Answer to Question #303838 in Physics for Wisdom

Question

A curling stone is given an initial velocity of 18.4m/s across ice rink. If the coefficient of friction between the stone and the ice s 0.054, how far will the stone travel before it comes to a stop.

Answer 1

The distance can be found as follows:

$d = \dfrac{W}{F}$

where $W$ is the work done by friction force $F$. According to the energy-work theorem, the work is equal to the change in kinetic energy $W = \Delta K$. Since the stone starts from $v = 18.4m/s$ and comes to the rest, the change in kinetic energy is the following:

$\Delta K = \dfrac{mv^2}{2}$

where $m$ is the mass of the stone.

The friction force is given as follows:

$F = \mu m g$

where $\mu=0.054,g = 9.8m/s^2$. Thus, find:

$d = \dfrac{mv^2}{2\mu mg} = \dfrac{v^2}{2\mu g} = \dfrac{18.4^2}{2\cdot 0.054\cdot 9.8} \approx 320m$

Answer. 320m.