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Q.

Two blocks A and B masses $m_{A} = 1 k g$ and $m_{B} = 3 k g$ are kept on the table as shown in figure. The coefficient of friction between A and B is 0.2 and between B and the surface of the table is also 0.2 The maximum force F ( in newton) that can be applied on B horizontally, so that the block A does not slide Over the block B is : $[T a k e g = 10 m / s^{2}]$

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answer is 16.

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Detailed Solution

Taking $(A + B)$ as system
$F - μ (M + m) g = (M + m) a \Rightarrow a = \frac{F - μ (M + m) g}{(M + m)} a = \frac{F - (0.2) 4 \times 10}{4} = (\frac{F - 8}{4}) ... (i)$
But, $a_{\max} = μ g = 0.2 \times 10 = 2$
$∴ \frac{F - 8}{4} = 2$
$\Rightarrow F = 16 N$

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Two blocks A and B masses mA=1kg and mB=3kg are kept on the table as shown in figure. The coefficient of friction between A and B is 0.2 and between B and the surface of the table is also 0.2 The maximum force F ( in newton) that can be applied on B horizontally, so that the block A does not slide Over the block B is : [Take g=10 m/s2]