B How does an object gain potential energy?

[adjurovich]

In the scenario you asked about, work is done on the object by you, which equals the increase in the system's potential energy.

No one is saying work is done on the system.

I think I already said that in the post #23.

$W_{ext} = \Delta{U}$​

 

[kuruman]

Also I would like to note that in my high school textbook, potential energy is studied before systems are even mentioned and at high school classes I was taught that the potential energy is basically the energy body has due to its position.

You were taught incorrectly. That incorrect teaching is what led you to this question

What my reasoning was is that if two forces do work on the same object in opposite directions, energy is basically “disappearing” and that seems to be very incorrect. But my intuition stops here. Where does the energy go if there’s the net work is zero?

I have given you the correct way to consider this situation but I seem not to accept or understand what I am saying and go in that direction. Hence your statement

I am thankful for the time people invest in their answers here, but I feel like we are not going in any direction.

You are welcome. We are doing this as a service to you. The bottom line of all that has been said is that being clear about the boundaries of the system under consideration is good practice that protects against confusing yourself and others. It is a simple plea for clarity. Defining clearly what you are talking about is a simple idea that has nothing to do with quantum field theory.

I have said all I have to say on this thread.

 

[A.T.]

I was taught that the potential energy is basically the energy body has due to its position.

This is a commonly used approximation/shorthand, for cases where one body is much more massive than the other. Here, the PE of the system will almost exclusively go into the KE of the smaller body when they are released. Therefore, the system's PE is being often associated with the smaller body.

Unfortunately this shorthand leads to confusion, once you consider work done.

 

[DrBanana]

I asked the same question here: https://www.physicsforums.com/threa...-in-potential-energy-with-0-net-work.1062154/

Plus there is a three part series on the pitfalls of the work-kinetic energy theorem on the Insights blog, that might illuminate some things as well.

 

[Herman Trivilino]

Something very simple is getting endlessly complicated somehow.

Yes. It's the notion that mechanical work can be generalized to a complete understanding of energy transfers. It cannot. The concept of internal energy must first be introduced. Later it can be followed by heat and the 1st Law of Thermodynamics.

This was one of the great intellectual accomplishments of the 19th century and led to what we now call the conservation of energy.

Also I would like to note that in my high school textbook, potential energy is studied before systems are even mentioned and at high school classes I was taught that the potential energy is basically the energy body has due to its position.

And that's almost correct. The energy a system has due to the relative position of its constituents. There are other textbooks that do this correctly, but they are in the minority.

What my reasoning was is that if two forces do work on the same object in opposite directions, energy is basically “disappearing” and that seems to be very incorrect.

It is incorrect because you are using mechanical work to draw conclusions about the more general concept of energy.

But my intuition stops here. Where does the energy go if there’s the net work is zero?

It goes into increasing the potential energy of the system. It comes from the person doing the lifting.

Do you have some literature recommendation (high school or less advanced college-level) that explains laws of conservation in detail, because I want to get deep understanding of this topic?

Development of energy concepts in introductory physics courses
Arnold B. Arons
Citation: Am. J. Phys. 67, 1063 (1999); doi: 10.1119/1.19182
View online: http://dx.doi.org/10.1119/1.19182
View Table of Contents: http://ajp.aapt.org/resource/1/AJPIAS/v67/i12
Published by the American Association of Physics Teachers

 

[adjurovich]

Yes. It's the notion that mechanical work can be generalized to a complete understanding of energy transfers. It cannot. The concept of internal energy must first be introduced. Later it can be followed by heat and the 1st Law of Thermodynamics.

This was one of the great intellectual accomplishments of the 19th century and led to what we now call the conservation of energy.

And that's almost correct. The energy a system has due to the relative position of its constituents. There are other textbooks that do this correctly, but they are in the minority.

It is incorrect because you are using mechanical work to draw conclusions about the more general concept of energy.

It goes into increasing the potential energy of the system. It comes from the person doing the lifting.

Development of energy concepts in introductory physics courses
Arnold B. Arons
Citation: Am. J. Phys. 67, 1063 (1999); doi: 10.1119/1.19182
View online: http://dx.doi.org/10.1119/1.19182
View Table of Contents: http://ajp.aapt.org/resource/1/AJPIAS/v67/i12
Published by the American Association of Physics Teachers

I think I almost understood it. The last question I will ask is: if the energy coming from external work is stored in the system as potential energy. What happens to the energy from work of gravity?

