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The coefficient of friction between 4kg and 5kg blocks is 0.2 and between 5kg block and ground is 0.1 respectively. The minimum force required for slipping between 4kg and 5kg block is $n \times 3$ N, the value of $n$ is $[g = 10 {ms}^{- 2}]$

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answer is 9.

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Detailed Solution

minimum force required to move system = 9N

$\begin{array}{l} f_{2} = 4, μ_{2} m g = 4 \times 0.2 \times g = 8 N \\ f_{1} = μ_{1} m g = 9 \times 0.1 \times g = 9 N \end{array}$

Let acceleration of 5kg block is $a_{1}$

$5 a_{1} = - (f_{1} + f_{2})$

$5 a_{1} = F - 17$

$a_{1} = \frac{F - 17}{5}$

For relative motion between 4kg and 5kg

$4 a_{1} \geq 8, 4 \times \frac{F - 17}{5} \geq 8$

$F - 17 \geq 10$

$F \geq 27 N$

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