Question

A conveyor belt is moving with constant speed of `6 m//s`. A small block is just dropped on it.Coefficient of friction between the two is `mu = 0.3`. Find
(a) The time when relative motion between them will stop
(b) Displacement of block upto that instant `(g = 10 m//s^(2))`.

Answer 1

`a = (f)/(m) = mu g = 3m//s^(2)`

(a) Relative motion will stop when velocity of block also becomes `6 m//s` by the above acceleration
`v= at`
`t = (v)/(a) = (6)/(3) = 2 s`
(b) `S = (1)/(2) at^(2) = (1)/(2) (3) (2) = 6 m`