Question

If a `5 kg` object moving at `15 m/s` slows to a halt after moving `12 m`, what is the coefficient of kinetic friction of the surface that the object was moving over?

Answer 1

`μ=0.956`

Explanation:

We know that the work done by friction is equal to the change in kinetic energy of the object.

`ΔΚ=W_f =>K_"after"-K_"initial"=F_f*d =>0-1/2*5*15^2=`

`-Nμd` because the force of friction is `F_f=-Nμ`

where `N` is the upward force from the surface applied to the object and it is equal to the gravitational force because there is no movement in the upward or downward direction.

so `N=mg`

and `μ` is the coefficient of friction

`0-1/2*5*15^2=-Nμd=>1125/2=mgμd=>`

`μ=1125/(2mgd)=1125/(2*5*9,81*12)=0.956`