Question

A 16.5 kg block is dragged over a rough, horizontal surface by a constant force of 187 N acting at an angle of 27° above the horizontal. The block is displaced 31.3 m, and the coefficient of kinetic friction is 0.22. What is the work done by the force in moving the block?

1) 0 J
2) 1,000 J
3) 5,000 J
4) 10,000 J

Answer 1

The work done by the force in moving the block is 5,000 J. Therefore, the correct option is 3.

Step-by-step explanation:

The work done by the constant force in moving the 16.5 kg block over a rough surface is calculated using the formula:

Work = Force × Distance × cos(θ)

Given a force of 187 N at an angle of 27° above the horizontal and a displacement of 31.3 m, the work done is expressed as:

Work = 187 N × 31.3 m × cos(27°)

Considering the angle, the cosine of 27° is approximately 0.8988. Therefore, the work done is:

Work = 187 N × 31.3 m × 0.8988 ≈ 5,000 J

Thus, the correct answer is 3) 5,000 J, representing the work done by the force in moving the block over the given distance.