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Homework Help: Equation: work done by friction

  1. Jun 22, 2010 #1
    1. The problem statement, all variables and given/known data
    How much work is done by the friction force and by vector F ? (Don't forget the signs.) (Use mu_k for µk, q for θ, and m, g, x, and F as appropriate.)
    (a.) work done by friction
    (b.) work done by vector F

    3. The attempt at a solution

    a. mu_k(mg-Fsinq)x
    b. Fxcosq
    Last edited: Jun 22, 2010
  2. jcsd
  3. Jun 22, 2010 #2


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    Those look correct.
  4. Jun 22, 2010 #3
    a is incorrect
  5. Jun 22, 2010 #4
    a is correct.. as long as x is displacement
  6. Jun 22, 2010 #5
    it is, and it keeps telling me that it is incorrect.
  7. Jun 22, 2010 #6


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    Is there a figure for this? The answer sort of depends on how θ is defined with respect to F and the horizontal.

    (i) First I should ask, is θ actually the angle between the horizontal and F, or is it defined differently?

    (ii) Does F point in a somewhat downwards direction or in a somewhat upwards direction?


    Ask yourself this to make sure that your answer makes sense. You have given your answer as


    That implies that the normal reactive force is mg-Fsinq. Which implies that the figure is drawn such that as the magnitude of F gets bigger, the normal force becomes smaller (indicating that F has a somewhat upward direction). Does this make sense according to the figure (if there was a figure)? If the figure is drawn such that as F gets larger so does the normal force, how would you modify your equation to express this?
  8. Jun 22, 2010 #7


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    Oh, I think I see what's going on. Silly us. Don't feel bad, I missed it too. :redface: The hint is given in the problem statement. Don't forget the signs!

    Work done by a force is positive if the force vector component is parallel (same direction) to the displacement vector. But the work done by a force is negative if anti-parallel (opposite direction). (That's because if they are in opposite directions, the positive work is really done to the force, not by the force.)

    Work done by friction is pretty much always going to be such that the frictional force is in the opposite direction as the displacement vector. Which means the work done by friction is going to be... :wink:

    mandy9008, Do you see what needs to be changed in your mu_k(mg-Fsinq)x equation now?
  9. Jun 23, 2010 #8
    its just -ve.. i thought they were asking for the magnitude.. either way on an exam a question like this would be on a mc so you would see the signs =]

    you've got nothing to worry about.
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