The friction coefficient between the blocks is `mu`. If force `F` is applied on the lower block , analyse motion of two blocks.

Explanation

When force `F` is applied on the lower block , two possibilities , either two blocks move together with same acceleration or the upper block will move in backward direction with respect to the lower block.
Let two blocks move together with acceleration `a`.
`a = (F)/(m_(1) + m_(2))`
Since the upper block is moving in the forward direction , this is due to the friction force acting on it the forward direction.
When force `F` is applied on the lower block. the upper block has a tendency to move backward but the friction will try to prevent relative motion hence it will act on the upper block in the forward direction . But the friction has its maximum value `mu N = mu m_(1) g` , up to a certain stage it will prevent the backward motion of the upper block with respect to the lowe block
`f = m_(1) a = (m_(1) F)/(m_(1) + m_(2))`
`f_(max) = mu N = mu m_(1)g`
Both the blocks will move together if
`f le f_(max)`
`(m_(1) F)/(m_(1) + m_(2)) le mu m_(1) g`
`F le mu(m_(1) + m_(2)) g`
The maximum value of `F` so that both the blocks move together
`F_(max) = mu(m_(1) + m_(2))g`
If `F gt F_(max)` , the upper block will move backward with respect to the lower block.
When `F gt F_(max)` ,
`m_(1) : mu m_(1) g = m_(1) a_(1) rArr a_(1) = mu g` (i)
`m_(2) : F - mu m_(1) g = m_(2) a_(2)` (ii)
`rArr a_(2) = (F)/(m_(2) - (mu m_(1) g)/(m_(2))`
`a_(1) lt a_(2)` , acceleration of the upper block w.r.t. the lower block is in the backward direction.
After sometime the upper block will separate from the lower block.
To solve problems , assume that the blocks are moving together , find the common acceleration `a = (F)/(m_(1) + m_(2))`.
(a) If `a le mu g` , the blocks move together.
(b) If `a gt mu g` , the upper block will move in the backward direction with respect to the lower block.
( c) The maximum value of `F` so that the blocks move together,
`F_(max) = mu (m_(1) + m_(2))g`