Question

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

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

Answer 1

The correct option is B

$2\sqrt{2}m{s}^{-1}$

Let x be the extension in the string when 2 kg block leaves the contact with ground. Then tension in the spring should be equal to weight of 2 kg block:

$Kx=2gorx=\frac{2g}{K}=\frac{2\times 10}{40}=\frac{1}{2}m$

Now from conservation of mechanical energy,

$mgx=\frac{1}{2}K{x}^2+\frac{1}{2}m{v}^2$

$\Rightarrow =\sqrt{2gx-\frac{K{x}^2}{m}}=\sqrt{2\times 10\times \frac{1}{2}-\frac{40}{4\times 5}}=2\sqrt{2}m{s}^{-1}$