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Vertical mass-spring system - force or energy?

  1. Dec 7, 2011 #1
    A lot of problems I see have vertical mass-spring systems, where a mass m hangs from a spring with spring constant k stretching it a distance D, and usually all but one of those quantities is known. But would you equate the forces or the energies, i.e. which is the correct equation to use: mg=kD or mgD = 1/2kD^2, because the two give different answers.

    Which is correct? Why is the other incorrect?

    Thank You
  2. jcsd
  3. Dec 7, 2011 #2
    The first one is correct in general (i mean, unless the spring itself has mass, but that would be ridiculous!). It's just Hooke's Law.

    The second one is more complex. Imagine quickly hooking on the weight: the weight is going to start oscillating on the spring, with the center of oscillation being at the point given by mg = kD. However, at that point, the weight's going to have kinetic energy, which you haven't accounted for. Your equation has no kinetic energy: the only way that could happen would be if you gently lowered the block into place. If you do that, you'd be doing negative work on the block, so conservation of energy would not be valid.

    If you want to work it out, conservation of energy still holds (in a way) for vertical springs! In fact, you can completely ignore GPE and just write KE + EPE = constant. The only difference is that you have to measure x in (1/2)kx^2 from the new equilibrium point, given by mg = kD. If you want, you can work that out and see why it's true; the new way of defining x accounts exactly for the GPE chance.
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