Why does a conservative force return the work done against it by a body to that body?

Question

Newton's 3rd law of motion:

Newton's third law of motion, or the law of action and reaction, implies that there is no isolated force in nature. Whenever there is any force at all, there must be at least two forces. There can be no force without an opposing force. In fact,

Every force results from the interaction of two bodies. The two bodies involved experience equal and opposite forces.

The "WORDS" to which I refer are as follows:

When two bodies interact, and one body exerts a force on the other and does negative work, the other body simultaneously exerts a reactive force on the former and does positive work on it. (This statement is extracted from an answer written by Floris.)

Non-conservative force follows the "WORDS":

When a body moves on a rough floor, kinetic friction $F_k$ opposes the motion and does negative work on the body which is given by $W=-F_{k}\cdot d~ $, where $d$ is the displacement. Simultaneously, by Newton's third law, the body exerts the same force but opposite in direction on the rough floor and does positive work on the floor: partly to rotate it (though very very much less, since it is a part of the earth, which is a large sphere) and mostly to increase the kinetic energy of the molecules of the floor, (i.e., thermal energy). Thus, the positive work done by the body is given by $W=F_{k}\cdot d$. Thus, non-conservative forces like friction here follow the "WORDS". The agent of the friction, the rough floor, gains the positive work done on it, most of it going to heat. Thus, the body loses energy by the same amount.

Conservative forces don't follow the "WORDS":

When a body moves against conservative forces (gravitation, electric fields, etc.), conservative forces do negative work on the body. Here also, the body will also exert a force (as forces are mutual, and are attributed to Newton's 3rd law, the law of gravitation, Coulomb's law...) on the agents of the forces (earth, charge) & will do positive work on them. But instead of gaining that work, it is stored as potential energy. Thus, though the body loses energy, the agent doesn't gain that energy.

My question:

Why don't, unlike the rough floor, agents of the conservative force gain that work, but rather store it as potential energy???

Answer 1

A conservative force only returns the energy back when the object moves in a closed path, that is, it returns to the initial position (it doesn't matter if he returns due to other forces). This can be demonstrated as a theorem, but the intuitive explanation is that a conservative force depends only on the spatial coordinates, and not in the direction of motion (such as friction), and thus eventually when the body moves back the field force is in the opposite directions and makes work of the opposite sign, such that in a closed loop the overall work of the field over the particle is zero. The additional property for a field is that this happens regardless of the path taken (that is, you do not need to go back to the original position using the same path you used when moving forward. This can be shown to be the case when the force is described as the gradient of a potential.

Update: I reread your question and I realized that I didn't actually answer your specific question (I misinterpreted the question). Let us start with the easy case: a mass attached to a spring. Here your body is the mass and the agent is the spring. Describing the mass moving in a conservative field instead of describing it in terms of mutual forces is much easier, as we do not have to take into account the details of the internal forces within the spring. The mass losses energy and this energy is stored in the spring, not in the mass. However we can give an alternative description ignoring the spring and saying that the mass accumulated potential energy. This is very convenient, but only if your are sure that the spring will always be attached to the mass (so you can describe the force from the spring as conservative). But if you cut the spring when the mass is at rest, the mass will "suddenly lose" its potential energy (the force is no longer conservative), the energy was actually in the spring an will be dissipated as heat (assuming the spring will eventually stop moving, as in a real spring).

In the case of long distance forces that cannot be switched off, such as gravity, the description is a little more complex. If you have a very large mass as the agent, you can approximate it as not moving due to the reaction force from the object, and describe the object as moving in a potential field where it stores potential energy. A more accurate description would be that the agent actually moves and gains this as kinetic energy (when the object moves "up",
the earth will follow it and also move up too so it gains kinetic energy). But this motion is so small that you do not take it into account in your description. For all practical purposes the object sees the agent as being at rest. Of course, I am assuming that you can move the object with a force that doesn't interact with the agent, otherwise it is the same but this time the agent might be storing the energy in some different way, however the details do not matter for all practical purposes.

Answer 2

I think there is some confusion about the terms and the relationship between conservative forces and energy conserving systems. I will make a few general statements:

1) A conservative force can be described by a potential function $\Phi(x)$ with $\vec F=-\nabla{\Phi}$, a non-conservative force can not be reduced to such a function.

