To solve this problem, we need to analyze the forces acting on both blocks.$\\$Consider block A on the table:$\\$• The force of tension $T$ acts horizontally.$\\$• The force of kinetic friction $f_k = \mu_k m_1 g$ opposes the motion.$\\$For block B hanging vertically:$\\$• The gravitational force $m_2 g$ acts downward.$\\$• The tension $T$ acts upward.$\\$Since the system is accelerating, we apply Newton's second law to each block.$\\$For block A:$\\ T - \mu_k m_1 g = m_1 a \quad (1) \\$For block B:$\\ m_2 g - T = m_2 a \quad (2) \\$Add equations (1) and (2) to eliminate $T: \\ (T - \mu_k m_1 g) + (m_2 g - T) = m_1 a + m_2 a \\ m_2 g - \mu_k m_1 g = (m_1 + m_2) a \\$Solve for acceleration $a: \\ a = \frac{m_2 g - \mu_k m_1 g}{m_1 + m_2} \\$Substitute $a$ back into equation (1) to find $T: \\ T = m_1 a + \mu_k m_1 g \\ T = m_1 \left(\frac{m_2 g - \mu_k m_1 g}{m_1 + m_2}\right) + \mu_k m_1 g \\ T = \frac{m_1 m_2 g - \mu_k m_1^2 g}{m_1 + m_2} + \frac{\mu_k m_1 m_2 g}{m_1 + m_2} \\ T = \frac{m_1 m_2 g - \mu_k m_1^2 g + \mu_k m_1 m_2 g}{m_1 + m_2} \\ T = \frac{m_1 m_2 g (1 + \mu_k) - \mu_k m_1^2 g}{m_1 + m_2} \\ T = \frac{(m_1 m_2) (1 + \mu_k) g}{m_1 + m_2} \\$Therefore, the tension in the string is given by Option 2.