What is the coefficient of kinetic friction for the incline?

  1. A 12kg block is released from rest on an inclined plane with angle 35. Acceleration of the block is 1.23457 The acceleration of gravity is 9.8m/2^2.

    What is the coefficient of kinetic friction for the incline?

    I found the normal force (12*9.8*cos35)=96.33228
    So, I thought mu=Fnet/N=0.8462, but that answer is apparently wrong. Any ideas?
  2. jcsd
  3. [tex]F_f+F_g+N=m\vec{a}[/tex]

    You know that the mass is 12kg, the and the acceleration 1.23457 m/s2. You know that the product of those two is equal to the frictional force + the downward force. The downward force is the sum of the gravitational force and the normal force. Make a few substitutions and you should have your answer.

  4. Ok, so gravitational force is 9.8*12, right? That would be 117.6N. So the downward force is 117.6+96.33228=213.93228. But that means ma does not equal the product of the two. Where am I messing up?
  5. This will be much easier for you if you write it out with vectors. For this type of problem, try the following:

    1. Draw a free-body diagram.
    2. Write Newton's Second Law in vector form (important!).
    3. Write an expression for each of the forces.
    4. Write a scalar equation for each.
    5. Express any constraints on your equations.
    6. Solve the final system(s) of equations.
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