Question

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

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

Answer 1

The correct option is B $2\sqrt{2}$

Given,

Masses, $m=2kgandM=5kg$

Let x be the extension in spring when $m=2kg$ mass leaves contact with the ground.

$kx=mg=2g$

$x=\frac{2g}{k}=\frac{2\times 10}{40}=\frac{1}{2}$$m$

By law of conservation of energy for $M=5kg$ mass

$Mgx=\frac{1}{2}k{x}^2+\frac{1}{2}M{v}^2$

$v=\sqrt{\frac{2Mgx}{M}-\frac{k{x}^2}{M}}$

$v=\sqrt{2\times 10\times \frac{1}{2}-\frac{40}{5}\times {\left(\frac{1}{2}\right)}^2}$

$v=2\sqrt{2}m/s$

Hence, Velocity of $5kg$ is $2\sqrt{2}m{s}^{-1}$