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Homework Help: Pogo Stick/Spring Question

  1. May 28, 2007 #1
    A child's pogo stick (Fig. P8.56) stores energy in a spring with a force constant of 2.50 104 N/m. At position A (xA = -0.120 m), the spring compression is a maximum and the child is momentarily at rest. At position B (xB = 0), the spring is relaxed and the child is moving upward. At position C, the child is again momentarily at rest at the top of the jump.

    The mass of the child is 25.5 kg. (Ignore the mass of the pogo stick.)

    (a) Calculate the total energy of the child-Earth system if both gravitational and elastic potential energies are zero for x = 0.

    (b) Determine xC.

    (c) Calculate the speed of the child at x = 0.

    (d) Determine the value of x for which the kinetic energy of the system is a maximum.

    (e) Calculate the child's maximum upward speed.


    I have gotten parts ABC but am having real trouble with part D....I know how to do part E once I get part D but just cant get it...heres all the work I have so far

    (a) Etot = KE + PEg + PEspr (choose reference at x = 0)
    = KE + 0 + 0

    So we need to find the KE of the child at the ref position. Apply conservation of energy between the ref point and the initial position of the child.

    Eref = Einitial
    KEref = KEi + PEi + PEs i
    = 0 + mgx + 1/2kx2
    = 25.5 kg * 9.8 * -0.120 m + 0.5 * 2.50 x 104 * (-0.120 m)2
    = - 30.0 J + 180 J
    = 150 J

    In other words, the total energy is all converted into kinetic energy in the child when he passes through the ref point.

    (b) To find xC, we apply cons of energy between the ref point and point C. The pogo stick is irrelevant in this process as it doesn't change shape when the kid moves up from the ref point.

    KEref = KEC + PEC
    150 J = 0 + PEC At max height, the kid doesn't move, all the KE converts to
    (grav) PE
    150 J = mgxC
    150 J / 25.5 kg * 9.8 = xC = 0.600 m Note: This is measured from the ref point to the
    max point, where the kid's feet are. Presumably
    this is the length of the pogo stick.

    (c) Get this from the KE we found in part a.

    KE = 150 J = 1/2 mv2
    v = ?(2*150 J / 25.5 kg)
    = 3.43 m/s

    (d) I'm having trouble with this one ... I used calculus to get an answer of xA = - 1.00 mm but I plugged it back into the initial equation and it tells me the KE is negative, or the kid can't leave the ground.

    KE = mgx + 1/2 kx2
    dKE / dx = 0 = mg + kx
    - mg = kx
    - mg / k = x = - (25.5 kg * 9.8 / 2.50 x 104 m)
    = - 1.00 mm

    What to do? Well, it's a bit of a fudge, but as the spring compresses, PEs >> PEg until only the spring's PE contributes to total energy. In this way, the energy is maxed when the spring's compression is maxed, which occurs if the stick is totally compressed.

    We got an answer of 0.600 m for the pogo stick length, so let's assume it can compress this far.

    (d) --> 600 mm then

    KE = 25.5 * 9.8 * -0.6 + 0.5 * 2.5 x 104 * 0.62
    = 4350 J

    (e) v = ?(2KE / m) = 18.5 m/s

    That's my attempt anyway ... other responses?
  2. jcsd
  3. May 29, 2007 #2


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