Originally Posted by
JD-EE
henra, grity mentioned the aircraft's thrust. That's a stable system. So the real answer is probably between the two calculations you give. I'd suspect it is closer to the square of the delta velocity. (In fact I did a long ways back.)
In fact if you want to be exact, grity is right. The equilibrium of drag and thrust is only valid for one speed. So when the AC slows down in the climb, the 1g drag decreases. Therefore the thrust of the engines will add some energy to the equation.
However, this Delta of energy is negligable compared to the Deltas in energy state we are talking here. The effects of increased drag by maneuvering and changing bank angles will by far exceed the Excess thrust due to the speed decrease.
Between the two calculations I compared there is no doubt about which one is the physically correct one:
it is 1/2(m) v1^2 (Kinetic Energy before the climb) minus 1/2 (m) v2^2 (Kinetic Energy after the climb).