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