Work-energy theorem: A 5.00-kg box slides 4.00 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?

Question

Quizplus · Accepted Answer

The work-energy theorem states that the work done on an object is equal to its change in kinetic energy. In this problem, the work done on the box is equal to the force of friction multiplied by the distance it travels. Therefore, we have:

work done on box = force of friction x distance traveled

We can also express the work done on the box as its change in kinetic energy:

work done on box = (final kinetic energy) - (initial kinetic energy)

Since the box comes to rest at the end of its motion, its final kinetic energy is zero. So we have:

0.5 x 5.00 kg x (3.00 m/s)^2 = force of friction x 4.00 m

Solving for the force of friction, we get:

force of friction = 22.5 N

The force of friction is equal to the coefficient of kinetic friction multiplied by the normal force:

force of friction = coefficient of kinetic friction x (mass of box x acceleration due to gravity)

The normal force is equal to the weight of the box:

normal force = mass of box x acceleration due to gravity

Substituting these equations into our original expression for the force of friction, we get:

coefficient of kinetic friction = force of friction / (mass of box x acceleration due to gravity)

Plugging in the values we found earlier, we get:

coefficient of kinetic friction = 22.5 N / (5.00 kg x 9.81 m/s^2) = 0.115

Therefore, the answer is C: 0.115.

Work-Energy Theorem: a 5