Coefficient of kinetic friction

missyjane

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
Fnet=ma=12*1.23457=14.81484
Ff=N-Fnet=81.51744
So, I thought mu=Fnet/N=0.8462, but that answer is apparently wrong. Any ideas?

amcavoy

[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.

Alex

missyjane

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?

amcavoy

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.