Question

System shown in figure is released from the rest. Pulley and string is massless and friction is absent everywhere. The speed of $5kg$ block when $2kg$ block leaves contact with ground is : (Take force constant of spring $k=40N/m$ and $g=10m/s^2$)

• A. $\sqrt{2}m/s$
• B. $2\sqrt{2}m/s$
• C. $2m/s$
• D. $4\sqrt{2}m/s$

Answer 1

The correct option is B $2\sqrt{2}m/s$

$\begin{array}{c}Letxbetheextensioninspringwhen2kg\\ massleavescontactwiththeground.\\ kx=2g\\ x=\frac{2g}{k}\\ =\frac{2\times 10}{40}\\ =\frac{1}{2}m\\ Bylawofconservationofenergyfor5kgmass\\ mgx=\frac{1}{2}{k}^2+\frac{1}{2}m{v}^2\\ v=\sqrt{2gx-\frac{k{x}^2}{m}}\\ v=\sqrt{2\times 10\times \frac{1}{2}-\frac{40}{5}\times \frac{1}{4}}\\ v=2\sqrt{2}m/s\\ Hence,theoptionBisthecorrectanswer.\end{array}$