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Two unequal masses of 1kg and 2kg are connected by an inextensible light string passing over a smooth pulley as shown in fig. A force F = 20 N is applied on 1 kg block. The acceleration of either block [ ${g=10 m/s}^{2}$ ]

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\frac{10}{3} {m/s}^{2}

\frac{13}{3} {m/s}^{2}

\frac{5}{3} {m/s}^{2}

\frac{20}{3} {m/s}^{2}

answer is A.

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Detailed Solution

F + m_{1} g - g = m_{1} a \to (1)

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

$F + m_{1} g - g + g - m_{1} a + m_{2} a$

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

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

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

$a = \frac{20 + 10 (1 - 2)}{3}$

$= \frac{20 + 10 (- 1)}{3}$

$= \frac{20 - 10}{3} \Rightarrow \frac{10}{3} {m/s}^{2}$

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