A body of mass $0.5 \mathrm{kg}$ travels in a straight line with velocity $v=k{x}^{3/2}$ where $k=5{ \mathrm{m}}^{-1/2}{ \mathrm{s}}^{-1}$. The work done by the net force during its displacement from $x=0$ to $x=2 \mathrm{m}$ is $............... \mathrm{J}.$

Given : $m=0.5 \mathrm{kg}, v=k{x}^{3/2} where, k=5{ \mathrm{m}}^{-1/2}{ \mathrm{s}}^{-1}$ 

Acceleration, $a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx} \left(\because v=\frac{dx}{dt}\right)$

As $v^2=k^2{x}^3$

Differentiating both sides with respect to x, we get:

$$\begin{gathered} 2v\frac{dv}{dx}=3k^2{x}^2 \\ \therefore \text{ Acceleration, }a=\frac{3}{2}{k}^2{x}^2 \\ \text{ Force, }F=\text{ Mass }\times \text{ Acceleration }=\frac{3}{2}m{k}^2{x}^2 \\ \text{ Work done, }W=\int Fdx={\int }_0^2\frac{3}{2}m{k}^2{x}^{2}dx \\ W=\frac{3}{2}m{k}^2{\left[\frac{x^3}{3}\right]}_0^2=\frac{3}{6}\times 0.5\times 5^2\times \left[2^3-0\right]=50 \mathrm{J} \end{gathered}$$