Question

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

Answer 1

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.

Answer:

The value for static friction is
`\mu_s = 0.60`

The value for static friction is
`\mu_k = 0.70`

Step-by-step explanation:

From the question we are told that

The mass of the clock is
`m = 95 \ kg`

The first horizontal force is
`F _s = 560 \ N`

The second horizontal force is
`F _k  =  650  \  N`

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

So

`F _s  =  m  *  g  *  \mu_s`

=>
`560  =  95  *  9.8  *  \mu_s`

=>
`\mu_s =  0.60`

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

So

`F _k  =  m  *  g  *  \mu_k`

`650 =  95  *  9.8  *  \mu_k`

`\mu_k =  0.70`