# Work energy KE theorem for a book being lifted up in a gravitional field

• ChiralSuperfields
In summary, according to the text, the work-energy theorem does not apply to a system which can convert some of the work performed on it into internal energy. However, this contradicts the fact that the work done on the system is shown by equation 7.18. Can someone please explain this paradox?
ChiralSuperfields
Homework Statement
Relevant Equations
For this,

From the work kinetic energy theorem, if we assume that the book and the earth is the system, and that the finial and inital speed of the system is zero, then is the work KE theorem there is no net work done on the system. However, clearly there is work done on the system is shown by equation 7.18. Can someone please explain this paradox?

Many thanks!

ChiralSuperfields said:
is the work KE theorem there is no net work done on the system
As the text explains, sort of, the work KE theorem applies to rigid bodies. It does not apply to a system which can convert some of the work performed on it into internal energy.

MatinSAR, PhDeezNutz and ChiralSuperfields
haruspex said:
As the text explains, sort of, the work KE theorem applies to rigid bodies. It does not apply to a system which can convert some of the work performed on it into internal energy.

Sorry I not sure if I understand. If I assume that the book is a rigid object, then according to textbook, the work done on the book is ##W_{ext} = mg \Delta h##. Assuming that the finial and inital KE of the book is zero, then according the the Work KE theorem, the ##W_{ext} = \Delta {KE}##. However, this contradicts ##W_{ext} = mg \Delta h##. Do you please know why that is the case?

Many thanks!

The text is missleading. The work energy theorem works for this case as well for systems of objects, even if the system is not rigid. It just has to be formulated properly. In this case, if the change in kinetic energy is zero, net work is zero, as predicetd by the work energy theorem. For non-rigid objects, the net work means all the work done by both external and internal forces (for rigid bodies the work done by internal forces is zero). The potential energy may or may not be brought into the picture, but the work-energy theorem holds anyway. If you want to consider potential energy, then you may use some other relationship and not the work-energy (some books calls is conservation of energy even if there are non-conservative forces) but not because the W-E does not "work". It is just a matter of choice.

ChiralSuperfields
ChiralSuperfields said:
If I assume that the book is a rigid object, then according to textbook, the work done on the book is ##W_{ext} = mg \Delta h##. Assuming that the finial and inital KE of the book is zero, then according the the Work KE theorem, the ##W_{ext} = \Delta {KE}##. However, this contradicts ##W_{ext} = mg \Delta h##. Do you please know why that is the case?
But the system defined in the text is Earth+book, which is not a rigid body.
If we redefine the system as just the book then gravity acts on it, making the net force zero.
nasu said:
For non-rigid objects, the net work means all the work done by both external and internal forces
Yes, it could be formulated that way, but it sits awkwardly with the principle that internal forces of a system do not count as forces acting on the system.
All the online sources I could find easily specify either a rigid body or a particle.

ChiralSuperfields and MatinSAR
Consider just the book. Initially it is placed at rest at height ##h=0##. Then an upward force ##F## (say, from a piece of string or something) is acting on the book (let positive direction be upwards). If ##F > mg## then the book will start to travel upwards, i.e. its displacement will be positive. Let us assume this discplacement is ##h##. Total work performed on the book is ##W_\text{tot} = (F-mg)h## (because there are two external forces acting on the book).

Now, as you said, the work-energy theorem states that the total work performed on a rigid object is the change in kinetic energy. We thus have ## (F-mg)h = \dfrac{m}{2}\left( v(h)^2 - v(0)^2 \right)##. We can rewrite this as ## Fh = mgh + \dfrac{mv(h)^2}{2} ## (using ##v(0) = 0##) and identify ##mgh## with gravitational potential energy. The work performed by the force ##F## has in this case increased the potential energy of the book, and also its kinetic energy (if ##F = mg## then ##v(h) = 0## too and we have just increased the potential energy.)

