When a spring is stretched by a distance x, it exerts a force given by F = (–5x – 16x³)N. The work done, when the spring is stretched from 0.1m to 0.2m is 

The amount of work performed is equal to the variation in potential energy for the two positions of spring.

$\mathrm{F} = (–5\mathrm{x} – 16{\mathrm{x}}^3)\mathrm{N}=-(5+16{\mathrm{x}}^2)\mathrm{x} =-\mathrm{kx}$

Now work done is,

$\mathrm{W}=\frac{1}{2}{\mathrm{k}}_2{{\mathrm{x}}_2}^2-\frac{1}{2}{\mathrm{k}}_1{{\mathrm{x}}_1}^2$

$\mathrm{W}=\frac{1}{2}[5+16{(0.2)}^2]{(0.2)}^2-\frac{1}{2} [5+16{(0.1)}^2]{(0.1)}^2$

$\mathrm{W}=2.82\times 4\times {10}^{-2}-2.58\times {10}^{-2} =0.087 \mathrm{J}$