3. The attempt at a solution
    I have had a gruesome time trying to understand spring related questions.
    First I took F=-kx and solved for k, leading to -k=F/x leading to k=mg/x. This is wrong however, because the force of the trampoline is greater than the force exerted by gravity on the person. so in the equation we have 2 unknowns, k and f.
    So I took formula KE=-PE
    KE=(1/2)mv2
    SpringPE=(1/2)kx2
    leading to (1/2)mv2=-(1/2)kx2
    thus -mv2/x2=k
    This didn't give the correct answer and I'm not sure why. Is KE also supposed to inclued the KE of the falling person?
     
  2. jcsd
  3. Mar 22, 2014 #2
    seems like you forgot to include gravitational potential energy in your calculation
     
  4. Mar 22, 2014 #3

    Simon Bridge

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    When the student reaches maximum height, is she in contact with the surface of the trampoline?

    They can be pretty gruesome.

    Yes: mg=kx when the person is just standing on the trampoline - this is the equilibrium condition.
    When she jumps, she passes through the equilibrium.

    Where else would you get the KE from?
    Why is your value of k negative? Does that make sense?

    Your calculation assumes that all the KE from the person falling a height h goes into the trampoline - and nothing else. What about gravitational PE dropping additional distance d?
     
  5. Mar 22, 2014 #4
    Thank you guys very much. With your help and suggestions, I have come up with the equation (1/2)mv^2=0.5kx^2+mgh.
    I thank you guys very much.
     
  6. Mar 22, 2014 #5

    Simon Bridge

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    Um...

    You should be careful to use the variables from the problem statement in your formulas.
    i.e. (1/2)mv^2 is the kinetic energy gained falling a distance h ... which is mgh
    so you wrote: mgh=0.5kx^2+mgh which means that k=0.

    I'm sure that is not what you meant?

    Note: how much of that initial GPE ends up stored in the spring?
     
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