Question

If an object with a mass of `10 kg` is moving on a surface at `15 ms^-1` and slows to a halt after `4 s`, what is the friction coefficient of the surface?

Answer 1

The acceleration of the object is `3.75 ms^-2`, which means the frictional force acting is `37.5 N` and the frictional coefficient is `mu = 0.38`.

Explanation:

First we calculate the acceleration of the object:

`v=u+at`

where `v` is final velocity `(ms^-1)`, `u` is initial velocity `9ms^-1)`, `a` is the acceleration `(ms^-2)` and `t` is the time `(s)`.

Rearranging:

`a=(v-u)/t=(0-15)/4 = -3.75 ms^-2`

The negative sign just shows that it is a deceleration, and we can ignore it for the rest of this calculation.

Now we find the force acting to slow the object, using Newton's Second Law :

`F=ma=10*3.75=37.5 N`

This is simply the frictional force acting. Frictional force is related to the normal force acting on the object, which in this case is the weight of the object, `F_N=mg=10*9.8 = 98 N`:

`F_f = muF_N`

Rearranging and substituting in the force:

`mu = F_f/F_N = 37.5/98 = 0.38`

Note that `mu` is a dimensionless number, that is, it doesn't have any units.