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Two plates each of mass $m$ are connected by a massless spring as shown below. A weight $w$ is put on the upper plate which compresses the spring further. When $w$ is removed, the entire assembly jumps up. The minimum weight $w$ needed for the assembly to jump up when the weight is removed is just more than ...........$ \,m$
$1$
$2$
$3$
$4$
Solution
(b)
For block $m_2$ to lift off,
Spring force $=$ Weight of block
$\Rightarrow \quad k x=m g$ or $x=\frac{m g}{k}$
Now, from energy conservation (position ( $1$ and $11$ marked in the diagram),
Initial stored energy (due to $h$ compression) = Energy of extended spring + Potential energy of mass of upper block
$\Rightarrow \quad \frac{1}{2} k h^2=m g(h+x)+\frac{1}{2} k x^2$
$\Rightarrow \quad \frac{1}{2} k h^2=m g\left(h+\frac{m g}{k}\right)+\frac{1}{2} k\left(\frac{m g}{k}\right)^2$
$\Rightarrow \quad \frac{1}{2} k h^2=m g h+\frac{m^2 g^2}{k}+\frac{m^2 g^2}{2 k}$
$\Rightarrow \quad \frac{k h^2}{2}=m g h+\frac{3 m^2 g^2}{2 k}$
$\Rightarrow \quad k^2 h^2-(2 m g k) h-3 m^2 g^2=0$
$\Rightarrow h=\frac{2 m g k \pm \sqrt{4 m^2 g^2 k^2-4\left(k^2\right)\left(-3 m^2 g^2\right)}}{2 k^2}$
$=\frac{2 m g k \pm 4 m g k}{2 k^2}=\frac{3 m g}{k}$
or $h k=3 m g$
Now, considering equilibrium in position $1$ ,
Spring force $=$ Weight of upper block $+$ Weight of mass $M$
$\Rightarrow k h =m g+M g$
$\Rightarrow 3 m g =m g+M g$
$\text { or } M =2 m$
