In the Fig force `F = alpha t ` is applied on the block of mass `m_(2)`. Here `alpha` is a constant and `t` is the time .Find the acceleration of the block Also draw the acceleration time graph of both the block

Explanation

If `F le mu (m_(2))/(m_(1))(m_(1) + m_(2)) g` , then both blocks will move togather (illustration 7.28)
Here , `1 le mu (m_(2))/(alpham_(1))(m_(1) + m_(2))g`
During this time, acceleration of both blocks
`a_(1) = a_(2) = (F)/((m_(1) + m_(2)))= (alpha t)/((m_(1) + m_(2)))`
If `t ge (mu m_(2))/(am_(1))(m_(1) + m_(2)) g` friction will be of kinetic nature

Free - body diagrams :
Acceleration of `m_(1) : a_(1) = (mu N)/(m_(1)) = (mu m_(2)g)/(m_(1))`
Acceleration of `m_(2) : a_(2) = (alpha t - mu m_(2)g)/(m_(2))`
Acceleration time graph for `m_(1)` :
`t_(0) =( mu m_(2))/(alpha m_(1))(m_(1) + (m_(2))g`
` a_(t_(0)) = (alpha t_(0))/(m_(1) + (m_(2)) = (mu m_(2) g)/(m_(2))`

Acceleration time graph for `m_(2)` Before time `t_(0)` . slope of the graph is `(alpha)/(m_(1) + (m_(2))` and after `t_(0)`slope of the graph becomes `alpha//m_(2)` .Slope increase after `t_(0)` hence the graph