Sand particles drop vertically at the rate of `2kgs^-1` on a conveyor belt moving horizontally with a velocity of `0.2ms^-1`.
The time rate of change of kinetic energy of sand particles is

Options

A. (a) `0.4Js^-1`
B. (b) `0.08Js^-1`
C. (c) `0.04Js^-1`
D. (d) `0.2Js^-1`

Answer

C

Explanation

Force required to keep the belt moving `=F`
`F=v(dm)/(dt)=0.2xx2=0.4N`
Extra power required: `P=Fv=0.4xx0.2=0.08W`
Rate of change of kinetic energy is
`1/2v^2((dm)/(dt))=1/2(0.2)^2xx2=0.04Js^-1`