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 :
[tex]v^2-u^2=2ax[/tex]
v = 0 (as it stops)
[tex]-u^2=2ax[/tex]
[tex]a=\dfrac{-u^2}{2x}[/tex]
[tex]a=\dfrac{-(8.1)^2}{2\times 16.8}[/tex]
[tex]a=-1.95\ m/s^2[/tex]
The frictional force is given by :
f = m a
[tex]f=0.112\ kg\times 1.95\ m/s^2[/tex]
f = 0.218 N
(b) Also the frictional force is given by :
[tex]f=\mu N[/tex]
And N = mg (normal force)
[tex]f=\mu mg[/tex]
[tex]\mu=\dfrac{f}{mg}[/tex]
[tex]\mu=\dfrac{0.218}{0.112\times 9.8}[/tex]
[tex]\mu=0.198[/tex]
Hence, this is the required solution.
