As shown in the figure, an elevator is descending with an...

Let the normal force between block A and block B be $N$.

Considering the free body diagram of block A, the forces acting on it are its weight ($m_A g$) downwards and the normal reaction ($N$) upwards.

Since the elevator is accelerating downwards with an acceleration $a = 2 \text{ m/s}^2$, the equation of motion for block A is:

$m_A g - N = m_A a$

$\Rightarrow N = m_A(g - a)$

Given $m_A = 0.5 \text{ kg}$, $a = 2 \text{ m/s}^2$, and taking $g = 10 \text{ m/s}^2$:

$N = 0.5 \times (10 - 2)$

$N = 0.5 \times 8 = 4 \text{ N}$

By Newton's third law, the force exerted by block A on block B is equal to the normal reaction $N$, which is $4 \text{ N}$.