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Compressed spring (trampolin)

  1. Apr 4, 2007 #1
    1. The problem statement, all variables and given/known data
    A 60kg person is at rest with a vertical distance 1 m above the groud, compressed against the a spring. The spring is compressed a distance of .2m from its equilibrium position, with a spring constant of 1000 N/m. The person lands into a springy trampoline with a spring constant of 4000 N/m and the spring is maximally compressed a distance X2 below equilibrium position (ground). The reference is to be taken below the trampolin (ground), so the distance will have to be positive. How far below equilibrium will the spring be compressed?

    2. Relevant equations
    mgy + 1/2KX^2 = 1/2KX^2

    3. The attempt at a solution The fact that the reference is below the trampolin in confusing me, I tried soving the equation for X2, but I don't know how to acout for the vertical distance (1m + X) because we need to solve for X.
  2. jcsd
  3. Apr 4, 2007 #2


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    Gold Member

    Is the person being thrown up by the first spring then landing on the second one ? It's not clear from your description.

    The easiest way to do this problem is to work out how much energy is stored in the first spring, then equate that to the energy in the second spring after the leap.
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