Three solids of masses m_1, m_2 and m_3 are c | vedclass

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(C) The total mass of the system is $M = m_1 + m_2 + m_3$.Since the system is pulled by a force $T$ on a frictionless surface,the acceleration $a$ of

Vedclass · Accepted Answer

$\frac{m_1 + m_2}{m_1 + m_2 + m_3} T$

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(C) The total mass of the system is $M = m_1 + m_2 + m_3$.Since the system is pulled by a force $T$ on a frictionless surface,the acceleration $a$ of the system is given by $a = \frac{T}{m_1 + m_2 + m_3}$.To find the tension $T'$ in the string between $m_2$ and $m_3$,we consider the system consisting of masses $m_1$ and $m_2$ being pulled by this tension $T'$.Applying Newton's second law to this subsystem: $T' = (m_1 + m_2) a$.Substituting the value of $a$: $T' = (m_1 + m_2) 	imes \frac{T}{m_1 + m_2 + m_3}$.Thus,the tension is $\frac{m_1 + m_2}{m_1 + m_2 + m_3} T$.

Three solids of masses $m_1, m_2$ and $m_3$ are connected with weightless strings in succession and are placed on a frictionless table. If the mass $m_3$ is dragged with a force $T$,the tension in the string between $m_2$ and $m_3$ is

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