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Work and energy and that sort of rot. :/

  1. Jan 8, 2009 #1
    Sally applies a horizontal force of 520 N with a rope to drag wooden crate across a floor with a constant speed. The rope tied to the crate is pulled at an angle of 53.0°.

    (a) How much force is exerted by the rope on the crate?
    Solved and correct.

    (b) What work is done by Sally if the crate is moved 28.5 m?
    Solved and correct

    (c) What work is done by the floor through force of friction between the floor and the crate?
    This one I have no idea how to do, unfortunately :(
    FxD=Work
    Trig (cos, sine, tan)
    Work=F(d)+(friction force)(d)
    work= change in Kinetic Energy
    Potential Energy=mass(g)(Height)


    (a) How much force is exerted by the rope on the crate?
    I managed to figure this one out.
    cos53=520N/x
    cos53x=520
    x=864.05 N

    (b) What work is done by Sally if the crate is moved 28.5 m?
    All I did with this one was multiply force (520) by distance (28.5) and got 14820 J

    (c) What work is done by the floor through force of friction between the floor and the crate?
    This one I have no idea how to do, unfortunately :(


    I've unfortunately procrastinated and it's due tomorrow but I'm sure I'll manage if someone doesn't reply....:/
     
  2. jcsd
  3. Jan 8, 2009 #2

    berkeman

    User Avatar

    Staff: Mentor

    Work is Force X Distance, right? What is the Force of friction? How is it defined in terms of the normal Force on the floor, and the coefficient of (dynamic, sliding) friction?
     
  4. Jan 9, 2009 #3
    Also keep in mind that the Fn isnt going to just equal the Fw because it is being pulled at an angle and some of that force is going into the y-component.
     
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