Question

a hockey player hits a 0.20kg puck across a frozen lake with an initial speed of 12m/s. How far does the puck move if the coefficient between the puck and ice is 0.10?

Answer 1

72 meters

Step-by-step explanation:

The puck is acted upon by a force of friction between it and the ice. The strength of that force is relatively low on ice (coefficient of 0.1), that's why we have so much fun sliding over frozen puddles.

The force is determined as follows:

(friction) = (norm force of surface) x (kinetic coefficient of friction)

and it acts perpendicularly to the norm force, i.e., horizontally/parallel to ice surface. (the norm force is the opposite to gravity here).

(friction) = 0.20kg*9.8m/s^2*0.10 = 0.20N

So the puck experiences a deceleration of 0.20N/0.20kg=1 m/s^2.

The distance the puck makes is then calculated using the kinematic equation for decelerated motion with initial velocity v0 (and final velocity of 0):

`v_1^2=0 = v_0^2-2ad\\\implies d = (v_0^2)/(2a)=(12^2 (m^2)/(s^2))/(2(m)/(s^2))=72m`

The puck traveled 72m, unless it hit another hockey player, a tree, or a hole in ice.