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A horizontal force of 20 N is required to keep a 5.0 kg box traveling at a constant speed up a frictionless incline for a vertical height change of 3.0 m. (a) What is the work done by gravity during this change in height? (b) What is the work done by the normal force? (c) What is the work done by the horizontal force?

(a) 0 J
(b) 0 J
(c) 60 J
(d) 100 J

User BSP
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1 Answer

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Final answer:

The work done by gravity is 0 J, the work done by the normal force is 0 J, and the work done by the horizontal force is 60 J.

Step-by-step explanation:

(a) The work done by gravity during the change in height can be calculated using the formula:

work = force × distance × cos(angle).

In this case, the force of gravity is equal to the weight of the box, which is given by mass × acceleration due to gravity = 5.0 kg × 9.8 m/s² = 49 N. The distance is 3.0 m and the angle between the force and displacement is 90°. Therefore, the work done by gravity is 49 N × 3.0 m × cos(90°) = 0 J.

(b) The work done by the normal force is zero because the normal force is perpendicular to the displacement of the box.

(c) The work done by the horizontal force can be calculated using the same formula as in (a). The force is 20 N, the distance is 3.0 m, and the angle between the force and displacement is 0°. Therefore, the work done by the horizontal force is 20 N × 3.0 m × cos(0°) = 60 J.

User Cmfolio
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