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Mass Dropped on Vertical Spring

  1. Jul 1, 2011 #1
    1. The problem statement, all variables and given/known data
    A mass, m, is dropped from a height, h, on an initially uncompressed spring of length L.

    (a) Determine the amount by which the spring is compressed before the mass comes to rest.
    (b) Determine the mass's maximum speed.

    2. Relevant equations
    U(e)=.5k*x^2, U(g)=mgh, KE=.5mv^2

    I think I am supposed to do this all symbolically, since I don't have k.

    3. The attempt at a solution
    For part (a):
    So, take the bottom of compression to be the 0 for gravitational potential energy. Then, we can say that the mass is dropped from a distance d above the top of the unstretched spring, where d=h-L. If the spring compresses by a distance, x, then let us call L-x the 0 for gravitational potential energy. I end up with this conservation of energy equation:
    E0=Ef... so: mg(h)= .5k*x^2+ mg(L-x)
    This simplifies to: 0=.5k*x^2-mgx-mg(L-h)

    I can use the quadratic equation to solve this, but is that too complicated for this type of question? Which solution to that equation would I use, the + or the - section?

    For part (b), I believe I find the new equilibrium position of the mass (x=mg/k) and then plug that into the conservation of energy as the x, solving for v at that point. Correct?
    Last edited: Jul 1, 2011
  2. jcsd
  3. Jul 1, 2011 #2
    Yes, solving the quadratic equation would be the correct way to solve this. As for whether it's too complicated or not, if you were taught to solve quadratic equations, I guess it's reasonable to expect you to use that knowledge in solving problems.

    It seem to me that since you've taken L-x as the length of the spring after compression, x would be positive.

    I agree with your method of solving part (b).
  4. Jul 1, 2011 #3
    Ok... we weren't explicitly taught to solve the quadratic, but I see no way around it. Is this a standard sort of problem, which would then include that formula?

    And I was trying to figure out whether the choice of x as positive means I choose the positive root (well, not the positive root but the one that involves addition).

    And for (b), I just realize that I actually need to find all of that in terms of a 0 for gravitational PE. Should I set that 0 at the new equilibrium point?
  5. Jul 1, 2011 #4
    That I don't know about, sorry. But it seems like a perfectly valid problem to me, to which a solution can be found.

    I tried solving the quadratic equation. One of the two solutions for x is negative, provided h>L. And btw, there is a small algebraic mistake in your final expression for the quadratic equation.

    It doesn't matter where you choose to have zero gravitational PE. As long as you apply energy conservation correctly, you will get the solution.
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