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A block A of mass m, 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 m2 is suspended. The coefficient of kinetic friction between the block and the table is µ.. When the block A is sliding on the table, the tension in the string is

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$\begin{aligned} \frac{(m_{2} - μ {km}_{1}) g}{(m_{1} + m_{2})} \end{aligned}$

$\begin{array}{r} \frac{m_{1} m_{2} (1 + μ_{k}) g}{(m_{1} + m_{2})} \end{array}$

$\begin{array}{r} \frac{m_{1} m_{2} (1 - μ_{k}) g}{(m_{1} + m_{2})} \end{array}$

$\frac{(m_{2} + μ_{k} m_{1}) g}{(m_{1} + m_{2})}$

answer is B.

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Detailed Solution

For the motion of both the blocks $m_{1} a = T - μ_{k} m_{1} g$ $m_{2} g - T = m, a$

$\begin{array}{l} a = \frac{m_{2} g - μ_{k} m_{1} g}{m_{1} + m_{2}} \\ m_{2} g - T = (m_{2}) (\frac{m_{2} g - μ_{k} m_{1} g}{m_{1} + m_{2}}) \end{array}$

solving we get tension in the string

$T = \frac{m_{1} m_{2} g (1 + μ_{k}) g}{m_{1} + m_{2}}$

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