Question

Two identical 5.0-kg boxes are stacked one on top of the other. The stack of boxes is accelerated at a rate of 2.0 m/s^2 without any slipping. The coefficient of friction between the two boxes is 0.4. Which of the following can we say about the friction component of the force the bottom box exerts on the top box?

a. The friction force is 10 N.

b. The friction force is 20 N.

c. The friction force is 30 N.

d. The friction force is 40 N.

Answer 1

The friction force between the two boxes is 19.6 N.

Step-by-step explanation:

The friction component of the force the bottom box exerts on the top box can be found using the equation:

Friction force = coefficient of friction × normal force

Since the two boxes are identical and have the same mass, the normal force exerted by the bottom box on the top box is equal to the weight of the top box:

Normal force = mass × gravity

Therefore, the friction force is equal to:

Friction force = coefficient of friction × mass × gravity

Plugging in the values:

Friction force = 0.4 × 5.0 kg × 9.8 m/s²

Friction force = 19.6 N

Therefore, the correct answer is: a. The friction force is 19.6 N.