Curious about the work - energy theorem

[John Constantine]

TL;DR Summary

Why can't we add the work done by forces that originate from the same source?

I'm sorry my question seems too strange.

The work done by a net force is equal to the change in the kinetic energy of an object.This is very simple, but I suddenly encountered a confusing problem. If you hang an object on a weightless string and apply a force of 10N in a straight line, moving it 1 meter, the increase in the object's kinetic energy will be 10J, right?The work I did on the object = The work done by the tension on the object = 10J. Very simple.My question might seem too obvious and natural, but I have one question.Why can't we add the work I did on the object and the work done by the tension on the object? Why is it incorrect to say that the increase in the object's kinetic energy is the work I did + the work done by the tension = 20J? If I understand it as "the cause of the force is that I pulled the object, and the force the string applied to the object was merely transmitted through the string," it seems easy, but it's not very clear to me. Is there a more physically rigorous explanation?

 

[Dale]

You need to be clear about what is the “system” that you are analyzing. Is the system the object or is it the object plus the string?

If the system is chosen to be the object, then you do no work on it. Your force is exerted on the string which is not part of the system. Only the tension does work on the system.

If the system is chosen to be the object plus the string, then you do work on it. The tension is now an internal force, and internal forces cannot transfer energy in or out of the system. So the tension does no work.

In neither case do both forces do work on the system.

 

[John Constantine]

If the system is chosen to be the object plus the string, then you do work on it. The tension is now an internal force, and internal forces cannot transfer energy in or out of the system. So the tension does no work.

When considering a string and an object as a single system, is it impossible to determine the work done on the object by a particular force? Are you saying that the work done by the string on the object is 10J and the reaction, which is the work done by the object on the string, is -10J, thus canceling each other out?

 

[A.T.]

When considering a string and an object as a single system, is it impossible to determine the work done on the object by a particular force?

If you are interested in the work between string and mass, why would you choose to consider them both a single object? It is you choice after all.

Are you saying that the work done by the string on the object is 10J and the reaction, which is the work done by the object on the string, is -10J, thus canceling each other out?

Those amounts of work are done on two different objects, so they don't "cancel out" in the sense of energy inflow/outflow into a particular object. They "cancel out" in the sense that no mechanical energy is generated or dissipated in the interaction between the two objects.

The work done by you on the string 10J and the work done by the mass on the string -10J "cancel out" in the sense that there is no net energy inflow into the string.

 

[jbriggs444]

When considering a string and an object as a single system, is it impossible to determine the work done on the object by a particular force?

You are muddying the water by considering a composite system in one breath and then considering only part of the system in the next.

If you want to know how much mechanical energy the object has gained due to the force that you applied on the string and its ensuing consequences then you can ask that question. You already know the answer.

That answer is the same as the work done by the force on the string. (non-zero)
That answer is not the same as the work done by the force on the object. (zero)

Are you saying that the work done by the string on the object is 10J and the reaction, which is the work done by the object on the string, is -10J, thus canceling each other out?

Yes. Those two works are equal and opposite. If both object and string are part of the system then the net energy imparted by that internal force pair is zero.

Internal contact forces between mating surfaces that are not in relative motion do no "real" work.

 

[Dale]

When considering a string and an object as a single system, is it impossible to determine the work done on the object by a particular force?

You can certainly determine the work done by a particular force.

The whole point of specifying a system is so that you can ignore the internal forces and ignore the forces not acting on the system. So, if you want to determine the work done by a particular force you simply choose your system such that the particular force of interest is an external force acting on the system.

If you want to know the work done by the tension, then you choose your system to be the object so that the tension is an external force acting on the system. Otherwise, tension is an internal force, so it is not part of the analysis.

If you want to know the work done by you, then you choose your system to be the object and the string so that your force is an external force acting on the system. Otherwise, your force is not acting on the system, so it is not part of the analysis.

Are you saying that the work done by the string on the object is 10J and the reaction, which is the work done by the object on the string, is -10J, thus canceling each other out?

You could think of it that way if you really want. In my view, it is simpler than that. Internal forces simply don't even show up on the free body diagram for a system. They are not part of the analysis.

 

[kuruman]

All that has been said is appropriate and to the point. I should a calculation to show exactly what the work done by the string and by the pulling force are. I will assume that the string has mass $\mu$ (we can always set it eqaul to zero in the end) and the object mass $m$. A "disembodied hand" is pulling on the string with constant force $~F_{\text{hand}}~$ and the mass is sliding on a horizontal, frictionless surface. The string + mass system is displaced by distance $x$.

