Question Number 27111 by Tinkutara last updated on 02/Jan/18
Answered by mrW1 last updated on 02/Jan/18
$${T}={tension}\:{in}\:{string} \\ $$$${a}_{{A}} =\mathrm{2}{a}_{{B}} \\ $$$${m}_{{B}} {g}−\mathrm{2}{T}={m}_{{B}} {a}_{{B}} \\ $$$${T}={m}_{{A}} {a}_{{A}} =\mathrm{2}{m}_{{A}} {a}_{{B}} \\ $$$$\Rightarrow{m}_{{B}} {g}=\left(\mathrm{4}{m}_{{A}} +{m}_{{B}} \right){a}_{{B}} \\ $$$$\Rightarrow{a}_{{B}} =\frac{{m}_{{B}} }{\mathrm{4}{m}_{{A}} +{m}_{{B}} }×{g}=\frac{\mathrm{8}}{\mathrm{4}×\mathrm{2}+\mathrm{8}}×\mathrm{10}=\mathrm{5}\:{m}/{s}^{\mathrm{2}} \\ $$
Commented by Tinkutara last updated on 02/Jan/18
Thank you very much Sir! I got the answer.