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

The maximum friction force between the two blocks is given by f_{\text\max} = \mu m_1 g

f_{\text\max} = 0.20 \times 2 \times 10 = 4 \text{ N}

Assuming both blocks move together, their common acceleration would be $a = \dfrac{F}{m_1 + m_2}$

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

The force required to move the lower block with this acceleration is $f = m_2 a = 4 \times 2 = 8 \text{ N}$

Since f > f_{\text\max}, the blocks will slip over each other. The friction force acting between them will be $f_k = 4 \text{ N}$.

For the upper block, the equation of motion is $F - f_k = m_1 a_1$

$12 - 4 = 2 a_1 \Rightarrow a_1 = 4 \text{ m/s}^2$

For the lower block, the equation of motion is $f_k = m_2 a_2$

$4 = 4 a_2 \Rightarrow a_2 = 1 \text{ m/s}^2$