In the given figure, calculate the acceleration of the block $m_{2} i n {m/s}^{2}$ , if the coefficient of friction between the blocks and the surface is 0.2. Take $m_{1} = 2 k g, m_{2} = 6 k g, F = 93 N, g = 10 m / s^{2},$ the pulley is smooth and weightless.

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Solution

Let acceleration of $m_{2}$ be $a$
$m_{1} = 2 k g$
$f = μ N = μ m g$
$f_{1} = 0.2 \times 2 \times 10 = 4 N$
$f_{2} = 0.2 \times 6 \times 10 = 12 N$
force equation for $m_{1}$
$F - 2 T - f_{1} = m_{1} a_{!}$
$93 - 2 T - 4 = 2 \times a / 2$
$89 - 2 T = a . . . . . . . (i)$
Force equation for $m_{2}$
$T - f_{2} = m_{2} a$
$T - 12 = 6 a$
$T = 6 a + 12....... (i i)$
After solving both the equations,
$89 - 2 (6 a + 12) = a$
$65 = 13 a$
$a = \frac{65}{13}$
$a = 5 \frac{m}{s e c^{2}}$ .

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