Question

Boxes are transported from one location to another in a warehouse by means of a conveyor belt that moves with a constant speed of 0.50 m/s. At a certain location the conveyor belt moves for 2.0 m up an incline that makes an angle of 10° with the horizontal, then for 2.0 m horizontally, and finally for 2.0 m down an incline that makes an angle of 10° with the horizontal. Assume that a 2.0 kg box rides on the belt without slipping. At what rate is the force of the conveyor belt doing work on the box as the box moves (a) up the 10° incline, (b) horizontally, and (c) down the 10° incline?

Solution

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Answered 6 months ago
Answered 6 months ago
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Givens:
The speed of the conveyor belt, v=0.5v=0.5 m/s.
The displacement of the conveyor belt up the incline, x1=2x_1=2 m.
The angle between the incline and the horizontal, θ=10\theta = 10^\circ.
The displacement of the conveyor belt horizontally x2=2x_2=2 m.
The displacement of the conveyor belt down the incline, x3=2x_3=2 m.
The mass of the box riding the belt, m=2m=2 kg.
Fg\vec{F_g} is the weight of the box.

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