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Pogo Stick Elastic Energy: Substituting Correct Variables

  1. May 1, 2012 #1
    1. The problem statement, all variables and given/known data
    A 35 kg child is jumping on a pogo stick. If the spring has a constant of 4945 N/m and it is compressed 25 cm, how high will the child bounce?

    Answer: 0.45 m

    Need help substituting the right variable.

    2. Relevant equations

    ETi= ETf
    Egi + Eki + Eei = Egf + Ekf + Eef
    mghi + 0.5mvi2 + 0.5ksi2 = mghf + 0.5mvf2 + 0.5ksf2


    3. The attempt at a solution

    The real problem I have is setting the defined variables; everything I can solve for. I used trial and error and got 0.45 but I NEED TO KNOW WHY I use the variables (magnitudes in the right place)!

    m = 35 kg
    k = 4954 N/m
    g = 9.8 m/s²
    si = 0 m
    sf = 0.37 m
    hi = 0 m
    hf = ?
    vi = 0 m/s²
    vf = 0 m/s²

    I rearranged and solved and I got -0.45 m, some please tell me where what variables I have to substitute correctly! Thank you!
     
  2. jcsd
  3. May 1, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    There are a number of different ways to answer this question but from the "relevant equations" you list, you appear to be using "conservation of energy". That's what your first equations says- initial total energy equals final total energy. There are three types of energy involved in such a jump- the gravitational potential energy, kinetic energy, and the stored mechanical energy of the spring. Those are your "[itex]E_g[/itex]", "[itex]E_k[/itex]", and "[itex]E_e[/itex]". The additional "i" and "f" subscripts "initially", at the beginning of the jump, and "finally", at the highest point of the jump. The third equation replaces those with the standard formulas.

    We can take the ground as our base for potential energy so the potential energy is 0 there and mgh at h feet at h meters above the ground. At the start and at the highest point, the child is motionless so kinetic energy is 0 at both points. On the ground the stored mechanical energy is, as you say, [itex]0.5ks_f^2[/itex]. That is, on the ground the total energy is 0+ 0+ 0.5(4945)(.25)^2 and at the highest point of the jump the total energy is (35)(9.88)(h)+ 0+ 0. Set those equal and solve for h.
     
    Last edited: May 1, 2012
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