Q. When a spring is extended by 2 cm energy stored is 100 J. When extended by further 2 cm, the energy increases by
A) 400 J
B) 300 J
C) 200 J
D) 100 J
Explanation
(b) Potential energy stored in the spring is given by
$U = \frac{1}{2}kx^{2}$
$\therefore\ \frac{U_{1}}{U_{2}} = \left( \frac{x_{1}}{x_{2}} \right)^{2}$
Or $\frac{100}{U_{2}} = \frac{(2)^{2}}{(4)^{2}}$
Or $U_{2} = 400J$
$\therefore$ Potential energy increases by
400-100=300J
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.