Consider the situation as shown in the figgure. The coefficent of friction between the blocks is `mu`. Find the acceleration of each block.

Explanation

(a) Consider the motion of the block of mass `m_(1)`.
`N = m_(1) g`
`f_(max) = mu N = mu m_(1) g`
If `F lt f_(max) (= mu m_(1) g)`, the block of mass `m_(1)` will not sure move over `m_(2)`. In fact both blocks will move together as a single block.
`F = (m_(1) + m_(2)) a rArr a = (F)/(m_(1) + m_(2))`
Note : A body on rough surface moves only when the applied force `gt f_(max)`.
(b) `F gt f_(max) ( = mu m_(1) g)` , the two blocks will move with different accelerations.
`F - mu m_(1) g = m_(1) a_(1)`
`a_(1) = (F)/(m_(1)) - mu g`
`mu m_(1) g = m_(2) a_(2)`
`a_(2) = (mu m_(1) g)/(m_(2)`