Two blocks are connected over a massless pulley as shown in figure. The mass of block A is 10 kg and the coefficient of kinetic friction is 0.2. Block A slides down the incline at constant speed. The mass of block B (in kg) is

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5.4

3.3

4.2

6.8

answer is B.

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Detailed Solution

Net pulling force on the system should be zero, as velocity is constant. Hence,

$m_{A} g \sin 30 ° = μ m_{A} g \cos 30 ° + m_{B} g$

$∴ m_{B} = (\frac{m_{A}}{2}) - (\frac{μ m_{A} \sqrt{3}}{2})$

$= 10 [\frac{1}{2} - 0.2 \times \frac{\sqrt{3}}{2}] = 3.3 k g$

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