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Two blocks of masses $2 kg$ and 4Kg are connected by a light string and kept on horizontal surface. A force of $16 N$ is acted on 4kg block horizontally as shown in figure. Besides, it is given that coefficient of friction between 4kg and ground is 0.3 and between 2kg block and ground is 0.6. Then, frictional force between 2kg block and ground is

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4 N

6 N

12 N

Zero

answer is C.

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Detailed Solution

${(f_{2 k g})}_{\max} = μ_{2} m_{2} g = 0.6 \times 2 \times 10 = 12 N {(f_{4 k g})}_{\max} = μ_{4} m_{4} g = 0.3 \times 4 \times 10 = 12 N$

Net pulling force $F = 16 N$ and

Net resistive force

$F^{'} = {(f_{2 kg})}_{\max} + {(f_{4 kg})}_{\max} = 24 N$

Since, $F < F^{'}$ , system will not move and free body diagrams of two blocks are as shown below.

$4 kg T + 12 = 16 \Rightarrow T = 4 N 2 kg f_{2} = T = 4 N$

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