A constant force $F = \frac{m_{2} g}{2}$ is applied on the block of mass $m_{1}$ as shown in the figure. The string and the pulley are light and the surface of the table is smooth. The acceleration of $m_{1}$ is

\frac{m_{2} g}{2 (m_{1} + m_{2})}

towards right

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\frac{m_{2} g}{2 (m_{1} - m_{2})}

towards left

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\frac{m_{2} g}{2 (m_{2} - m_{1})}

towards right

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\frac{m_{2} g}{2 (m_{2} - m_{1})}

towards left

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Solution

The correct option is A $\frac{m_{2} g}{2 (m_{1} + m_{2})}$ towards right
Let $a$ be the acceleration of mass $m_{2}$ in the downward direction. Then
$T - (\frac{m_{2} g}{2}) = m_{1} a$ ... (i)
and $m_{2} g - T = m_{2} a$ ... (ii)
Adding eqs. (i) and (ii), we get
$(m_{1} + m_{2}) a = m_{2} g - m_{2} (\frac{g}{2}) = \frac{m_{2} g}{2}$
$∴ a = \frac{m_{2} g}{2 (m_{1} + m_{2})}$

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