Question

If a `3kg` object moving at `1 m/s` slows down to a halt after moving 2 m, what is the friction coefficient of the surface that the object was moving over?

Answer 1

`mu= 0,0255`

Explanation:

initial kinetic energy
`E_k = 1/2 m xx v^2 = 1/2 xx 3 kg xx 1 m^2/s^2 = 1,5 J`
if the object will stop after 2 m this energy is lost with the friction work
`W_f = F xx s = mu xx g xx m xx s= mu xx 9,8 m/s^2 xx 3 kg xx 2 m = mu xx 58,8J`
since
`E_k= W_f`
`1,5 J = mu xx 58,8 J`
friction coefficient
`mu = 0,0255`

by the second dynamic principle
`F= m xx a`
`a= F/m = (mu xx g xx m)/ m = mu xx g = 0,25 m/s^2`
by the cinematic equation
`v= -a xx t + v°`
`t= v°/a = (1 m/s)/(0,25 m/s^2)= 4 s`
`s= -1/2 a xx t^2+ v° xx t= 1/2 xx 0,25 m/s^2 xx (4s)^2 + 1m/s xx 4 s=-2m + 4 m = 2m`
and the datas are verificated