A 5.00-kg box slides 5.50 m across the floor before coming to rest. What is the coefficient of kinetic friction between the floor and the box if the box had an initial speed of 3.00 m/s?

μk = 0.08

Explanation:

Applying the work-energy theorem, we know that the change in kinetic energy of the box, is equal to the total work done on it.The only force acting on the box capable of doing work, is the dynamic friction force.This force is just the product of  the coefficient of kinetic friction, and the normal force.In this case, if the floor is level, the normal force is equal and opposite to the gravity force, Fg.So we can write the following expression:

       

Replacing by the givens, we can solve for μk, as follows: