Question

If a `10kg` object moving at `3 m/s` slows down to a halt after moving `5/2 m`, what is the friction coefficient of the surface that the object was moving over?

Answer 1

`=0.18`

Explanation:

Here work done against force of friction = KE of the moving body
`mumgxxd=1/2mv^2`
given
`m =10kg`
`d=5/2m=2.5m`
`v =3m/s`
`mu=?`
`:.mumgxxd=1/2mv^2`
`=>mucancelmgxxd=1/2cancelmv^2`
`=>mu =v^2/(2gd)=3^2/(2xx10xx2.5)=0.18`