2) Friction is a non-conservative force that also doesn't conserve energy, but not every non-conservative force is a friction force. There are non-conservative forces that conserve energy and can return all of it, but that can't be described by a scalar potential function of the mechanical coordinates. One such example would be a torsion spring, where the total work performed on the spring is not uniquely defined by the coordinates of the particle (rather it's defined as a potential of the total angle that the spring has been wound up). Time dependent forces may be energy conserving, but may not return all of the energy to the mass they act on. Another example is an energy conserving velocity dependent force like a magnetic field acting on a charged particle.

3) Even a conservative force may never return the work that was performed against it. A simple example is a particle that rolls up on an incline and keeps moving at constant velocity on the flat top of the potential.

The easiest way to avoid confusion is to apply the potential field definition. If a potential can be defined, we are looking at a conservative force. If it can not be, then we have a more complex physical situation to deal with, and then there are multiple different things that can happen, some of which are energy-conserving and some which are not (at least from the reduced point of view of the mass that the force is acting on).

Answer 3

My Interpretation of your question :

Why don't agents of the conservative force gain internal energy (i.e. heat) as in the case of friction, instead of gaining just potential energy.

In any Conservative force, Mechanical Energy is conserved. $$\therefore~~ E_{k~(\text{initial})} + E_{p~(\text{initial})} = E_{k~(\text{final})} + E_{p~(\text{final})}~~~~~~~~~(1)$$

In non-conservative forces, mechanical energy is not conserved, and some energy is lost: $$E_{k~(\text{initial})} + E_{p~(\text{initial})} > E_{k~(\text{final})} + E_{p~(\text{final})}$$ To accomodate this lost energy, we can reform the equation as : $$E_{k~(\text{initial})} + E_{p~(\text{initial})}-E_{dissipated} = E_{k~(\text{final})} + E_{p~(\text{final})}~~~~~~~~~(2)$$

Back to your question Why don't agents of the conservative force gain internal energy (i.e. heat) as in the case of friction

EXAMPLE 1

Consider this deliberately made example :

In this example, the body has some potential energy because of its position. As the body starts moving towards the Earth, this $E_{p}$ starts converting into $E_{p}$, until $E_{p}~\text{becomes 0}$. So, $$E_{k~(\text{initial})} + E_{p~(\text{initial})} = E_{k~(\text{final})} + E_{p~(\text{final})}$$ $$0+E_{p(\text{initial})}=E_{k(\text{final})}+0$$ $E_{k}$ just before the body reaches the Earth is equal to $E_{p}$ There is no need for storing energy internally, as energy is already conserved.

EXAMPLE 2

Consider another example, this time of a non-conservative force.

If we push the block at A, it moves (e.g.) till B, and stops. Applying equation 1, we get $$E_{k~(\text{initial})} \stackrel{?}{=} 0$$ but this is not true, as we push the block with some velocity, and it has some initial kinetic energy. Therefore to account for the "lost" energy, we add another term $E_{dissipated}$ to RHS, to comply with Conservation of Energy. And hence we get $eq^{n}~2$ for non-conservative forces.

NOTE

Non-conservative forces like friction and air resistance set in motion the particles the come in contact with. Energy goes into heating up the particles. At microscopic level, this thermal energy includes both the kinetic energy and potential energy of a system's constituent particles, which may be atoms, molecules or electrons etc.

This heating up does not occur in case of conservative forces as these are much more macroscopic and action of these forces is at a distance. Therefore there is no dissipation of energy. Also, most forces in real life are non-conservative, as there is always some energy that gets dissipated.

EDIT

As @t.c stated :

... its path dependence means that work done to an object cannot be successfully retrieved by letting it return to its original position - defeating the whole meaning of "potential" energy altogether.

The Energy possessed by the body in case of non-conservative force is not retained and contained within the system. It gets dissipated. If in example 2, you move the body directly from A to B, there is less dissipation of energy than if it follows an indirect path (eg. a zig-zag or a haphazard path)

Following is a simple demonstration of dissipation of energy (although it is not very accurate)

Source: mw.concord.org

In Example 1, when the body reaches the ground, then the energy will get transferred to the earth. When the body collides with the earth, friction force comes into play, which is non-conservative. Most energy will be transferred in the form of kinetic energy, and most of it will go on to change the entropy of Earth rather than changing its velocity.

