Question

If an object with a mass of `10 kg` is moving on a surface at `4 m/s` and slows to a halt after `6 s`, what is the friction coefficient of the surface?

Answer 1

`mu = 1/15`

Explanation:

The initial kinetic energy is lost against the friction losses.

`1/2mv_0^2 = mg mu Delta x`.

Here `mu` is the kinetic friction coefficent and `Deltax` is the path until repos.

Also the movement along the path `Delta x` is dictated by

`Delta x = v_0 t - 1/2(mu g)t^2`

Solving for `Delta x` and `mu` the system

#{ (1/2mv_0^2 = mg mu Delta x), (Delta x = v_0 t - 1/2(mu g)t^2) :}#

we obtain

`Delta x = (v_0 t)/2` and
`mu = v_0/(g t)`

Assuming `g = 10` we have

`mu = 4/60 = 1/15`