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Standard XII

Physics

Net Force

A constant fo...

Question

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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