Q. When a $1.0\ kg$ mass hangs attached to a spring of length $50\ cm$, the spring stretches by $2\ cm$. The mass is pulled down until the length of the spring becomes $60\ cm$. What is the amount of elastic energy stored in the spring in this condition, if $g = 10\ m/s^{2}$
A) $1.5\ joule$
B) $2.0\ joule$
C) $2.5\ joule$
D) $3.0\ joule$
Explanation
(b) Force constant of a spring
$k = \frac{F}{x} = \frac{mg}{x} = \frac{1 \times 10}{2 \times 10^{- 2}} \Rightarrow k = 500\ N/m$
Increment in the length $= 60 - 50 = 10\ cm$
$U = \frac{1}{2}kx^{2} = \frac{1}{2}500\left( 10 \times 10^{- 2} \right)^{2} = 2.5\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.