Question

A force of 123 N acts on a 10.0kg object for 45.6 meters. If the objects starts at rest, and the surface is flat and frictionless, how fast is the object moving after the force is removed.

I got acceleration to be 12.3, but I am not sure how to move on with the problem.

Answer 1

d = (1/2) a t^2 = 45.6
You know a, solve for t

v = a t

Answer 2

the work (energy) is equal to the KE of the object

f d = 1/2 m v^2

123 * 45.6 = 1/2 * 10.0 * v^2