Two plates each of mass m are connected by a | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 conser

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(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

$\Rightarrow \quad \frac{1}{2} k h^2=m g(h+x)+\frac{1}{2} k x^2$

$\Rightarrow \quad \frac{1}{2} k h^2=m g\left(h+\frac{m g}{k}ight)+\frac{1}{2} k\left(\frac{m g}{k}ight)^2$

$\Rightarrow \quad \frac{1}{2} k h^2=m g h+\frac{m^2 g^2}{k}+\frac{m^2 g^2}{2 k}$

$\Rightarrow \quad \frac{k h^2}{2}=m g h+\frac{3 m^2 g^2}{2 k}$

$\Rightarrow \quad k^2 h^2-(2 m g k) h-3 m^2 g^2=0$

$\Rightarrow h=\frac{2 m g k \pm \sqrt{4 m^2 g^2 k^2-4\left(k^2ight)\left(-3 m^2 g^2ight)}}{2 k^2}$

$=\frac{2 m g k \pm 4 m g k}{2 k^2}=\frac{3 m g}{k}$

or $h k=3 m g$

Now, considering equilibrium in position $1$ ,

Spring force $=$ Weight of upper block $+$ Weight of mass $M$

$\Rightarrow k h =m g+M g$

$\Rightarrow 3 m g =m g+M g$

$	ext { or } M =2 m$

Similar Questions

Two springs have their force constant as ${k_1}$ and ${k_2}({k_1} > {k_2})$. When they are stretched by the same force

Two springs have their force constant as $k_1$ and $k_2 (k_1 > k_2)$. when they are stretched by the same force

A $2\ kg$ block slides on a horizontal floor with a speed of $4\ m/s$. It strikes a uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is $15\ N$ and spring constant is $10,000\ N/m$. The spring compresses by ............. $\mathrm{cm}$