Question

If a `1 kg` object moving at `1 m/s` slows down to a halt after moving `1 m`, what is the friction coefficient of the surface that the object was moving over?

Answer 1

The coefficient of friction is `=0.051`

Explanation:

The coefficient of kinetic friction is `mu_k=F_r/N`

The normal force is `N=1gN`

Initial velocity is `u=1ms^-1`

Final velocity is `v=0ms^-1`

Distance is `s=1m`

We apply the equation of motion

`v^2=u^2+2as`

`0=1+2a`

`a=-1/2ms^-2`

The negative sign indicates that it is a deceleration.

According to Newton's Second Law,

`F=ma`

`F_r=1*1/2=1/2N`

Therefore,

`mu_k=(1/2)/(1g)=1/(2g)=0.051`