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Two blocks of masses $m_{1} = 1 k g$ and $m_{2} = 2 k g$ are connected by a non-deformed light spring. They are lying on a rough horizontal surface. The coefficient of friction between the blocks and the surface is $0.4$ . What minimum constant force $F$ has to be applied in horizontal direction to the block of mass $m_{1}$ in order to shift the other block $(g = 10 m s^{- 2})$

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$10 N$

$15 N$

$8 N$

$25 N$

answer is A.

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Detailed Solution

According to work energy theorem

$F x - μ m_{1} g x - \frac{1}{2} k x^{2} = 0$

$F x - μ m_{1} g x - \frac{1}{2} μ m_{2} g x = 0$

$F = μ (m_{1} + \frac{m_{2}}{2}) g = 0.4 (1 + \frac{2}{2}) 10 = 8 N$

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