Question

A hockey stick strikes a hockey puck of mass 0.17 kg. If the force exterted on the hockey puck is 35.0 N and there is a force of friction of 2.7 N opposing the motion. what is the acceleration of the hockey puck?
Choices:
0.0054 m/s^2
0.005 m/s^2
200 m/s^2
190 m/s^2

Answer 1

Answer:

$a=190\ m/s^2$

Explanation:

Mass of a hockey puck, m = 0.17 kg

Force exerted by the hockey puck, F' = 35 N

The force of friction, f = 2.7 N

We need to find the acceleration of the hockey puck.

Net force, F=F'-f

F=35-2.7

F=32.3 N

Now, using second law of motion,

F = ma

a is the acceleration of the hockey puck

$a=\dfrac{F}{m}\\\\a=\dfrac{32.3}{0.17}\\\\a=190\ m/s^2$

So, the acceleration of the hockey puck is $190\ m/s^2$.