Q. When a spring is stretched by a distance $x$, it exerts a force, given by $F = \left( - 5x - 16x^{3} \right)N$. The work done, when the spring is stretched from $0.1\ m$ to $0.2\ m$ is
A) $8.7 \times 10^{- 2}J$
B) $12.2 \times 10^{- 2}J$
C) $8.7 \times 10^{- 1}J$
D) $12.2 \times 10^{- 1}J$
Explanation
(a) $F = - 5x - 16x^{3} = - \left( 5 + 16x^{2} \right)x = - kx$
$\therefore k = 5 + 16x^{2}$
Work done, $W = \frac{1}{2}k_{2}x_{2}^{2} - \frac{1}{2}k_{1}x_{1}^{2}$
$= \frac{1}{2}\left\lbrack 5 + 16(0.2)^{2} \right\rbrack(0.2)^{2} - \frac{1}{2}\left\lbrack 5 + 16(0.1)^{2} \right\rbrack(0.1)^{2}$
$= 2.82 \times 4 \times 10^{- 2} - 2.58 \times 10^{- 2} = 8.7 \times 10^{- 2}J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.