A hockey puck (mass = 3.5 kg) leaves the players stick (moving to the left) with
ID: 1579641 • Letter: A
Question
A hockey puck (mass = 3.5 kg) leaves the players stick (moving to the left) with a speed of 28 m/s and slides on the ice before coming to rest.
The coefficient of friction between the puck and the ice is 0.9.
What is the normal force on the puck?
N
What is the friction force exerted on the puck due to the ice?
N
What is the magnitude of the acceleration of the puck due to this friction force?
m/s2
How long will the puck slide after leaving the players stick? (what amount of time will it be sliding)
s
How far will the puck slide after leaving the players stick?
m
Explanation / Answer
Normal force N = m*g
N = 3.5*9.8
N = 34.3 Newtons
Friction force = u*N where u is cofficient of friction
Friction force = 0.9*34.3 = 30.87 Newton'
To calculate acceleration use Newton's second law.
F = m*a
a = 30.87 / 3.5 = 8.82 m/s2
Amount of time it will be sliding
t = v / a
t = 28 / 8.82
t = 3.1746 seconds
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