The Lion MUST be caught ?!?

Car pulls ( engine pulls rather ) with force F.
Car moves with constant velocity V.
:. Net force on car = 0
Force by dissipiative forces such as friction = -F.

Now ; P = δW/δt = δ(F.s)/δt = F.δs/δt
Now since velocity is always relative, δs/δt = u is always relative!

However, since there are two powers acting, one by friction & one by engine, & net F = 0 , the net power on car is 0. This is from ground frame.

If a cycle follows at velocity U, in frame of cycle, the cars velocity = V-U. It is wrong to say that cyclist would say the power delivered by engine is F.(V-U) because the cyclist finds the car not accelerating & hence he caculates the work done by engine to be 0. The cyclist has no direct means to find from his frame what the work done by the engine alone is. ( Cyclist can't measure F, the internal force )

Besides, whenever engine applies a POWER, it makes the car accelerate. And acceleration is independant of relative refrence frames.Both someone on the car and on the cycle would measure the same Acceleration & hence find out the engine's power ( in absence of dissipiative forces ).

To get my point, try to sit on a cycle & follow a car (running at constant velocity) & think of an experiment to determine the power of its engine! You can do so by guessing at its friction, but not directly! This is the key to the problem. Work done is relative.

Therez another realated interesting thought. WHEN WE CLIMB STAIRS UPSTAIRS OR DOWNSTAIRS, THE NET CHANGE IN POTENTIAL E IS THE SAME. THEN WHY DO WE GET TIRED WHILE CLIMBING UPSTAIRS TO A GRATER DEGREE ???