Question

A skier with a mass of 62 kg is sliding down a snowy slope. Find thecoefficient of kinetic friction for the skier if friction for the skier isknown to be 45 N.

Answer 1

Given:

The angle of inclination is,

`\theta=25^(\circ)`

the frictional force is,

`f=45\text{ N}`

The mass of the skier is,

`m=62\text{ kg}`

according to the diagram the frictional force is,

`\begin{gathered} f=\mu mg\cos \theta \\ \mu=\frac{f}{\text{mgcos}\theta} \end{gathered}`

Substituting the values we get,

`\begin{gathered} \mu=(45)/(62*9.8*\cos 25^(\circ)) \\ =0.08 \end{gathered}`

Hence, the coefficient of friction is 0.08.