Consider this: you start out with a certain mechanical energy, $E_i$. As you move along, work ($W_x$) is done on the system, changing the mechanical energy of the system. At the end, the energy of the system is $E_f.$
Because energy is conserved, and we've restricted ourselves to mechanical interactions, the final energy must be the initial plus the work:
$$E_f = E_i + W_x.$$
$$W_x = E_f - E_i = mgh-mgH$$
Your mistake is you added work to the final mechanical energy rather than the initial mechanical energy.
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.
I asked my Physics teacher and he explained it to me, but I am still confused on how to use it.
Best Answer
The net work $W_\text{net}$ can be broken down into two pieces:
Thus, $$W_\text{net} = W_\text{conservative} + W_\text{external}.$$
You are correct that the work-energy theorem states $W_\text{net} = K_f - K_i$ where $K = \frac{1}{2} mv^2$ is the kinetic energy (this is nothing more than an algebraic rearrangement of the kinematics formula $v_f^2 = v_i^2 + 2a(x_f - x_i)$ paired with the defnition of net work $W_\text{net} = \mathbf F_\text{net} \cdot \Delta \mathbf x$). Substituting this in we have the following.
$$\Delta K = W_\text{conservative} + W_\text{external}$$
The work done by conservative forces (the energy transferred into the object by the conservative forces) is exactly the amount of energy lost from the potential energy (this is basically the definition of potential energy. As an equation, this is $W_\text{conservative} = - \Delta U$ where $U$ represents kinetic energy. Substituting this fact in and then rearranging gives the following.
$$\Delta K = - \Delta U + W_\text{external}$$
$$\Delta K + \Delta U = W_\text{external}$$
The quantity $K + U$ is what we as physicists define as total mechanical energy, or $E_\text{total} := K+U$. This allows us to write the following.
$$\Delta E_\text{total} = \Delta K + \Delta U = W_\text{external}$$
So, in short, yes, this is just another form of the work-energy theorem. It's just written in such a way so that $W_\text{external}$ means all the work done by forces acting on the object that do not have a potential energy $U$ associated with them.