Question

If a `2 kg` object moving at `5 m/s` slows down to a halt after moving `100 m`, what is the friction coefficient of the surface that the object was moving over?

Answer 1

`0.0125`

Explanation:

Here,if constant retardation occurring due to frictional force is a then we can use the equation `v^2=u^2-2as`(all symbols are bearing their conventional meaning) where, `v=0` ,`u=5m/s` and `s=100`

So, `a = 0.125 m/s^2`

So,here amount of force acting against the motion of the object is `m×a` or `(2×0.125) N`

And this will be equal to the amount of frictional force acting at their interface i.e `mu×N` = `mu×mg` (where, `mu`= coefficient of frictional force)

So,we can write,

`mu×2×10 = 2 × 0.125` or `mu` = `0.0125`