Question

A 0.171-kilogram hockey puck is struck by a hockey stick. If the force from the hockey stick is 30.2 Newtons, but the puck also experiences a force of friction of 3.5 Newtons at the same time, what is the acceleration of the puck?

Answer 1

Answer:

a = 156.14 m/s^2

Explanation:

Using the laws of newton:

∑F = ma  

where ∑F is the sumatory of forces, m the mass and a the aceleration.

so:

F -$f_k$ = ma

where F is the force from the hockey stick and $f_k$ is the force of friction.

Replacing values, we get:

(30.2N) - (3.5N)= (0.171kg)a

Finally, solving for a:

a = 156.14 m/s^2