Question

The work done in stretching a spring of force constant $k$ from length $l_1$ to $l_2$ is

• A. $k(l_2^2-l_1^2)$
• B. $\frac{1}{2}k(l_2^2-l_1^2)$
• C. $k(l_2-l_1)$
• D. $k/2(l_2-l_1)$

Answer 1

Answer: (B)

$\frac{1}{2}k(l_2^2-l_1^2)$

Potential energy in a stretched spring $U=\frac{1}{2}k{x}^2$
where, $x$ is the extension in the spring.
So, $U_2=\frac{1}{2}k{l}_2^2$ and $U_1=\frac{1}{2}k{l}_1^2$
Work done $W=U_2-U_1$
$W=\frac{1}{2}k{l}_2^2-\frac{1}{2}k{l}_1^2=\frac{1}{2}k(l_2^2-l_1^2)$.