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Block on a rough plane Coefficient of Kinetic Friction

  1. Mar 20, 2013 #1
    1. A 3kg block slides down a rough plane inclined at an angle of 27 degrees. If the acceleration of the block is 2.5 m/s/s find the coefficient of kinetic friction.



    2. solving for components
    Force Net = M x A
    forces in the y direction: Fn - Fg cos 27 degrees = 0
    forces in the x direction: M x G sin 27 degrees - Fk = M x A
    Fk = coef-k x m x g cos 27 degrees




    3. isolating Fk resolves to:
    9.8 sin 27 degrees - 2.5 m/s = Fk
    substituting for fk to resolve for coef of kinetic friction resolves to this:
    9.8 sin 27 degrees - 2.5 m/s = mu-k x 3 kg x 9.8 cos 27 degrees
    1.949 = mu-k x 26.196
    mu-k = .074
     
  2. jcsd
  3. Mar 20, 2013 #2
    is this correct..? I am wondering if all the masses should cancel out in this equation or if I have it set up correctly
     
  4. Mar 21, 2013 #3
    Your work is section 2 is fine.

    You've divided the left-hand side by m, but not the right-hand side.
     
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