If we consider air resistance (non-conservative force), some of the kinetic energy of the body will get transferred to the particles of air, and $E_{k~(body)}$ will decrease, and $E_{k~(air~particles)}$ will increase. The entropy of air will therefore increase.

Therefore, non-conservative forces cause a change in the entropy of the system

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EDIT 2 - Response to the comment by @Alb

OP : When a body moves against conservative forces ... on the agent of the forces(earth,charge) & will do +ve work on them. But instead of gaining that work , it is stored as potential energy. Thus though the body loses energy ,the agent doesn't gain that.

The agent doesn't gain that energy because it already gets conserved. When a ball is thrown upwards, and it reaches a max. height $h$, the $E_{k}~(\frac{1}{2}mv^{2})$ when the ball was thrown fully converts to $E_{p}~(mgh)$ at the height $h$. If the agent (say, Earth) of a conservative force (gravity) started gaining the energy lost by the body, it will mean energy sprouted out of nowhere, which'll be a defiance to the law of conservation of energy.

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PS: As @Simha pointed out, if earth's negligible motion towards the body is taken into account the agent (Earth) will gain negligible amount of $E_{k}$, but the agent certainly will not gain any internal thermal energy. Consider this :

In the above-given image, consider both bodies A and B have equal mass. B starts to move towards the right, away from A with a constant velocity. Now when B reaches maximum distance before returning back to A, A will have gained some $E_{k}$ and has moved a little bit towards B. Therefore $E_{p}$ of B has reduced, as h (distance) between A and B has reduced. The $E_{p}$ of B as it moves back towards A starts converting to $E_{k}$, and with each moment, $E_{p}$ of B reduces due to two factors - Firstly, $E_{p}$ is converting to $E_{k}$ and Secondly, $h$ is reducing every instant. But if we look at the sum total of all the energies - $E_{k~(A)}+E_{p~(A)}+E_{k~(B)}+E_{p~(B)}$ remains constant.

As far as $E_{p}$ is concerned , it can be argued that it is possessed both by earth and by the body, relative to each other. $$E_{p}~earth=m_{earth}g_{body}h~~and~~E_{p}~body=m_{body}g_{earth}h$$ $g_{body}$ is acceleration of earth due to body and $g_{earth}$ is acceleration of body due to earth. These turn out to be exactly equal:- $$F_{gravity}=\frac{G m_{body} m_{earth}}{h^2}$$ Reshuffling the equation, we get the values of $g_{body}$ and $g_{earth}$ : $$g_{body}=\frac{G m_{body}}{h^2}~~and~~g_{earth}=\frac{G m_{earth}}{h^2}$$ Putting value of $g_{earth}$ and $g_{body}$ in above equations of $E_{p~earth}$ and $E_{p~body}$, we get : $$E_{p~earth}=\frac{G m_{body} m_{earth}}{h}~~~and~~~E_{p~body}=\frac{G m_{body} m_{earth}}{h}$$ $\therefore E_{p~earth}=E_{p~body}$ where h is the height or distance b/w the two bodies. Therefore, it can't be said that potential energy is possessed by one body. It is possessed by the system, due to the virtue of position of its constituent bodies. However, we can take it to be possessed by either and go ahead with our calculation.

Answer 4

I reminded you elsewhere that you know very well when 3rd law is applicable, and that you even teach it to others for example here

Newton's third law of motion is not applied on a single body. - user36790

but then, inexplicably, you forget to use it properly in your own posts:

Let a body move and a conservative force oppose its motion. Therefore, by Newton's third law...

I regret I cannot yet draw pictures, but I'll borrow some familiar ones from your other questions so that you know exactly what I am referring to: here the skater is pushing on the rail to move on the ice:

and here a foot is pushing backward on the ground and the ground is providing an opposing force in order to walk forward.

