• # question_answer System shown in the figure is released from rest. Pulley and spring are massless and friction is absent everywhere. The speed of 5 kg block when 2 kg block leaves the contact with ground is (Take force constant of spring K = 40 N/m and$g\text{ }=\text{ }10\text{ }m/{{s}^{2}}$) A) $\sqrt{2}\,m/s$                      B) $2\sqrt{2}\,m/s$        C)   $2\,m/s$         D)   $4\sqrt{2}\,m/s$

Hence, the correction option is (a).              Let $x$be the extension in the spring when 2 kg block leaves contact with the ground. $\therefore$$Kx=2g$or $x=\frac{2g}{K}=\frac{2\times 10}{40}=\frac{1}{2}m$ From conservation of mechanical energy (for m = 5 kg block),             $mgx=\frac{1}{2}K{{x}^{2}}+\frac{1}{2}m{{v}^{2}}$             $5\times 10\times \frac{1}{2}=\frac{1}{2}\times 40\times {{\left( \frac{1}{2} \right)}^{2}}+\frac{1}{2}\times 5{{v}^{2}}$             $25=5+\frac{5}{2}{{v}^{2}}$             $v=\sqrt{8}=2\sqrt{2}\,m/s$ Hence, the correction option is (b).