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1)A 25 kg air compressor is dragged up a rough incline from r⃗ 1=(1.3ı^+1.3ȷ^)m to r⃗ 2=(8.3ı^+2.6ȷ^)m, where the y-axis is vertical. How much work does gravity do on the compressor during this displacement?

2)A 25 kg box sliding to the left across a horizontal surface is brought to a halt in a distance of 55 cm by a horizontal rope pulling to the right with 16 N tension.How much work is done by tension?How much work is done by gravity?

User Returneax
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Final answer:

The gravity does -318.5 J of work on the air compressor as it is dragged uphill, and tension does 8.8 J of work on the box to bring it to a halt; gravity does no work on the horizontally sliding box.

Step-by-step explanation:

Work Done by Gravity and Tension

The work done by gravity on a 25 kg air compressor being dragged up an incline can be calculated by determining the change in height and using the formula for gravitational work, which is W = mgh, where m is mass, g is the acceleration due to gravity, and h is the change in vertical height. To find the change in height, we subtract the initial y-coordinate from the final y-coordinate: 2.6 m - 1.3 m = 1.3 m. Therefore, the work done by gravity is (25 kg)(9.8 m/s2)(1.3 m), which is -318.5 J. The negative sign indicates that gravity does work against the direction of the displacement.

For the 25 kg box being halted by tension, the work done by the tension force can be determined by the formula Work = force × distance. Since the tension is 16 N and the box stopped over a distance of 55 cm (or 0.55 m), the work done by tension is (16 N)(0.55 m), yielding 8.8 J. Gravity does zero work on the box as it is moving horizontally and there is no vertical displacement.

User Kimichang
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