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System shown in figure is released from rest. Pulley and spring are massless and the friction is absent everywhere. The speed of $5\, kg$ block, when $2\, kg$ block leaves the contact with ground is (take force constant of the sprign $k = 40\, N/m$ and $g = 10\, m/s^2$)
$\sqrt 2 \,m/s$
$2\sqrt 2 \,m/s$
$2\, m/s$
$4\sqrt 2 \,m/s$
Solution
Let $x$ be the extension in the spring when $2 \,kg$ block leaves the contact with ground. Then,
$\mathrm{Kx}=2 \mathrm{g}$
$\mathbf{o r}$ $\mathrm{x}=\frac{2 \mathrm{g}}{\mathrm{K}}=\frac{2 \times 10}{40}=\frac{1}{2} \mathrm{m}$
Now from conservation of mechanical energy,
$\operatorname{mg} \mathrm{x}=\frac{1}{2} \mathrm{Kx}^{2}+\frac{1}{2} \mathrm{mv}^{2}$
$\mathbf{O r}$ $v=\sqrt{2 g x-\frac{K x^{2}}{m}}$
$=\sqrt{2 \times 10 \times \frac{1}{2}-\frac{40}{4 \times 5}}$
$=2 \sqrt{2} \mathrm{m} / \mathrm{s}$
