A constant force F = $m_{2} g / 2$ is applied on the block of mass $m_{1}$ as shown in 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 left

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$\frac{m_{2} g}{2 (m_{1} + m_{2})}$ towards right

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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 B
$\frac{m_{2} g}{2 (m_{1} + m_{2})}$ towards right

So if $m_{1}$ has an acceleration of a towards left $m_{2}$ will have to have an acceleration of a upwards - constraint relation.

lets draw Free Body Diagram of $m_{1}$ & $m_{2}$

Net force in direction of acceleration T - $m_{2} g = m_{2} a$ -----------------(II)

F - T = $m_{2} a$

We know $F = \frac{m_{2} g}{2}$

$\Rightarrow \frac{m_{2} g}{2} - T = m_{2} a$ ----------------(I)

Adding equation (I) & (II) we get

$\frac{m_{2} g}{2} - m_{2} g = (m_{2} + m_{2}) a$

$a = \frac{- m_{2} g}{2 (m_{1} + m_{2})}$

so accleration a of $m_{2} = a = \frac{- m_{2} g}{2 (m_{1} + m_{2})}$

The negative sign infers that acceleration is opposite in direction to that of assumed positive i.e., rightwards.So acceleration of $m_{1} = \frac{- m_{2} g}{2 (m_{1} + m_{2})}$ rightwards.

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A constant force F = m2g/2 is applied on the block of mass m1 as shown in figure. The string and the pulley are light and the surface of the table is smooth. The acceleration of m1 is

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