Sand is droped vetically downward at the constant rate `mu kg//s` on a conveyor belt which is moving horizontally with velocity `'v'`. Assume that sant particles comes to rest with respect to belt immediately after landing.
The rae at which work is done by the external force applied on the belt.

Options

A. `(3)/(2)muv^(2)`
B. `muv(2)`
C. `(muv^(2))/(2)`
D. `-(muv^(2))/(2)`

Explanation

Let `'F'` be the force to be applied on belt to move with same `'V'`
`F=(dp)/(dt)=(d(mv))/(dt)=m(dv)/(dt)+(dm)/(dt)v=muv`
`(dw)/(dt)=(d)/(dt)(Fvdt)=F.v=muv^(2)`
By work energy theorem
`(dw_(F))/(dt)+(dw_("fr"))/(dt)=(dK)/(dt)`
`k=(mv^(2))/(2)`
`(dk)/(dt)=(dm)/(dt)(1)/(2)v^(2)=(muv^(2))/(2)`
`rArr muv^(2)+(dw_("fr"))/(dt)=(muv^(2))/(2)`
`(dw_("fr"))/(dt)=(muv^(2))/(2)`