Explanation

(a) Given $F = - 5x - 16x^{3} = - \left( 5 + 16x^{2}\ \right)x = - kx$
where $k( = 5 + 16x^{2})$is force constant of spring ,Therefore , work done in stretching the spring from position $x_{1}$ to position $x_{2}\ is$
$w = \frac{1}{2}k_{2}x_{2}^{2} - \frac{1}{2}k_{1}x_{1}^{2}$
We have,$x_{1} = 0.1\ m$ and$x_{2} = 0.2m.$
$\therefore W = \frac{1}{2}\left\lbrack 5 + 16(0.2)^{2} \right\rbrack(0.2)^{2} - \frac{1}{2}\left\lbrack 5 + 16(0.1)^{2} \right\rbrack(0.1)^{2}$
=$2.82 \times 4 \times 10^{- 2} - 2.58 \times 10^{- 2} = 8.7 \times 10^{- 2}J$