Aconveyor belt, run by a motor, is moving at a constant speed of 5 m *s^(-1) . If 5 kg of sand is sprinkled per second on the belt, then what extra power should the motor supply to the belt?

Aconveyor belt, run by a motor, is moving at a constant speed of 5 m *

1.	Aconveyor belt, run by a motor, is moving at a constant speed of 5 m *s^(-1) . If 5 kg of sand is sprinkled per second on the belt, then what extra power should the motor supply to the belt?
Answer» Solution :The initial horizontal speed fosand before falling on the conveyor belt =0. So, the rate of SUPPLY of kinetic energy of the sand by the belt `=1/2 xx("mass"xx"velocity"^2)/("second")=1/2 xx(5xx5^2)/(1) =62.5 Js^(-1)` If there is no loss of energy otherwise, then EXTRA power to be supplied to the beltby the motor `=62.5 Js^(-1) or 62.5 W`.