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Work Done by Gravity

  1. Feb 28, 2012 #1
    1. The problem statement, all variables and given/known data

    A 265-kg load is lifted 24.0 m vertically with an acceleration =0.190 g by a single cable.

    Determine the work done by gravity on the load.

    2. Relevant equations

    Force of Gravity = mg
    W = F * d * cos (theta)

    3. The attempt at a solution
    I think I know why I did this incorrectly but I want to make sure.

    First attempt: Force gravity = (265 kg)(9.8 m/s^2) = 2.6 * 10^3 N.
    W = (2.6 x 10^3 N) (24.0 m) * cos (0°)
    W = 6.24 x 10^4 J.

    The correct answer is the same magnitude but the opposite sign. Is that because the angle should have been 180° because gravity always points down? The displacement is up so therefore the angle between the net force and the displacement would have been 180°, not 0°?
  2. jcsd
  3. Feb 28, 2012 #2


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    That looks fine. What's your question?
  4. Feb 28, 2012 #3
    I did it incorrectly the first time but I wanted to make sure I knew why it was wrong.
  5. Feb 28, 2012 #4
    Yes, if your displacement is in the positive direction, then in this case gravity must be in the opposite direction (180 degrees from positive direction).
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