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The system shown is released at rest. Speed of block A (in m/s), after B has descended by 2 cm is

Option: 1
0.4

Option: 2
0.5

Option: 3
0.6

Option: 4
0.8

Answers (1)

best_answer

Solution :

$\\ \text{ For block A :}$

$T - \mu m g = m_A a ......(1)$

$\\ \text{For Block B :}$

$m_{B}g - T = m_Ba .......(2)$

$\\ \text{Adding (1) and (2) }$

$a = \frac{m_Bg - \mu m_Ag }{m_A+m_B}= \frac{30-0 .5 \times 2\times 10}{5}=4 m/s^2$

$\\ \text{Now,for block B : }$

$\\ u=0, \text{and} \ s=2 cm=\frac{2}{100}m$

$\\ \text{Now,} \$

$s=\frac{1}{2}at^2$

$\frac{2}{100}=\frac{1}{2}\times 4\times t^2$

$\\ t=\frac{1}{10}sec$

$\\ \text{now for block A :}$

$u=0, t=\frac{1}{10} \sec$

$\\ \text{So,}$

$\\ v=u+at \\ \Rightarrow 0+(4\times \frac{1}{10})=0.4 \\ \\ \therefore v_{a}=0.4 \ m/s$

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Divya Prakash Singh

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