Question

If a `3/2 kg` object moving at `4/3 m/s` slows to a halt after moving `1/3 m`, what is the coefficient of kinetic friction of the surface that the object was moving over?

Answer 1

`mu_k = 0.272`

Explanation:

The original kinetic energy was `KE = 1/2*3/2 kg*(4/3 m/s)^2 = 4/3 J`.

Work of 4/3 J needs to be done by the friction to stop it.

`"work" = 4/3 J = F*1/3 m`

`F = (4/cancel(3) J)/(1/cancel(3) m) = 4 J/m = 4 N`

To find the value of `mu_k`, we solve

`F_"fk" = 4 N = mu_k*m*g = mu_k*3/2 kg*9.8 m/s^2 = mu_k*14.7 N`

`mu_k = (4 N)/(14.7 N) = 0.272`

I hope this helps,
Steve