Answer
$-2.51\;J$
Work Step by Step
The water flows steadily through the pipe. Therefore, the flow within any tube of flow obeys the equation of continuity:
$A_{left}v_{left}=A_{mid}v_{mid}$
or, $v_{left}=\frac{A_{mid}v_{mid}}{A_{left}}$
or, $v_{left}=\frac{R^2v_{mid}}{4R^2}$
or, $v_{left}=\frac{v_{mid}}{4}$
and
$A_{right}v_{right}=A_{mid}v_{mid}$
or, $v_{right}=\frac{A_{mid}v_{mid}}{A_{right}}$
or, $v_{right}=\frac{R^2v_{mid}}{9R^2}$
or, $v_{right}=\frac{v_{mid}}{9}$
Now, the net work done on $0.400\;m^3$ of the water as it moves from the left section to the right section is equals to the change in kinetic energy
$W=E_{right}-E_{left}$
or, $W=\frac{1}{2}m(v^2_{right}-v^2_{left})$
or, $W=\frac{1}{2}\rho V(v^2_{right}-v^2_{left})$
or, $W=\frac{1}{2}\rho V v^2_{mid}(\frac{1}{9^2}-\frac{1}{4^2})$
or, $W=\frac{1}{2} \times 1000\times 0.4\times 0.5^2_{mid}\times(\frac{1}{9^2}-\frac{1}{4^2})\;J$
or, $W=-2.51\;J$
Therefore, the net work done on $0.400\;m^3$ of the water as it moves from the left section to the right section is $-2.51\;J$
