B How does an object gain potential energy?

[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.