Answer
Please see the work below.
Work Step by Step
We know that
$v=\sqrt{\frac{F}{\mu}}$
We plug in the known values to obtain:
$v=\sqrt{\frac{550}{0.280}}=44.32\frac{m}{s}$
Now we can find the average power as
$P=(\frac{1}{2})\mu \omega^2A^2v$
We plug in the known value to obtain:
$P=(\frac{1}{2})(0.280)(2\times 3.1416\times 3.3)^2(0.061)^2(44.32)$
$P=9.9W$
