Initially friction between mass A & table is just sufficient to prevent the motion. If an additional 1.25 kg is added to the 3.75 kg mass, find the acceleration of the masses

As initially system is in equilibrium, applying conditions of equilibrium.

Block A

$T - µ 25 g=0 \\ 3.75g - µ 25(9.81)=0 \\ µ=0.15 $

Block B

$3.75(g)-T=0 \\ T= 3.75g $

Now, after addition of mass to block B, system starts motion

Let acc of blocks be 'a'

Applying NSL to block A

$\sum Fx= max \\ T - 0.15(25g)=25a \text {____(i) }$

Applying NSL to block B

$\sum Fy= may \\ 5g - T=5a\text {____(ii) } $

Solving (i) & (ii)

$T=47 N $ & $a=0.409 m/s^2 $

Acceleration of masses is $0.409 m/s^2$