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Showing work is path independent

  1. May 27, 2013 #1
    1. The problem:
    A 5-kg block is moved from the ground to a height of 1 m via two different routes, vertically
    and along an inclined plane with angle of 30° to the horizontal. How much work is done on
    the block against gravity in each case?

    2. Relevant equations
    W=Fcosθd where θ is between the displacement and force, Fg=mg


    3. The attempt at a solution
    I realize the answer is that the work done is the same because it is path independent, but I don't understand why I can't show this in my calculations. Any help is greatly appreciated.

    For the vertical path, I did
    W=Fcosθd
    W=mgcos(0)(1m)
    W=(5kg)(9.8m/s^2)(1)(1m) ≈ 50 J

    For the inclined path I did
    W=Fcosθd where the force required to move the object along the plane is =mgsin(30°) and the angle between the displacement and force would now be 60° so
    W=mgsin(30°)cos(60°)(1m) ≈12.5 J
    Where I get only a fourth of the work I found above. What am I doing wrong?

    (I assumed there was no friction because the question only asks about the work done against gravity)
     
  2. jcsd
  3. May 27, 2013 #2

    TSny

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    For the inclined path, the displacement is not 1 m vertically. It's a certain distance along the incline.
     
  4. May 27, 2013 #3
    Thank you! However wouldn't that distance be 1/tan(30) ? Because then the value I get is around 20...
     
  5. May 27, 2013 #4

    TSny

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    That's not quite right. Make a right triangle with one side horizontal and the other vertical. Let the hypotenuse represent the distance along the incline, and the vertical side represent the height of 1 m.
     
  6. May 27, 2013 #5
    Oops, it should be 1/sin(30), right? But then I'm still left with a cos(60) that doesn't cancel
     
  7. May 27, 2013 #6
    Oh it should be cos(0)- I forgot about that when you corrected my displacement. Thanks so much for your help!
     
  8. May 27, 2013 #7

    TSny

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    You're welcome. Good work!
     
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