Work-Energy theorem and System

In the book Classical Mechanics by David Morin,while discussing Work Energy theorem he gives an example of importance of choosing System. As he considers lifting the book upward then by WE theorem:

$$W_{external}$$= ∆K+ ∆U +∆Internal Energy

Now,if we consider lifting the book and choosing only book as System,we apply WE to get:

$$W_{person}+W_{earth}=0+0+0$$

Now,I understood that change in K.E and Internal energy of System is 0 but why change in potential energy is 0 if we keep book as system.

• If the book is the only part of the system where did W_person come from?
– user196418
Dec 28 '18 at 17:09
• Because on this system there are 2 external forces. One by the person which is upwards and other by gravity downwards Dec 28 '18 at 17:11
• The change in potential IS the work done on the book by gravity (modulo a sign).
– user196418
Dec 28 '18 at 17:19

You sure this is the exact statement. Because $$\Delta U = -W_{Earth}$$ is the definition of gravitational potential energy. You should never include both of them in the same expression.

Either this or the fact that since you are only considering the book as your system the potential energy of the system(book only) does not get changed at all. What is getting changed is the potential of the system containing earth and book.

The book & Earth as a system has gravitational potential energy whereas the book by itself as a system cannot have gravitational potential energy.

When you have only the book as the system and an external force (you?) lifting the book then the book has two external forces acting on it; the contact force that you exert on the book and the gravitational attractive force due to the Earth on the book.
If these two external forces are equal in magnitude and opposite in direction then the net force on the book is zero and the net work done by the external forces is zero.
Thus the change in the kinetic energy of the book is zero.

If the book (the system) of mass $$m$$ is thrown upwards with a speed $$v$$ in a downward gravitational field of magnitude $$g$$ and the book reaches a height $$h$$ before it stops then the external work done on the book by the external (gravitational attractive) force is $$-mg\times h$$ and the change in kinetic energy of the book is $$0 - \frac 12 m v^2$$.
Using the work energy theorem gives $$-mg\times h =0 - \frac 12 m v^2 \Rightarrow mgh = \frac 12 mv^2$$
Notice that in this derivation no mention has been made of gravitational potential energy although you might be tempted to do so because of the term $$mgh$$ occurring in the final equation.

If the system is the book & the Earth then it is usually assumed that the mass of the book is much, much less than the mass of the Earth and that the Earth does not move.
In this case an upward external force, equal in magnitude to the weight of the book $$mg$$, moves a distance $$h$$ upwards and does $$mgh$$ amount of work.
If the speed of the book does not change then the change in kinetic energy of the book is zero but what does changed is the gravitational potential energy of the book & Earth system which in this case would be $$mgh$$.

If the book is thrown upwards with a speed $$v$$ and it rises by a height $$h$$ the change in kinetic energy of the book is $$0-\frac 12 mv^2$$.
As there are no external forces on the book during its flight the work done by external forces is zero.
However, there is a change in the gravitational potential energy of the system equal to $$mgh$$.
Applying the work energy theorem gives $$0 = \left (0-\frac 12 mv^2 \right ) +mgh\Rightarrow mgh = \frac 12 mv^2$$ as before.

Such an analysis for the book and Earth system could be extended without making the assumption that the Earth is much, much more massive than the Earth.

Are you sure the book is talking about the importance of choosing a "system" as opposed to a "frame of reference"? It sounds like its talking about a frame of reference. To explain:

If you say the change in kinetic energy is zero when the book has a change in velocity with respect to, say, the ground, you are using the book as the reference frame. It’s like if you are in a car traveling at a velocity $$v$$ with respect to the road. The kinetic energy of the car in your reference frame (the reference frame of the car) is zero. With respect to the reference frame of a person standing on the road, however, its $$\frac{mv^2}{2}$$.

Similarly, if you lift a book a height $$h$$ the change in potential energy of the book with respect to the reference frame of the book is zero. With respect to the floor it is $$mgh$$. Think of the book on a table and lifting the table a height $$h$$. The change in potential energy with respect to the table is zero.

But the equation you started with is the first law (without consideration of heat transfer) as it applies to the sum of the internal and external energy of the system (the book). The change in potential energy, $$\Delta U$$, and change in kinetic energy, $$\Delta K$$, are the changes in the external energy of the system, which are the changes with respect to an external (not the book) frame of reference (e.g. the floor).

The internal energy is the sum of changes in the internal energy of the system (the book). These changes are with respect to the reference frame of the book. If the book were, say, compressed by an external force or heated by an external heat source, there would be changes in its internal energy (e.g., a temperature increase). These considerations are more applicable to a system such as a gas in a cylinder.

Hope this helps.