Question

System shown in the figure is released from rest with mass $2kg$ in contact with ground. Pulley and spring are massless and the friction is absent everywhere. The speed of $5kg$ block when $2kg$ block leaves the contact with the ground is (force constant of spring $k=40N{m}^{-1}$ and $g=10m{s}^{-2})$

• A. $\sqrt{2}m{s}^{-1}$
• B. $2\sqrt{2}m{s}^{-1}$
• C. $2m{s}^{-1}$
• D. $4\sqrt{2}m{s}^{-1}$

Answer 1

Answer: (B)

Let $x$ be the extension in the string when $2kg$ block leaves the contact with ground. Then tension in the spring should be equal to weight of $2kg$ block.
$kx=2g$ or $x=\frac{2g}{k}=\frac{2\times 10}{40}=\frac{1}{2}m$
Now, from conservation of mechanical energy
$mgx=\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2$
Or $v=\sqrt{2gx-\frac{k{x}^2}{m}}$
$=\sqrt{2\times 10\times \frac{1}{2}-\frac{40}{4\times 5}}$
$=2\sqrt{2}m{s}^{-1}$