At a factory, a 300-kg crate is dropped vertically from a packing machine onto a conveyor belt (I apologize for not having a picture) moving at a speed of 1.2 m/s. The coefficient of kinetic friction between the belt and the crate is .40. After a short time, slipping between the belt and the crate ceases and the crate then moves along the belt. For the period of time during which the crate is being brought to rest relative to the belt, calculate, for a coordinate system at rest in the factory, (A) the work done by friction...
I can do the entirety of the problem except I am unsure of this first part and need a little help from the community.
W(friction) = E = (delta)KE = 1/2m(vf)^2 - 1/2m(vi)^2 = 216J.
This answer is correct however I am uneasy with the approach to the solution, if it is correct(doubtful) could someone explain how? otherwise a solution would be appreciated. -Thank you!