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,
$U = \int_{x_{1}}^{x_{2}}{F_{external}dx}$
Or $U = \int_{x_{1}}^{x_{2}}{kx\ dx}$
$= \frac{1}{2}k(x_{2}^{2} - x_{1}^{2})$
$= \frac{800}{2}\left\lbrack (0.15)^{2} - (0.05)^{2} \right\rbrack\left\lbrack K = 800\ (given) \right\rbrack$
$= 400\lbrack 0.2 \times 0.1\rbrack$
$= 8\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.