Work-energy theorem: can't figure out an apparent exception to it as stated in my textbook

Question

In this page of free College Physics text book by OpenStax, they state

this is different from the question Does work-energy theorem involve potentials? because that question is concerned with the theorem applied to objects which makes sense to me. my concern is with the validity of the statement in the college physics text book which applies generally to systems.

The net work done by all forces acting on a system equals its change in kinetic energy. In equation form, this is $W_{\text{net}}=\frac{1}{2}mv^2-\frac{1}{2}mv_0^2 = \Delta KE$.

Now, for most circumstances this makes sense to me.

However, if the system was the earth-object system and an external force to this was applied on the object:

Then if this force is equal to the object's weight, and the object moves with constant velocity. I realize we need to accelerate the object but let's consider the system only during the period it moves with constant velocity, then the kinetic energy of the system does not change although an external force is applied. This would be a contradiction to the statement of the work energy theorem in the textbook.

Now, I realize that the energy has gone into potential energy. My concern however is with the statement of the theorem. Is it correct, and if it is what am I missing that makes it correct?

Should the theorem be applied to something more specific than a system? e.g. an object or a particle? Or, should there be caveats in the types of systems that it applies to?

Answer 1

If you are going to consider the Earth and the object as a single system, then you need to include the KE of the Earth and the PE of the gravitational field between them in the work-energy equation. The PE of the gravitational field needs to be included because the gravitational attraction between the Earth and the object is now an internal force (if your system is the object alone, then gravity is an external force and you do not need to include its PE).

The external force may act on the object so as to keep its KE constant, but the object attracts the Earth and so the earth's KE changes and the separation between the Earth and the object also changes. The work done by the external force on the Earth-object system is equal to the KE gained by the Earth minus the PE lost as their separation decreases.

Answer 2

The statement in the textbook is restricted to systems which have no means of storing internal energy, or changing their internal energy content. For example, the statement can be applied to a simple particle but not to a spring. You can do work on a spring by squashing or stretching it, and then the work goes to internal energy of the spring.

In the case of a pair of bodies mutually attracting by gravitation, the gravitational energy associated with their attraction is a form of internal energy of the total "two-body plus field" system. You can think of further examples involving electric and magnetic fields.

Answer 3

Unfortunately, the Work Energy Theorem is one that is a very misunderstood theorem and in addition different authors use different conventions and details in the derivation and specification of the theorem. It is important to learn what the author of your textbook means and to recognize that it might be different from what other authors intend. In particular, some authors restrict the work energy theorem to only apply to point particles and others to arbitrary systems.

Assuming that your textbook uses the theorem to apply to arbitrary systems, then it is very important to pay attention to the meaning of the variables. The net work, $W_{net}$ is equal to the net force (the sum of all external forces) times the displacement of the center of mass of the system, $W_{net}=\vec F_{net}\cdot \Delta \vec r_{com}$. This version of the work energy theorem states that $$W_{net}=\frac{1}{2}m v_{com,final}^2-\frac{1}{2}m v_{com,initial}^2=\Delta KE_{com}$$

Now, because the earth-object system is very massive, the displacement of the center of mass, $\Delta \vec r_{com}$, is very small. So, the net work is also very small, and this very small work corresponds to a very small change in the KE of the system. There is some slight upwards acceleration of the Earth due to the unbalanced gravitational force between the Earth and the object, and it is this unbalanced force on the Earth that leads to the small increase in KE, although the object's KE is constant due to the balanced forces on the object.

Answer 4

In the case of a system of particles the work energy theorem applies to the change in kinetic energy of the center of mass of the system, not the particle subjected to the force. So

$$F_{ext}d=\frac{1}{2}(m+M)v^2$$

Where

$F_{ext}$ = the external force applied to the object

$d$ = displacement of the center of mass of the object/Earth system

$m,M$ = the mass of the object and the Earth, respectively

$v$= the velocity of the center of mass of the object/Earth system, which would be essentially the velocity of the center of the Earth if $m<<<M$.

thus, if $F_{ext}=mgh$ and $m<<<M$, the velocity of the center of mass would be infinitesimal.

It should be noted the work energy theorem applies to rotational kinetic energy as well as translational kinetic energy. In this case, however, since the direction of the external force is through the center of mass, rotational kinetic energy doesn’t apply.

Hope this helps.

Answer 5

We could write energy conservation as something like:

$$ \Delta W = \Delta KE + \Delta PE + \Delta IE +\Delta CE + \Delta RE + ...$$

where PE is potential energy, IE is internal energy, CE is chemical energy, RE is rest energy. There are various other forms of energy that can be added to this list such as elastic energy. The work energy theorem is a cut down version on the conservation of energy where changes in all forms of energy except KE are assumed to be zero. In this case $\Delta W \ne \Delta KE$, but in fact what is happening in this case is $\Delta W = \Delta PE$. The scenario has specified that we take the measurement when the velocity is constant so the $\Delta KE$ must be zero by definition. Even if we consider the velocity of the centre of mass of the system, it too will be constant and we still have $\Delta KE =0$. Since the work energy theorem assumes $\Delta PE = 0$ you are using the work energy theorem outside of its very limited area of application. Its much better to just use the conservation of energy rule, which is much more generally applicable.