Question

The system shown in the figure is released from rest with spring in natural length and string just taut. The pulley and the spring are massless and friction is absent everywhere. The value of spring constant is $k=40 \, N \, m^{- 1}$ . The speed of $5 \, kg$ block when $2 \, kg$ block leaves contact with the support below it, will be

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

Answer 1

Answer: (B)

$T=20 \, N$ and $kx=20 \, N$
$\therefore \, x=\frac{20}{40}=\frac{1}{2} \, m$
Now $mgx=\frac{1}{2}mv^{2}+\frac{1}{2}kx^{2}$
$\Rightarrow \, 5\times 10\times \frac{1}{2}=\frac{1}{2}\times 5v^{2}+\frac{1}{2}\times 40\times \frac{1}{4}$
$\Rightarrow \, \, \, 25=5+\frac{5}{2}v^{2}$
$\Rightarrow v=2\sqrt{2} \, m \, s^{- 1}$