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Work Confusion

  1. Nov 13, 2004 #1

    Rather confused about a concept with "Work". Say you're given an inclined plane problem where you're told to find the work energy required to move an object to the top of the plane (this is a TYPE of work problem, I should think).

    The answer (anyhow), in value is equal to the gravitational potential energy (mgh). Now... I ask. Why is that? (In other words, why isn't it the applied force (minus Fg parallel or Fg sin theta) - could you not consider finding work using the applied force PLUS using the length of the incline (hypotenuse) as delta d instead?)

    This problem has shown up in the past, the absolute BEST you could do is just clarify "conceptually". Or just help. Thanks!
  2. jcsd
  3. Nov 13, 2004 #2


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    Calculating mgh gives you the minimum amount of work needed to raise the object to that height. You could very well do more work to get it there, if you were so inclined (ohhh...terrible pun). For instance, if you wanted to accelerate the object, or if there were friction on the ramp, then you would end up doing more work to achieve the same result. However, if you do the calculation for a frictionless incline in the scenario in which there is no net force on the object (ie the applied force just counteracts gravity and the object moves up the ramp at a constant velocity) then you should get the same answer as mgh. Try it!

    That's the clearest explanation I could think of. For more detail...does this mean anything to you: if the net force on the object is zero, then the work you do on the object is equal to the negative of the work done by the graviational force on that object throughout its displacement. Since the graviational force is a conservative vector field, the work it does on an object moving from point a to point b is independent of the path taken by the object between those two points. If this does not mean anything yet, don't worry about it. :smile:
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