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Homework Help: Law of Conservation of Energy Question

  1. Feb 26, 2010 #1
    1. The problem statement, all variables and given/known data

    A 10.0 kg mass slides from rest down a frictionless inclined plane from a height of 0.500 m. After traveling 5.0 m along the ramp, it moves along a horizontal surface (frictionless again) where it makes contact with a spring. The force constant of the spring is 100.0 N/m. Determine the distance that the spring is compressed before the mass comes momentarily to a halt.

    m=10.0 kg
    k=100.0 N/m
    h=0.500 m

    2. Relevant equations

    KE (initial) + PE (initial) = KE (final) + PE (final)

    3. The attempt at a solution

    {Initial} [1/2(m)(v^2)]+[1/2(k)(x^2)] = {Final} [1/2(m)(v^2)]+[1/2(k)(x^2)]

    That's all I could do. I do not know if I have the correct equation here. But, if I do, I do not know how to get the (v^2) in the equation.

    Also, another thing that is troubling me is that I do not know what effect the incline has on the question, or if it does at all.

    Thanks, in advance, for your help!
  2. jcsd
  3. Feb 26, 2010 #2


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    Homework Helper

    Here the inclined plane is used to accelerate the mass.
    KE (initial) + PE (initial) = KE (final) + PE (final). That will give you the final KE of the mass when it starts moving on the horizontal surface.
    Again using the conservation of energy, you find the compression in the spring.
  4. Feb 26, 2010 #3

    Doc Al

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    Staff: Mentor

    You forgot about gravitational PE.
    Take your initial point to be when the mass is first released (what's its speed then?) and the final point to be when the spring is maximally compressed (what's its speed then?).
  5. Feb 26, 2010 #4
    It's the height of the mass at the start that matters. That's 0.500m
    Work out the PE. This is converted totally to KE as it slides down the frictionless track.
    There is no need to find its velocity, though!
    All this KE is then converted into the elastic PE in the spring.
  6. Feb 26, 2010 #5
    the block starts from rest (so it is not moving), then it has no kinetic energy, but it has gravitational potential energy. as it falls down , it gains kinetic energy, but since the total mechanical energy is conserved in this system, the gravitational potential energy starts reducing at the same rate (same as the gain of kinetic energy). Also, it loses gravitational potential energy simply because it is falling down. Then, it gains potential energy again as it gets in contact with the spring
  7. Mar 1, 2010 #6
    So, how about this:

    [0] + [(10 kg)(9.8)(0.5 m)] = [1/2(10 kg)(0^2)] = [1/2(100 N/m)(x^2)]
    0 + 49 = 0 + [50(x^2)]
    x= {sqrt[49/50]}
    x= 0.99m

    Did I do this right?

    Thanks for all the replies by the way!
  8. Mar 1, 2010 #7

    Doc Al

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    Staff: Mentor

    Yes. Perfect!
  9. Mar 1, 2010 #8
    Thanks very very much!
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