Physics: friction force: how far will the curling stone glide?

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?


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

Ff = μFN

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

Ff = μFN = 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/s2

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

a = -0.098m/s2

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

v22 = v12 + 2ad

where v1= initial velocity = 1.2m/s, v2 = 0m/s


02=1.22 + 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.


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.

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