Q. A spring of force constant $800\ N/m$ has an extension of $5cm$. The work done in extending it from $5\ cm$ to $15\ cm$ is
A) $16\ J$
B) $8\ J$
C) $32\ J$
D) $24\ J$
Explanation
(b) $W = \frac{1}{2}k\left( x_{2}^{2} - x_{1}^{2} \right) = \frac{1}{2} \times 800 \times \left( 15^{2} - 5^{2} \right) \times 10^{- 4} = 8J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.