Is the work I do on the object always equal in magnitude but opposite in sign to the work the object does on me?

Question

According to the definition of work, the work I do on an object is equal to the force I apply to the object multiplied by the distance the object moves. By Newton's third law, the reaction force (F21) that the object exerts on me is equal in magnitude and opposite in direction to the force (F12) that I exert on the object. Does Newton's third law also hold in the work-energy theorem? In other words, are the 'work I do on the object' and the 'work the object does on me' equal in magnitude but opposite in sign?

Answer 1

I think there is a difficulty in answering this question because it is not always possible to locate energy in one body or the other when they are interacting. Some of the energy (and it might even be all of it) is in the field by which the bodies are interacting.

Here is an example.

Two bodies of different masses $m_1$, $m_2$ approach one another with equal and opposite momenta $p$. Let's say they are charged and repel one another. They approach one another and come to rest (conservation of momentum). The initial kinetic energies were $p^2/(2 m_1)$ and $p^2 / (2 m_2)$ so the lighter body had more kinetic energy. Clearly both bodies lose kinetic energy in this interaction, and the lighter body loses more than the heavier one. But where has the energy gone? It is in the electromagnetic field. So really this is a system of three things not two.

Now take another case: just like the above but now the bodies do not repel. They collide and stick together. Again, the lighter body has lost more kinetic energy. And where has the energy gone? It has been converted into heat, and it is now internal energy of the stuck-together bodies.

The general conclusion is that energy is conserved, but this conservation does not give rise to a Newton-third-law-like formula except when there are strictly just the two things involved in an interaction and the energy is not converted to other forms. But under that restriction, then, yes, the work done by A on B ($W_{AB}$) means that energy passed from A to B so A has lost the energy and if the work done by B was the only means of energy transfer and no other party has picked up some energy, then $W_{BA} = - W_{AB}$. An example would be a reversible motor slowly and reversibly compressing a gas via a piston.

Answer 2

A mass $m$ which starts from rest has constant force $F$ acting on it for a time $\Delta t$.
The change in position of the mass is $\frac 12 \frac F m (\Delta t)^2$ and this can be used to determine the work done by the force in time $\Delta t$, and the change in kinetic energy of the mass.

So now consider the example of two masses, $m_1$ and $m_2$ each exerting a constant force $F$ on each other (Newton's third law) for an equal time $\Delta t$.
So the magnitude of the work done on the two masses is $\frac 12 \frac {F^2}{ m_1} (\Delta t)^2$ and $\frac 12 \frac {F^2}{ m_2} (\Delta t)^2$ and this not the same unless $m_1=m_2$ which answers your question, $\dots$ are the 'work I do on the object' and the 'work the object does on me' equal in magnitude but opposite in sign?

Possibly the most obvious example of differing amount of work being done on two objects, is dropping something on the Earth, where the dropped object gains kinetic energy and the Earth gains so little kinetic energy that it is usually ignored? In both cases the same magnitude gravitational attractive forces do work but the Earth moves so little as compared with the less massive object.

Answer 3

Newton's third law holds

If we deal with a system that can be described by the laws of classical mechanics, third law is a principle of such a theory and thus it holds. So, to answer your question

Does Newton's third law also hold in the work-energy theorem?

Yes, it does. I won't add anything after "Newton's third law holds".

Third principle and "reciprocal power"

In other words, are the 'work I do on the object' and the 'work the object does on me' equal in magnitude but opposite in sign?

This sentence is not equivalent to the Newton's third law: you can't replace the word "force" with "work" and hope for it to work. If $\mathbf{F}_{ik}$ and $P_{ik} = \mathbf{v}_i \cdot \mathbf{F}_{ik}$ are the force and the power on $i$ due to $k$, in general,

$$\begin{aligned} \mathbf{F}_{ik} & = - \mathbf{F}_{ki} \\ P_{ik} & \ne - P_{ki} \ . \end{aligned}$$

Using the definition $P_{ik} := \mathbf{v}_i \cdot \mathbf{F}_{ik}$ and the third principle, you can easily realize that the equality $P_{ik} = - P_{ki}$ implies

$$\mathbf{v}_i \cdot \mathbf{F}_{ik} = -\mathbf{v}_k \cdot \mathbf{F}_{ki} \qquad \rightarrow \qquad \mathbf{F}_{ik} \cdot ( \mathbf{v}_i - \mathbf{v}_k ) = 0 \ ,$$

i.e. equaity holds only if the bodies have velocity with the same projection along the exchanged force.

Theorem of the kinetic energy in classical mechanics

Let's derive the theorem of kinetic energy for a system of point mass to realize it. The governing equation of each particle is Newton's second law. For the $i$-th particle

$$m_i \dot{\mathbf{v}}_i = \mathbf{F}_i = \sum_{k\ne i} \mathbf{F}_{ik} + \mathbf{F}^{ext,i}_i \ .$$

Time derivative of the kinetic energy of the $i$-th particle, $K_i = \frac{1}{2}m_i |\mathbf{v}_i|^2$ reads

$$\frac{d K_i}{dt} = \mathbf{v}_i \cdot m_i \dot{\mathbf{v}}_i = \mathbf{v}_i \cdot \left( \sum_{k\ne i} \mathbf{F}_{ik} + \mathbf{F}^{ext,i}_i \right) \ ,$$

being $P_{ik}^{int} = \mathbf{v}_i \cdot \mathbf{F}_{ik}$ the power of the force produced by $k$-th particle on the $i$-th particle.

