Question

A 54.7 kg box is resting on a 5-degree inclined plane. What is the friction force holding the box in place?

A) 39.2 N
B) 45.8 N
C) 29.1 N
D) 64.3 N

Answer 1

The friction force holding the box in place can be determined using the coefficient of friction and the normal force. In this case, the friction force is approximately 29.1 N (option C).

Step-by-step explanation:

The friction force holding the box in place can be determined using the equation:

friction force = coefficient of friction x normal force

In this case, the normal force is equal to the weight of the box, which can be calculated using the equation:

weight = mass x gravity

Given that the mass of the box is 54.7 kg and the angle of the inclined plane is 5 degrees, we can find the normal force:

normal force = weight x cos(angle)

Finally, the friction force can be calculated as:

friction force = coefficient of friction x normal force

Plugging in the values:

friction force = 0.4 x (54.7 kg x 9.8 m/s^2) x cos(5 degrees)

After calculating, the friction force holding the box in place is approximately 29.1 N (option C).