Consider the arrangement depicted in the figure. Determine the...

Let the masses be $m_1 = 1.0\text{ kg}$ (left), $m_2 = 1.0\text{ kg}$ (middle), and $m_3 = 0.5\text{ kg}$ (hanging).

The maximum friction force on the two blocks on the horizontal surface is:
f_{\text\max} = \mu m_1 g + \mu m_2 g = 0.2 \times 1.0 \times 10 + 0.2 \times 1.0 \times 10 = 2 + 2 = 4\text{ N}

The driving force due to the hanging block is:
$F_{\text{drive}} = m_3 g = 0.5 \times 10 = 5\text{ N}$

Since F_{\text{drive}} > f_{\text\max}, the system will accelerate. Let the common acceleration be $a$.

The equation of motion for the entire system is:
m_3 g - f_{\text\max} = (m_1 + m_2 + m_3) a
$5 - 4 = (1.0 + 1.0 + 0.5) a$
$1 = 2.5 a \implies a = 0.4\text{ m/s}^2$

Let $T$ be the tension in the string connecting the two $1.0\text{ kg}$ blocks. Considering the free body diagram of the left block ($m_1$):
$T - \mu m_1 g = m_1 a$
$T - 2 = 1.0 \times 0.4$
$T = 2.4\text{ N}$