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Gravitational Potential Energy, Elastic Potential Energy, and Kinetic Energy

  1. Mar 1, 2012 #1
    1. The problem statement, all variables and given/known data

    A 4.0 kg mass is pressed down on a vertical spring of spring constant 400 N/m, compressing it to 0.250 m. After it is released, the amount of kinetic energy this mass would have when it leaves the spring is ___.

    2. Relevant equations

    mgy(final) + 1/2 kx^2 (final) + 1/2 mv^2 (final) = mgy (initial) + 1/2kx^2 (initial) + 1/2 mv^2 (initial)

    3. The attempt at a solution

    Ok so I think what I'm solving for is 1/2 mv^2 (final)

    (4.0 kg)(9.8m/s^2)(0m) + 1/2 (400 N/m)(0m)^2 + 1/2(4.0 kg)(v)^2 = (4.0 kg)(9.8 m/s^2)(-0.250 m) + 1/2(400 N/m)(-0.250m)^2 + 1/2(4.0 kg)(0 m/s)^2

    That was sort of long so to simplify it a bit:

    1/2(4.0 kg)(v)^2 = (4.0 kg)(9.8m/s^2)(-.250m) + 1/2(400 N/m)(-0.250m)^2 + 1/2(4.0 kg)(0 m/s)^2

    My main concern: My "x" and "y" are the same. Is that because it's a vertical spring? I'm calling the end of the release x=0 m and y=0 m and the compression -0.250 m.
  2. jcsd
  3. Mar 1, 2012 #2


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    Homework Helper

    you have done it all correctly. Technically speaking, x is the displacement from the equilibrium point. So you can think of it as Δx = x2-x1, where the x1 is the equilibrium point. We also know that the displacement Δx is purely vertical, so what is the (nice and simple) relationship between Δx and Δh?
  4. Mar 1, 2012 #3
    Ok, so Δx and Δh in this case are the same; that makes sense.
  5. Mar 1, 2012 #4


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    Homework Helper

    yep, that's right. Often you'll find that its the change in height that is important in questions with uniform gravity.
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