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Two blocks (m$_1$) and (m$_1$), where m$_1 \gt\;$ m$_2$ are connected as shown on frictionless surface. Find the common acceleration of the masses, ignoring any friction effects.

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Answer: $ \large\frac{m_2g}{m_1+m_2}$
We can draw the free body diagrams for our problem as follows:
When the system is released, let the acceleration of the blocks be aa.
From the Free Body Diagrams, for the bigger block $m_1 \rightarrow T = m_1a$ and for the lighter block $m_2 \rightarrow m_2g - T = m_2a$
Solving for $a$, we get $a = \large\frac{m_2g}{m_1+m_2}$
answered Aug 20, 2014 by balaji.thirumalai
edited Aug 20, 2014 by balaji.thirumalai

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