Sand is being dropped on a conveyor belt at the rate of `Mkg//s` . The force necessary to kept the belt moving with a constant with a constant velocity of `vm//s` will be.

Options

A. `(Mv)/(2)` newton
B. zero
C. Mv newton
D. 2 Mv newton

Answer

C

Explanation

`F_("ext.")=(dp)/(dt)=(d(mv))/(dt),`
m = mass of system as conveyor belt with sand drops at time t.
`F_("ext.")=m(dv)/(dt)+v(dm)/(dt)`
`"but v "rarr "constant so, "(dv)/(dt)=0`
`F_("ext")=v(dm)/(dt)=Mv`
`F_("ext")` is in direction of `vec v` of belt.
OR
`"Since "m.(dv)/(dt)=F_("ext")+F_("reaction")" ......(i)"`
Consider belt as a system with variable mass
`(dv)/(dt)=0,m = "mass of systme at time t"`
`F_("reaction")=v_("rec").(dm)/(dt)=-v(dm)/(dt)" ......(ii)"`
`rArr" "F_("ext.")=v.(dm)/(dt)=Mv.`
`vecF_("ext")` is in direction of `vec v` to keep belt moving with constant velocity.