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

Let the masses of the blocks on the horizontal surface be $m_1 = 1.0 \text{ kg}$ and $m_2 = 1.0 \text{ kg}$.

Let the mass of the hanging block be $m_3 = 0.5 \text{ kg}$.

The driving force for the system is the weight of the hanging block:

$F_{\text{driving}} = m_3 g = 0.5 \times 10 = 5 \text{ N}$

The frictional force opposing the motion acts on both blocks on the horizontal surface. The total frictional force is:

$f_k = \mu m_1 g + \mu m_2 g = \mu (m_1 + m_2) g$

$f_k = 0.2 \times (1.0 + 1.0) \times 10 = 4 \text{ N}$

Since the driving force ($5 \text{ N}$) is greater than the total frictional force ($4 \text{ N}$), the system will accelerate.

The net force acting on the system is:

$F_{\text{net}} = F_{\text{driving}} - f_k = 5 - 4 = 1 \text{ N}$

The total mass of the system moving together is:

$M_{\text{total}} = m_1 + m_2 + m_3 = 1.0 + 1.0 + 0.5 = 2.5 \text{ kg}$

The acceleration $a$ of the system is given by Newton's second law:

$a = \dfrac{F_{\text{net}}}{M_{\text{total}}} = \dfrac{1}{2.5} = 0.4 \text{ m/s}^2$