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Block B of mass 100 kg rests on a rough surface of friction coefficient $μ = 1/3$ . A rope is tied to block B as shown in figure. The maximum acceleration with which boy A of 25 kg can climbs on rope without making block move is :

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\frac{g}{3}

\frac{4g}{3}

\frac{g}{2}

\frac{3g}{4}

answer is B.

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Detailed Solution

T - 25 g = 25 a

$T cos37° = F$

$N + Tsin θ = 100g$

$N = 100g - T \frac{3}{5}$

$N = \frac{500 g - 3 T}{5}$

$T \cos 37 ° = μ N$

$T = \frac{1}{3} \times \frac{5}{4} \times \frac{(500 g - 3 T)}{5}$

$T = \frac{500 g - 3 T}{12}$

$12 T = 500 g - 3 T$

$500g = 15T$

$T = \frac{100 g}{3}$

Sub in equation (1)

$\frac{100 g}{3} - 25 g = 25 a$

$a = \frac{g}{3}$

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