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Homework Help: Spring causing a mass to jump

  1. Jul 25, 2011 #1
    This is not a homework/coursework problem, but I came across it and wanted to check my answer.

    1. The problem statement, all variables and given/known data
    As in the diagram, there are two blocks of mass M and m. The mass m is suspended above M by a spring of spring constant k.

    Initially, the block m is pushed down with a force 3mg until the system reaches equilibrium, then the force is released.

    What is the maximum value for M for which the bottom block will jump off the ground?

    2. Relevant equations
    [tex] F = kx [/tex]
    [tex] F = mg [/tex]
    [tex] U = \frac{1}{2}kx^2 [/tex]

    3. The attempt at a solution

    I think this is the right way to do it, but I'm not sure.

    To find the equilibrium offset:
    [tex] 3mg = kx [/tex]
    [tex] x = \frac{3mg}{k} [/tex]

    Finding the potential energy due to the spring:
    [tex] \frac{1}{2}kx^2 = \frac{9m^2g^2}{2k}[/tex]

    To find the maximum upward offset (d) of the spring after release:
    [tex] \int_{-3mg/k}^{d} (-kx-mg)dx = -\frac{9m^2g^2}{2k}[/tex]

    Evaluating this returns : (If I did it right)
    [tex] d = \frac{mg(\sqrt{7}-1)}{k} [/tex]

    If kx > Mg, then the block will lift off, so I end up with:
    [tex] M \lt m(\sqrt{7}-1) [/tex]

    I'm just curious if I got this right. If anybody is willing to try the problem or check my logic, that would be great. Thank you!

    Attached Files:

  2. jcsd
  3. Jul 26, 2011 #2


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    Homework Helper

    Your logic works quite well, but:

    The block is pushed with a force of 3mg, but there is gravity, too, so the net force is 4mg.

    The integral is the work done by all forces between its initial and final position which is equal to the change of KE according to the work-energy theorem. d is the minimum upward displacement above the height of the relaxed spring where the spring force is Mg/k. As that, the velocity of the upper block is zero there. So the integral has to be zero.

  4. Jul 26, 2011 #3
    Ok, I see.

    Easy mistake:
    [tex] x = \frac{4mg}{k} [/tex]

    It seems everything else I did was unnecessary, I don't need to deal with the potential energy.

    So evaluating the integral:
    [tex] \int_{\frac{-4mg}{k}}^{d} (-kx - mg) dx = 0[/tex]

    [tex] d =\frac{2mg}{k}[/tex]
    [tex] M \lt 2m [/tex]

    This makes sense, Thanks!
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