Classically, the net work done on an initially stationary object equals the final kinetic energy of the object. Verify that this also holds relativistically. Consider only one-dimension motion. It will be helpful to use the expression for p as a function of u in the following:

$W=\int Fdx=\int \frac{dp}{dt}dx=\int \frac{dx}{dt}dp=\int udp$

The one-dimensional motion net work is given by the following.

$$\begin{gathered} W={\int }_0^{u_f}udp \\ ={\int }_0^{u_f}ud\left(\frac{mu}{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}\right) \end{gathered}$$

Let’s get the expression for $dp$first.

$$\begin{gathered} dp=d\left(\frac{mu}{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}\right) \\ =m\left[\frac{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}du-u\left({\displaystyle \frac{1}{2}}{\left(1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right)}^{{\displaystyle \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\left(-{\displaystyle \frac{2udu}{c^2}}\right)\right)}{\left(1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right)}\right] \\ =m\left[\frac{du}{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}+\frac{u^{2}du}{{\left(1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right] \\ =m\frac{du}{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}\left[1+\frac{{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}{\left(1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right)}\right] \end{gathered}$$$$\begin{gathered} dp=d\left(\frac{mu}{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}\right) \\ =m\left[\frac{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}du-u\left({\displaystyle \frac{1}{2}}{\left(1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right)}^{{\displaystyle \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\left(-{\displaystyle \frac{2udu}{c^2}}\right)\right)}}{\left(1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right)}\right] \\ =m\left[\frac{du}{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}+\frac{u^{2}du}{{\left(1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right] \\ =m\frac{du}{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}\left[1+\frac{{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}{\left(1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right)}\right] \end{gathered}$$

Hence, the final expression for the differential change in momentum is

$dp=\frac{mdu}{{\left(1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$

Putting the expression for $dp$in expression of net work done.

$$\begin{gathered} W={\int }_0^{U_f}u\left(\frac{mdu}{{\left(1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right) \\ =m{\int }_0^{u_f}\frac{udu}{{\left(1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ ={\left[\frac{m{c}^2}{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}\right]}_0^{u_f} \\ ={\left[{\gamma }_{u}m{c}^2\right]}_0^{u_f} \end{gathered}$$

Hence, the final expression for net work done on the stationary object is

$W=\left({\gamma }_{u_f}-1\right)m{c}^2$

Thus, it is proved that relativistically, the net work done on an initially stationary object is equal to the final kinetic energy of the object.