A particle of mass 0.2 kg is moving along x-axis and its displacement is given by $\mathrm{x}=2{\mathrm{t}}^2\mathrm{m}$ , where t is time in second. Work done on the particle in 4 sec is

$\mathrm{x}=2{\mathrm{t}}^2$

$\therefore \mathrm{V}=\frac{\mathrm{dx}}{\mathrm{dt}}=4\mathrm{t}\mathrm{m}/\mathrm{s}.\mathrm{At} \mathrm{t}=4\mathrm{s},\mathrm{V}=16\mathrm{m}/\mathrm{s}$

By work energy theorem, work done = ${\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}}=\left[\frac{1}{2}\times 0.2\times {\left(16\right)}^2-0\right]\mathrm{J}$

                                                                            = 256 Joule