Why does power increase as a constant force accelerates a body? [duplicate]

Question

If a constant force is being applied to a body, without any other external forces, F = ma says that that body will accelerate at a constant rate. This acceleration will continuously increase the body's velocity.

According to P = Fv, since the force is constant and the velocity is continuously increasing, the power required by the force will continuously increase.

I understand all the maths, but am trying to get a better intuitive understanding of this. I cannot seem to come to terms with the fact that a constant force will need to supply an increasing power. What is this power being represented by, if the force is constant? What typical inner-workings of such a force would require its power to increase, even though its ultimate "output" is the same? What actually constitutes "power" and "force" at the "force-side" of things?

Answer 1

Suppose you are pushing a box on a surface and want to apply a constant force, F. As you point out, the box will accelerate. But as its speed increases, you have to run faster in order to continue supplying the same force F. You must use more energy per unit time (power) in order to keep the force at F. So the answer to your question is that you have to examine how that constant force F is able to be maintained over the range of velocities for which it is in effect.

Answer 2

Suppose the body starts from rest at time $0$ and accelerates at a constant rate $a$. At time $t$ it has speed $v=at$ and kinetic energy $E=\frac 1 2 ma^2t^2$.

At time $t+\delta t$ it has speed $a(t+\delta t)$ and energy $E + \delta E = \frac 1 2 ma^2(t+\delta t)^2$. So

$\delta E = m a^2 t \delta t + \frac 1 2 ma^2 (\delta t)^2$

So the power applied to the body is

$\frac {\delta E}{\delta t} \rightarrow ma^2t$ as $\delta t \rightarrow 0$

So power increases linearly with time because kinetic energy increases as the square of speed.

Answer 3

There is no internal mechanism, imagine a ball falling in a constant gravitational field. It basically means that the same force acts for a longer distance (not a longer time) if the speed is larger. Intuitively, for a conservative force like gravity this means that you get potential energy faster

Answer 4

You have already accepted an anwser, but in a comment you wonder whether a closer intuition is possible.

For a thought demonstration: imagine a rocket that is being refueled at regular distances. Think of that as similar to aerial refueling of aircrafts.

Except, in the case of aerial refuelling the tanker aircraft only needs to match the velocity of the receiving aircraft. But the spacecraft is continously accelerating, so each subsequent refueling mission has to travel to a location that is quadratically farther away from the home base.

That is, in the case of refueling a spacecraft from a single homebase the energy cost of transporting propellent all the way out to the accelerating spacecraft increases quadratically.

In the case of for example gravity this aspect of acceleration is so non-obvious that it is easily overlooked. It's really quite remarkable that the amount of gravitational acceleration that a gravitating body causes is independent of the velocity relative to that gravitating mass. That is, gravity is an accelerating influence that never runs out of capacity to accelerate more.

A spacecraft that is being accelerated by gravity is like a spacecraft that is continuously receiving propellent supply.

Back to the car

So, how to translate this to propulsion on Earth; a car being propelled by an engine.

The counterpart of the home base is the ground that the wheels are resting on. The traction of the car is with respect to the ground. The reason that this matters is the following: the kinetic energy of an object is not something intrinsic to that object. Kinetic energy is Galilean invariant, which means that for kinetic energy galilean relativity holds good

The kinetic energy that you attribute to a car is the kinetic energy associated with the velocity relative to another object. In a collision the amount of damage inflicted is proportional to the kinetic energy.

A marble that is penetrating clay

Probably the simplest visualization of damage inflicted on impact is a marble penetrating a lump of clay. For the sake of simplifying, let's say that an experimenter finds that twice the velocity give 4 times the depth of penetration, and that in general the penetration depth is kwadratically proportional to the velocity.

As we know: uniform acceleration has the following property: to get double the velocity you have to quadruple the height; the relation is quadratic.

The experimentor will find that the relation between penetration into the clay and height of the fall is 1-on-1; linear.

With that in place I return to the car. What does it take to cause uniform acceleration? For simplification: let's say the car is an electric car, and air resistance has been removed (let's say the car is in a vacuum tunnel) We disregard rolling resistance too, of course.

The acceleration from 0 unit of velocity to 1 unit of velocity take a certain amount of energy from the battery pack. Let's call that 1 unit of energy. The acceleration from 1 unit of velocity to 2 units of velocity takes 3 units of energy.

The car increases its own velocity by having traction with respect to the ground.

The only way for that car to increase its velocity is to increase its kinetic energy with respect to the ground, the only traction that the car has is with respect to the ground. Relative to the car the ground is moving backwards ever faster, so it is harder and harder for the car to increase its velocity.

The electric engine of the car maintains a constant force. To maintain that constant force every revolution of the engine draws a particular amount of energy from the battery pack. But this constant force must now act over a longer distance. In order to maintain the same traction the wheels must spin faster, the engine must spin faster and thus the engine will draw more units of energy from the battery pack.