A 300 kg crate is dropped vertically onto a conveyor belt that is moving at 1.20 m/s. A motor maintains the belt's constant speed. The belt initially slides under the crate, with a coefficient of

Question

A 300 kg crate is dropped vertically onto a conveyor belt that is moving at 1.20 m/s. A motor maintains the belt's constant speed. The belt initially slides under the crate, with a coefficient of friction of 0.400. After a short time, the crate is moving at the speed of the belt. During the period in which the crate is being accelerated, find the work done by the motor which drives the belt :-

TestHub · Accepted Answer

Kinetic friction on the crate: f = μ m g = 0.4 × 300 × 10 = 1200 N , giving crate acceleration a = μg = 4 m/ s 2 Time to reach belt speed: t = 4 1.2 ​ = 0.3 s ; belt displacement meanwhile: x = 1.2 × 0.3 = 0.36 m The motor holds the belt at constant speed against the friction the crate drags on it: W = f x = 1200 × 0.36 = 432 J Ans. (A)

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