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Question Number 168982 by Tawa11 last updated on 22/Apr/22

Commented by mr W last updated on 22/Apr/22

$$allanswersseemtobewrong.$$

Answered by mr W last updated on 22/Apr/22

$$l=\sqrt{2}R$$$$mgR+\frac{k{(2R-l)}^2}{2}=\frac{{mv}^2}{2}+\frac{k{(\sqrt{2}R-l)}^2}{2}$$$$\frac{{mv}^2}{R}=2mg+k(2-\sqrt{2})(2R+\sqrt{2}R-2l)$$$$N=\frac{{mv}^2}{R}-\frac{k(\sqrt{2}R-l)}{\sqrt{2}}$$$$N=2mg+k(2-\sqrt{2})(2R+\sqrt{2}R-2l)-\frac{k(\sqrt{2}R-l)}{\sqrt{2}}$$$$N=2mg+k(2-\sqrt{2})(2R+\sqrt{2}R-2\sqrt{2}R)-\frac{k(\sqrt{2}R-\sqrt{2}R)}{\sqrt{2}}$$$$N=2mg+k{(2-\sqrt{2})}^{2}R$$$$N=2\times 1\times 10+(2+\sqrt{2}){(2-\sqrt{2})}^2\times 5\approx 26N$$

Commented by mr W last updated on 23/Apr/22

$$amIright?$$

Commented by Tawa11 last updated on 22/Apr/22

$$\mathrm{Ohh}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.$$

Commented by Tawa11 last updated on 22/Apr/22

$$\mathrm{Thanks}\mathrm{for}\mathrm{your}\mathrm{time}.$$

Commented by Tawa11 last updated on 23/Apr/22

$$\mathrm{They}\mathrm{choose}30\mathrm{sir}.$$

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