A 2 kg block is positioned over a 4 kg block, and both are placed...

Let $m_1 = 2 \text{ kg}$ be the mass of the upper block and $m_2 = 4 \text{ kg}$ be the mass of the lower block.

The maximum static friction between the two blocks is given by f_{\text\max} = \mu m_1 g = 0.20 \times 2 \times 10 = 4 \text{ N}.

Assuming both blocks move together as a single system, their common acceleration $a$ is:

$a = \dfrac{F}{m_1 + m_2} = \dfrac{12}{2 + 4} = 2 \text{ m/s}^2$

The frictional force required to provide this acceleration to the upper block is:

$f_{\text{req}} = m_1 a = 2 \times 2 = 4 \text{ N}$

Since the required friction $f_{\text{req}}$ is equal to the maximum available static friction f_{\text\max}, the blocks will not slip relative to each other.

Therefore, both the upper and lower blocks will move with an acceleration of $2 \text{ m/s}^2$.