For a constant force, $P=Fv$. I understand the mathematical derivation of this, but this seems to me, intuitively, to be nonsense. I feel that my discomfort with this comes from a fundamental misunderstanding of force and Newton's Second Law, so I'm not really looking for any mathematical explanation. So, to begin:
How is it that a constant force does not add energy to a system at a fixed rate? Consider a rocket burning a fuel at a constant rate. The chemical potential energy should be converted to kinetic energy at a constant rate, that is, $(1/2)mv^2$ should be increase linearly. The magnitude of the velocity of the rocket would then increase at a less than linear rate, implying a nonconstant acceleration and therefore, a nonconstant force/thrust (F=ma).
If force is indeed a "push or a pull," shouldn't that constant rate of burning of fuel yield a constant "push or pull" as well? Clearly not, so I would have to think that, somehow, a given force applied to a certain object at rest would in some way be different than that a force of the same magnitude being applied to that same object in motion. In this sense, is force merely a mathematical construct? What does it tangibly mean, in physical terms? Would a given force acting upon me "feel" differently to me (in terms of tug) as I am moving at differing velocities?
Force being defined as a "push or pull," which is how it has been taught in my high school class, seems rather "handwavy," and maybe that's the issue. It's been troubling me for a couple of weeks and my teacher hasn't really been able to help, so thanks!
Best Answer
There's nothing wrong with any of these other answers, but for another perspective, if you have a constant force acting on an object starting with zero velocity, then it will accelerate with constant acceleration $\frac{F}{m}$, and thus, after $t$ time, will have velocity $v=\frac{F}{m}t$. This means that the kinetic energy that it has acquired will be given by $\frac{1}{2}mv^{2} = \frac{F^{2}t^{2}}{2m}$.
Since the power is the rate of energy consumption, we have:
$$P = {\dot E} = \frac{F^{2}t}{m}$$
so, it should be obvious that the power increases with time. It should also be clear that our expression for $P$ is equal to $Fv$.