 

[Herman Trivilino]

The work is not external. It's internal to the Earth-object system.

 

[adjurovich]

The work is not external. It's internal to the Earth-object system.

Let’s be a little more precise: work done by gravity or work done by some force $F$?

 

[Herman Trivilino]

Let’s be a little more precise: work done by gravity or work done by some force $F$?

I answered the question you asked:

What happens to the energy from work of gravity?

As I told you before, the energy expended by the agent separating the object from Earth is stored as potential energy in the object-Earth system.

 

[adjurovich]

I answered the question you asked:

Thanks a lot!! I get it now.

 

[A.T.]

if the energy coming from external work is stored in the system as potential energy. What happens to the energy from work of gravity?

Work is transfer of mechanical energy. If an external force does positive work on one of the bodies then it transfers energy to that body. Some of that energy might go into KE of that body, and some into the PE of the system. The later transfer is represented by the negative work by gravity on that body.

It might help to replace the gravitational field with a spring. Here it's more clear where the PE is stored, while the whole energy transfer scheme is the same.

 

[PeroK]

I think there is a general problem here of starting with the equations and then trying to interpret them. Whereas, the meaning of an equation is inherent in the definitions and assumptions that led to the equation in the first place.

In the case of $F =ma$ it is often forgotten that $F$ is the total external force and that this equation is not valid independent of the definition of the three quantities $F, m$ and $a$.

The same applies to the work energy theorem. That equation holds under a precise set of hypotheses. And if it appears to fail, then one or more of the hypotheses must not apply.

 

[cianfa72]

Here are three examples. Say that "We" do work and add 50 joules to the object + Earth system.

• If the object moves vertically up at constant speed, 50 joules are added to the system's energy in the form of potential energy of the system. There is no change in the kinetic energy of the system.
• If the object moves horizontally, 50 joules are added to the system's energy in the form of kinetic energy of the object. There is no change in the potential energy of the system.
• If the object moves vertically up at increasing speed, 50 joules are added to the system's energy in the form of potential energy of the system and kinetic energy of the object.

In all three cases, the change in potential energy is the negative of the work done by gravity. It has nothing to do with the 50 joules that we put in but a lot to do with where the object starts and where it ends up in space.

I came late to this thread: let's assume "Earth + book" as the system. As pointed out by @kuruman in the post before the referenced one, when "We" lift the book raising its kinetic energy, the Earth kinetic energy changes as well. However one can disregard the latter (reasonably assuming that the location of system's center of mass doesn't change) hence all the system's kinetic energy actually goes into the book's kinetic energy.

 

[kuruman]

I came late to this thread: let's assume "Earth + book" as the system. As pointed out by @kuruman in the post before the referenced one, when "We" lift the book raising its kinetic energy, the Earth kinetic energy changes as well. However one can disregard the latter (reasonably assuming that the location of system's center of mass doesn't change) hence all the system's kinetic energy actually goes into the book's kinetic energy.

Good point. One should keep in mind that work done on a single object by "something outside it" is always accompanied by equal and opposite work done by the single object object on the "something outside it." This is guaranteed by Newton's third law.

 

[jbriggs444]

Good point. One should keep in mind that work done on a single object by "something outside it" is always accompanied by equal and opposite work done by the single object object on the "something outside it." This is guaranteed by Newton's third law.

This is correct if we add the proviso that there is no relative motion between the thing inside the system which is subject to the external force and the thing outside the system on which the third law partner force acts.

In that case, the relevant displacements are equal. So the work done: $F_\text{action} \Delta s + F_\text{reaction} \Delta s = 0$

If there can be relative motion (e.g. kinetic friction, gravity, electrostatic force) then we can find energy being absorbed into or emerging from the force pair.

 

[cianfa72]

This is correct if we add the proviso that there is no relative motion between the thing inside the system which is subject to the external force and the thing outside the system on which the third law partner force acts.

In that case, the relevant displacements are equal. So the work done: $F_\text{action} \Delta s + F_\text{reaction} \Delta s = 0$

Just to give an example in the realm of Newtonian physics: let's take an isolated binary system made of two bodies A and B rotating around their common Center of Mass (CoM). Now suppose an external thing suddenly interacts with B doing "external" work on the system (where system = binary system). Generally speaking, the system will change both its potential energy and kinetic energy of its components (A and B) w.r.t. their common CoM.

Assuming A much more massive than B, the displacement of A w.r.t. CoM can be neglected, so basically the change in system's kinetic energy is accounted to B kinetic energy (w.r.t. CoM) alone.

