So here also energy of block doesn't change, then why we say that potential energy of block increases.
This is a common confusion and is due to poor communication, not a failure in your understanding which appears to be correct.
Potential energy is energy that is available by virtue of the configuration of a system. In this case the potential energy is $mgh$. Now, consider what system this potential energy describes. We have $m$, which is a property of the block, but $g$ is a property of the earth, and $h$ is a relationship between the block and the earth. So the system that has potential energy $mgh$ is the system of both the block and the earth.
So, when we say "the potential energy of the block increases", we are actually making a mistake. We should say "the potential energy of the block+earth system increases". Thus, when you say:
Therefore $mgh$ goes from me to block and $mgh$ from block to earth
If we taught potential energy correctly from the beginning you would have said "Therefore $mgh$ goes from me to block and the block+earth system's potential energy is increased by $mgh$"
Thus, if we consider the book to be our system, the net force and by extension, the net work comes out to be 0. Moreover, the work done by gravity is defined as the negative of the change in potential energy of the book-Earth system. Hence, we can say that the change in potential energy ฮ๐=๐๐โ
First, in this question letโs completely discard the concept of โnet workโ which is simply the net force times the displacement of the center of mass. It is typically not useful and is often even counter-productive when looking at conservation of energy.
Now, the key to such problems is to carefully define the system. In this case, the system is the book. Note that in $mgh$ the $h$ is not a property of the book. It is a relationship between the book and the earth. Since it is not part of the book the PE $\Delta U$ does not belong to the book.
So the correct energy analysis for the book as the system is that the contact force does positive work on the book and gravity does negative work on the book, so the system consisting only of the book has no change in energy.
Usually we consider the earth as being so massive that it doesnโt move. In that case, treating the potential energy as โbelongingโ to the book does not cause any problems. But it is technically wrong and can cause confusion like yours.
Now I want to analyze the same situation by considering the book and the Earth to be our system.
Now, in this case $h$ is a property of the system. It is a โdegree of freedomโ that can be used to change the configuration of the system and store energy. For energy, we are only interested in the external forces. The internal forces just shuffle energy around inside the system, only external forces change the energy of the system.
The work done by the contact force on the book is positive, thus increasing the energy of the system by $\Delta U= mgh$. However, since the Earth doesnโt move, the contact force on the earth does no work. So the external work done is equal to the change in energy, so energy is conserved.
Now, if you do choose to analyze the work done by the internal forces, you will find that it is negative. Since that is an external force, that is not energy lost, it is just energy that is converted to internal potential energy rather than kinetic energy.
From the book's perspective, the earth is moving downward, while the gravity is acting upward, towards the book. So, shouldn't the book do a work of โ๐น๐๐๐๐/๐๐๐๐กโ.โ=โ๐๐โ on the Earth ? I'm told this is not true, and I don't seem to understand why. Is there a special choice of origin that we stick to, throughout the entire analysis ?
You can pick any inertial reference frame. There is no special inertial frame, but you do need to stick to that frame throughout the analysis. You cannot calculate part of the energy in one frame and the rest of the energy in another frame. This is because the energy will be different in different frames (but it will be conserved in all frames).
However, even though you can, in principle, use any inertial frame, the math is much simpler in the frame where Earth is at rest. If you want to do the analysis in a frame where the Earth moves then you must consider the conservation of momentum, and you must consider the earth to have a finite mass.
Best Answer
TL;DR There are few things wrong in how you understand the work-energy theorem:
The work-energy theorem
The work-energy theorem says that the total work done on an object equals change in its kinetic energy
$$\Delta K = K_2 - K_1 = W \qquad \text{or} \qquad \boxed{K_1 + W = K_2}$$
where $W$ is total work done on an object. This means work done by all forces acting on the object! Note that by definition, work is a scalar value that can be either positive or negative.
One thing people usually forget is that the work done by gravitational and elastic (spring) forces equals negative change in the potential energy:
$$W_g = -\Delta U_g \qquad \text{and} \qquad W_e = -\Delta U_e$$
where $U_g = mgy$ is gravitational potential energy and $U_e = \frac{1}{2} k x^2$ is elastic potential energy. Since difference $\Delta$ always means final minus initial value, the work by gravitational and elastic forces is defined as
$$W_g = U_{g,1} - U_{g,2} \qquad \text{and} \qquad W_e = U_{e,1} - U_{e,2}$$
If we use this in the work-energy theorem we get
$$\boxed{K_1 + U_{g,1} + U_{e,1} + W_\text{other} = K_2 + U_{g,2} + U_{e,2}}$$
where $W_\text{other}$ is work done by forces other than gravitational and elastic forces.
Example: A freely-falling ball
As you have correctly identified, the change in kinetic energy can be negative, but the kinetic energy itself cannot. In example of dropping the ball from a height $h$, the energy diagram would be
$$\underbrace{\frac{1}{2} m 0^2}_{K_1} + \underbrace{mgh}_{U_{g,1}} = \underbrace{\frac{1}{2} m v_1^2}_{K_2} + \underbrace{mg0}_{U_{g,2}}$$
or in a general case
$$\frac{1}{2} m v_1^2 + mgy_1 = \frac{1}{2} m v_2^2 + mgy_2$$
which might be more familiar in this form
$$\boxed{v_2^2 = v_1^2 - 2g(y_2 - y_1)}$$
You must be careful what you use for $y_1$ and $y_2$ - subscript 1 denotes the starting point and 2 denotes the ending point.
Work by gravitational force equals negative $\Delta U_g$
As to why work done by gravitational force is negative change of potential energy, we start from the definition of work:
$$W = \vec{F} \cdot \vec{x} \qquad \text{or} \qquad W = \int \vec{F} \cdot d\vec{x}$$
where both force and displacement are vectors, which means they have both magnitude and direction. The work is defined as a scalar product of force and displacement, which means that only portion of force parallel to displacement does work, whereas perpendicular portion does no work. The scalar product finds that portion of work parallel to displacement. If force is constant and displacement is a straight line you can use the (simpler) left-hand side equation, but in all other cases you should use the right-hand side equation.
Let positive $\hat{\jmath}$ axis point upwards (away from Earth's center), then the gravitational force is
$$\vec{F}_g = -mg\hat{\jmath}$$
where $\hat{\jmath}$ is just a unit vector (magnitude equals to one) which defines direction of the force. The work done by gravitational force $\vec{F}_g$ over some distance $\Delta \vec{y}$ is
$$W_g = \vec{F}_g \cdot \Delta \vec{y} = -mg\hat{\jmath} \cdot (y_2 - y_1) \hat{\jmath} = -(mgy_2 - mgy_1) = -\Delta U_g$$
where $\hat{\jmath} \cdot \hat{\jmath} = 1$ is a scalar product between two unit vectors.