Question

A block of mass $2\text{kg}$ is suspended by a spring of force constant $k=10{\text{Nm}}^{-1}$. Another identical spring is fixed below $1\text{m}$ from mass $2\text{kg}$. Initially both the springs are unstretched, and the mass released from rest, then
$($Take $g=10{\text{ms}}^{-2})$

• A. the maximum extension in the upper spring is $2.82\text{m}$.
  
• B. the 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:
$m=2\text{kg};k=10{\text{Nm}}^{-1};g=10{\text{ms}}^{-2}$

Let, compression of the bottom spring is $x$ and assuming that initial position of the block as datum.


Using conservation of mechanical energy,

$\text{Initial energy}\left(E_i\right)=\text{Final energy}\left(E_f\right)$

$\Rightarrow 0=\frac{1}{2}k(1+x{)}^2+\frac{1}{2}k{x}^2-mg(1+x\left)\right[\because E_i=0]$

$\Rightarrow \frac{1}{2}k(1+x{)}^2+\frac{1}{2}k{x}^2=mg(1+x)$

$\Rightarrow \frac{1}{2}\times 10\times (1+x{)}^2+\frac{1}{2}\times 10\times x^2=20(1+x)$

$\Rightarrow 5(1+x^2+2x)+5x^2=20+20x$

$\Rightarrow 10x^2+10x+5=20x+20$

$\Rightarrow 2x^2-2x-3=0$

$\Rightarrow x=\frac{2\pm \sqrt{4+24}}{4}=\frac{2\pm \sqrt{28}}{4}$

Since we're considering elongation in upper spring, so, taking positive value of $x$ only.

$\therefore x=1.822\text{m}$

$\therefore$ Maximum elongation of the upper string is,

$x_{\text{upper}}=1+1.822=2.82\text{m}$

To find the equilibrium position, drawing the FBD of $2\text{kg}$ block as shown, we get,

So, from the FBD, we have

$kx+k(1+x)=mg$

$\Rightarrow 2kx=mg-k$

$\Rightarrow 2\times 10\times x=20-10$

$\therefore x=0.5\text{m}$

Equilibrium position of block from released position,

$x_{eq}=1+0.5=1.5\text{m}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, options $\left(A\right)$ and $\left(D\right)$ are the correct answer.