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Physics problem involving work done by friction

  1. Mar 25, 2007 #1
    hi, I am learning about how to solve energy problems using the law of conservation of energy. I am okay with most of the problems, but I am confused about how to solve the ones involving the work done by frictional force. For example:

    A clerk pushes a filing cabinet of mass 22.0 kg across the floor by exerting a horizontal force of magnitude 98N. The magnitude of the force of kinetic friction acting on the cabinet is 87N. The cabinet starts from the rest. Use the law of conservation of energy to dertermine the speed of the cabinet after it moves 1.2m.

    First, I should state that due to the law of conservation of energy, total energy before=total energy after, which can be restated as:
    E applied + W done by friction = E_k

    but is it right? I know if i do it like this, i get the right answer, but that's the only reason I ended up with the statement, E applied + W done by friction = E_k, after many trials and errors!
    Is the W done by friction always included in the "total energy before"? I thought it was supposed to be a part of the "total energy after" cause doesn't friction occur after the object is given some work???
    plz try to explain this stuff!
  2. jcsd
  3. Mar 25, 2007 #2
    First, there is cosine involved since you are directly dealing with a horizontal component. Second the addition is wrong. The frictional force will oppose the movement, so the net force applied on the cabinet is 90N - 87N = 3N. Now since, [tex]F_{net}d = \frac{1}{2} v^{2} [/tex], you can calculate the velocity.
  4. Mar 25, 2007 #3
    ya, but im supposed to solve the problem using the concept of conservation of ENERGY! and i know that you can't subtract energy factors.
    read what I've asked for.
  5. Mar 25, 2007 #4
    Conservation of energy? I don't see how is that practical. What happens is that the big part the energy (87*1.2 J) given to the cabinet gets dissipated by friction (this energy is transformed into heat between the floor and the surface of the cabinet).
  6. Mar 25, 2007 #5
    i believe your approach is sound, giving V^2=1.2 In fact, except for the sequence of algebraic steps, appears to be the same as Wergs (tho he accidentally put 90 instead of 98).
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