Question

A block of mass $2\text{kg}$ is suspended by a spring of force constant $K=10\text{N/m}$. Another identical spring is fixed below $1\text{m}$ from mass $2\text{kg}$. Initially, both the springs are unstretched and the mass is released from rest. Then, which of the following options are correct? (Take $g=10{\text{m/s}}^2$)

• A. Maximum extension in the upper spring is $2.82\text{m}$
• B. Maximum extension in the upper spring is $1.41\text{m}$
• C. Equilibrium position of the block from initial released position is $0.75\text{m}$
• D. Equilibrium position of the block from initial released position is $1.5\text{m}$

Answer 1

Answer: (A), (D)

Equilibrium position of the block from initial released position is $1.5\text{m}$

Given: $m=2\text{kg},k=10\text{N/m},g=10{\text{m/s}}^2$
Let the compression of the bottom spring be '$x$'. Then elongation in upper spring will be $1+x$.

Applying work-energy theorem:
$W_{ext}=\Delta U+\Delta K.E$
At maximum elongation of upper spring (and maximum compression of lower spring), kinetic energy of block will be zero.
$\therefore \Delta K.E=0$ and there is no work done by any external force.
Hence, $\Delta U=0$ i.e loss in gravitation potential energy of the block is converged to elastic potential energy of the springs
$\frac{1}{2}k(1+x{)}^2+\frac{1}{2}k{x}^2=mg(1+x)$
$\Rightarrow \frac{1}{2}\times 10\times (1+x{)}^2+\frac{1}{2}\times 10\times x^2=20(1+x)$
$\Rightarrow 5(1+x^2+2x)+5x^2=20+20x$
$\Rightarrow 10x^2-10x-15=0$
$\Rightarrow 2x^2-2x-3=0$

So, $x=\frac{2\pm \sqrt{4+24}}{4}=\frac{2+\sqrt{28}}{4}=1.822\text{m}$
$\therefore$ Maximum extension in upper spring
$=1+1.822=2.82\text{m}$

Consider FBD of $2\text{kg}$ block at equilibrium (let $x$ be the compression of lower spring)


$kx+k(1+x)=mg$
$\Rightarrow 2kx=mg-k=20-10$
$\Rightarrow 2\times 10\times x=10$
$\Rightarrow x=0.5\text{m}$
So, equilibrium position of block from initial position
$=1+0.5$
$=1.5\text{m}$