Question

The work done in stretching a spring of force constant k from $y_1$ and $y_2$ from mean position will be

• A. $\frac{k}{2}(y_2^2-y_1^2)$
• B. $\frac{k}{2}(y_1^2-y_2^2)$
• C. $k(y_2-y_1)$
• D. zero

Answer 1

The correct option is A $\frac{k}{2}(y_2^2-y_1^2)$

Initial potential energy = $\frac{1}{2}k{y}_1^2$
Final PE = $\frac{1}{2}k{y}_2^2$
Thus work done = change in potential energy = $\frac{1}{2}k(y_2^2-y_1^2)$