In the system shown in figure pulleys and str | vedclass

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Question

$\Rightarrow \mathrm{m}_{1} \mathrm{g}-2 \mathrm{T}=\mathrm{m}_{1} \mathrm{a}_{1}$        $...(1)$

${\mathrm{m}_{2} \mathrm{g}-

Vedclass · Accepted Answer

$6\,m{s^{ - 2}} \uparrow $

Vedclass · Answer

$\Rightarrow \mathrm{m}_{1} \mathrm{g}-2 \mathrm{T}=\mathrm{m}_{1} \mathrm{a}_{1}$        $...(1)$

${\mathrm{m}_{2} \mathrm{g}-\mathrm{T}=\mathrm{m}_{2} \mathrm{a}_{2}}  {\ldots \ldots .(2)}$

${2 \mathrm{a}_{1}+\mathrm{a}_{2}=0}  {\ldots \ldots .(3)}$

$\Rightarrow 2\left(\frac{\mathrm{m}_{1} \mathrm{g}-2 \mathrm{T}}{\mathrm{m}_{1}}ight)+\left(\frac{\mathrm{m}_{2} \mathrm{g}-\mathrm{T}}{\mathrm{m}_{2}}ight)=0$

$\Rightarrow 2 \mathrm{g}-\frac{4 \mathrm{T}}{\mathrm{m}_{1}}+\mathrm{g}-\frac{\mathrm{T}}{\mathrm{m}_{2}}=0$

$\Rightarrow 3 \mathrm{g}=\mathrm{T}\left(\frac{4}{2}+\frac{1}{2}ight)$

$\Rightarrow 30=\mathrm{T}\left(\frac{5}{2}ight)$

$\Rightarrow \mathrm{T}=12 \mathrm{N}$

$	herefore \mathrm{a}_{1}=\frac{\mathrm{m}_{1} \mathrm{g}-2 \mathrm{T}}{\mathrm{m}_{1}}$

$a_{1}=\frac{20-24}{2}=-2 m / s^{2}$

$a_{2}=4 \mathrm{m} / \mathrm{s}^{2}$

$\vec{\mathrm{a}}_{1}=2 \hat{\mathrm{j}} ; \quad \vec{\mathrm{a}}_{2}=-4 \hat{\mathrm{j}}$

$\vec{\mathrm{a}}_{1 / 2}=6 \hat{\mathrm{j}}=6 \mathrm{m} / \mathrm{s}^{2} \uparrow$

In the system shown in figure pulleys and strings are ideal. Acceleration of $m_1\ w.r.t.\ m_2$ is $(m_1 = 2\ kg\ ; m_2 = 2\ kg)$

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