If m_1 = 4m_2 and the acceleration of m_1 is | vedclass

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(D) From the constraint relation,if $m_1$ moves up with acceleration $a$,then $m_2$ moves up with acceleration $2a$.For block $m_1$: $2T - m_1g = m_1a

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$\frac{3}{2}m_2g$

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(D) From the constraint relation,if $m_1$ moves up with acceleration $a$,then $m_2$ moves up with acceleration $2a$.For block $m_1$: $2T - m_1g = m_1a$ (Assuming upward motion is positive).For block $m_2$: $T - m_2g = m_2(2a)$.Given $m_1 = 4m_2$,substitute $m_1$ in the first equation: $2T - 4m_2g = 4m_2a \implies T - 2m_2g = 2m_2a$.Now we have two equations:$1$) $T - 2m_2g = 2m_2a$$2$) $T - m_2g = 2m_2a$Subtracting $(1)$ from $(2)$: $(T - m_2g) - (T - 2m_2g) = 2m_2a - 2m_2a \implies m_2g = 0$,which implies the system is in equilibrium or the acceleration $a$ must be determined differently. Re-evaluating the forces: $m_1g - 2T = m_1a$ and $T - m_2g = m_2(2a)$.Substitute $m_1 = 4m_2$: $4m_2g - 2T = 4m_2a \implies 2m_2g - T = 2m_2a$.Adding the two equations: $(2m_2g - T) + (T - m_2g) = 2m_2a + 2m_2a \implies m_2g = 4m_2a \implies a = g/4$.Substitute $a$ into $T - m_2g = m_2(2a)$: $T = m_2g + 2m_2(g/4) = m_2g + m_2g/2 = \frac{3}{2}m_2g$.

If $m_1 = 4m_2$ and the acceleration of $m_1$ is $a$,find the tension $T$ in the string.

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