Physics question

Question

A 5.93 kg block is placed at the top of a frictionless inclined plane angled at 22.4 degrees relative to the horizontal. When released (from rest), the block slides down the full 4.99 meter length of the incline. Calculate the speed (magnitude of the velocity) of the block at the bottom of the incline. [Start by drawing a free-body diagram for the block.] Note that all the information provided may not be necessary to solve the problem.

Answer 1

Initial height =4.99sin(22.4)=1.91m. Initial energy = mgh = final energy =.5mv². Solve to find v=√2gh = 20.3 m/s