Question

Block of mass $m$ is released from rest when spring is in its natural length. Assume pulley is ideal and block does not strike on ground during its motion in vertical plane. Then:

• A. Maximum elongation in spring is $\frac{4mg}{k}$
• B. Maximum elongation in spring is $\frac{2mg}{k}$
• 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)

Maximum speed of block is $2g\sqrt{\frac{m}{k}}$

When block is released, the spring will stretch by $x$ and block will move downwards by $2x$.

By energy conservation principle,

Energy gain by spring = Loss in gravitational potential energy by block
$\frac{1}{2}k{x}_{\text{max}}^2=mg\left(2x_{\text{max}}\right)$

$\Rightarrow x_{\text{max}}=\frac{4mg}{k}$ ( at maximum elongation)

The block will attain maximum velocity when net force on the pulley is zero.
Let net force on block be zero when $x=x^{\prime }$

At equilibrium, $k{x}^{\prime }=2mg$

$\Rightarrow x^{\prime }=\frac{2mg}{k}$

The kinetic energy of the block is given by
$\frac{1}{2}m{v}^2=mg\left(2x\right)-\frac{1}{2}k{x}^2$

$\therefore \frac{1}{2}m{v}_{\text{max}}^2=mg\left(2x^{\prime }\right)-\frac{1}{2}k{x}^{\prime 2}$

Substituting $x^{\prime }$ to get maximum velocity ($v_{\text{max}}$)

$\frac{1}{2}m{v}_{\text{max}}^2=2mg\left(\frac{2mg}{k}\right)-\frac{1}{2}k{\left(\frac{2mg}{k}\right)}^2$
$\Rightarrow \frac{1}{2}m{v}_{\text{max}}^2=\frac{2m^2{g}^2}{k}$
$\Rightarrow v_{\text{max}}=2g\sqrt{\frac{m}{k}}$

Hence, options (a) and (c) are the correct answers.