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Energy Lost due to Friction

  1. Feb 28, 2012 #1
    1. The problem statement, all variables and given/known data
    A 15.6 kg block is dragged over a rough, horizontal surface by a 68.9 N force acting at 19.7 degrees above the horizontal. The block is displaced 4.59 m, and the coefficient of kinetic friction is 0.286. Find the work done by the 68.9 N force.

    How much energy is lost due to friction?


    2. Relevant equations
    W=F (Δr) cos∅
    Friction (kinetic)= μk* m*g


    3. The attempt at a solution
    I found the work done by the 68.9 N force using the equation above to get 297.74 J.
    The energy lost due to friction should be simply the Force of kinetic friction multiplied by the displacement, right?
    I did 0.286*15.6 kg* 9.81 to get 43.77 N as the force of kinetic friction.
    Energy lost= Force friction * displacement i.e. 43.77 N * 4.59 m = 200.90 J, which is wrong.

    What am I missing?
     
  2. jcsd
  3. Feb 28, 2012 #2

    Delphi51

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    The normal force is reduced by the "lift" of the vertical component of the 68.9 N pulling force.
     
  4. Feb 28, 2012 #3
    So, normal force should be mg-68.9sin19.7 ?
     
  5. Feb 28, 2012 #4

    Delphi51

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    Yes, right on.
     
  6. Feb 28, 2012 #5
    Got it. Thanks so much!
     
  7. Feb 28, 2012 #6

    Delphi51

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    Most welcome! Good luck on the next one.
     
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