A horizontal converyor belt moves with a constant velocity V. A small block is projected with a velocity of 6 mis 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, g= 10 m//s^(2)`
Find V

Explanation

`|vec(V)_(b,c)|=V_(b)+V_(c )=6+V`
`f=mu mg=0.3xx m xx10=3m`
Retardation `a=(3m)/(m)=3m//s^(2)`
`u_(r )=6+V, V_(r )=0, t=4 sec`
`a_(r )=-3 ms^(-2)`
`V_(r ) =u_(r )+a_(r )t`
`0=(6+V)-3xx4`
`V=6m//s`