A conveyor 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?

Explanation

The initial horizontal speed fo sand 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 J*s^(-1)`
If there is no loss of energy otherwise, then extra power to be supplied to the belt by the motor `=62.5 J*s^(-1) or 62.5 W`.