# Potential Energy of a System and work–kinetic energy theorem

If you lift a book from the floor to a shelf, $K_f-K_i=0$,because the final and initial speed are 0.

So from the work–kinetic energy theorem the work done by external forces should be $0$, so there is no energy transferring mechanism here, since work done by external forces is $0$, and there is no other energy transferring mechanism here, so how does this the system gain energy in form of potential energy?

• The total work done by external forces is zero. But each force can still do a non-zero amount of work. Aug 31, 2017 at 8:00

. . . so how does this the system gain energy in form of potential energy?

The system which gains the gravitational potential energy is the book and the Earth not the book alone.
You do work increasing the separation between the book and the Earth and the result is that the book and the Earth have more gravitational potential energy.

The book alone as the system has two forces acting on it.

• The gravitational attractive force due to the Earth.
• The upward force that you exert.

So the net force on the book is zero, the net work done on the book is zero and so the change in kinetic energy of the book is zero.

Put another way the positive work done by you in lifting the book is equal to the negative work done by the gravitational attractive force.

You lifted the book. You are the external force supplier.

$W_{gravity} + W_{ext} = \Delta KE = 0$

$W_{ext}=-W_{gravity}$

The relation given when mechanical energy is conserved is a special modification of the work energy theorem when there are only conservative forces acting on the system.

$W_{conservative} + W_{non-conservative}=\Delta KE$,

When $W_{non-conservative}=0$,

$W_{conservative} =\Delta KE$,

Positive work done by conservative force decreases the potential energy of the system and negative work done increases the potential energy.

Therefore, $W_{conservative}=-\Delta U$,

$-\Delta U = \Delta KE$

$\Delta KE + \Delta U=0$

The work done by external forces is zero. Your hand which lifts the book does positive work on it. The force of gravity does negative work on the book while you lift it. Since there is no change in kinetic energy, they exactly cancel each other out. The reason that you might be confused is because the work done by gravity is actually defined as the negative change in the gravitational potential energy (i.e. $W_{g}=-\Delta PE$). Therefore, gaining gravitational potential energy is exactly equivalent to having negative work done on you by gravity.

Edit: Here is a brief derivation of the Work-Kinetic Energy Theorem which relies only on a couple of basic definitions.

$$\Delta E=\Delta KE \ + \Delta PE \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(Definition of Mechanical Energy) (1)}\\ W_{total}=W_{conservative}+W_{non-conservative} \ \ \ \ \ \ \ \ \text{(Mutually exclusive categories) (2)}\\ W_{conservative}=-\Delta PE \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(Definition of Potential Energy) (3)}\\ W_{non-conservative}=\Delta E \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(Definition of Non-conservative Forces) (4)}\\W_{total}=\Delta E-\Delta PE \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{((2)+(3)+(4)) (5)}\\W_{total}=\Delta KE \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{((1)+(5)) (6)}$$

As you can see, this derivation (and therefore the W-KE Theorem) is entirely agnostic with regards to whether the forces are internal or external to the system. The only thing that matters is whether ther forces at play are conservative or non-conservative.

• but the only external force acting on the system is my force, so if i do positive work on it, how is the external work zero? isn't the force of gravity an internal force? Aug 31, 2017 at 6:31
• @forpointing Not necessarily, and certainly not in the way that this problem was phrased. The number of external forces acting on the system depends entirely on what you take to be your system. Implicitly, you chose your system to be only the book. In that case, gravity is external. But this distinction doesn't matter very mug anyway since Conservation of Energy and the Work-Kinetic Energy Theorem do not distinguish between external and internal forces - they only care about whether the forces are conservative or non-conservative. I'll edit my answer with a short derivation that might help. Aug 31, 2017 at 6:41
• @forpointing The edit is in. I hope this helps! Aug 31, 2017 at 7:02

By work - kinetic energy theorem you presumably mean something like:

The work-energy theorem states that the work done by all forces acting on a particle equals the change in the particle's kinetic energy

But this is true only when there is no change in the potential energy. A more complete statement would be:

The work-energy theorem states that the work done by all forces acting on a particle equals the change in the particle's kinetic energy + the change in the particle's potential energy.

In the example you give there is no change in kinetic energy since the particle is stationary to start with and stationary at the end of its motion. In that case all the work goes into changing the potential energy.

• Are you sure about that, John? It seems to me that the work done by conservative forces is the negative change in potential energy. And since the change in total mechanical energy is the work done by non-conservative forces, then the total work must be the change in kinetic energy. As follows: $W=W_{NC}+W_{C}$, $W_{C}=-\Delta U$, $W_{NC}=\Delta E$, and $\Delta E= \Delta T + \Delta U$; therefore, $W=\Delta T$. Aug 30, 2017 at 9:39
• @Geoffrey: I'm not sure I see the relevance of non-conservative forces? What non-conservative forces are acting here? Aug 30, 2017 at 11:04
• It is relevant for two reasons. First, I was trying to address that your second "corrected" definition of the Work-Kinetic Energy Theorem appears to be wrong. It reads as though you think $W=\Delta E$, which it does not. If there were no non-conservative forces, then $\Delta E=0$, but in this situation $\Delta E>0$ since the potential energy of the book increases. Second, his hand which raises the book is a non-conservative force. There is no potential energy associated with the force of his hand, and the work that his hand does is path dependent. Therefore, it is non-conservative. Aug 30, 2017 at 17:11
• I must agree with @Geoffrey, we can derive $W_{tot} = \Delta T$ via Newton II, and then we can replace $W_c = -\Delta U$ for any conservative forces that are included in the net work term. If we raise a book against gravity at a constant speed, the total work done by all forces equals zero, however the potential energy is definitely increasing! Then we might say "the work done by all non-conservative forces acting on a particle equals the change in the particle's kinetic energy + the change in the particle's potential energy" Apr 1, 2020 at 11:51

There is a mistake in your approach: the final velocity is not zero. If you are measuring that movement, according, of course, with the point particle model, the module of the velocity vector is such that $K_{f}-K_{i}>0$ in the moment of the contact between the particle and the shelf. In that instant of time the trajectory is complete, and then we stop our clock, and no later.