Question

If an object with a mass of `10 kg` is moving on a surface at `15 m/s` and slows to a halt after `4 s`, what is the friction coefficient of the surface?

Answer 1

The coefficient of friction is `=0.38`

Explanation:

We start by calculating the deceleration.

We apply the equation of motion

`v=u+at`

The initial velocity is `u=15ms^-1`

The time is `t=4s`

The final velocity is `v=0ms^-1`

Therefore,

`0=15+4a`

The acceleration is

`a=-15/4ms^-2`

According to Newton's Second Law, the frictional force is `F_r=ma`

So,

`F_r=10*15/4=75/2N`

The normal force is

`N=mg=10*9.8=98N`

The coefficient of friction is

`mu_k=F_r/N=37.5/98=0.38`