## Related questions with answers

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

Verified**Givens:**

The speed of the conveyor belt, $v=0.5$ m/s.

The displacement of the conveyor belt up the incline, $x_1=2$ m.

The angle between the incline and the horizontal, $\theta = 10^\circ$.

The displacement of the conveyor belt horizontally $x_2=2$ m.

The displacement of the conveyor belt down the incline, $x_3=2$ m.

The mass of the box riding the belt, $m=2$ kg.

$\vec{F_g}$ is the weight of the box.

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