A horizontal conveyor belt moves with a constant velocity `V.A` small block is projected with a velocity of `6m//s` on it in a direction opposite to the direction of motion of the belt The block comes to rest relative to the belt in a time `4s. mu =0.3, g =10 .m//s^(2)` Find `V` .

Explanation

`|vecV_("b,c")|=V_(b)+V_(c)=6+V`
`f=mumg=0.3xxmxx10=3m`
`"Retardation "a=(F)/(m)=(3m)/(m)=3m//s^(2)`
`u_(r)=6+V, V_(r)=0, t=4sec, a_(r)=-3ms^(-2)`
`V_(r)=m_(r)+a_(r)t, 0=(6+V)-3xx4, V=6m//s`