Question

If a `9` `kg` object moving at `9` `ms^-1` slows down to a halt after moving `81` `m`, what is the coefficient of kinetic friction of the surface that the object was moving over?

Answer 1

The frictional coefficient `mu=0.102`

Explanation:

First step is to find the acceleration:

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

`a=(v^2-u^2)/(2s)=(0^2-9^2)/81=-1` `ms^-2`

(the negative sign just indicates that the acceleration is in the opposite direction to the initial velocity)

Second step is to find the force causing the acceleration:

`F=ma=9xx-1=-9` `N`

This is the magnitude of the frictional force. The third step is to find the normal force, which is just the weight force on the object:

`F=mg=9xx9.8=88.2` `N`

The final step is to find the friction coefficient:

`mu=F_"fric"/F_"norm"=9/88.2=0.102`

Frictional coefficients do not have units.