Question

If a `2` `kg` object moving at `10` `ms^-1` slows to a halt after moving `25` `m`, what is the coefficient of kinetic friction of the surface that the object was moving over?

Answer 1

`\mu=F_"frict"/(mg)=4/(2xx9.8)=0.20`

Explanation:

The object's final velocity will be `v=0` `ms^-1`, the initial velocity `u=10` `ms^-1` and the distance `25` `m`. We can use this to find the object's acceleration (deceleration):

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

`a=(v^2-u^2)/(2s)=(0^1-10^2)/(2xx25)=(-100)/50=-2` `ms^-2`

The frictional force will be given by `F_"frict"=ma=2xx-2=-4` `N`

That is, the force is `4` `N` in the direction opposite to the initial motion.

The relationship between the frictional force and the object's weight force is `F_"frict"=\muF_w=\mumg`.

Rearranging:

`\mu=F_"frict"/(mg)=4/(2xx9.8)=0.20`

(frictional coefficients have no units)