Boxes are transported by a conveyor belt with a velocity v0 to a fixed incline at A, where they slide and eventually fall off at B. Knowing that μk = 0.40, determine the velocity of the conveyor belt if the boxes leave the incline at B with a velocity of 10 ft/s.

The velocity of the conveyor belt if the boxes leave the incline at B with a velocity of 10 ft/s = 16.3 ft/s

Explanation:

The image for the question is attached to this solution as the first file, the second file represents the free body diagram for one of the boxes when it's on AB

For this inclined plane, the normal reaction, N = W cos 15°

Fr = - μN = - μW cos 15°

Length of AB = 20 ft

Work done by the force of gravity = W × 20 sin 15° = 5.176 mg

Work done by the frictional force = - μW cos 15° × 20 = - 7.73 mg

Total workdone between A and B = 5.176mg - 7.73mg = - 2.554 mg

kinetic energy at B = (1/2)m(10²) = 50m

Kinetic energy at A = (1/2) m(v₀²) = mv₀²/2

Using the energy - work principle

(Change in kinetic energy between A and B) = (work done between A and B)

(50m - mv₀²/2) = (- 2.554 mg)

g = 32.2 ft/s²

50 - (v₀²/2) = - 2.554 × 32.2

v₀²/2 = 50 + 82.24 = 132.24

v₀² = 264.48

v₀ = 16.3 ft/s