Last edited:
ChiralSuperfields and MatinSAR
haruspex said:
Yes, it could be formulated that way, but it sits awkwardly with the principle that internal forces of a system do not count as forces acting on the system.
All the online sources I could find easily specify either a rigid body or a particle.
But this is the problem. Not all systems are rigid or a particle. And this is the case in the OP problem, if the book-Earth system is considered.
I cannot say about "all" the mechanics books but in the ones I know the chapter on dynamics of systems have the work-energy theorem formulated in the general form, with the change in KE equal to the work done by borh internal and external forces. So it is not than it can be formulated this way, it is actually how it is done.

sits awkwardly with the principle that internal forces of a system do not count as forces acting on the system.
This should not be a problem because there is no such principle. Do you have a reference for it, for a general mechanical system? Enforcing this idea is the source of some errors we seen even on this forum.
The source of confusion may be the fact that the internal forces cannot change the momentum of the system and somehow this is extended to the KE. Which is not right.
If you have a projectile at rest relative to the observer and then the projectile explodes into fast moving pieces (relative to the same observer), what external force counts as producing the change in kinetic energy of the system?

ChiralSuperfields
I think the matter can be set straight if one considers total energy conservation which has been formulated as the first law of thermodynamics, $$\Delta E_{int}= W+Q.$$For a multicomponent system, on the left hand side of the equation, ##\Delta E_{int}~## is the change in internal energy of the components and includes changes in their kinetic energy, changes in their potential energy as a result of changes in their spatial configuration when the components are interacting, chemical and biochemical changes, etc.

On the right hand side of the equation ##W## and ##Q## are work and heat that cross the boundary of the system. It's a simple expression of the idea that "to change what's inside something must come in from the outside." I will apply this to the physical situation of a man lifting a book of mass ##m## from floor level to height ##h## above his head at constant speed. I will consider three cases of different systems. In all three ##Q=0.##

Case I: The system is the book.
In the equation ##\Delta U_{int}=0## because ##\Delta K=0## by assumption and there is no change in potential energy because this is a one-component system. Potential energy is defined when the system has at least two components interacting via conservative forces. On the right hand side ##W=W_g+W_{man}## and $$\Delta E_{int}=0=W_g+W_{man}=-mgh+W_{man}\implies W_{man}=mgh.$$This is the same result that one can also obtain with the work-energy theorem.

Case II: The system is the book + Earth
This is the system that the textbook chose. As in case I, the change in kinetic energy of the system is zero. However there is a change in gravitational potential energy because the spatial configuration of the two component system changes as the book is moved farther away from the center of the Earth. The change in gravitational potential energy is, by definition, the negative of the work done by gravity, ##\Delta U_g=+mgh.## Thus, $$\Delta E_{int}=\Delta U_g=W_{man} \implies W_{man}=mgh.$$ This is exactly the same result as in the previous case.

Case II: The system is the book + Earth + man
Here the system is isolated, no work crosses the boundary of the system, therefore the change in internal energy must be zero. As before, ##\Delta K=0## but the potential energy case is a bit more complicated for the three-component system. We have
##\Delta U_{b-e}=mgh =~## the change in potential energy of the book-Earth pair.
##\Delta U_{m-e}=Mg\Delta y_{cm} =~## the change in potential energy of the man-Earth pair as the man's CM is displaced by ##\Delta y_{cm}.##
##\Delta U_{m-b}\approx 0,~## we ignore the change in gravitational potential energy between man and book.

In addition we have the change in internal energy of the man. When he lifts the book, he spends biochemical energy that he needs to replenish by eating. The first law becomes
$$0=\Delta E_{int}=\Delta U_{b-e}+\Delta U_{m-e}+\Delta E_{biochem.}\implies \Delta E_{biochem.}=-mgh-Mg\Delta y_{cm}.$$The expression shows how the biochemical energy expended by the man is split into raising the book and his center of mass.
nasu said:
If you have a projectile at rest relative to the observer and then the projectile explodes into fast moving pieces (relative to the same observer), what external force counts as producing the change in kinetic energy of the system?
In this case, if the projectile initially at rest explodes into ##N## pieces, we have an isolated ##N##-component system. No external force acts on the system and the center of mass remains at rest after the explosion. Assuming that the pieces are non-interacting, the change in internal energy of the system is $$0=\Delta E_{int}=\Delta K+\Delta E_{chem} \implies \Delta E_{chem}=-\frac{1}{2}\sum_{i=1}^Nm_iv_i^2.$$It's a pity that textbooks and instructors don't introduce the first law of thermodynamics in the special case ##Q=0## when they start talking about "energy conservation."

