A body of mass 10 kg moves according to the relation $\mathrm{x} = {\mathrm{t}}^2 + 2{\mathrm{t}}^3$. The work done by the force in the first 2s is.

We know that,

$\mathrm{x}={\mathrm{t}}^2+2{\mathrm{t}}^3$

$\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=2\mathrm{t}+6{\mathrm{t}}^2$

$\mathrm{Work} \mathrm{done}=∆\mathrm{KE}$

At t=0

$$\begin{gathered} v_i=2\times 0+6\times 0 \\ {v}_i=0 \end{gathered}$$

At t=2

$$\begin{gathered} v_f=2\times 2+6\times 2^2 \\ {v}_f=28m/s \end{gathered}$$

Thus work done is given by 

$\mathrm{Work} \mathrm{done}=∆\mathrm{KE}=\frac{1}{2}\mathrm{m}({{\mathrm{v}}_{\mathrm{f}}}^2-{{\mathrm{v}}_{\mathrm{i}}}^2)$

$\mathrm{work} \mathrm{done}=\frac{1}{2}\times 10[{28}^2-0^2]=3920\mathrm{J}$

Hence the correct answer is 3920J.