Question

If a `6 kg` object moving at `1 m/s` slows down to a halt after moving `3 m`, what is the coefficient of kinetic friction of the surface that the object was moving over?

Answer 1

`mu=1/(80)`

Explanation:

We know that,
`v^2=u^2+2as`

`1/4 = 1 +2xx3xxa`

`-3/4 = 6xxa`

`a= -1/8`

By Newton's law,

`F=ma`

`f= 6xx1/8`

`mu N=3/4`

`mu=3/(4xx60)`

`mu=1/(80)`