In unloading grain from the hold of a ship, an elevator lifts the grian through a distance of 12 m. Grain is discharged at the top of the elevator at a rate of 2 kg each second and the discharge speed of each of grain particle is 3 `ms^(-1)`. Find the minimum horsepower of the motor that can elevate grain in this way. (Take g=10 `ms^(-2)`)

Explanation

The work done by the motor each second ,i.e.,
Power`=mgh+(1)/(2)mv^(2)`, as t=1 s.
Given, m=2kg, `v=3 ms^(-1)` and h=12 m.
`:. "Power"=2xx(10)xx(12)+(1)/(2)(2)(3)^(2)=249 W=(249)/(746)=0.33 hp`
The motor must have an output of at least 0.33 hp.