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

`|vecV_(b.c)| = V_(b) + V_( c) = 6 + v`
`f= mumg = 0.3 xx m xx 10 = 3m`
Retardation `a=(3m)/m = 3 m//s^(2)`
`mu_(f) = 6 + V, V_(r) = 0, t= 4` sec
`a_(r) = -3 ms^(-2), V_(r) = mu_(r) + a_(r)t`
`0 = (6+V)- 3 xx 4, V = 6 m//s`