A bag is gently dropped on a conveyor belt moving at a speed of 2m...

When the bag is dropped on the conveyor belt, the acceleration of the bag will be $a = \mug = 4m{s}^{- 2}$

The bag will stop moving on the conveyor when the velocity of the bag and the velocity of the conveyor becomes equal.

$u = 0, v = 2m{s}^{- 1}$

Using the equation of motion for constant acceleration,

${v}^{2} = {u}^{2} + 2as$

$⇒ s = 0 . 5m$