Question

If a `4 kg` object moving at `1 m/s` slows down to a halt after moving `90 m`, what is the friction coefficient of the surface that the object was moving over?

Answer 1

`5 * 10 ^-4`

Explanation:

`2as = v^2 - u^2`

`a =( Deltav^2)/(2s) = 1 / 180`

`f = mu N`

`N = mg = 40 N`

`D_f = mu N/m = mu g`

`mu g = 1/180`

`mu = 1/1800 = 0.00055555555 approx 5 * 10 ^-4`