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Question
A constant force F = m2g/2 is applied on the block of mass m1 as shown in the following figure. The string and the pulley are light and the surface of the table is smooth. Find the acceleration of m1.
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Solution
The free-body diagrams for both the blocks are shown below:
From the free-body diagram of block of mass m1,
m1a = T − F ...(i)
From the free-body diagram of block of mass m2,
m2a = m2g − T ...(ii)
Adding both the equations, we get:
\[a\left( m_1 + m_2 \right) = m_2 g - \frac{m_2 g}{2} \left......... [\text{ because F }= \frac{m_2 g}{2} \right]\]
\[ \Rightarrow a = \frac{m_2 g}{2\left( m_1 + m_2 \right)}\]
So, the acceleration of mass m1,
\[a = \frac{m_2 g}{2\left( m_1 + m_2 \right)}, \text{ towards the right }.\]
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