Q. A spring is extended by 30 mm when a force of 1.5 N is applied to it. Calculate the energy stored in the spring when hanging vertically supporting a mass of 0.20 kg if the spring was instructed before applying the mass.
A) 0.01 J
B) 0.02 J
C) 0.04 J
D) 0.08 J
Explanation
(c) $K = \ \frac{1.5\ N^{2}}{30\ \times 10^{- 3}} = 50\ Nm^{- 1}$
$l\ \ = \ \frac{0.2 \times 10}{50}m = 0.04\ m$
Energy stored = $\frac{1}{2}\ \times 0.20\ \times 10 \times 0.04\ J = 0.04\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.