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Work of Friction Done Opposing Force at an Angle

  1. Feb 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Ben does 600J of work pushing a 55kg cart 5m in the +x direction by applying a force at a downward angle of 30° to the direction of displacement. If the coefficient of kinetic friction μκ is equal to 0.1, find:

    • The magnitude of the applied force.
    • The work done by the force of friction on the cart.


    2. Relevant equations

    W = Fd
    Work of friction = -μκ(mg + Fsinθ)d
    Work in x direction = Fcosθ

    3. The attempt at a solution

    The magnitude of the applied force:

    W = Fd
    600 = F(5)
    F = 120N

    The work done by friction:

    Wopp = -μκ(mg + Fsinθ)d
    Wopp = -0.1(55*10 + 120sin30°)(5)
    Wopp = -305J



    ----------------------------------------------

    Is it right? I have an undying feeling that it is correct; however, the student teacher that teaches my class substituted the initial W = Fd with W = Fdcosθ. That would only give you the force in the x direction, and then the force of friction found in the second part would be incorrect, right?

    Help me out, please. Thanks.
     
  2. jcsd
  3. Feb 8, 2012 #2

    PhanthomJay

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    Science Advisor
    Homework Helper
    Gold Member

    Teacher is correct, there is no work being done in the y direction because there is no displacemnt in the y direction. For constant force, Work is equal to the product of the force times the displacement in the direction of the force (W=Fdcostheta). (Or you could say, Work is equal to the component of the force in the direction of the displacement time the displacement...W = Fcostheta(d)).
     
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