Work is defined as

$$W = \vec{F}\cdot\vec{s}$$

But what what exactly is $\vec{s}$? Is it the displacement of the body on which the force is being applied? Or is it the displacement of the point of application of the force?

If you look at the derivation for the definition of work from kinetic energy (the way many textbooks do it), it would seem to be the former, but I'm not sure since I can think of many examples that contradict the fact.

Or is it something else entirely?

  • 2
    $\begingroup$ You could give one of those examples you can think of that seems to contradict so it's easier to point out where the problem is in your understanding. $\endgroup$ – Ignacio Vergara Kausel Jun 26 '14 at 13:28
  • $\begingroup$ @IgnacioVergaraKausel: Say you are rubbing a surface. When you rub, the surface doesn't move, but the point of application does. By the former view, no work is being done. But the surface gets warm, so clearly, energy was transferred. $\endgroup$ – Gerard Jun 26 '14 at 13:53
  • $\begingroup$ Work is done at the point where the force is applied - not at some remote point that is connected. Rotation of the object means that the two points need not move at the same velocity. Thus total energy of object = linear + rotational energy, and as soon at rotational energy is nonzero, you expect there to be a difference between the two calculations / definitions of $\vec{s}$. $\endgroup$ – Floris Jun 26 '14 at 14:04
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    $\begingroup$ @Gerard There is a difference between, motion of point of application and changing the point of application. $\endgroup$ – Isomorphic Jun 26 '14 at 14:08

First of all, I appreciate your doubt. It is conceptual and involves critical thinking regarding the topic.

In order to understand work, you need to understand the 2 physical quantities namely - Force and Displacement .

Note that using only one single word Displacement in order to define work is not appropriate. (Same is for Force ) . As, Displacement depends on the direction of Force . Now, again, it will not be appropriate to just write Force in order to define Work/Displacement. I will better write it as : As, Displacement depends on the direction of Force applied on the point of application.

Forces acting on a moving Car

Whenever a body is moving, there are many forces acting upon it. While you are walking, the body is applying force on you, the ground is applying force on you, the air is resisting you from moving etc. So, basically it is much more complicated than we think.

Coming to the main point, displacement! Let us just take a point object in order to understand the situation well. Now, before I go on, I will first define what is point-object.

The concept of a point object can be understood here : Point Particle

So, as you're clear with the basic concepts, I can go on. Each object is made up of millions of particles...Uncountable! Now, when you push an object, suppose a car, then the car is moving but the force has acted upon on a specific point. That point may be referred to point of application of the force. When the car moves (if the other opposing forces become lesser than the force applied by you) , then it surely displaces itself. Thus, it does some Work. While, you have also done the work to push the object, but we usually calculate the work done by the object experiencing the force.

The displacement here, will be based on the point particle though many simply refer it to the complete object. It is not wrong either, but just a little mis-conceptional to refer the displacement to a whole object. You must keep this in mind that basically, the force(external) is acting upon the point particle which later on affects the nature of the whole object, that's why the object displaces itself.

This can be explained in 1-2 para also, but, what I'm trying to explain is critically-thinking a topic. Whenever you try to follow a question or topic, first think about it. Try to compare the conditions with nature, notice the similarities and differences. This will make your concepts much stronger!

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    $\begingroup$ +1 For encouraging critical thinking. It is important to break down or analyze questions like this, but it is a difficult skill to learn. Taking to an extreme in this example: I push, and do work, on the first layer of molecules of the paint on the car. That layer pushes, and does work, on the next layer ... until the entire car is moving. Side benefits of this critical thinking / analytical picture: we see that internal interactions play an important role, that deformations are important, and we get an idea of how the temperature of the car will rise. $\endgroup$ – garyp Jun 26 '14 at 14:27
  • $\begingroup$ You're pretty right Garyp! +1 $\endgroup$ – Kushashwa Ravi Shrimali Jun 26 '14 at 14:30
  • $\begingroup$ "Thus, it does some Work. While, you have also done the work to push the object, but we usually calculate the work done by the object experiencing the force". I didn't quite understand what you meant by that. How can the object do work? We are doing work on the object and transferring energy to it, not the other way round. $\endgroup$ – Gerard Jun 26 '14 at 15:49
  • $\begingroup$ Gerard, yes we are doing work on the object but what I'm referring to this condition as relative. I've taken a fixed reference frame. It is doing relative displacement, or the body is moving, also the object will apply the reaction force on the person (newton's third law) . Thus, we can say (relatively), the object is doing work. $\endgroup$ – Kushashwa Ravi Shrimali Jun 26 '14 at 15:59

