A body of mass 2   kg is initially at rest. It starts moving unidirectionally under the influence of a…

Energy provided will be used to increase kinetic energy of the body. Therefore,

$\frac{1}{2}m{V}^2=Pt$

$V=\sqrt{\frac{2Pt}{m}}$

$$\begin{gathered} \Rightarrow \frac{dx}{dt}=\sqrt{\frac{2Pt}{m}} \\ \Rightarrow {\int }_0^{x}dx=\sqrt{\frac{2P}{m}}{\int }_0^4{t}^{\frac{1}{2}}dt \end{gathered}$$

$\Rightarrow x=\sqrt{\frac{2P}{m}}\frac{2}{3}{\left[t^{\frac{3}{2}}\right]}_0^4$

$\Rightarrow x=\frac{16\sqrt{P}}{3}=\frac{1}{3}\times 16\sqrt{P}$

Hence, required value of $\alpha =4$.