Total Work versus External Work

[Lost1ne]

Is there an important difference between total work and external work?

My knowledge would be that total work a.k.a. net work on a system would be equal to the change in kinetic energy of that system and equal to the line integral of the net force on the system dotted with the differential displacement vector over some path that would either track a single particle, if that was our defined system, or a center mass of a system of numerous particles/bodies.

Now please tell me where my thinking goes wrong:
Analyzing the spring image, we can see that the net force (always shorthand for the net external force as all internal forces in a system cancel out by Newton's 3rd Law) would equal zero. Thus, our line integral would result in a value of zero, and thus we may conclude that the spring system does not experience a change in kinetic energy. If this system was at rest before, it is at rest now. I guess I should be specific and say that the center of mass of this spring system had and maintains a non-zero speed. Sure, some of the composing parts of the spring move, but the center of mass of the spring does not accelerate.

But there are two external forces acting on this system, vectors F_1 and F_2, and when examining both of these forces, we can see that they both do positive work on this system according to the book.

Does this mean that if we were to look at the "net work" on this system without using the line-integral approach that we would find that the net work doesn't simply equal the work done by F_1 + the work done by F_2, meaning that "net work" does not necessarily equal "external work" and that there is something else that must be considered? (Internal work?)

(Now after re-examining this, I have another question: how do we define the work done by F_1 and F_2 on the spring? If the spring is our system, composed of many particles, aren't we supposed to examine the center of mass displacements of the spring? If the center of mass doesn't move, how can we claim that these forces "do positive work on the spring as they compress it"?)

I'll also post another example that may also be discussed.

 

[sophiecentaur]

My knowledge would be that total work a.k.a. net work on a system would be equal to the change in kinetic energy of that system

You seem to be ignoring Potential Energy. A significant factor when deforming a spring.

 

[Lost1ne]

You seem to be ignoring Potential Energy. A significant factor when deforming a spring.

But wouldn’t that line that you were referencing directly come from the work-kinetic energy theorem where the line integral is equal to the net work which is equal to ΔK and to -ΔU? So if ΔK = 0 then -ΔU = 0? Am I interpreting this theorem incorrectly? I can see and do concede that the spring system does gain spring potential energy, but how would this mathematically be explained?

If I read a couple of pages later in this chapter of the textbook, I’m introduced to the relation W_external = ΔE_system = ΔE_mech + ΔE_therm + ΔE_chem + ΔE_other. I presume that this equation would help explain the spring gaining spring potential energy. If so, then am I correct in stating that there is an important difference between “net work” on a system and “net external work” on a system, and how should I then carefully interpret the work-KE theorem I discussed above?

 

[sophiecentaur]

work-kinetic energy theorem

Can you quote the actual equation (reference) that you are using?
Your idea has to be wrong. You can do Work quasi-statically and transfer a vanishingly small amount of KE to a spring but the Potential Energy is still there.

 

[nasu]

The total work is not the work of the net force but the sum of the works done by each force on the system.
And this work is equal to the change in the kinetic energy of the system. The kinetic energy of the system can change even when the velocity of the center of mass does not change. You may be confusing the second law applied to a system with the work-energy theorem. A zero net force indeed means that the COM does not accelerate. But it does not mean that the KE of the system is constant. Both internal and external forces can change the KE of the system without necessarily changing the velocity of the COM.

 

[Lost1ne]

The total work is not the work of the net force but the sum of the works done by each force on the system.

My understanding was always that the net work is the work done by the net force as you could always "decompose" that F_net in the equation into a sum of the (external) forces acting on our system of interest (external because the internal forces would cancel out as vectors by Newton's 3rd Law) and then use the distributive property of the dot product and the additive integral property to reveal this as the sum of the work of all of the forces. (But now that I'm thinking about it, wouldn't this make it the net external work?)

I think I can see your kinetic energy argument because if we were to take the sum over i of 1/2 * m_i * (v_i)^2 for the particles composing the spring, we would retrieve a non-zero kinetic energy value even if the center of mass of the spring system still maintains zero speed.

But overall, I'm still a bit confused with things.

 

[ZeusPerseus]

The vector sum of all the external forces on a system (regardless as to where on the system they are applied) will equal the sum of the individual masses in the system times the vector acceleration of the masses.
$$\sum\limits_i {{{\bf{F}}_i}} = \sum\limits_j {{m_j}} {{\bf{a}}_j}$$
The vector sum of the time integral of the external forces on a system will equal the sum of the individual masses in the system times the vector change in the velocity of the individual masses:
$$\sum\limits_i {\int\limits_0^{\Delta t} {{{\bf{F}}_i}(t)\,dt = } } \sum\limits_j {{m_j}} \Delta {{\bf{v}}_j}$$
The sum of the integrals of the dot product of the external forces and the vector change in the position of application of the external forces will equal the total change in energy (kinetic, potential, thermal, etc) of the system:
$$\sum\limits_i {\int\limits_0^{\Delta t} {{{\bf{F}}_i}({{\bf{r}}_i},t) \cdot \,d{{\bf{r}}_i} = } } \Delta E$$

This corresponds to the total work performed on the system by the external world.

Key points to note about these equations:
1) All motions are measured relative to an inertial coordinate system.
2) There may be one external force, but many masses in the system.
3) These equations say nothing about how the individual masses move, only what the sums are equal to.
4) In the third equation, each force has a unique point of application ${\bf{r}}_i$ whose position in space varies in time, according to the motion of the (point) mass at which the force is applied.

For the system involving the (massless) spring, forces applied at the end points p1 and p2, the forces can be assumed to be linearly increasing in time in equal amounts but in opposite directions. If F is the final force applied to the spring at each end, where $F = k\Delta ({r_1} - {r_2})$, then the work performed on the spring will equal $\frac{1}{2}k\Delta {({r_1} - {r_2})^2}$. This result also holds for a spring with mass if the forces are applied very gradually.

 

[ZeusPerseus]

In your original post, you wrote "Analyzing the spring image, we can see that the net force (always shorthand for the net external force as all internal forces in a system cancel out by Newton's 3rd Law) would equal zero. Thus, our line integral would result in a value of zero, and thus we may conclude that the spring system does not experience a change in kinetic energy." However, the line integrals are not zero because you have to use the motion of the endpoints at which each force is applied, not the center of mass of the system (unless it is a completely rigid object). If the forces are applied slowly, the resulting change in the energy of the spring system will be in the form of potential energy. If the forces are applied quickly, the spring will vibrate along its length, alternating between internal kinetic and internal potential energy. If the forces at the two endpoints are unbalanced, then the spring system will also gain kinetic energy in terms of the motion of its center of mass.

 

[Lost1ne]

I have an additional question that I forgot to ask earlier. Let's say we have a ball-earth system. Although it's common to say something like "the ball has gravitational potential energy," is it more technically correct to say that the ball-earth system has a gravitational potential energy, implying that this potential energy is shared? Or are we supposed to go as far to say that both the ball and the Earth have separate gravitational potential energies because this system features two equal and opposite gravity forces that do internal, conservative work on the ball and Earth separately?