Question

$K$ is the force constant of a spring. The work done in increasing its extension from $l_1$ to $l_2$ will be:

• A. $K(l_2-l_1)$
• B. $\frac{K}{2}(l_2+l_1)$
• C. $K(l_2^2-l_1^2)$
• D. $\frac{K}{2}(l_2^2-l_1^2)$

Answer 1

The correct option is D $\frac{K}{2}(l_2^2-l_1^2)$

Work done = Change in potential energy

At extension $l_1$, the stored energy =$\frac{1}{2}k{l}_1^2$

At extension $l_2$, the stored energy =$\frac{1}{2}k{l}_2^2$

Thus, Work done in increasing its extension from $l_1$ to $l_2$ is $\frac{1}{2}K(l_2^2-l_1^2)$