The stopping distance is 143.1 m
Explanation:
First of all, we have to find the acceleration of the hockey puck. This can be done by using Newton's second law of motion:
[tex]\sum F =ma[/tex]
where
[tex]\sum F = F_f = -0.14 N[/tex] is the net force acting on the puck (the force of friction, negative because it acts in a direction opposite to the direction of motion)
m = 0.12 kg is the mass of the puck
a is the acceleration
Solving for a,
[tex]a=\frac{\sum F}{m}=\frac{-0.14}{0.12}=-1.17 m/s^2[/tex]
The motion of the puck is a uniformly accelerated motion, therefore we can use the following suvat equation:
[tex]v^2-u^2=2as[/tex]
where:
v = 0 is the final velocity (the puck comes to a stop)
u = 18.3 m/s is the initial velocity
[tex]a=-1.17 m/s^2[/tex] is the acceleration
s is the stopping distance
And solving for s, we find
[tex]s=\frac{v^2-u^2}{2a}=\frac{0-(18.3)^2}{2(-1.17)}=143.1 m[/tex]
Learn more about accelerated motion:
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