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(B) The acceleration $a$ of an Atwood machine with masses $m_1$ and $m_2$ is given by the formula:$a = \left( \frac{m_1 - m_2}{m_1 + m_2} \right) g$Gi

Vedclass · Accepted Answer

$9:7$

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(B) The acceleration $a$ of an Atwood machine with masses $m_1$ and $m_2$ is given by the formula:$a = \left( \frac{m_1 - m_2}{m_1 + m_2} \right) g$Given that the acceleration $a = \frac{g}{8}$,we substitute this into the equation:$\frac{g}{8} = \left( \frac{m_1 - m_2}{m_1 + m_2} \right) g$Dividing both sides by $g$:$\frac{1}{8} = \frac{m_1 - m_2}{m_1 + m_2}$Cross-multiplying gives:$m_1 + m_2 = 8(m_1 - m_2)$$m_1 + m_2 = 8m_1 - 8m_2$Rearranging the terms to group the masses:$9m_2 = 7m_1$Therefore,the ratio of the masses is:$\frac{m_1}{m_2} = \frac{9}{7}$

$A$ light string passing over a smooth light pulley connects two blocks of masses $m_1$ and $m_2$ (vertically). If the acceleration of the system is $\frac{g}{8}$,then the ratio of the masses is

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