A block of mass 40kg slides over a surface, when a mass of 4kg is...

For hanging block,

$40 - T = 4a ⋯ \left(1\right)$

Now, maximum value of friction force will be,

{f}_\max = \muN = 0 . 02 \times 40 \times 10 = 8N

For block on table, $T - 8 = 40a ⋯ \left(2\right)$

From equation(1) and equation (2), we get$32 = 44a ⇒ a = \dfrac{32}{44} = \dfrac{8}{11}m{s}^{- 2}$