Question

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?

Answer 1

Answer:

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:

       $\Delta K = -\frac{1}{2} * m* v_{0} ^{2} = \mu_{k} *m*g* cos (180\º)$

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

       $\mu_{k} = \frac{v_{0}^{2} }{2*g*d} = \frac{9 (m/s)2}{2*9.8 m/s2*5.50m} = 0.08$