Question

In unloading grain from the hold of a ship, an elevator lifts the grain through a height of 12 m. Grain is discharged at the top of the elevator at a rate of 2 kg/s, and the discharge speed of each grain particle is 3m/s. Then,the minimum power of the motor that can elevate grain in this way is

• A. 149 W
• B. 249 W
• C. 100 W
• D. 549 W

Answer 1

The correct option is B 249 W

The work done by motor = Change in Mechanical energy
The power output of the motor is
$Power=\frac{\Delta K+\Delta U}{t}=\frac{\frac{1}{2}m(v_f^2-v_i^2+gh)}{t}=\frac{m}{t}[\frac{1}{2}{v}_f^2+gh](\because v_i=0)=2\left[\frac{1}{2}\right(3{)}^2+\left(10\right)\left(12\right)]=249W$