Question

The work done by an external agent in stretching a spring of force constant $K$ from length $l$ to $3l$ is

• A. $8K{l}^2$
• B. $4K{l}^2$
• C. $2Kl$
• D. $\frac{K}{4l}$

Answer 1

The correct option is B $4K{l}^2$

Potential energy in a stretched spring:
$U=\frac{1}{2}k{x}^2$, where $x$ is the elongation or compression in the spring.

Work done by external agent,
$W=\Delta U=U_2-U_1$
$U_1=\frac{1}{2}K{l}^2=\frac{K{l}^2}{2}$
& $U_2=\frac{1}{2}K(3l{)}^2=\frac{9K{l}^2}{2}$
$\Rightarrow W=U_2-U_1=4K{l}^2$