The final form of the theorem of kinetic energy is obtained remembering that the kinetic energy is an additive quantity and summing the $\frac{d K_i}{dt}$-equations over all the particles, and obtaining

$$\begin{aligned} \frac{d K}{d t} = \sum_i \frac{dK_i}{dt} & = \sum_{i} \left[\mathbf{v}_i \cdot \left( \sum_{k\ne i} \mathbf{F}_{ik} + \mathbf{F}^{ext,i}_i \right) \right] = \\ & = \sum_{\{i,k\} , \ i\ne k} \left(\mathbf{v}_i - \mathbf{v}_k \right) \cdot \mathbf{F}_{ik} + \sum_i \mathbf{v}_i \cdot \mathbf{F}^{ext,i}_i \\ & = P^{int} + P^{ext} = \\ & = P^{tot} \ , \end{aligned}$$

i.e. the time derivative of the kinetic energy of a system equals the total power (sum of internal and external) acting on it.

Little remark about "complex systems" and the work-energy theorem

Dealing with complex systems made of lots of components (like almost every object of everyday life), some fraction of the "macroscopic energy" is converted into "microscopic energy" like molecular random motion. Using the macroscopic description of the system with classical thermodynamics, you need to:

• consider the total energy $E^{tot} = K + U$, with $K$ the macroscopic energy and $U$ the internal energy of the system;
• the first principle of thermodynamics to describe the total energy balance,

$$\dot{E}^{tot} = P^{ext} + \dot{Q}^{ext}$$

Answer 4

In other words, are the 'work I do on the object' and the 'work the object does on me' equal in magnitude but opposite in sign?

Sometimes yes, sometimes no.

For the work to be equal and opposite the acceleration of each must be the same. That, in turn, depends on the mass of you and the object being the same per Newton's 2nd law.

Does Newton's third law also hold in the work-energy theorem?

It depends on what you mean by "hold".

In addition to the above, you need to keep in mind that that not all of the work done on an object equals its change in kinetic energy. Just the net work.

If I lift an object of mass $m$ from rest on the ground and bring it to rest a height $h$ above the ground, the force I exert up on the object is equal and opposite to the force the object exerts down on me, per Newton's 3rd law. I do work equal on the object of $mgh$.

However, the change in kinetic energy of object is zero. That's because gravity did an equal amount of negative work taking the energy I transferred to the object and storing it as gravitational potential energy in the Earth-object system.

In the case of a perfectly elastic collision where two objects A and B with different masses collide, one object will lose energy and the other will gain energy. If B gains energy, the work done by A on B is positive, and since A loses energy, the work done by B on A is negative. When two objects with different masses collide, is the amount of work they do on each other different as well?

A perfectly elastic collision between A and B where there are no external forces acting on a A or B is a special case. Assuming the only source of energy available to each do work on the other is the kinetic energy each had prior to the collision (i.e, there is no release of energy stored in A or B prior to the collision), then both kinetic energy and momentum are conserved.

Keeping in mind that the net work done on an object equals its change in kinetic energy, an increase in the kinetic energy of A due to positive work done by B on A, must be matched by a decrease in the kinetic energy of B, due to negative work done by A on B, and vice versa, regardless of their masses.

Hope this helps.

Answer 5

Does Newton's third law also hold in the work-energy theorem? In other words, are the 'work I do on the object' and the 'work the object does on me' equal in magnitude but opposite in sign?

Sometimes yes, but not in general. This is because while the two forces have the same magnitude and opposite directions, the relevant displacements of both bodies (those along the direction of force) are not necessarily of the same magnitude.

Let's consider motion of two compact bodies having different masses $m_1 > m_2$ - so the first body is heavier - as they are released from some distance. They experience attractive gravity forces of same magnitude but opposite directions, along the joining line.

Because the first body is heavier, it will move less, so displacement along the joining line $\Delta \mathbf r_1$ will be of smaller magnitude than displacement of the lighter body $\Delta \mathbf r_2$. So work done on the body 1 will be smaller than work done on the body 2. This can be pushed to extreme, when mass $m_1$ is very many times greater than $m_2$, like in case of Earth and a man. In such case, when the man makes a jump, the man moves a lot, and big work is done on it by force due to Earth; but the heavy Earth body almost does not move, so the force acting on Earth due to man (which has the same magnitude) does only very very small work on the Earth.

The two works are of the same magnitude (and opposite sign) when the body material point experiencing the force makes the same displacement, whether in body 1, or body 2, like when the two bodies "move together". For example, when a lift floor raises a weight, or a train locomotive pulls the railwagon behind it.

Friction forces often do not comply with this condition; friction happens when two bodies touch and move with respect to each other. Then again, displacement of the material points involved in force interaction is not the same, and also, they do not even stay the same material points, but different material points participate at different times. So, for example, when you scratch a hard table surface with your nail, without making a dent, you do zero work on the table, because no part of it moves, but the table does negative work on your nail, because the nail moves against the friction force it experiences.