ChiralSuperfields and MatinSAR
What is the krooked thing that you want to set straight? The work-energy theorem works as it is formulated, for mechanical systems. You don't need thermodynamics and you don't need potential energy to use "fix" the work-energy theorem. There is nothing to be fixed. Maybe just some misconceptions from introductory physics, where statement valid for points or rigid objects are extended to be absolutely true.

For the projectile, it is true that the "source" of the KE of the parts is in the chemical (internal) energy of the explosive. But the transfer of this energy to the parts is done by the work of the internal forces. There is nothing wrong with internal forces doing net work that results in the increase in the KE of the system.

malawi_glenn and ChiralSuperfields
I thought this (from another thread) was enlightening.

nasu said:
I said "net work". The work done by gravity is negative and can be described by an increase in potential energy. But this not the net work. It must be another force acting against gravity. This second force does positive work on the sledge. The sum of the two is the net work. If the sledge accelerates upwards, the net work is positive. If it moves up with constant velocity the net work is zero. In both cases the work done by gravity is the same, for a given height. The work done by the other force depends on how do you do the lifting.

ChiralSuperfields
nasu said:
The work-energy theorem works as it is formulated, for mechanical systems.
As I understand it, the formulation of the work-energy theorem for mechanical systems is $$\Delta K=W_{net}$$ where ##W_{net}## is the sum of all the works done on the components of the system by external forces, i.e. the total work that crosses the system boundaries.

The point that the textbook example wants to make is that application of the work-energy theorem to the two-component Earth + book system doesn't work and here is why. The change in kinetic energy is zero while the only work that crosses the boundary is that done by the man on the book which means that ##W_{net}=W_{man}.## So according to the work-energy theorem as formulated $$0=\Delta K=W_{man} \implies W_{man}=0.$$ which is certainly an absurd result.

The work-energy theorem works fine for single-component systems but fails with multi-component systems in which there are internal forces between pairs of components, i.e. deformable systems, an example of which is the Earth + book system1. That's what I think is "krooked" and can be set straight by invoking the higher authority of total energy conservation of which the first law of thermodynamics is an example.

1@haruspex said as much in post #2.

ChiralSuperfields and gmax137
Thank you @haruspex, @nasu, @malawi_glenn , @kuruman, and @gmax137 for your replies!

So it turns out work energy theorem is actually valid in this case! However, we just have to remember that
##W_{net} = W_{applied} + W_{conservative}## as found on Physics SE.

Anybody else agree/disagree?

Many thanks!

ChiralSuperfields said:
Thank you @haruspex, @nasu, @malawi_glenn , @kuruman, and @gmax137 for your replies!

So it turns out work energy theorem is actually valid in this case! However, we just have to remember that
##W_{net} = W_{applied} + W_{conservative}## as found on Physics SE.
View attachment 326632

Anybody else agree/disagree?

Many thanks!
I also do not understand why non-conservative internal forces are being ignored.
If I strike a cold squash ball with a racket, a lot of the work goes into warming the ball.

ChiralSuperfields
haruspex said:
I also do not understand why non-conservative internal forces are being ignored.
If I strike a cold squash ball with a racket, a lot of the work goes into warming the ball.

haruspex said:
If I strike a cold squash ball with a racket, a lot of the work goes into warming the ball.
A lot? Do you have a rough estimate?

ChiralSuperfields
ChiralSuperfields and malawi_glenn
haruspex said:
According to https://www.researchgate.net/figure...tution-for-two-types-of-squash_fig1_253254383, the coefficient of restitution can be as little as 0.43, implying 80% of the KE is lost.
However, that might apply more to the bounce off a wall than the bounce off the racket strings.
Yeah, that was my thought as well.
Reason I am asking is because I had students working on squash balls as their high school science project and they examined how the coefficient of restitution against the schools lab floor depended on how many bounces the ball has made.
Pretty tedious experiment, several thousands of drops... from 3 m height.
I remember there was some argument that their result could not be applied to "real" game situations because 1) the impact speed is much higher there then in their experiment
2) you also hit the ball with a racket.

ChiralSuperfields
malawi_glenn said:
you also hit the ball with a racket.
Yes, but then it bounces off a wall, maybe several, and perhaps the floor.

ChiralSuperfields
haruspex said:
Yes, but then it bounces off a wall, maybe several, and perhaps the floor.
which is why I wrote "also" and not "instead"

ChiralSuperfields and SammyS

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