A 30.0-kg block is initially at rest on a horizontal surface. A horizontal force

Question

A 30.0-kg block is initially at rest on a horizontal surface. A horizontal force of 76.0 N is required to set the block in motion, after which a horizontal force of 58.0 N is required to keep the block moving with constant speed.

(a) Find the coefficient of static friction between the block and the surface.

(b) Find the coefficient of kinetic friction between the block and the surface.

.......

I used this equation F=umg and I got 0.25 for part (a) and 0.19 for part (b) but it marked me wrong !!

Please Help..

Explanation / Answer

The relationship between the coefficient of friction (?), the friction force (Ff) and the normal force (Fn) {The normal force is the force at right angles to the friction surface.}

? = Ff/Fn

The surface is horizontal, so the normal force is just the weight force (mg)
Fn =30.0 * 9.81 = 294.3 N

(a) You are told that you need a force of 76.0 N to overcome the (static) friction force, so we can assume that Ff = 76.0 N

? (static) = 76.0 / 294.3 = 0.258

(b) The block is moving with constant speed, so the net force on the block = 0. That must mean that the kinetic friction force is equal and opposite to the applied 58.0 N force.

? (kinetic) = 58.0 / 294.3 = 0.197

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