So the title's a bit irrelevant. And I'm having a tough time thinking how to describe my doubt but I'll try my best. Apologies if it is vague, do let me know in the comments.

So I just finished reading the work-energy theorem and was moving on to the concept of potential energy (which my book explains poorly, or I'm missing something).

I understand that by the work-energy theorem the change in the kinetic energy of a body is the work done by an external force force on the body. ✓Got it, the change in the kinetic energy of a body is equal to the work done by the net force

So I moved on to potential energy, where my book tries to explain stuff by starting off with an example in which we're elevating the book of mass $m$ by doing work $mgh$ on it, where $h$ is the height to which it is raised.
My first doubt is, since the force we applied is $mg$ (equal to the gravitational force of attraction in magnitude yet opposite in direction), the resultant force on the book is 0. Since the resultant force is zero, the work done by the resultant force on the book should be 0, and there should be no change in the kinetic energy.

Secondly, the force done by us (the external agnecy), is along the direction of the displacement, the work done is positive and so the change in kinetic energy should be positive (by the work energy theorem?)

What am I messing up? Mixing the two concepts (potential energy and work energy theorem)? Yet after the work we do, potential energy increases, not the kinetic energy? What am I confusing?


So the given answers helped me understand where I was going wrong (somewhat). The change in the kinetic energy of a particle is equal to the work done on it by the net external force (if the particle is the system)

Now I'm having trouble interpreting the work energy theorem for a system of particles. A book explains it by considering a system of two charges (one positive and one negative; attracting each other). It says:

Because of mutual attraction, the particles are accelerated towards each other and the kinetic energy of the system increases

Firstly, how can we consider something which isn't fixed as our system (the charges move...).

Secondly, there is no net force on the system, so how is the kinetic energy even increasing.

PS: I'm still all confused about potential energy. Before my book starts about potential energy it goes on to talk about the work energy theorem for a system of particles, and that's where I'm confused now.

  • $\begingroup$ One thing t0 note is that if you system is deformable and has potential energy , the the work-kinetic energy theorem does not apply. $\endgroup$
    – garyp
    Mar 30, 2017 at 17:43
  • $\begingroup$ It is probably best to restrict the work-energy theorem to a single particle or a rigid body.However you can use the idea that the external work done on a system is equal to the change in kinetic energy plus the change in potential energy.So in your two charge case there are no external forces so the external work done is zero.The system starts with no kinetic energy & some electric potential energy.A little later the system (charges) has some kinetic energy & less electric potential energy.The sum of the change in KE (positive) and the change in electric potential energy (negative) is zero. $\endgroup$
    – Farcher
    Mar 31, 2017 at 9:41
  • $\begingroup$ For one of the charges as the system the work is done by the external force due to other charge and that results in a increase of the kinetic energy of the charge. $\endgroup$
    – Farcher
    Mar 31, 2017 at 9:55
  • $\begingroup$ @Farcher So you mean to say that the work energy theorem doesn't hold for a system of particles? $\endgroup$ Mar 31, 2017 at 17:03
  • $\begingroup$ It can but then you need to consider the internal work as well as external work. So even if there are no external forces the kinetic energy of parts is the system can increase. Often the summation of the changes of kinetic energy and the potential energy of the system is considered. In thermodynamics this is called the internal energy of the system and the change in internal energy it is equated to the work done on the system and the heat input into the system. $\endgroup$
    – Farcher
    Mar 31, 2017 at 18:55

4 Answers 4


The first thing you must do is define your system.
If the system is the book alone the the external forces on the book are the force that you exert on the book and the gravitational attraction on the book by the Earth.
If the book starts and finishes at rest then there is no change in the kinetic energy.
The work done by you on the book is positive as the direction of the force that you exert on the book is the same as the displacement of the book.
The work done by the gravitational force due to the Earth is negative because the gravitational force is in the opposite direction to the displacement of the book.

If the two external forces are equal in magnitude and opposite in direction then the net work done on the book is zero (equal to the change in kinetic energy).
Of course one could reason that the net external force on the book is zero so the net work done by external forces on the book is zero.

