A 95kg clock initially at rest on a horizontal floor requires a 560N horizontal force to set it in motion. After the clock is in motion, in horizontal force of 650N

Complete Question

A 95 kg clock initially at rest on a horizontal floor requires a 650 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 560 N keeps it moving with a constant velocity. Find the coefficient of static friction and the coefficient of kinetic friction.

The value for static friction is

The value for static friction is

Explanation:

From the question we are told that

The mass of the clock is

The first horizontal force is

    The second horizontal force is    

Generally the static frictional force is equal to the first  horizontal force

So

=>  

=>    

Generally the kinetic frictional force is equal to the second horizontal force

So

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr