Question

The system shown in the figure is released from rest with the mass $2\text{kg}$ in contact with the ground. The pulley and spring are massless, and friction is absent everywhere. Then, find the speed of the $5\text{kg}$ block when the $2\text{kg}$ block leaves contact with the ground (force constant of the spring $k=40{\text{Nm}}^{-1}$ and $g=10{\text{ms}}^{-2}$)

• A. $\sqrt{2}{\text{ms}}^{-1}$
• B. $2\sqrt{2}{\text{ms}}^{-1}$
• C. $2{\text{ms}}^{-1}$
• D. $2\sqrt{3}{\text{ms}}^{-1}$

Answer 1

The correct option is B $2\sqrt{2}{\text{ms}}^{-1}$

Let $x$ be the extension in the string when $2\text{kg}$ block leaves contact with the ground.


For $2\text{kg}$ block just breaking contact with the ground,
Normal reaction $N=0$
i.e $T=mg=20\text{N}$

Since the string is joined with one end of the spring and both spring and string are massless, so tension will be the same everywhere
i.e $T=\text{spring force}$
$T=kx...\left(i\right)$
$\Rightarrow kx=T=20\text{N}$
$\therefore x=\frac{20}{k}=\frac{20}{40}=0.5\text{m}$

Applying mechanical energy conservation on the system (blocks+spring+string):
(since the internal forces are non-dissipative and work done by external force from ceiling zero, and external gravitational force is also conservative)

$\therefore \text{Loss in gravitational PE of 5 kg block}=\text{Gain in elastic PE of spring}+\text{Gain in KE of 5 kg block}$

i.e $Mgx=\frac{1}{2}k{x}^2+\frac{1}{2}M{v}^2$
Here $M=5\text{kg}$
$\Rightarrow 5\times 10\times 0.5=\frac{1}{2}\times 40\times {0.5}^2+\frac{1}{2}\times 5\times v^2$
$\Rightarrow 50=10+5v^2$
$\therefore v=\sqrt{8}=2\sqrt{2}\text{m/s}$