If I run and push a car, the force considered is the one felt in my arms not in my legs? The distance the one marked on the field? If the energy outputs is constant and linear with time, why isn't work also linear with the time and increase so exponentially? If I stop applying the force on the car, stop doing work on it, d is still increasing because of the velocity. then the formula F.d is integrating something superfluous that it shouldn't count.
Consider a rocket trajectory in space for 10 sec from distance d0 to d. From the earth I know it has traveled a greater distance in the last seconds because of the stored momentum. If I apply the work formula F.d to measure the energy transfer. It means the engine has done more work from second 9 to 10 than between second 0 to 1?
My physic book doesn't seem to care whether it correlates with the chemical energy output or whether I understand the formula. It says the force is the same all over the distance and we multiply it by the distance traveled. Imagine the flow rate of the propellant is constant, it outputs X watts (I guess this is the condition for the force be the same), does the ergol have more potential when it goes faster?
I've seen momentum, impulse so far, I can't find any calculus that lead us to F.d, that would justify we decided to scale all our energy quantification with the work and kinetic energy formula, I understand the relation between those two. But I'd like to see some calculus starting from the momentum theorem which I understand, and by changing the variable from t to d, or by integrating the impulse.
What bothers me is that if work is linear to the distance, it can't be linear to time (which is the distance traveled by my hand in my watch but which has been used as a reference to measure the acceleration of my rocket and the power of my propellant). (I see that using d instead of t may be convenient with torques but even there I see no proof that the time to torque represents the total time of my impulses on both sides, maybe they are just less or more frequent depending on the time the normal force to come back from the center)
I think I would have understood if the force had be defined with m.t/d² from the beginning.