Question:
The acceleration of block \( B \) in the given pulley system is
The acceleration of block \( B \) in the given pulley system is
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In pulley systems, first find displacement relation, then apply Newton’s laws.
Updated On: Apr 23, 2026
- \(\dfrac{m_2 g}{4m_1 + m_2}\)
- \(\dfrac{2m_2 g}{4m_1 - m_2}\)
- \(\dfrac{2m_2 g}{m_1 + 4m_2}\)
- \(\dfrac{2m_2 g}{m_1 + m_2}\)
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The Correct Option is A
Solution and Explanation
Concept:
Pulley constraint $\Rightarrow$ displacement/acceleration relation.
Step 1: Constraint
If block \(B\) moves down by \(x\), block \(A\) moves horizontally by \(2x\). \[ a_A = 2a_B \]
Step 2: Apply Newton’s law
For \(m_1\): \[ T = m_1 a_A = 2m_1 a_B \] For \(m_2\): \[ m_2 g - 2T = m_2 a_B \]
Step 3: Substitute
\[ m_2 g - 4m_1 a_B = m_2 a_B \] \[ a_B = \frac{m_2 g}{4m_1 + m_2} \] Conclusion: \[ {(A)} \]
Step 1: Constraint
If block \(B\) moves down by \(x\), block \(A\) moves horizontally by \(2x\). \[ a_A = 2a_B \]
Step 2: Apply Newton’s law
For \(m_1\): \[ T = m_1 a_A = 2m_1 a_B \] For \(m_2\): \[ m_2 g - 2T = m_2 a_B \]
Step 3: Substitute
\[ m_2 g - 4m_1 a_B = m_2 a_B \] \[ a_B = \frac{m_2 g}{4m_1 + m_2} \] Conclusion: \[ {(A)} \]
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