Q. A spring of force constant $800\ Nm^{- 1}$ has an extension of 5 cm. 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) The work is stored as the PE of the body and is given by,
$U = \int_{x_{1}}^{x_{2}}{F_{ext}\ dx}$
or $U = \int_{x_{1}}^{x_{2}}{kx\ dx}$
$= \frac{1}{2}k(x_{2}^{2} - x_{1}^{2})$
$= \frac{800}{2}\lbrack(0.15)^{2} - (0.05)^{2}\rbrack$
$\lbrack 0.02 - 0.002\rbrack\lbrack\therefore k = 800Ns^{- 1}\rbrack$
$= 400 \times \lbrack 0.2 \times 0.1\rbrack$
$= 8\ J$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.