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Work due to friction?

  1. May 19, 2007 #1
    While tutoring a friend of mine, I encountered a problem I was startled to discover I was unable to find the correct answer to. I am far from being a stranger to Work, yet the solution eludes me. I don't remember the exact values of the coefficient of friction et all, so I'll provide the problem in terms of the variables.

    1. The problem statement, all variables and given/known data

    A block of mass M is being pulled across a rough surface by a rope at angle theta above the block, providing a tension force T. The coefficient of kinetic friction between the block and the surface is MuK. What is the work done by friction over the distance D.

    2. Relevant equations

    F(friction) = Mu(k) * N
    T(y) = TSin(theta)
    W = F * D

    3. The attempt at a solution

    Under the understanding that the work is equal to the force multiplied by the distance, we went to solve for the force, first. Force of friction is Mu(k) * N, seemingly simple. So, since there is actively a force pulling up on the block, the block would not be pressing down on the surface as much thus the normal force would be less. We solved for the normal force under the assumption that the N would be equal to the difference between the force of gravity downward, and the upward pull due to tension. N = Fg - Ft. Then, solving for both values, expanded the problem to: N = Mg - TSin(theta). We then plugged this into the friction force equation, F = Mu(k) * N, resulting in F = Mu(k) * (Mg - TSin(theta)). Finally, after finding that, we plugged that force into the equation for the Work: W = F * D. Thus, our final equation was: Mu(k) * (Mg - TSin(theta)) * D. For some reason, this was incorrect, and I am at a lack to explain why that may be. Any help you all could provide would be greatly appreciated.

  2. jcsd
  3. May 19, 2007 #2


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    I know you know this, but in case you didn't, always draw a free body diagram. That having been said, I have drawn a free body diagram and have come to the same conclusions as you have. I have one downward acting force Mg, one upward acting force:

    [tex] T_y = T\sin \theta [/tex]

    one rightward force [itex] T_x [/itex] and one leftward force (the lowercase f stands for friction):

    [tex] F_f = \mu_k N [/tex]

    where N is the net vertical force on the block (ignoring the ground, which exactly counteracts this by pushing up with force N, so that the net vertical force on the block is actually zero).

    You can see that I have tried to be careful and have stated the obvious in some places. My question would be: what do you mean when you say "this was incorrect?" I assume that you mean that you did not get the correct numerical answer. Since I cannot see anything wrong with the physics, are you sure there was not a mistake with arithmetic or units?
  4. May 19, 2007 #3
    Hello Cepheid, thanks for your time.

    I'm afraid there is nothing wrong with the units as after arriving at the same, incorrect, answer twice we were very confused at what the problem was. I'll try to get the actual data as soon as I can so I can post it here. Though I can't quite believe it to be simply an arithmetic error, it seems that's the only possible issue.
  5. May 27, 2007 #4
    I finally found the reason why I kept getting it wrong...

    The work due to friction is, of course, negative.
  6. May 27, 2007 #5
    8 days later... now this is someone that doesn't give up!
  7. May 27, 2007 #6


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    Was it one of those damned nitpicky online quizzes (which are stupid way to test physics knowledge, IMO)...?
  8. May 27, 2007 #7


    Not so bad of a website, but every now and then you find problems with answers like that one. You're confidant you have the right answer, but it's wrong and it's not gonna tell you why.

    On the other hand, sometimes there are multiple choice questions with fewer selections than you have chances to get the right answer. :rofl:
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