Question

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

• 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 $2\sqrt{2}m/s$

Let y be the extension of the spring.

Given, stiffness of the spring K= 40 N/m

Also $g=10m/s^2$

$Ky=2kg\times gm{s}^{-2}y=\frac{2\times 10}{40}=\frac{1}{2}m$

According to Law of conservation,

$mgy=\frac{1}{2}K{y}^2+\frac{1}{2}m{v}^{2}v=\sqrt{\frac{2mgy}{m}-\frac{K{y}^2}{m}}=\sqrt{2gy-\frac{k{y}^2}{m}}=\sqrt{2\times 10\times \frac{1}{2}-\frac{40}{5}\times \frac{1}{4}}=\sqrt{8}=2\sqrt{2}m/s$