Question Number 178746 by Acem last updated on 21/Oct/22
Answered by mr W last updated on 21/Oct/22
$${s}={total}\:{length}\:{of}\:{chain}=\mathrm{3}{L}=\mathrm{3}\:{m} \\ $$$$\frac{\mathrm{2}{mg}}{\mathrm{3}}×\frac{{s}}{\mathrm{3}}−\frac{{mg}}{\mathrm{3}}×\frac{{s}}{\mathrm{6}}=\frac{{mv}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow{v}=\sqrt{\frac{{gs}}{\mathrm{3}}}=\sqrt{{g}}\:{m}/{s} \\ $$
Commented by Tawa11 last updated on 21/Oct/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by Acem last updated on 21/Oct/22
$${Morning}\:{friend},\:{its}\:{length}=\:\mathrm{1}{m} \\ $$
Commented by mr W last updated on 21/Oct/22
$${that}\:{is}\:{not}\:{important}.\:{when}\:{you}\: \\ $$$${understand}\:{that}\:{L}=\mathrm{1}{m}\:{is}\:{the}\:{total} \\ $$$${length},\:{it}'{s}\:{ok}.\:{but}\:{i}\:{understand}\:{that} \\ $$$${L}=\mathrm{1}\:{m}\:{is}\:{only}\:{one}\:{third}\:{of}\:{the}\:{total} \\ $$$${length}. \\ $$
Commented by Acem last updated on 21/Oct/22
$${Yes}\:{friend},\:{i}\:{know}\:{that}\:{isn}'{t}\:{important},\:{what}'{s} \\ $$$$\:{interested}\:{in}\:{is}\:{knowing}\:{how}\:{to}\:{deal}\:{with}\:{the}\:{C}_{{gr}} \\ $$$$\:{for}\:{the}\:{free}\:{section}\:{of}\:{that}\:{chain},\:{thanks} \\ $$