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Homework Help: Conservation of Energy and a spring

  1. Oct 27, 2008 #1
    A 2.4kg block is dropped onto a spring with a force constant of 3.96*10^3 N/m from a height of 5m. When the block is momentarily at rest, the spring is compressed by 25cm. Find the speed of the block when the compression of the spring is 15cm.

    Relevant equations:
    External Work = change in mechanical energy = change in potential energy of spring + change in potential energy of gravity + change in kinetic energy

    External Work = 0

    for change in potential energy of the spring I did
    final - initial
    1/2 K (.15m)^2 - 1/2 K (.25m)^2 = -79.2J

    for change in potential energy of gravity I did
    final - initial
    ?? - mgh

    for change in KE I did
    final - initial
    1/2 m v^2 - 0

    Im not sure what the final potential energy of gravity is... should it be 0 since it is converted to potential energy of the spring?

    Would it be
    0 = -79.92 - mgh + 1/2 m v^2
    and solve for v?
  2. jcsd
  3. Oct 27, 2008 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    Hello maniacp08,

    What I would do is look at what happens to the total energy of the system *in sequence*. Initially, because the object has some height, it has gravitational potential energy. Let's use the coordinate y for vertical position. Let's say y = 0 occurs at the spring. Then the object's initial height is y0 = 5, and its gravitational potential energy is

    [tex] E_p = mgy_0 [/tex]

    Then, as it falls, all of that gravitational potential energy is converted into kinetic energy, so that at the moment that it hits the spring, it has kinetic energy:

    [tex] E_k = \frac{1}{2}mv^2 = mgy_0 [/tex]

    It compresses the spring until all of its kinetic energy has been converted into elastic potential energy, at which point its velocity is zero and therefore it has stopped (let's call this point y1 = -25 cm:

    [tex] E_p = \frac{1}{2}ky_1^2 = mgy_0 [/tex]

    As you can see, the net result of this is to turn the gravitational potential energy the object initally had before being dropped into elastic potential energy. Now, at some earlier spring compression (let's call it y2 = -15 cm), NOT ALL of the object's kinetic energy has been converted into elastic potential energy yet. It still has some kinetic energy, which means it still has some velocity v', and we want to figure out what that is. The difference between its *initial energy* (which is the total energy of the system), and the elastic energy stored in the spring at y2 gives us the leftover kinetic energy:

    [tex] mgy_0 - \frac{1}{2}ky_2^2 = E_k = \frac{1}{2}mv^{\prime}^2 [/tex]

    Edit: I just realized that I forgot to add in the (negative) gravitational potential energy that the object would have at height y2. It's necessary to do that in order for things to work out properly.
    Last edited: Oct 27, 2008
  4. Oct 27, 2008 #3
    Wow, thanks for the very detailed explanation.
    I understand it now. Thanks again cepheid.
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