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Finding work done by F given coefficient of kinetic friction

  1. Oct 25, 2008 #1
    1. The problem statement, all variables and given/known data
    A 2.55 kg block is pushed 1.54 m up a vertical wall with constant speed by a constant force of magnitude F applied at an angle of 69.9 degrees with the horizontal. The acceleration of gravity is 9.8 m/s^2. If the coefficient of kinetic friction between the block and wall is 0.613, find work done by F.

    2. Relevant equations
    W = F*x

    3. The attempt at a solution
    t = theta
    u = coefficient of kinetic friction

    Fsint - uN = ma + mg

    ma = 0 (since it's constant speed)
    N = Fcost

    F = (mg + uFcost)/(sint)

    W = F * x

    I think my method is wrong because I end up with a F in the equation which I don't know. Please help me asap.
  2. jcsd
  3. Oct 26, 2008 #2


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    Gold Member

    just factor the F out and then divide:

    [tex]Fsin \theta- \mu F cos \theta = mg [/tex]

    [tex]F= \frac{mg}{sin \theta-{\mu} cos \theta }[/tex]


    you know m, g, u, x, and the angle, so you can plug in and solve for F, then plug in and solve for W.
  4. Oct 26, 2008 #3
    I tried to solve the problem before your response, and I got the same equation. I solved for F and got 34.307 N. Then I put that into the second equation to solve for W and got 52.832 J as my final answer. But this is still wrong (online submit)! I don't understand what I'm doing wrong.
  5. Oct 26, 2008 #4
    The angle is given with respect to the *horizontal*
  6. Oct 26, 2008 #5
    So would the equation for work be:
    W= Fcost * x

    Because I did that and I got the answer 18.156 J but that is also wrong! Do you understand what I'm doing wrong?
  7. Oct 26, 2008 #6
    W=Fcos t *x is almost correct (Fcost t is the normal force)
    Last edited: Oct 26, 2008
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