For the block A:

Here the net downward (along the incline) force acting on the block Fa = m*g*sinθ + m*aT (component of weight and psedo force)

So Fa = 1*10*sin30 + 1*5 = 10 N

Also the normal reaction on the block will be Ra = m*g*cosθ = 1*10*cos30 = 8.66 N

Hence the maximum frictional force possible on the block fmax(a) = μs(a) * Ra = 0.4*8.66 = 3.46 N < fa

Hence the block A will move down the incline with acceleration aA = (fa - fk(a))/m = (10 - 0.2*8.66)/1 = 8.27 m/s2 (This acceleration is with respect to the truck. Also while the block is moving then only the kinetic friction force fk = μk(a) * Ra will act)

For the block B:

Normal reaction Rb = 2*m*g*cosθ = 17.32 N so the maximum frictional force possible fmax(b) = 0.5*17.32 =