Question

An 18.5 kg block is dragged over a rough, horizontal surface by a constant force of 132 N acting at an angle of 27.2�� above the horizontal. The block is displaced 17.3 m, and the coefficient of kinetic friction is 0.117. What is the work done by the force in moving the block?

1) 0 J
2) 2,288.5 J
3) 2,288.5 J
4) 2,288.5 J

Answer 1

The work done by the force can be calculated using the formula Work = Force * displacement * cos(theta). In this case, the work done is 2,288.5 J.

Step-by-step explanation:

The work done by a force can be calculated by multiplying the magnitude of the force by the displacement in the same direction as the force. In this case, the force is 132 N and the displacement is 17.3 m. However, we need to take into account the angle at which the force is acting. The work done by the force can be calculated as:

Work = Force * displacement * cos(theta)

where theta is the angle between the force and the displacement. In this case, theta is 27.2 degrees.

Using the given values, the work done by the force is calculated as:

Work = 132 N * 17.3 m * cos(27.2)

Calculating this expression gives us the answer of 2,288.5 J.