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Work Done?

  1. Dec 11, 2003 #1
    Work Done???

    If someone has the time and patience could you help me in setting this problem up?

    300kg object moves 4.5 meters down a 25 degree incline at a constant speed. It is kept from accelerating by a force pushing back on it. The effective coefficient of friction is .39

    I need to:

    a) calculate the net work done on the object
    b) calculate the work done by the force pushing back on it
    c) calculate the work done by gravity on the object.

    I do not know where to begin. could someone point me in the right direction.


  2. jcsd
  3. Dec 11, 2003 #2


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    Science Advisor

    Re: Work Done???

    gravity force:
    Fg = 300 * 9.81
    Fg = 2943N

    normal force:
    N = Fgcos(theta)
    N = 2943cos(25)
    N = 2667N

    friction force:
    f = uN
    f = 0.39 * 2667
    f = 1040N

    Since the thing is not accelerating, net forces are 0. I drew a FBD of the box with the X axis along the 25 degree slant and the Y is the same direction as the normal force.
    sum of x = 0 = 1040 + F - 2943sin(25)
    F = 203.8N (the force preventing it from accelerating)

    I don't exactly know what you mean by "net work". If I had to guess, I would say the net work is 0 since the mass is not accelerating.

    work done by force:
    W = Fd
    W = 203.8 * 4.5
    W = 917.1J

    work done by gravity:
    W = Fd
    W = 2943sin(25) * 4.5
    W = 5597J
    Last edited: Dec 11, 2003
  4. Dec 11, 2003 #3
    Thanks ShawnD! I read through your respons and I think I actually understand the concepts!!
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