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In the arrangement shown, the blocks A and B are oscillating horizontally without any relative motion between them. Blocks A and B have masses 2 kg and 1 kg respectively. Spring constant $k = 600 N / m$ . There is no friction between block A and the bottom surface but friction is there between A and B. If amplitude of oscillation is 10 cm, the maximum frictional force acting on B is

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20 N

30 N

60 N

15 N

answer is A.

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Detailed Solution

$ω = \sqrt{k / (M + m)}$

Maximum pseudo force on block B $= m ω^{2} A = \frac{m k A}{(M + m)}$

$∴ f = \frac{m k A}{M + m} = \frac{1 \times 600 \times 0.1}{(2 + 1)} N = 20 N$

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