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Work Done By a Gravitational Force

  1. Sep 17, 2007 #1
    Hi, I'm having trouble understanding this concept. A refrigerator is on a ramp of length d and it is being pushed up the ramp. What is the work done by the gravitational force? In the book it says mgdcos180 which is -mgd. I thought that the horizontal component of gravity was mgsin(theta) so the work done would by (mgsin(theta))(d)(cos180).

    Also, is work done by the gravitational force difference than gravitational potential energy?

    Thanks in advance.
    Last edited: Sep 17, 2007
  2. jcsd
  3. Sep 17, 2007 #2

    Doc Al

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    Staff: Mentor

    Assuming that d is the distance along the ramp and not the height, then you're correct. In terms of height (h), the work done by gravity is just F*S = -mgh, which is equivalent to -mgdsin(theta).

    Same thing (just opposite sign).
  4. Sep 17, 2007 #3
    Oh, I see it now. Thank you for clearing that up for me.
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