In both cases the opposing forces are applied on two different bodies: a person and the rail, a person and the ground. These are case where 3rd law is applicable:

action $\rightarrow$PERSON || GROUND/RAIL $\leftarrow$ opposite reaction

If any two (opposing) forces act on the same body the two forces are not related and no one of them is a reaction to the other. If a gun shoots a bullet up, the force of the explosion will give it acceleration and a muzzle velocity v and KE k.

the body will also exert reactive force on the conservative force.

the body (the bullet) exerts a reactive force only whwn it is shot: on the gun and after that cannot exert any action nor reaction. The KE of the (bullet) will bring it up and will increase the distance of the body from the ground. In a conservative force *the distance $d$ from the ground by $F = ma$ corresponds to the value of the KE a body loses or will acquire in its journey to the ground. If it were ever to fall back to the ground it would regain same KE, because of its distance from the ground.

In a non-conservative force the increased distance is meaningless.

Update after the edit by OP:

The WORDS which I follow: "When two bodies interact and one body exerts force on the other and does negative work, the other body simultaneously exerts reactive force on the former and does positive work on it".

Your new version is completely different from the original,

Let a body move and a conservative force oppose its motion. Therefore, by Newton's third law...

If ONE body moves 3rd law cannot be involved, as everybody has been telling you, and as you taught to sobebody else. Until you do not realize this contradiction you cannot move forward

I do not want to add to your confusion. There is no limit to the questions you can ask in this site, I suggest you reset your notions and ask one single question at a time, after you have assimilated the previous one.

Answer 5

But in the real world,instead of storing that work, the agent of the conservative force returns it to the body itself which will be stored as potential energy.

I don't believe this is the case.

For example, consider the simple mass-spring system.

When the spring is compressed and the mass is motionless, all of the energy of the mass-spring system is in the form of potential energy.

Where is this potential energy stored? Is this potential energy a property of the mass? The spring? The system as a whole?

Consider a static electric charge distribution. There is an associated stored energy. This stored energy equals the work done to assemble the charge distribution. Where is this energy stored? The charge? The electric field?

Answer 6

The earth also gains positive work the same way the rough floor does, only it's not apparent at first.

To understand how, let's go to the microscopic level to see what actually happens when you rub an object against a rough surface. You have the molecules and atoms of the object above, and those of the surface below - as in the image shown.

The molecules of the object are held in place (due to it being a 'solid') by Coulombic forces (which are a 'conservative' force) which act between the molecules of the object. The same force also holds the molecules of the rough surface in place. The very reason both the surface and the object are called 'solids' is because (almost) every molecule in both surface and object has found a neat little place where the net force due to all the neighbouring molecules is zero. Thus, these molecules are each in equilibrium. Therefore, they will remain in that position unless an external agent comes and displaces them.

Thus essentially, any interaction between the molecules of the object and the molecules of the rough surface must ultimately be due to conservative (Coulombic) forces. When the molecules of the object 'rub' against the molecules of the rough surface, the equilibrium between the molecules of the surface (and the object) is disturbed. (By 'rubbing', I mean that the molecules of the object and those of the surface are forced to 'press' against one another, due to which there will be Coulombic repulsions from the molecules within the individual surfaces.) However, there is no actual 'touching' of molecules. This repulsive force is what causes the equilibrium to be disturbed. Once equilibrium is disturbed, a net Coulombic force acts on each of these molecules due to the neighbouring molecules. The (conservative) forces do work on the molecule(s) and give it a kinetic energy. The molecules start oscillating about their equilibrium position (see simple harmonic motion), consequently oscillating between potential and kinetic energy. Many such molecules gain kinetic energy and this net increase in kinetic energy of the surface is seen as an increase in 'temperature'.

Therefore, the molecules 'store' potential energy the same way the earth does. However, this potential energy is quickly converted to kinetic energy due to the oscillations. The difference between the earth and the rough surface case is that the earth does not have a mechanism by which it can oscillate between the energies. Thus, it can only 'store' that potential energy.

The main purpose of this answer is to point out that fundamentally, both the friction case and the earth case are caused due to conservative forces.

Hope it helped!

Note: The description of the molecular structure I have given of the surface and object are hypothetical. There is no such 'ideal' surface which is completely homogeneous, uniform and neutral. However, even in real surfaces the 'mechanism' by which heat is produced is the same as what I've given. Even there, the heat is produced due to molecules which oscillate under the action of Coulombic forces.

