Question

You observe a hockey puck of mass 0.12 kg, traveling across the ice at speed 18.3 m/sec. The interaction of the puck and the ice results in a frictional force on the puck, f = 0.14 N.
Calculate: the stopping distance for this puck.

Answer 1

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:

$\sum F =ma$

where

$\sum F = F_f = -0.14 N$ 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,

$a=\frac{\sum F}{m}=\frac{-0.14}{0.12}=-1.17 m/s^2$

The motion of the puck is a uniformly accelerated motion, therefore we can use the following suvat equation:

$v^2-u^2=2as$

where:

v = 0 is the final velocity (the puck comes to a stop)

u = 18.3 m/s is the initial velocity

$a=-1.17 m/s^2$ is the acceleration

s is the stopping distance

And solving for s, we find

$s=\frac{v^2-u^2}{2a}=\frac{0-(18.3)^2}{2(-1.17)}=143.1 m$

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