If a block with a mass of 10kg moves down an inclined plane what is the coefficient of kinetic friction?

Question

If a block with a mass of 10kg moves down an inclined plane (inclination is 30 degrees above the horizontal) with a constant velocity (i.e., no acceleration), what is the coefficient of kinetic friction? If the block moves a distance of 1 meter along the slope, what is the work on the block due to friction?

Answer 1

The block slides down the inclined plane at constant velocity so the frictional force is equal to the parallel component of the mg.

Resolve mg parallel to and normal to the inclined plane
1. Parallel to plane, F// = mg sin30 = 0.5mg
2. Normal to plane Fn = mg cos30

Frictional force Fr = Fn x u = u mg cos30

So mg sin 30 = u mg cos 30, ie, u = tan 30 (independent of mass)

Work done = Force along the plane x distance
= mg sin30 x 1 meter (you can use mg sin 30 or u mg cos 30, they are the same)

Answer 2

What is the resistance value of the surface of the plane. Without that, no one can calculate your answer unless they give a resistance value of zero to the plane inclined at 30 degrees.