Answer 7

Newton's 3rd law is nothing but the conservation of momentum. Recall that force can be expressed as the time rate of change of momentum,

$$\vec{F} = \frac{d\vec{p}}{dt}$$

In order for the conservation of energy to hold, when Object 1 pushes against Object 2,

$$m_1d\vec{v}_1 = -m_2d\vec{v}_2$$

Since the mass does not change with time (conservation of mass),

$$d\vec{p}_1 = -d\vec{p}_2$$

Applying the definition of a force as the time rate change of momentum,

$$\vec{F}_1dt = -\vec{F}_2dt$$

The time in which the force-pair act would be the same for both objects. Therefore,

$$\vec{F}_1 = -\vec{F}_2$$

This shows that the two bodies will experience an equal but opposite force - satisfying Newton's third law.

Imagine two skaters skating on ice. Newton's third law implies that in order to conserve momentum, when person A applies a force to person B, the momentum gain by person A must be equal and opposite in magnitude to the momentum gain by person B. Similarly, if a person walks on the floor, as he gains momentum, something has to lose momentum (or gain momentum in the opposite direction) in order to satisfy the law of conservation of momentum. In that case, it's the Earth.

A few more examples: When a person jumps into the sky by pushing against the Earth, as the person gains upward momentum, the Earth gains downward momentum. (Newton's law of universal gravitation also satisfies momentum conservation - as a free-falling object gets pulled into Earth's gravity, Earth gets pulled towards the object as well.)

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For the second part, the law of conservation of energy must hold whether or not a force is conservative. However, the definition of a conservative force is as follows:

A conservative force is a force with the property that the work done in moving a particle between two points is independent of the taken path.

In other words,

$$W = \oint_C \vec{F}_{gravity} \cdot \mathrm{d}\vec r = 0.\,$$

Gravity is a conservative force. This means that the work done due to gravity is zero if you start and end up at the same position, regardless of the path travelled. If you, from position A, climb up the stairs to position B, then walk to the lift lobby at position C, take the lift down to position D, and walk back to position A, the work done by gravity on you would be $0$.

Friction is a non-conservative force. In other words,

$$W = \oint_C \vec{F}_{friction} \cdot \mathrm{d}\vec r \neq 0.\,$$

Of course, when a person drags a box from position A to position B, and back to position A, the work done against friction would not be zero. However, the conservation of energy must be satisfied. Therefore, the frictional work that is done on the box is converted irreversibly to heat - the box now feels hot after you drag it around a closed loop.

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Lastly, the concept of potential energy simply means energy that can be stored. Usually this applies to the case of forces with a potential function (or equivalently, conservative forces), but there are some exceptions (e.g. chemical potential energy from food).

It makes sense to relate conservative forces, such as gravity, to the gravitational potential energy, because work done to an object is independent of path and solely dependent on position. If you do work to an object, you can expect to get back the same amount of work by letting it revert to its original position. Conversely, it does not make sense to relate non-conservative forces, such as friction, to the concept of potential energy, because (1) you cannot store friction, and (2) its path dependence means that work done to an object cannot be successfully retrieved by letting it return to its original position - defeating the whole meaning of "potential" energy altogether.

Answer 8

A conservative force is gradient of a potential energy,

$\mathbf F=\nabla U=\dfrac{\partial U}{\partial r}\tag1$

Work done by a force from $A$ to $B$ is,

$\displaystyle\int_A^B dW=\int_A^B \mathbf F\cdot d\mathbf r=W_{AB}\tag2$

As potential energy is function of position, so

$dU=\dfrac{\partial U}{\partial r}\cdot d\mathbf r\tag3$

From (1), (2) and (3),

$\displaystyle\int_A^B dW=\int_A^B dU= U(B)-U(A)\tag4$

From (4), it's clear that work done by or upon a conservative force is depends upon value of potential energy at point and independent of path.

Now in a closed path, where work done upon a conservative force from point $B$ to $A$ along path $BA$ is $W_{BA}$ and work done in back return to same point $B$ by path $AB$ is work done by force is $W_{AB}$. Now work done for closed path in conservative force is zero because initial and final position is same. Therefore,

$\oint dW=\oint \mathbf F\cdot d\mathbf r=0=W_{BA}+W_{AB}$

$\Rightarrow W_{BA}=-W_{AB}$

This shows that in conservative force, work done upon the system is equals to work done by the system, negative sign is convention for work done by the system. So a conservative force returns work done upon it from work done by a system because this is nature of conservative force as we see it above.