True enough, but that's only the
force due to drag. To compute the
power loss due to this force we must recognize that power=force*velocity. The drag force scales with v^2 as you showed, and to compute the power loss we simply multiply by an additional factor of v. v^2 * v = v^3, and therefore, power loss due to drag scales with v^3.
Let's check the units. From
Wikipedia, the drag equation is
[Note: the term 'u' represents the velocity of the body relative to the fluid (air); this corresponds to the quantity I've been referring to as 'v']
rho has units of kg / m^3
u has units of m / s
C_D is unitless
A has units of m^2
(kg / m^3) * (m / s)^2 * m^2 = kg * m / s^2 =
newton, a unit of
force
Multiplying by a factor of u (units of m / s, as above) to get power, we have
kg * (m / s^2) * (m / s) = kg * m^2 / s^3 =
watt, a unit of
power
See the field "SI base units" on both Wikipedia entries for the units and you will find that my calculations are correct.