Here's a way to argue it isn't the displacement of the (center of mass) of the body.

Take a spring that's at rest. Apply two forces on either end so it compresses. You can do this in such a way that the center of mass of the spring doesn't move. However, the energy of the spring system has changed (the potential energy increased). In order to satisfy the work energy theorem $$W_\text{net, external} = \Delta E_\text{total},$$ it would seem that the work must be non-zero. If you used the center of mass displacement, then $W=0$, which is inconsistent. The solution is to use the displacement of the point of application of the (two) force(s): $$W=W_1^\text{app} + W_2^\text{app}.$$


It's the displacement of the point of application of the force. Elementary textbooks initially model objects as point particles. They have exactly two properties: position and mass. No others. So the displacement of the object is necessarily the same as the displacement of the point of application.

Later, extended or compound objects may (or may not) be introduced. Now the object has more properties: a size, an internal structure, maybe others. And it may not be rigid. We usually define the position of an extended object as the location of the center of mass. A force can be applied to some part of the object, and the displacement of that part is not necessarily the same as the displacement of the center of mass. In those cases, the work done is the force times the displacement of the point of application, not the displacement of the position of the object (center of mass). Note the Kinetic Energy theorem fails to be applicable in such systems!

  • $\begingroup$ +1 for pointing out that the KE theorem is not valid for the systems you mentioned. I was having trouble there. So, just confirming, the KE theorem is valid only for single particles? $\endgroup$ – Gerard Jun 26 '14 at 13:55
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    $\begingroup$ Yes, and the work must be considered to be the work done by the net force. So if you push and object up, you have to consider the applied force of your hand, and the force of gravity. The KE theorem is deceptively limited in applicability. $\endgroup$ – garyp Jun 26 '14 at 14:02
  • $\begingroup$ (-1) : The KE theorem is always valid if you define work correctly. $\endgroup$ – Isomorphic Jun 26 '14 at 14:07
  • $\begingroup$ @Iota How does it apply to the case of a deformable body that has internal interactions and internal potential energy? $\endgroup$ – garyp Jun 26 '14 at 14:14
  • $\begingroup$ It will be mathematically difficult to apply , and might involve a term, due to internal forces that leads to conversion of Kinetic Energy to Potential Energy, but still. When you integrate the work done on a point, it will still give, the change in kinetic energy of the body. $\endgroup$ – Isomorphic Jun 26 '14 at 14:18

It is the displacement of the point on which the force is applied. This is it. That is the real definition, period. The books which do it otherwise are wrong. I would like you to read Resnick Halliday Krane , physics Vol. 1, it has a separate chapter devoted to this problem with most of the books available.

As an example to show that this is what the definition should be, you write an infinitesimal work done as, you could not use things like $m\frac{d^2x}{dt^2}dx$, the $d^2x$ and $dx$ would then be for different points and the integration wouldn't proceed the way it does.

Other simplifications arise from constraints, newton's 3rd law and viewing from different reference frame when applied to complex bodies.

I would still advise you to at least once go through that marvellous book. It is the best book on introductory physics that exists.

  • $\begingroup$ I do have Resnick Halliday Walker. Are the two any different? $\endgroup$ – Gerard Jun 26 '14 at 13:56
  • $\begingroup$ @Gerard Extremely.It is a lot more clearer compared to Resnick Haliday Walker and more detailed. It is a different book altogether and has no link to RHW's book. $\endgroup$ – Isomorphic Jun 26 '14 at 14:04

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