Question

Block is released from rest when spring is in its natural length (assume pulley is ideal and block does not strike on ground during it's motion in vertical plane then

• A. Maximum elongation in spring is $4mg/k$.
• B. Maximum elongation in spring is $2mg/k$.
• C. Maximum speed of block is $2g\sqrt{\frac{m}{k}}$.
• D. Maximum speed of block is $g\sqrt{\frac{m}{k}}$.

Answer 1

Answer: (A), (C)

The correct options are
A Maximum elongation in spring is $4mg/k$.
C Maximum speed of block is $2g\sqrt{\frac{m}{k}}$.

If the maximum elongation in spring is $x_{max}$, block will move by a distance of $2x_{max}$.
By energy conservation,
$\frac{1}{2}k{x}_{max}^2=mg\left(2x_{max}\right)$
$\Rightarrow x_{max}=\frac{4mg}{k}$
At equlibrium position,
$kx=2mg\Rightarrow x=\frac{2mg}{k}$
So,$(K.E.{)}_{max}=mg(2x)-\frac{1}{2}k{x}^2=2mg(\frac{2mg}{k})-\frac{1}{2}k(\frac{2mg}{k}{)}^2$
$\frac{1}{2}m{v}_{max}^2=\frac{2m^2{g}^2}{k}$
$\Rightarrow v_{max}=2g\sqrt{\frac{m}{k}}$