FBD of the block is shown in the fig.

Let 'N' be the Normal reaction.

Since, it is under equilibrium, $\sum F_Y=0;$

$N-5g\cos40=0$

$N = 5g\ cos40$

$Dynamic\ Friction\ force= F_D= \mu N$

$= 0.2 \times 5g\cos 40 = g\ cos40$

Let 'ma' be the D'Alemberts force acting on the block.

Due to D'Alemberts force, block 'A' is under equilibrium hence $\sum F_X=0, $

$ma + F_D-5g\sin40=0$

$5a + g\cos40 - 5g\sin40=0$

$a = \dfrac {5g\sin40 - g\ cos40} 5$

$Hence,\ a=4.8011\ m/s^2$

$Acceleration\ of\ the\ block\ using\ D'Alemberts\ priniciple\ =a=4.8011\ m/s^2$