A constant force of $F = \frac{m_{2} g}{2}$ 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 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 and $T$ be the tension in the string.

From the FBD of block of mass $m_{1}$ ,
$\begin{matrix} T - \frac{m_{2} g}{2} = m_{1} a \\ (1) \end{matrix}$
From the FBD of block of mass $m_{2}$ ,
$\begin{matrix} m_{2} g - T = m_{2} a \\ (2) \end{matrix}$
Adding equation $(1)$ and $(2)$ , we get
$(m_{1} + m_{2}) a = m_{2} g - \frac{m_{2} g}{2} = \frac{m_{2} g}{2}$
$\Rightarrow a = \frac{m_{2} g}{2 (m_{1} + m_{2})}$

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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

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(P) Acceleration of mass $M_{1}$	(1) $\frac{F}{M_{2}}$
(Q) Acceleration of mass $M_{2}$	(2) $\frac{F}{M_{1} + M_{2}}$
(R) Acceleration of mass $M$	(3) zero
(S) Tension in the string	(4) $\frac{M_{2} F}{M_{1} + M_{2}}$

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A constant force of F=m2g2 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 of $F = \frac{m_{2} g}{2}$ 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