Question

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

Answer 1

`0.20`, rounded to second decimal place.

Explanation:

To find out the retardation due to kinetic friction let us use the equation
`v^2-u^2=2as`,

where `v,u,a,s` are the final velocity, initial velocity, acceleration and distance moved respectively.
`0^2-1^2=2xxaxx1/4`
or `a=-2ms^-2`

Now Force `F=m.a`
or Retarding force which is the force `|F_f|=2xx2=4N`
We know that ths `F_f=muxx"Normal Reaction"`, where `mu` is the coefficient of kinetic friction.

Assuming that the surface is a level surface, `:. "Normal Reaction"=mg`
Taking `g=9.81ms^-2`, we obtain
`4=muxx2xx9.81`, solving for `mu`
`mu=4/(2xx9.81)=0.20`, rounded to second decimal place.