# The tutor gives an example with friction force.

I’ve never been curling. Even so, I can imagine the following question resonates with many curlers and spectators:

A curling stone is released at 1.2m/s. If the coefficient of friction between the ice and the stone is 0.010, how far will it travel before coming to rest?

Solution:

From my previous post, the force of friction is calculated by

F_{f} = μF_{N}

In this case, the stone is on level ice, so

F_{f} = μF_{N} = μmg

The force of friction is the only unbalanced force acting on the stone, so its acceleration is due only to friction. From Newton’s Second Law,

a = F/m = μmg/m = μg = 0.010(9.8) = 0.098m/s^{2}

Since the acceleration opposes the velocity, its sign is negative:

a = -0.098m/s^{2}

Now, we can calculate the distance the stone travels using kinematics:

v_{2}^{2} = v_{1}^{2} + 2ad

where v_{1}= initial velocity = 1.2m/s, v_{2} = 0m/s

Therefore,

0^{2}=1.2^{2} + 2(-0.098)d

0 = 1.44 – 0.196d

We add 0.196d to both sides:

0.196d = 1.44

We divide both sides by 0.196:

d = 1.44/0.196 = 7.3469…m (7.3m in sig figs)

Apparently the stone will glide 7.3m before stopping.

Source:

Heath, Robert W et al. __Fundamentals of Physics__. D.C. Heath Canada Ltd., 1981.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.