Answer to Question #104980 in Physics for SARUUL

Question

On the surface of the ice s = 20.4m for sliding stones at initial velocity v = 3m / s
stopped on the road Find the coefficient of friction of stones on ice

Answer 1

The change of kinetic energy is equal to work done by a friction force

$$\frac{mv^2}{2}=\mu mgs.$$

Hence

$$\mu=\frac{v^2}{2gs}=\frac{3^2}{2\times 9.8\times 20.4}=0.023$$