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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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Two bodies of masses m$_1$ (4 kg) and m$_2$ (3 kg) are suspended at the ends of a string of negligible mass passing over a friction-less pulley as shown here.The acceleration of the system is


\[(a)\;4.9 m/s^2 \quad (b)\;2.45 m/s^2 \quad (c)\;9.5 m/s^2 \quad(d)\;1.4 m/s^2 \]

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Answer: 1.4 m/s$^2$
When the bodies are released, the heavier one moves downward and the lighter one moves upward.
Net force in the direction of motion of $m_1$ is $F_1 = m_1g - T$ which is equal to $m_1a$ as per Newton's second law.
$\Rightarrow m_1g - T = m_1 a \quad (i)$
Net force in the direction of motion of $m_2$ is $F_2 = T-m_2g $ which is equal to $m_2a$ as per Newton's second law.
$\Rightarrow T-m_2g = m_2a \quad (ii) $
Solving $(i)$ and $(ii)$ for acceleration, $a \rightarrow a = \large\frac{m_1-m_2}{m_1+m_2}$$g$
In our case, $m_1 = 4, \;m_2 = 3 \rightarrow a = \large\frac{4-3}{4+3}$$\times 9.8 = 1.4\; m/s^2$
answered Jul 4, 2013 by meena.p
edited Aug 19, 2014 by balaji.thirumalai
 

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