Question

A 112 g hockey puck sent sliding over ice is stopped in 16.8 m by the frictional force on it from the ice. (a) If its initial speed is 8.1 m/s, what is the magnitude of the frictional force? (b) What is the coefficient of friction between the puck and the ice?

Answer 1

Explanation:

It is given that,

Mass of the hockey puck, m = 112 g = 0.112 kg

The hockey puck is stopped at a distance of 16.8 m.

(a) Initial speed of the puck, u = 8.1 m/s

We need to find the magnitude of the frictional force. Firstly, calculating the acceleration of the puck using third equation of kinematics as :

$v^2-u^2=2ax$

v = 0 (as it stops)

$-u^2=2ax$

$a=\dfrac{-u^2}{2x}$

$a=\dfrac{-(8.1)^2}{2\times 16.8}$

$a=-1.95\ m/s^2$

The frictional force is given by :

f = m a

$f=0.112\ kg\times 1.95\ m/s^2$

f = 0.218 N

(b) Also the frictional force is given by :

$f=\mu N$

And N = mg (normal force)

$f=\mu mg$

$\mu=\dfrac{f}{mg}$

$\mu=\dfrac{0.218}{0.112\times 9.8}$

$\mu=0.198$

Hence, this is the required solution.