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Inclined plane with spring underneath.

  1. Dec 11, 2011 #1
    1. The problem statement, all variables and given/known data

    A body of mass m = 2.5kg is on a frictionless inclined plane (by 40°). At an height of h = 1.6m.
    On the bottom of the plane there's a spring, which at rest has an extension of 30cm.
    If the spring is compressed maximum of 12cm, what's the "k" of the spring?

    2. Relevant equations



    3. The attempt at a solution

    Alright, there is conservation of energy because there is no friction.
    mgh = 1/2kx^2

    But there is an inclined plane so ...
    mg(h/sen40) = 1/2kx^2

    Now, the only step I don't fully get, what is X?
    I thought x would be:
    30cm - 12cm = 18cm.
    So I could solve for k.

    Is this right? Thanks.
     
  2. jcsd
  3. Dec 11, 2011 #2
    Assuming that the spring is lying on the ground and not sitting up against the slope then the angle of inclination doesn't matter, all you need is
    [tex]mgh = \frac{1}{2}kx^2[/tex]
    Where x is the distance it is compressed (given as 12cm in the problem)

    As a side note I reread your problem and you may be correct in saying that the [itex]x[/itex] is 18cm, I'm not exactly sure what your wording means.
     
  4. Dec 11, 2011 #3
    It means that when it is not compressed, the spring has a lenght of 30cm's.
    Then when the block pushes it, it's compressed of a maximum of 12 cm's.
    Now I would like to know what I must plug into x.
     
  5. Dec 11, 2011 #4
    A spring is comprensed or stretched along only one axis. That said, it seems that the confusion here is what must we use as x. If 12 cm (0.12 m) is the compressed lenght, then x = 0.12 m. But if 0.12 m is the final lenght of the spring, x would be instead 18 cm.
    One question: is the spring lying along the inclined plane or is parallel to the floor at the bottom?
    Note: you didn't have to use E = mg(h/sen 40°), since h must be and it is already parallel to g.
     
  6. Dec 11, 2011 #5
    the spring would be perpendicular to the inclined plane, something like this (check attachment).

    So, a question, when we have an inclined plane, and we want to find the potential energy.

    mgh

    h is always the height from the ground or it should be relative to the inclination of the plane?
     

    Attached Files:

  7. Dec 11, 2011 #6
    The height h doesn't depends of inclination; it is measured from the ground.
    From the drawing you made, it is clear that not all gravitational potential energy goes into compressing the spring, if you take ground as a point of 0 gravitational potential energy. Try to find the difference in height between the upper point and the lower point (when the spring is compressed): that will give the amount of gravitational energy that goes in the spring.
    Hint: draw a triangle whose hypotenuse is 18 cm at the position of the spring, in order to find the height of the lower point.
     
    Last edited: Dec 11, 2011
  8. Dec 11, 2011 #7
    I don't think I follow.
     
  9. Dec 11, 2011 #8
    To find the difference in height, try 1.6 m - (0.18 m) sen 40°. Calculate gravitational energy and equate it to elastic potential energy.
     
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