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The coefficient of static friction, μₛ, between block A of mass 2 kg and the table as shown in the figure is 0.2. What would be the maximum mass value of block B so that the two blocks do not move? The string and the pulley are assumed to be smooth and massless.

1. 0.4 kg
2. 2.0 kg
3. 4.0 kg
4. 0.2 kg

Correct answer: 0.4 kg

Solution

The maximum force of static friction is μₛ × mₐ × g = 0.2 × 2 × 9.8 = 3.92 N. For the system to remain stationary, the weight of block B (m_b × g) must not exceed this force. Solving m_b × 9.8 = 3.92 gives m_b = 0.4 kg.

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