Question

If a `11 kg` object moving at `45 m/s` slows down to a halt after moving `900 m`, what is the coefficient of kinetic friction of the surface that the object was moving over?

Answer 1

By conservation of mechanical energy
Work done against Frictional force = KE of the object
`mu_kxxmxxgxxd=1/2xxmv^2`
`mu_k=v^2/(2gd)=45^2/(2xx10xx900)=0.1125`