Two blocks of masses $m_{1} = 4 k g$ and $m_{2} = 6 k g$ are connected by a string of negligible mass passing over a frictionless pulley as shown in the figure. The coefficient of friction between the block $m_{1}$ and the horizontal surface is $0.4$ . When the system is released, the masses $m_{1}$ and $m_{2}$ start accelerating. What additional mass $m$ should be placed over $m_{1}$ so that the masses $(m_{1} + m)$ slide with a uniform speed ?

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$12 k g$

$11 k g$

$10 k g$

$2 k g$

answer is B.

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Detailed Solution

For block of mass $m_{1}$

$f = T$

But $f = μ N = μ (m + m_{1}) g$

$∴ T = μ (m + m_{1}) g$

For uniform motion of of masses

$m_{2} g - T = 0$ or $m_{2} g = T$

$∴ μ (m + m_{1}) g = m_{2} g$

or $m + m_{1} = \frac{m_{2}}{μ}$ or $m = \frac{m_{2}}{μ} - m_{1}$

or $m = \frac{6}{0.4} - 4 = \frac{60}{4} - 4 = 15 - 4 = 11 k g$

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