Question

A ski hill is inclined at 30 degrees. When a skier goes down the hill, his speed at the bottom is half the value expected in the absence of friction. What is the coefficient of kinetic friction between the skier and the hill?

Answer 1

Without friction: potential energy at top of the hill h meters above the ground fully converts to kinetic energy:

$$mgh=\frac{mv^2}{2},\\ v=\sqrt{2gh}.$$

With friction this potential energy converts partly to work against friction, partly to kinetic energy:

$$mgh=\mu mgh\text{ cos}30°+\frac{mu^2}{2},\\\space\\ \mu=\frac{1}{\text{cos}30°}\bigg(1-\frac{u^2}{2gh}\bigg).$$

And since $v=\sqrt{2gh}$ and $v=2u,$

$$\mu=\frac{1}{\text{cos}30°}\bigg(1-\frac{2gh}{2\cdot2gh}\bigg)=\\\space\\ =\frac{1}{2\text{cos}30°}=0.577.$$