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A ring of mass 1 kg can slide on smooth certical rod shown in fig. The ring is connected to a spring of spring constant R = 40 N/m and natural length of spring is 0.2 m. The ring is taken at a height of 1.5 m and released .Find the velocity of ring when it reaches point C (in m/s)

$lengthofthespringforpositionp\phantom{\rule{0ex}{0ex}}l=\sqrt{6.25}=2.5m\phantom{\rule{0ex}{0ex}}lengthofthespringforthepointC\phantom{\rule{0ex}{0ex}}l\text{'}=2m\phantom{\rule{0ex}{0ex}}\frac{m{v}^{2}}{2}=\frac{40}{2}\left(2.{3}^{2}-1.{8}^{2}\right)+mg\left(1.5\right)\phantom{\rule{0ex}{0ex}}\frac{{v}^{2}}{2}=41+15\phantom{\rule{0ex}{0ex}}v=\sqrt{112}m/s\phantom{\rule{0ex}{0ex}}bestofmyknowledge$

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