Q. A spring with spring constant k is extended from $x = 0\ \ \ to\ \ x = x_{1}$.The work done will be
A) $\kappa x_{1}^{2}$
B) $\frac{1}{2}\kappa x_{1}^{2}$
C) $2\kappa x_{1}^{2}$
D) $2\kappa x_{1}$
Explanation
(b) Work done $W = \int_{x_{0}}^{x_{1}}{F \bullet dx}$
$= \int_{0\ }^{x_{1}}{kx\ dx}$
$= k\left\lbrack \frac{x^{2}}{2} \right\rbrack_{0\ }^{x_{1}}$=$\frac{1}{2}kx_{1}^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.