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Work with friction

  1. Feb 15, 2017 #1
    Hi. The problem is really simple I think, but I feel like the text is wrong even though the result seems to be correct.

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

    An object of known mass m (71.5 Kg) is pushed on a [flat] surface with a known friction coefficient of μ (0.272). How much does the object move with a work W (642 J) ?

    2. Relevant equations
    W = F*s
    Friction force = Normal force * μ

    3. The attempt at a solution
    The way I solved it, and the result seems to be correct at 3.36 meters, is like this.


    Friction force = m*g*μ (since it's a flat surface, the Normal force = Weight force) = 191 N

    W = F*s
    642 = 191 * s => s = 642/191 = 3.36

    The problem is, I feel like it's missing a piece of the puzzle, and more specifically the Force F that pushes the object in the first place (called F? in the drawing).

    What I mean is, the Work should be given by F*s, but "F" here should be the total Forces along the "movement axis" of the object...right? So I thought that the F in W = F*s in this case should have been:

    F = -Friction Force + Force_that_pushes_the_object

    I figured I'd try solving the exercise assuming that F = Friction Force, and it turns out that the result is in fact correct. However, I feel like saying that the result "matches the result of the workbook" is more accurate than saying that it's "correct", because I really believe that the F should be = Force_push - Friction Force, and not F = Friction Force.

    So, in my opinion, the text should have given the additional information on how much F? is worth, and then the solution should have been:

    s = Work / (F? - Friction)

    Am I wrong?
  2. jcsd
  3. Feb 15, 2017 #2


    User Avatar
    Science Advisor

    You are expected to read the problem statement as specifying the work done by the unknown force which you call F. You are also expected to assume that the object moves in a straight line (or that you measure distance along its path), that it begins and ends the scenario with the same speed, that air resistance is negligible and that this is done on earth where the acceleration of gravity is 9.8 meters/sec2.

    If the object starts and ends at rest and if the 642J is taken as the net work done by all forces on the object then this would violate the work-energy theorem. So that interpretation cannot be correct.

    In the interest of brevity, almost all physics problems are stated without spelling out all the details. Often this means that they become ambiguous and sometimes hard to read. With practice, you can develop the skill to see which interpretations make sense and test the subject matter you have been learning.

    One key rule applies: If you have to make assumptions about the problem to solve it, state those assumptions clearly and then work the problem. Even if the assumptions you make turn out not to be the ones intended, this doing this should at least earn you partial credit.
  4. Feb 15, 2017 #3
    To be precise, the question should have been worded "how much work is the friction force doing on the mass, as reckoned from the laboratory frame of reference?" We don't know how much work the force F is doing on the mass, because we don't know if the mass is accelerating. The friction force F is doing positive work on the mass, while the frictional force is doing negative work (the displacement is opposite to the direction of the frictional force). If the force F just matches the friction force, then the work of the two forces cancel, and the kinetic energy of the mass stays constant. If the force F is larger than the friction force, there is net positive work done on the mass, and its kinetic energy (velocity) increases.
  5. Feb 15, 2017 #4


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    Staff: Mentor

    Correct (provided you drop that negative sign). But the push required to keep a body moving at a steady velocity is zero Newtons, in the absence of friction.
  6. Feb 15, 2017 #5
    Ok, thanks all for the answers.

    Just a little doubt.
    You meant "the force F?" there?
  7. Feb 15, 2017 #6
    Yes. Sorry.
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