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Q.

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

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a

$\frac{m_{2} g}{2 m_{1}}$

b

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

c

$\frac{3 m_{2} g}{2 (m_{1} + m_{2})}$

d

$\frac{3 m_{2} g}{2 m_{1}}$

answer is B.

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Detailed Solution

$\begin{array}{l} T - F = m_{1} a \dots \dots \dots (1) \\ m_{2} g - T = m_{2} a \dots \dots . (2) \\ (1) + (2) \\ m_{2} g - F = (m_{1} + m_{2}) a \\ m_{2} g - m_{2} \frac{g}{2} = (m_{1} + m_{2}) a \\ \Rightarrow \frac{m_{2} g}{2} = (m_{1} + m_{2}) a ∴ a = \frac{m_{2} g}{2 (m_{1} + m_{2})} \end{array}$
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A constant force F = m2g/2 is applied on the block of mass m1 as shown. The string and the pulley are light and the surface of the table is smooth. Find the acceleration of m1.