Question

An 8.5 kg crate is pulled 5.3 m up a 30° incline by a rope angled 19 ° above the incline. The tension in the rope is 120 N and the crate's coefficient of kinetic friction on the incline is 0.27. What is the work done by the tension force in pulling the crate up the incline?

1) 636 J
2) 318 J
3) 212 J
4) 106 J

Answer 1

The work done by the tension force in pulling the crate up the incline is 636 J.

Step-by-step explanation:

To calculate the work done by the tension force in pulling the crate up the incline, we need to consider the different forces acting on the crate.

The work done by the tension force can be calculated using the formula:

Work = Force * Displacement * cos(theta)

In this case, the force is the tension in the rope, which is 120 N. The displacement is the distance the crate is pulled up the incline, which is 5.3 m. The angle between the force and displacement vectors is the angle between the rope and the incline, which is 19° + 30° = 49°.

Now we can calculate the work:

Work = 120 N * 5.3 m * cos(49°) = 636 J

Therefore, the work done by the tension force in pulling the crate up the incline is 636 J.