Question

When a small mass $m$ is suspended at lower end of an elastic wire having upper end fixed with ceiling. There is loss in gravitational potential energy, let it be $x$, due to extension of wire, mark correct option

• A. The lost energy can be recovered
• B. The lost energy is irrecoverable
• C. Only $\frac{x}{2}$ amount of energy is recoverable
• D. Only $\frac{x}{3}$ amount of energy is recoverable

Answer 1

Answer: (C)

$\Delta U$ (loss in gravitational potential energy) $=m g \times \Delta l $
$\Delta U=x $ (given)
So $ x=m g \times \Delta$
Where
$m=$ mass suspended
$\Delta l =$ elongation in wire
Elastic potential energy gained $=\frac{1}{2} \times $ Force $ \times$ Elongation
$=\frac{1}{2} \times M g \times \Delta l $
$=\frac{1}{2} M g \times \Delta \quad[\because M g \Delta l=x] $
$=\frac{1}{2} x$
So only $\frac{x}{2}$ amount of energy is recoverable which is stored as elastic potential energy in wire.