If there can be relative motion (e.g. kinetic friction, gravity, electrostatic force) then we can find energy being absorbed into or emerging from the force pair.

Sorry, might you be more explicit about this point?

 

[kuruman]

This is correct if we add the proviso that there is no relative motion between the thing inside the system which is subject to the external force and the thing outside the system on which the third law partner force acts.

I don't think that a proviso is needed. By "single object" I meant a system with no internal structure.

 

[jbriggs444]

Sorry, might you be more explicit about this point?

It goes back to the definition of mechanical work.

The work done by a force is the dot product of the applied force and the displacement of the material at the point of application of that force: $$W = F \cdot \Delta s$$Newton's third law says that if there is an applied force of $F$ then there will be a third law partner force $-F$ applied on some other body. If we apply the definition of work again, we need to worry about the displacement of the material on that other body.

@kuruman would have us believe that because $F_2 = -F_1$, it follows that $F_1 \cdot {\Delta s}_1 + F_2 \cdot {\Delta s}_2 = 0$

But that is clearly not the case if $\Delta s_1 \ne \Delta s_2$.

I don't think that a proviso is needed. By "single object" I meant a system with no internal structure.

Let us try a concrete example.

We have a car. Model it like a block of wood with no internal structure. It is skidding to a stop. Our system of interest is the block of wood. There is an external force of friction between road and block.

@kuruman would assert that the work done by road on block is equal and opposite to the work done by block on road. Let us test that assertion.

We will adopt the frame of reference of the road. It does not matter. Any inertial frame will deliver a similar result.

Let us say that the block has mass $m$, velocity $v$ and kinetic energy $\frac{1}{2}mv^2$. The coefficient of kinetic friction is $\mu$. The local acceleration of gravity is $g$. The frictional force between road and block is $F = -\mu m g$.

The block will skid to a stop over a distance $\Delta s$ which will be given by $\frac{ \frac{1}{2} mv^2} {\mu m g}$.

The work done by road on block is $F \Delta s = -\frac{1}{2}mv^2$. [Force and displacement are oppositely directed. The block is losing KE. This should indeed be negative]

Let us consider the work done by block on road. The road does not move. The work done is zero.

The two works are not equal and opposite. There is a deficit of $\frac{1}{2}mv^2$. This is an invariant. It is the amount of kinetic energy dissipated via kinetic friction.


Let us revisit our choice of inertial frame...

We can adopt a frame where the block begins at rest and accelerates to match speeds with the moving road. We will find that the work done by road on car is $\frac {1}{2}mv^2$. We will find that the work done by car on road is $-mv^2$. Again, there is an invariant deficit of $\frac{1}{2}mv^2$.

We can adopt a frame where both block and road are moving at $\frac{v}{2}$ toward each other. This time, the work done by road on car will be 0 (it begins and ends with the same speed) and the work done by car on road will be $-\frac{1}{2}mv^2$ (half the velocity of the previous case, so half the work done). Again, the invariant deficit is $\frac{1}{2}mv^2$

 

[A.T.]

One should keep in mind that work done on a single object by "something outside it" is always accompanied by equal and opposite work done by the single object object on the "something outside it." This is guaranteed by Newton's third law.

Newton's 3rd is about conserving momentum: equal but opposite impulse.

It doesn't imply conservation of mechanical energy: the work doesn't have to be equal but opposite.

Macroscopic mechanical energy can be converted to/from other forms of energy.

 

[kuruman]

We have a car. Model it like a block of wood with no internal structure. It is skidding to a stop. Our system of interest is the block of wood. There is an external force of friction between road and block.

You can also model a cow as a sphere, but is it realistic? By my reckoning, a car skidding to a stop has internal structure and hence internal energy. The heat dissipated by friction raises the internal (thermal) energy of the molecules in the brake pads, the tires, etc. You are limiting your argument to mechanical work when the law that rules here is the first law of thermodynamics. If the system is the car, then when it is skidding to a stop both mechanical work and heat cross the system's boundary. What happens to the heat entering the system if it doesn't go into raising the system's internal energy?

@kuruman would assert that the work done by road on block is equal and opposite to the work done by block on road.

I would not assert that. I would get around the issue by considering the system to consist of the car and the Earth. I would figure out all the energy transformations and use total energy conservation to analyze the situation. I have already written about this here. See example II.2 which includes friction and is the closest example to a skidding car.

