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:
![\sum F =ma](https://tex.z-dn.net/?f=%5Csum%20F%20%3Dma)
where
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,
![a=\frac{\sum F}{m}=\frac{-0.14}{0.12}=-1.17 m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B%5Csum%20F%7D%7Bm%7D%3D%5Cfrac%7B-0.14%7D%7B0.12%7D%3D-1.17%20m%2Fs%5E2)
The motion of the puck is a uniformly accelerated motion, therefore we can use the following suvat equation:
![v^2-u^2=2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
where:
v = 0 is the final velocity (the puck comes to a stop)
u = 18.3 m/s is the initial velocity
is the acceleration
s is the stopping distance
And solving for s, we find
![s=\frac{v^2-u^2}{2a}=\frac{0-(18.3)^2}{2(-1.17)}=143.1 m](https://tex.z-dn.net/?f=s%3D%5Cfrac%7Bv%5E2-u%5E2%7D%7B2a%7D%3D%5Cfrac%7B0-%2818.3%29%5E2%7D%7B2%28-1.17%29%7D%3D143.1%20m)
Learn more about accelerated motion:
brainly.com/question/9527152
brainly.com/question/11181826
brainly.com/question/2506873
brainly.com/question/2562700
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