d = (1/2) a t^2 = 45.6
You know a, solve for t
v = a t
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.
2 answers
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
f d = 1/2 m v^2
123 * 45.6 = 1/2 * 10.0 * v^2