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 rate which work is done by friction force on sand is

Options

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