Neglect friction & inertia of pulley

Let length of strings associated with A & B be l1 & l2 respectively.

$As 2r2 = r1 \\ 2l2 = l1 $

Now, as total length of string remains constant

$2l2 - l1 = c \text {_____ }$[as l2 is increasing & l1 is decreasing]

Differentiating w.r.t. t

$2v2 = v1 \text {_____(i) } $

Differentiating w.r.t. t

$2a2 = a1 \text {_____(ii) }$

Now, considering block B

Applying newton's second law

$\sum Fy= may \\ T - 60 = (60/9.81)*a2 \\ T - 6.17a2 = 60 \text {_____(iii) }$

Considering block A

$\sum Fy= may \\ 40 - T = (40/9.81)*a1 \\ T + 4.08a1 = 40 \text {_____(iv) }$

Using equations (ii),(iii) & (iv)

We get, $a1= -2.79 m/s^2 , a2 = 1.395 m/s^2$ & $T=51.39 N$

Downward acceleration of A is $-2 .79 m/s^2 $