Why is Kinetic energy not an explicit function of acceleration?

Question

A few days ago a high schooler asked me this question to which I couldn't give an answer.

His main question was that acceleration is also a property of a moving body so why is Kinetic energy which keeps track of energy associated with motion doesn't include the acceleration explicitly? Of course there would be some dimensional correction needed but still Is there any reason for both instantaneous velocity and acceleration to be not there simultaneously in the formula?

My reply was that the instantaneous acceleration would actually affect the kinetic energy but its contribution would be taken care of when we consider the velocity after an small time interval. I am not sure if it is the answer or not. So someone please clarify once.

Answer 1

This is just another "why is this thing not the definition I think it is?" question. The thing we define to be kinetic energy is a function of speed. If you want to get to the heart of the question, you have to ask the student what idea they are "starting with"; you will most likely find it to be a qualitative idea that doesn't match up with the actual definition of energy.

There has to be something deeper than what is given in the OP; otherwise you could argue that really anything that is relevant could be included. e.g. "if an object is in a gravitational field, then why doesn't anything about gravity show up in the formula for kinetic energy?", etc.

Answer 2

Kinetic energy is not a generic property of a moving body. It is a property defined in such a way as to satisfy some essential relations directly originating from Newtonian Mechanics.

In particular, the reason it is called kinetic energy originates from the work-energy theorem, which is a direct consequence of the definition of work and the Newtonian equations of motion. For a single point-like particle: $$ \frac12 m \vec v^2(B) - \frac12 m \vec v^2(A) = \int_A^B\vec F \cdot d \vec l. $$

The form of the equation of motion $\vec F = m \frac{ d \vec v}{d t} $ is responsible for the limitation only to the velocity of the formula for the kinetic energy.

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Post Scriptum

Notice that from the historical point of view, the appearance of quantity like $\frac12 m \vec v^2$ in the energy-work theorem was one of the strongest motivations to give it a name and call in a way reminiscent of some relation with work (in ancient Greek ergon == work). The original name of kinetic energy was vis viva. In Latin, vis means force. That could hint at the conceptual elaboration underlying the introduction of the kinetic energy concept.

Answer 3

There is one assumption made in the way he framed the question:

His main question was that acceleration is also a property of a moving body so why is Kinetic energy which keeps track of energy associated with motion doesn't include the acceleration explicitly?

Emphasis mine, to highlight this assumption.

It is not correct, to the best of my understanding, to ascribe acceleration to the body undergoing the acceleration. If the body is being accelerated, it is usually taken to be a result of a dynamical process in which another body is exerting a force on it.

For example, if one charge pulls on another, accelerating it, but then suddenly disappears or, a bit more physically put, whisked away quickly towards infinity by some other agency, the acceleration goes away (as quickly as the field propagates towards the pulled charge), and the charge will indeed maintain a constant KE from that point onwards, in the absence of any other additional forces.

So your answer was correct, but one should add to it that acceleration, rather than being a property of the moving body, induces a change to a property of the moving body, which is its momentum.

In classical mechanics at least, the prevailing model is one in which, we only need knowledge of two properties: positions and momenta of the particles, to predict the evolution of the system, assuming of course that we are also given the forces, e.g. in the form of a potential so that, we have something like: $\vec{F}(\vec{x},t)=-\nabla U(\vec{x},t)$.

But, this added information from which we derive the forces is not to be seen as a property of the masses/bodies themselves, but rather a result of one or more interactions that occur between them (for example, as mentioned, an electric interaction of charges). Put differently, it is more correct to take the forces, as a property of the system, rather than of any particular part of it. These forces then generate accelerations, which therefore arise from these various mutual interactions, thus cannot be ascribed exclusively to a specific body that is accelerated (this is also embedded in the correct understanding of N3L).

For more about this last point see this related Phys.SE question -- there is an extensive list of links OP included in this question.

Answer 4

Most high school students have seen the kinematic equations. Specifically

$$v^2 = v_0^2 + 2ad$$

This is the explicit equation that shows $\Delta v^2$ is directly proportional to acceleration.

If you multiply both sides by the mass m and rearrange. You get the Work-Energy theorm

$$\frac 1 2 mv^2 - \frac 1 2 mv_0^2 = mad = Fd$$

Answer 5

His main question was that acceleration is also a property of a moving body so why is Kinetic energy which keeps track of energy associated with motion doesn't include the acceleration explicitly?

That's a non sequitur. Why would two things being a property of an object mean that one can be written in terms of the other? Moving bodies (all bodies, in fact) have some electrical charge (even if that charge is zero). Can charge be written in terms of KE, or vice versa?

Of course there would be some dimensional correction needed

And how do you make that correction? The units of KE are kg m^2/s^2. The units of acceleration are m/s^2. So to go from acceleration to KE you need a mass and a distance. For mass, you can use the mass of the object, but what do you use for distance? You'd need some distance that the object travels over, and that would then get you the change in KE that it experienced while it traveled over that distance, not the total KE, and even then that would only be for constant acceleration.

but still Is there any reason for both instantaneous velocity and acceleration to be not there simultaneously in the formula?

Velocity is change in position. Acceleration is change in velocity. They are both measures of change, so to get a total amount such as KE, you need to integrate from some zero point. If we know the velocity of an object is zero at time $t_0$, then we could write the KE at time $t$ as $\int_{t_0}^t ma dx$, but that's generally not a useful way of writing it.

Answer 6

A change in kinetic energy depends on acceleration and is independent of the choice of relatively inertial reference frames.

On the other hand, kinetic energy itself, which is a function of velocity, depends on the inertial frame of reference.

See

https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Classical_Mechanics_(Dourmashkin)/15%3A_Collision_Theory/15.02%3A_Reference_Frames_and_Relative_Velocities

Hope this helps.