Question

If a `3/2 kg` object moving at `3/4` `ms^-1` slows to a halt after moving `5/8 m`, what is the coefficient of kinetic friction of the surface that the object was moving over?

Answer 1

First find the acceleration, then use it to find the magnitude of the frictional force, and use that to find the coefficient of friction.

Explanation:

First find the acceleration:

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

Rearranging to make `a` the subject:

`a = (v^2-u^2)/(2s)`

`a = (0^2-(3/4)^2)/(2xx(5/8)) = -0.45` `ms^-2`

The negative sign just shows that the object is decelerating - its acceleration is in the opposite direction to the initial velocity.

Now to find the frictional force:

`F = ma = (3/2)xx(0.45) = 0.675` `N`

The frictional force is given by `F_"frict" = muF_"norm"`, where the normal force is `F_"norm"=mg`.

Rearranging to make the frictional coefficient the subject:

`mu=(F_"frict")/(F_"norm") = 0.675/(3/2xx9.8) = 0.022`

(frictional coefficients have no units, since they are ratios of forces)

(note: technically, a coefficient of friction is between two bodies, not a property of the surface alone)