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  • A block of mass   on a smooth inclined plane at an angle <img alt="30 \degree" src="http:/

A block of mass m_1 = 2Kg  on a smooth inclined plane at an angle 30 \degree  is connected to the second block of mass m_2 = 3Kg  by a cord passing over as a frictionless pulley as shown in the figure 

How much time (in seconds )  will be taken by block B to reach the ground if they are released from the rest 

Option: 1

2


Option: 2

5


Option: 3

4


Option: 4

3


Answers (1)

best_answer

Solution :

\\ \text{let tension be T and acceleration a of the blocks} \\ \text{For } m_{2} \ \text{block :} \\ m_{2}g-T=m_{2}a \\ T=m_{2}\left ( g-a \right ) \ \ \ \ \ \ \ -(1) \\ \text{For }\ m_{1} \ \text{block :} \\ T-m_{1}g\sin30^{\circ}=m_{1}a \\ T=m_{1}\left ( a+g\sin30^{\circ} \right ) \ \ \ \ \ \ \ -(2) \\ \text{equating above equations} \\ m_{1}\left ( a+g\sin30^{\circ} \right )=m_{2}\left ( g-a \right ) \\ \begin{aligned} &\mathrm{a}=\frac{\mathrm{m}_2 \mathrm{~g}-\mathrm{m}_1 \mathrm{~g} \sin 30^{\circ}}{\mathrm{m}_1+\mathrm{m}_2} \\ &\mathrm{a}=4 \mathrm{~m} / \mathrm{sec}^2 \end{aligned}

\\ \text{Now, Time taken by block} \ m_{2} \ \text{ to reach the ground (starting from rest u=0)}

\\ S=ut+\frac{1}{2}at^{2} \\ 8=0+\frac{1}{2}\times 4\times t^{2} \\ \\ \therefore t=2 \sec

Posted by

himanshu.meshram

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