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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Laws of Motion

Two blocks of masses $m_1=m$ and $m_2=2m$ are connected by a light string passing over a frictionless pulley.The mass $m_1$ is placed on a smooth inclined plane of inclination $\theta=30^{\large\circ}$ and mass $m_2$ hangs vertically as shown in figure below.If the system is released,the blocks move with an acceleration equal to

$\begin{array}{1 1}\large\frac{g}{4}\\\large\frac{g}{3}\\\large\frac{g}{2}\\g\end{array} $

1 Answer

Answer : $\large\frac{g}{2}$
Since the inclined plane is smooth and $m_2 > m_1$,block $m_1$ will up the plane and block $m_2$ will move vertically with a common acceleration $a$.If $T$ is the tension in the string,the free-body diagrams of masses $m_1$ and $m_2$ are as shown in figure.
The equations of motion of the blocks are
$T-m_1g\sin \theta=m_1a$-------(1)
$m_2g-T=m_2a$--------(2)
Equations (1) and (2) give
$a= \large\frac{(m_2-m_1\sin \theta)g}{(m_1+m_2)}$
$\;\;\;=\large\frac{(2m-m\times \sin 30^{\large\circ})g}{(m+2m)}$
$\;\;\;=\large\frac{g}{2}$
Hence the correct choice is $\large\frac{g}{2}$
answered Aug 28, 2014 by sreemathi.v
 

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