Question

If a `4 kg` object moving at `5/2 m/s` slows to a halt after moving `12 m`, what is the coefficient of kinetic friction of the surface that the object was moving over?

Answer 1

The coefficient of kinetic friction is `=0.027`

Explanation:

Apply the equation of motion

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

To calculate the acceleration

The initial velocity is `u=5/2ms^-1`

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

The distance is `s=12m`

The acceleration is

`a=(v^2-u^2)/(2s)=(0-(5/2)^2)/(2*12)=-0.26ms^-2`

According to Newton's Second Law, the force of friction is

`F_r=ma=4*0.26=1.042N`

The normal force is

`N=mg=4*9.8=39.2N`

The coefficient of kinetic friction is

`mu_k=F_r/N=1.042/39.2=0.027`