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Question Number 215975 by Tawa11 last updated on 23/Jan/25

Answered by Tawa11 last updated on 23/Jan/25

$$\mathrm{v}=1.98\mathrm{m}/\mathrm{s}\mathrm{at}\mathrm{the}\mathrm{back}\mathrm{of}\mathrm{the}\mathrm{book}.$$

Answered by mr W last updated on 26/Jan/25

Commented by mr W last updated on 26/Jan/25

$$l_0=1.42m$$$$l_1=1.68m$$$$\mathrm{\Delta }{l}_1=1.68-1.42=0.26m$$$$\frac{l_2}{2}=\sqrt{0.{35}^2+{(0.5\times 1.68)}^2}=0.91m$$$$l_2=2\times 0.91=1.82m$$$$\mathrm{\Delta }{l}_2=1.82-1.42=0.4m$$$$\sin \theta =\frac{0.35}{0.91}=\frac{5}{13}$$$$2T\sin \theta =mg$$$$\Rightarrow T=\frac{mg}{2\sin \theta }=\frac{2\times 9.81\times 13}{10}=25.506N$$$$\mathrm{\Delta }{l}_2=\frac{T}{k}$$$$k=\frac{T}{\mathrm{\Delta }{l}_2}=\frac{25.506}{0.4}=63.765N/m$$$$energyconservation:$$$$\frac{k\times \mathrm{\Delta }{l}_1^2}{2}+mg\times 0.35=\frac{k\times \mathrm{\Delta }{l}_2^2}{2}+\frac{{mv}^2}{2}$$$$\frac{k(\mathrm{\Delta }{l}_1^2-\mathrm{\Delta }{l}_2^2)}{m}+2g\times 0.35=v^2$$$$\Rightarrow v=\sqrt{\frac{63.765(0.{26}^2-0.4^2)}{2}+2\times 9.81\times 0.35}$$$$=1.98m/s$$

Commented by Tawa11 last updated on 25/Jan/25

$$\mathrm{Wow},\mathrm{thanks}\mathrm{sir}.$$$$\mathrm{I}\mathrm{really}\mathrm{appreciate}.$$

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