Originally Posted by
Goldenrivett
But will it?
Say you are walking towards the truck at 5 Km/Hr relative to the ground then stop. Your change in KE relative to the ground is 0.5*m*V^2 say 25 units.
Relative to the frame of reference of the truck moving uniformly at say 50 Km/Hr over the ground towards you, it observed you had a relative velocity of 55 Km/Hr initially, then 50 Km/Hr after you had stopped walking towards the truck.
Change in KE according to the truck’s frame of reference = (55*55) - (50*50) = 3025 - 2500 = 525 units.
Yes, you are correct, the change in kinetic energy would differ with the different frame of reference, I guess I had momentum in mind. Which makes the last sentence in my next-to-last post incorrect. to wit:
So, if the change in velocity is identical, then the change in kinetic energy must be identical.
In the scenario with non-zero wind the change in
velocity is still the same, independent of frame of reference, but the change in
kinetic energy will differ. However, what *is* true, is that the change in kinetic energy in the frame of reference of the air mass is identical in both the scenario with wind and the calm air scenario. And if we must consider this from the aspect of kinetic energy, that frame of reference (the air mass) is the only relevant frame of reference, because it is the air mass which is performing work on the airplane, ie; the airplane is only accelerated from +100 knots to -100 knots by the force the air is exerting on the airplane.