A body of mass 2kg is initially at rest. It starts moving...

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

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

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

$⇒ \dfrac{dx}{dt} = \sqrt{\dfrac{2Pt}{m}} ⇒ {∫}_{0}^{x}dx = \sqrt{\dfrac{2P}{m}}{∫}_{0}^{4}t\dfrac{1}{2}dt$

$⇒ x = \sqrt{\dfrac{2P}{m}}\dfrac{2}{3}{\left[t\dfrac{3}{2}\right]}_{0}^{4}$

$⇒ x = \dfrac{16\sqrt{P}}{3} = \dfrac{1}{3} \times 16\sqrt{P}$

Hence, required value of $\alpha = 4$.