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

The maximum frictional force acting on the two blocks on the horizontal surface is given by:

$f = \mu m_1 g + \mu m_2 g$

Substituting the given values:

$f = 0.2 \times 1.0 \times 10 + 0.2 \times 1.0 \times 10 = 2 + 2 = 4 \text{ N}$

The driving force due to the hanging block is:

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

Since the driving force is greater than the total frictional force ($5 \text{ N} > 4 \text{ N}$), the system will accelerate. Let $a$ be the common acceleration of the system. The net force on the system is:

$F_{\text{net}} = W - f = 5 - 4 = 1 \text{ N}$

The total mass of the system is:

$M = 1.0 + 1.0 + 0.5 = 2.5 \text{ kg}$

The acceleration of the system is:

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

Let $T$ be the tension in the string attached to the $0.50 \text{ kg}$ block. Considering the free body diagram of the $0.50 \text{ kg}$ block, the equation of motion is:

$m_3 g - T = m_3 a$

$5 - T = 0.5 \times 0.4$

$5 - T = 0.2$

$T = 5 - 0.2 = 4.8 \text{ N}$