Question

If a `13 kg` object moving at `4 m/s` slows down to a halt after moving `80 m`, what is the friction coefficient of the surface that the object was moving over?

Answer 1

The coefficient of friction is `=0.01`

Explanation:

Apply the equation of motion to find the acceleration

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

The initial velocity is `u=4ms^-1`

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

The distance is `s=80m`

The acceleration is

`a=(v^2-u^2)/(2s)=(0-4^2)/(2*80)=-16/160=-0.1ms^-2`

The mass of the object is `m=13kg`

According to Newton's Second Law of Motion

`F=ma`

The force of friction is `F_r=13*0.1=1.3N`

The acceleration due to gravity is `g=9.8ms^-2`

The normal reaction is `N=mg=13*9.8=127.4N`

The coefficient of kinetic friction is

`mu_k=F_r/N=1.3/127.4=0.01`