 

[cianfa72]

Just to give an example in the realm of Newtonian physics: let's take an isolated binary system made of two bodies A and B rotating around their common Center of Mass (CoM). Now suppose an external thing suddenly interacts with B doing "external" work on the system (where system = binary system). Generally speaking, the system will change both its potential energy and kinetic energy of its components (A and B) w.r.t. their common CoM.

Assuming A much more massive than B, the displacement of A w.r.t. CoM can be neglected, so basically the change in system's kinetic energy is accounted to B kinetic energy (w.r.t. CoM) alone.

To be more precise: in the binary system example above, what is true is that the external work done on the "system" changes the sum of system's potential energy plus the sum of system's components kinetic energies calcuated w.r.t. a given inertial frame. Suppose to pick as inertial frame the frame $F$ where the system's CoM is initially at rest. Upon the external thing interaction with the system (say with B body), the system's CoM will start accelerating w.r.t. $F$. However the system's total energy change calculated w.r.t. frame $F$ will be exactly equal to the amount of external work done.

Then, assuming body A much more massive than B, one can reasonably neglect the displacement/acceleration of system's CoM, getting back to the result that the system's change in kinetic energy basically goes in the change of B's kinetic energy w.r.t. system's CoM.

 

[A.T.]

You are limiting your argument to mechanical work ...

Because you justified your statement in post #44 about "equal but opposite work" with Newton's 3rd Law.

I would not assert that.

Then we have no disagreement.

 

[jbriggs444]

Retrieving the controversial claim:

One should keep in mind that work done on a single object by "something outside it" is always accompanied by equal and opposite work done by the single object object on the "something outside it." This is guaranteed by Newton's third law.

Nothing about energy conservation here. You are talking about equal and opposite work.

The statement above is false. I pointed this out. You objected.

You can also model a cow as a sphere, but is it realistic? By my reckoning, a car skidding to a stop has internal structure and hence internal energy. The heat dissipated by friction raises the internal (thermal) energy of the molecules in the brake pads, the tires, etc. You are limiting your argument to mechanical work when the law that rules here is the first law of thermodynamics.

Now you wish to change the subject to energy conservation. You wish to change the justification away from Newton's third law to the first law of thermodynamics.

Yes, energy is conserved. If there is a transfer of energy into the system of interest, there must be a transfer of energy out of the environment.

As you point out, this is guaranteed by the first law of thermodynamics. It is not guaranteed by Newton's third law.

If the system is the car, then when it is skidding to a stop both mechanical work and heat cross the system's boundary. What happens to the heat entering the system if it doesn't go into raising the system's internal energy?

Heat is not the same thing as work.

@kuruman would assert that the work done by road on block is equal and opposite to the work done by block on road. Let us test that assertion.

I would not assert that.

I retrieved your quote. It is at the top of this post. You did assert that. You asserted that the work done by third law partner forces is equal and opposite.

Note that kinetic friction is not the only situation where Newton's third law fails to ensure equal and opposite work done by third law partner forces.

 

[cianfa72]

Let us revisit our choice of inertial frame...

We can adopt a frame where the block begins at rest and accelerates to match speeds with the moving road. We will find that the work done by road on car is $\frac {1}{2}mv^2$. We will find that the work done by car on road is $-mv^2$. Again, there is an invariant deficit of $\frac{1}{2}mv^2$.

Let's check my understanding. First of all the frame $\mathcal A$ in which the block is initially at rest is a valid inertial frame since it moves with constant velocity $v_0$ w.r.t. the road's rest frame (assumed to be inertial).

In $\mathcal A$ frame, the block begins at rest and accelerates "backwards" up to velocity $-v_0$ ($v_0$ is the block's initial speed w.r.t. the road rest frame). In this frame, block gains kinetic energy up to $\frac 1 2 mv_0^2$, hence this is the work done by road on it via kinetic friction $f_k$.

What about the work done by block on road ? W.r.t. frame $\mathcal A$, the material point of contact on the road keeps constant velocity $-v_0$ along all the "journey". Therefore this work is exactly 2 times $\frac 1 2 mv_0^2$ with negative sign because in $\mathcal A$ frame Newton's 3rd law pair force from block on the road and road velocity $-v_0$ are opposite.

As you pointed out the "work difference" is invariant though (i.e. it is the same regardless the inertial frame chosen to do the calculation).

 

[jbriggs444]

Let's check my understanding...

Yes, yes and yes, all the way.