By Newton's second law:
The common acceleration of the system is $a=\dfrac{F_{\text{hand}}}{m+\mu}.$
The "internal" force exerted by the string on the object is also the net force on the object, $F_{\text{int.}}^{(\text{obj.})}=F_{\text{net}}^{(\text{obj.})}=m\dfrac{F_{\text{hand}}}{m+\mu}.$
The net work done on the object only is
$W_{\text{net}}^{(\text{obj.})}=F_{\text{net}}^{(\text{obj.})}x=m\dfrac{F_{\text{hand}}}{m+\mu}x.$

The net force acting on the string is
$F_{\text{net}}^{(\text{str.})}=F_{\text{hand}}-F_{\text{int.}}^{(\text{obj.})}=F_{\text{hand}}\left(1-\dfrac{m}{m+\mu}\right)=\mu\dfrac{F_{\text{hand}}}{m+\mu}.$
The net work done on the string only is
$W_{\text{net}}^{(\text{str.})}=F_{\text{net}}^{(\text{str.})}x=\mu\dfrac{F_{\text{hand}}}{m+\mu}x.$

Thus, the net work done on the two-component system of string + object is the sum
$$W_{\text{net}}^{(\text{sys.})}=W_{\text{net}}^{(\text{obj.})}+W_{\text{net}}^{(\text{str.})} =\dfrac{m}{m+\mu}(F_{\text{hand}}x)+\dfrac{\mu}{m+\mu}(F_{\text{hand}} x)=F_{\text{hand}}x.$$ Note that if you start with a massive string, some of the total work $F_{\text{hand}}x$ goes into the string and increases its kinetic energy. If you make the string massless ($\mu\approx 0$), all of the total work goes into increasing the kinetic energy of the object.

You can certainly determine the work done by a particular force.

I just did that here which I started composing before your post #6.

 

[jbriggs444]

You could think of it that way if you really want. In my view, it is simpler than that. Internal forces simply don't even show up on the free body diagram for a system. They are not part of the analysis.

There is a complexity lurking here. The same word ("work") can be used with two distinguishable meanings. The thing that distinguishes the two meanings is the specifics of what displacement is considered.

In what I like to call "center of mass" work, one multiplies an external applied force by the [parallel] motion of the center of mass of the system on which that force acts.

In what I like to call "real" work, one focuses in on a particular interface and multiplies an applied force (not necessarily external) by the [parallel] displacement of the material on which it acts at the place where it acts.

With this distinction in mind, you are correct that internal force pairs never do any "center of mass" work.

However, internal force pairs can do "real" work when the force pair acts between relatively moving internal entities. One example would be sliding friction. In that case, the force pair does negative work, draining internal mechanical energy into thermal energy.

 

[Dale]

There is a complexity lurking here. The same word ("work") can be used with two distinguishable meanings.

You are right that there is lurking complexity here that I glossed over, but it is actually a slightly different complexity from what you mentioned.

In what I like to call "center of mass" work, one multiplies an external applied force by the [parallel] motion of the center of mass of the system on which that force acts.

I use the term “net work” for this. Here “net” refers to the “net force” rather than the total work. I also like the term “center of mass work” as it is more descriptive. I just use “net work” because that was the term I was taught when I was a young physics student.

Note that in the above posts I never used the term “net work”. I was not referring to net work in my previous posts.

In what I like to call "real" work, one focuses in on a particular interface and multiplies an applied force (not necessarily external) by the [parallel] displacement of the material on which it acts at the place where it acts.

I call that “mechanical work”.

The term “work” is also used in a more general sense to refer to any transfer of energy into or out of a system by any means other than heat. It is in this sense that I use the word “work”.

With this distinction in mind, you are correct that internal force pairs never do any "center of mass" work.

Internal forces never do any work in the generalized sense I use the term. Internal forces can change one form of internal energy into another form or move it from one part of the system to another, but simply cannot transfer energy out of the system simply by virtue of being internal. That is one reason that I prefer the this meaning of work rather than the more restricted meaning of mechanical work.

However, internal force pairs can do "real" work when the force pair acts between relatively moving internal entities. One example would be sliding friction. In that case, the force pair does negative work, draining internal mechanical energy into thermal energy.

Sliding friction between internal parts moving at different speeds does not do work (generalized). No energy is transferred in or out of the system by this force pair. Instead, internal energy is converted from mechanical to thermal without being transferred in or out of the system.

One nice thing about this meaning of work is that you do not need to investigate internal forces. It doesn’t matter if they are mechanical or some other nature. It doesn’t matter if the parts are moving at the same speed or not. This usage simplifies the analysis.