Q. A height spring extends 40 mm when stretched by a force of 10 N, and for tensions up to this value the extension is proportional to the stretching force. Two such springs are joined end-to-end and the double- length spring is stretched 40 mm beyond its natural length. The total strain energy in (joule), stored in the double spring is
A) 0.05
B) 0.10
C) 0.80
D) 0.40
Explanation
(b) $k\ = \ \frac{10\ N}{40\ \times 10^{- 3}m} = \frac{1000}{4}\ Nm^{- 1} = 250\ Nm^{- 1}$
Spring constant of combination
=$\frac{250}{2}\ Nm^{- 1} = 125\ Nm^{- 1}\ $
Energy = $\frac{1}{2}\ \times 125\ \times \left( 40 \times 10^{- 3} \right)^{2}\ J = 0.1\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.