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In the arrangement shown, coefficient of friction between the block and the surface is 0.6 then

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acceleration of the block is 1 $m / s^{2}$

frictional force acting on the block is 10 N

acceleration of the block is 0.5 $m / s^{2}$

frictional force acting on the block is 12 N

answer is D.

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Detailed Solution

$f_{\max} = μ N = μ m g = 0.6 \times 2 \times 10 N = 12 N$

since $10 N < f_{\max}$ , the block will be in equilibrium

Hence frictional force acting on the block is 10 N.

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