Sand drops fall vertically at the rate of 3 kg/s on the conveyor belt moving horizontally with a speed of 0.25 m/s .the extra power required to keep it moving?

1 Answer
Jan 11, 2018

0.18 Watt

Explanation:

Force generated due to falling of sand on the conveyor belt is
#V(dm)/(dt)#
I.e #(0.25*3)#N or 0.75 N
So,this amount of force will be required per second to help per meter movement of the belt,hence extra power that would be required is #P=F*v=(0.75*0.25) # or 0.18 watt