Answer
This force does work on the rod at the rate of $~~1.44\times 10^{-7}~W$
Work Step by Step
In part (d), we found that the force on the rod is $2.88\times 10^{-8}~N$
We can find the power provided by this applied force:
$P = F~v$
$P = (2.88\times 10^{-8}~N)(5.00~m/s)$
$P = 1.44\times 10^{-7}~W$
This force does work on the rod at the rate of $~~1.44\times 10^{-7}~W$
