Newton's Laws of Motion?

Question

A 500g hockey puck experiences an applied horizontal force of 50N [W] as it slides across the ice with an acceleration of 2.3 m/s2 [W] against the force of friction. Determine the magnitude and direction of the frictional force acting on the hockey puck.

Answer 1

Since F=ma (Second Law) then a = F/m. So, the net force will be 50N - Frictional Force (which will be x)

(50N - x)/.5kg = 2.3m/s^2
50N - x = 1.15N
x = 48.85N

The direction of the frictional force is always opposite of the applied force.