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Pulling a block up an incline: work!

  1. Sep 24, 2007 #1
    What is the total work W done on the block by the applied force F as the block moves a distance L up the incline?

    http://session.masteringphysics.com/problemAsset/1006802/15/2550_a.jpg

    isn't it just FLcos(theta) minus the work of friction which is mew(mg)Lcos(theta)??

    what am i missing here????
     
  2. jcsd
  3. Sep 24, 2007 #2

    mgb_phys

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    Work is just force * distance.
    The friction opposing F is simply friction_coef * normal force
     
  4. Sep 24, 2007 #3
    first of all the answer must be expressed in terms of F, L, theta, mew, m, and g

    second, well... you're not taking into account the angle...
     
  5. Sep 24, 2007 #4

    mgb_phys

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    Sorry the diagram isn't very clear, I thought the force was acting in the direciton of the red arrow - if it actually acting horizontally as shown by the tiny arrow above the F, then the FLcos(theta) is correct.
     
  6. Sep 24, 2007 #5
    But FLcos(theta) is not the answer. shouldn't it be FLcos(theta) minus something like force of friction??
     
  7. Sep 24, 2007 #6

    mgb_phys

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    Normal force is m g cos(theta), so friction along the slope is coeff_friction * m g cos(theta).
    The force up the slope is F cos(theta) Assuming that the little arrow means F is horizontal.
    So the work done by the force is ( F cos(theta) - u m g cos(theta) ) times distance, or W = L ( F - u m g ) cos(theta) unless I made a mistake !
     
  8. Sep 24, 2007 #7
    that's the answer i originally came up with isn't it (look at first post). but i know for a fact that the answer is wrong.... i'm so confused. anyone know how to solve this thing??
     
  9. Sep 25, 2007 #8
    oh. haha. i get it. use sin instead of cos for the first part with the force of gravity. thanks for trying, mgb_phys!
     
  10. Feb 12, 2008 #9
    so what is the correct answer?
     
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