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A block A of mass $m_1$ rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass $m_2$ is suspended. The coefficient of kinetic friction between the block and the table is $\mu_k$. When the block A is sliding on the table, the tension in the string is
A
$\frac{(m_2+\mu_k m_1)g}{(m_1+m_2)}$
B
$\frac{(m_2-\mu_k m_1)g}{(m_1+m_2)}$
C
$\frac{m_1m_2(1+\mu_k)g}{(m_1+m_2)}$
D
$\frac{m_1m_2(1-\mu_k)g}{(m_1+m_2)}$
Detailed Solution
For B: $m_2g - T = m_2a$ ...(i)
For A: $T - \mu_k m_1 g = m_1a$ ...(ii)
Adding (i) and (ii): $a = \frac{(m_2 - \mu_k m_1)g}{m_1 + m_2}$
From (i): $T = m_2(g - a) = m_2g\left[1 - \frac{m_2 - \mu_k m_1}{m_1+m_2}\right]$
$T = \frac{m_1m_2(1+\mu_k)g}{m_1+m_2}$
For A: $T - \mu_k m_1 g = m_1a$ ...(ii)
Adding (i) and (ii): $a = \frac{(m_2 - \mu_k m_1)g}{m_1 + m_2}$
From (i): $T = m_2(g - a) = m_2g\left[1 - \frac{m_2 - \mu_k m_1}{m_1+m_2}\right]$
$T = \frac{m_1m_2(1+\mu_k)g}{m_1+m_2}$