There is no mention of gravitational potential energy because it is the energy associated with the book and the Earth as a system.
So now let's consider this system of the book and the Earth.
The external force is now the force that you apply on the book.
The force that the Earth exerts on the book is an internal force and its Newton third law pair is the force that the book exerts on the Earth.
When you do positive work separating the book and the Earth that work increases the gravitational potential energy of the book-Earth system.

If you released the book the separation between the book and the Earth will decrease and the gravitational potential energy of the system will decrease.
The book (and the Earth) would then have kinetic energy.
Usually only the motion and kinetic energy of the book is considered because the mass of the Earth is so much greater than the book.
This results in the speed and kinetic energy of the Earth being very much smaller that that of the book.

  • $\begingroup$ What is a system really? I know how books define it: 'part of the assembly under consideration'. But its physical significance? $\endgroup$ Mar 31, 2017 at 2:45
  • 1
    $\begingroup$ It is a way of putting up a boundary and reducing what is being looked at. Often this makes the analysis of situations easier. The work-energy theorem as you described in you question is an example of this. You define a system and then only need to consider one group of forces - the external ones. $\endgroup$
    – Farcher
    Mar 31, 2017 at 8:06
  • $\begingroup$ I've edited my question $\endgroup$ Mar 31, 2017 at 8:16

First question

There is no change in kinetic energy if force you apply is mg book will move with finitesimally small velocity but if you apply F>mg then there is kinetic energy

Second question yes if F> mg then K.E is positive

I hope you get it now


Your first doubt is correct! Since there is no net force on the object being raised in the gravitational field, it does not undergo any acceleration and hence its velocity and Kinetic Energy shall remain unchanged.

The work energy theorem stated in the earlier part of your post is slightly wrong. You need to make the following addition: "...by the work energy theorem the change in the kinetic energy of a body is the work done by an external net force on the body."

In other words, the work energy theorem you stated only works if the force applied (by you) is the only force on the body!

When you lift the object (using an external force) equal and opposite to the gravitational force (assumed constant = mgh), the net force is zero and hence we find no change in Kinetic Energy.

So why does the Potential Energy increases? Well potential is by definition the ability to do work. Now using the work energy theorem, if you stop exerting your own force on the object, it will start to accelerate downwards. This is because this time it will have a net force downwards which will do work on the object and thus (using the work energy theorem) increase its Kinetic Energy.

  • $\begingroup$ Still confused. Got everything but the last paragraph. I think I'm having trouble with how potential energy is defined. I've seen books where they mention internal forces and external forces while defining it. And I've seen books where they don't mention either while defining it. The problem is I've seen both. So it just doesn't make any sense (concept of potential energy). If it isn't much trouble, could you elaborate? On potential energy? Thanks $\endgroup$ Mar 30, 2017 at 17:11

Work is nothing but energy transferred from one body to another. That is why we say work is done on a body as opposed to work done by a body. I will use one of the SUVAT equations of motion which is v^2 = u^2 + 2as where v is the final velocity, u is the initial velocity a = acceleration and s is the distance travelled. Now we know force F= ma. Substituting a = F/m and Rearranging we get 1/2mv^2 - 1/2mu^2 = F x s which is work done, but 1/2mv^2 - 1/2mu^2 is nothing but the change in Kinetic energy which led to the definition of the work-energy theorem. The definition of the work-energy theorem is The net work done by the forces acting on a body is equal to the change in kinetic energy of the body.

if however there is a freely falling body then F = mg and this equation can be rewritten as 1/2mv^2 - 1/2mu^2 = mg(h2-h1) where s= h2-h1 . So change in kinetic energy = change in gravitational potential energy = F x s which is defined as work done on a body. Now this also proves the concept of conservation of energy which is energy is neither created nor destroyed but can be converted from one form to another. This shows any change in kinetic energy must be equal to the change in potential gravitational energy and vice versa. See in real life movement is not one dimensional.


Not the answer you're looking for? Browse other questions tagged or ask your own question.