Question

If a `5 kg` object moving at `2 m/s` slows to a halt after moving `50 m`, what is the coefficient of kinetic friction of the surface that the object was moving over?

Answer 1

`mu_k=0,004`

Explanation:

If frictional force is the only retarding force then it is the resultant force and so by Newton's 2nd Law of Motion (assuming motion along a flat surface in 1 direction)
`sumF=ma`
`therefore -f_k=ma`
`therefore -mu_kN=ma`
`therefore -mu_kmg=ma`

`therefore mu_k=-a/g`

We compute the acceleration `a` from the equations of motion for uniform acceleration in 1 direction as follows :

`v^2=u^2+2ax`

`therefore a=(v^2-u^2)/(2xxx)=-4/100=-0,04m//s^2`

Substituting back for `mu_k`, we get

`mu_k=0,04/9,8=0,004`