What is the final speed (in m/s) of the puck after being pushed for a time of .0721 seconds?
A 22.6-N horizontal force is applied to a 0.0710-kg hockey puck to accelerate it across the ice from an initial rest position.

The answer to this is 23.0. You should not think about air resistance at this point because you might only get confused. What you should focus on are the different forces that may cause the object to act that way such as the up and down force balance.

Another force to reckon with is the acceleration and the net force. The net force is 22.6 N which is equal to the applied force. Knowing this, you can follow one formula which is a = Fnet / m. This is based on Newton’s second formula. This will become 22.6 N, right / (0.0710 kg) = 318 m/s/s.

Now that you know the acceleration value, you can look for what the puck’s final speed is. You can use the formula vf = vi + atvf = 0 m/s which will translate to (318 m/s/s)(0.0721 s)vf = 23.0 m/s.

23.0 m/s

Upon neglecting air resistance, there are three forces acting upon the object. The up and down force balance each other and the acceleration is caused by the applied force. The net force is 22.6 N, right (equal to the only rightward force - the applied force). So the acceleration of the object can be computed using Newtons second law. a = Fnet / m = (22.6 N, right) / (0.0710 kg) = 318 m/s/s, right The acceleration value can be used with other kinematic information (vi = 0 m/s, t = 0.0721 s) to calculate the final speed of the puck. The kinematic equation, substitution and algebra steps are shown. vf = vi + at vf = 0 m/s + (318 m/s/s)(0.0721 s) vf = 23.0 m/s