Q. A spring with spring constant k when stretched through 1 cm the potential energy is U.If it is stretched by 4 cm ,the potential energy will be
A) 4U
B) 8U
C) 16U
D) 2U
Explanation
(c) Potential energy U$= \frac{1}{2}kx^{2}$
$\therefore\ \frac{U_{1}}{U_{2}} = \left( \frac{x_{1}}{x_{2}} \right)^{2}$
Or$\frac{U}{U_{2}} = \left( \frac{1}{4} \right)^{2}$
Or $U^{2} = 16\ U$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.