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A farmer after a successful harvest brings his product home using a crate. To the crate and the harvest with total mass of 50.0 kg, he exerts a force of 1.20 x 102 N by pulling on the rope.

A) How much work does he do on the crate if the rope is horizontal to the ground and he pulls the crate 5.00 m?

B) How much work does he on the crate if the rope makes an angle of 30.0 degrees with the horizontal and pulls the sled the same distance?

1 Answer

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A) When the rope is horizontal to the ground, the force the farmer exerts is in the same direction as the displacement of the crate. Therefore, the work done by the farmer on the crate can be calculated using the formula:

work = force x distance x cos(theta)

where force is the magnitude of the force applied, distance is the displacement of the crate, and theta is the angle between the force and the displacement. In this case, theta is 0 degrees because the force is in the same direction as the displacement.

Substituting the given values, we get:

work = (1.20 x 10^2 N) x (5.00 m) x cos(0 degrees)

work = 600 J

Therefore, the farmer does 600 J of work on the crate when the rope is horizontal to the ground.

B) When the rope makes an angle of 30.0 degrees with the horizontal, the force the farmer exerts is at an angle of 30.0 degrees to the displacement of the crate. Therefore, the work done by the farmer on the crate can be calculated using the same formula:

work = force x distance x cos(theta)

but this time, theta is 30.0 degrees.

Substituting the given values, we get:

work = (1.20 x 10^2 N) x (5.00 m) x cos(30.0 degrees)

work = (1.20 x 10^2 N) x (5.00 m) x 0.866

work = 519 J

Therefore, the farmer does 519 J of work on the crate when the rope makes an angle of 30.0 degrees with the horizontal.
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