Question

In an ice hockey game, a 250 g hockey puck is struck with a velocity of 108 km/h towards the goal located at a distance of 50 m. The speed of the puck recorded as it hits the back of the net of the goal is 72 km/h. Determine the frictional force offered by the surface of the ice rink.

• A. 1.3 N
• B. 1.5 N
• C. 1.25 N
• D. 4 N

Answer 1

The correct option is C 1.25 N

Given,
Mass of the hockey puck, m = 250 g = 0.25 kg
Initial velocity of the puck, u = 108 km/h = 30 m/s
Final velocity of the puck, v = 72 km/h = 20 m/s
Distance covered by the puck, s = 50 m
From the third equation of motion, we get,
$v^2$ - $u^2$ = $2as$
Calculating the acceleration of the puck,
$a$ = $\frac{v^2-u^2}{2s}$ ⇒ $\frac{{20}^2-{30}^2}{2✕50}$ = -5 $m/s^2$
From the second law of motion, F = ma or F = 0.25 ✕ (-5) = -1.25 N . This is eventually the frictional force exerted by the surface of the ice rink on the puck.