Question

If a `2 kg` object moving at `80 m/s` slows to a halt after moving `4 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 `=81.6`

Explanation:

Apply the equation of motion

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

to determine the acceleration `a`

The initial speed is `u=80ms^-1`

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

The distance is `s=4m`

The acceleration is

`a=(v^2-u^2)/(2s)=(0^2-80^2)/(2 xx 4)=-80*80/8=-800ms^-2`

The frictional force is `F_r=ma=2xx800=1600N`

Let the acceleration due to gravity be `g=9.8ms^-2`

The normal force is `N=mg=2*9.8=19.6N`

The coefficient of kinetic friction is

`mu _k=F_r/N=1600/19.6=81.6`