I don't understand the wordings of this theorem. Can someone please help me in understanding this? Secondly, on what basis are the sign conventions in this theorem applied? I get confused in positive and negative majorly while writing the the potential energy (due to earth or any other means) and the spring's work. When is it $-\frac{1}{2}kx^2$ and when $+\frac{1}{2}kx^2$?


3 Answers 3


The work-energy theorem can be derived from definition of work. $$W = \int \vec F\cdot d \vec s\\\text{since}\,\vec F=m\vec a=m\cdot \frac{d\vec v}{dt}\,\text{and}\,d\vec s=\vec v\,dt\\\text{we get:}\\W=\int m\frac{d\vec v}{dt}\cdot \vec v\space dt$$ $$\text{since}\space\space\space\ \vec v \cdot \vec v=v^2,\text{let us try take the derivative of it:}\\\frac{d(v^2)}{dt}=2\,\vec v\cdot\frac{d\vec v}{dt}\\\text{to obtain this result I used implicit differentiation, but the same result can be obtained by product rule}$$ Therefore: $$W_{ab}=\frac12\int_a^b m\frac{d(v^2)}{dt}dt=\frac12m(v_b^2-v_a^2)$$ We can see that total work done from a to b is equal to change in kinetic energy.

How does this relate to energy in general? We say that in isolated system the energy is constant. It implies that change of energy is zero. However in case of work the system is not isolated and we can see that there is indeed a change of energy (kinetic energy to be specific).

As you may already know the conservation of this quantity allows us to make specific calculations very simple (like for example determining final speed of isolated object).

I think one way you can safely think of this theorem is as extended principle of conservation. If we know work done on our object, we still can get information about useful quantities (like speed), even if conservation of energy no longer holds!

Later you may see, that work can be also described as some special function that is dependent only on initial and final position, if the specific force is applied.

To translate it into equation, we can say that:

$$\Delta E_{kinetic}+\Delta E_{potential}=W_{non-conservative}$$ *Please keep in mind above equation is valid only if mechanical energy is taken into account.

As for sign convention, it is usually defined that negative work is equal to potential energy. (think of it loosely as pushing in the direction against the force increases your potential energy).

Note that the negative sign is just for convenience, and does not arise from any mathematical law.


If you want to use the work-kinetic energy theorem you best stick to either a point like object or a particle, (or an object which can have both transnational kinetic energy and rotational kinetic energy) and then the theorem states that the work done by external forces on the particle (the system) is equal to the change in kinetic energy of the particle.

A particle (the system under consideration) is moving (ie has kinetic energy) along a rough surface so an external force due to the kinetic friction is acting on the particle.
Work done equals on the body is $\int \vec F_{\rm external} \cdot d\vec s$ where $F_{\rm external}$ is the frictional force and $d\vec s$ is the incremental displacement of the force.
In this case the direction of the friction force is opposite to the displacement of the particle so the work done on the particle is negative and so the change in kinetic energy of the particle is negative ie the kinetic energy of the particle is decreasing.
A particle (the system) is on a rough surface which is increasing in speed and the particle does not slip relative to the surface.
In this case the static frictional force is in the same direction as the displacement of the particle so the work done by the frictional force is positive and there is an increase in the kinetic energy of the particle.

Now suppose that you have a compressed horizontal spring fixed at one end and with a particle at rest at the other end.

Consider the particle as the system and so the spring exerts an external force on the particle.
Release the particle and the the force (external) on the particle due to the spring is in the sane direction as the motion of the spring so the spring does positive work on the particle and the kinetic energy of the particle increases.
If the compression of the spring was $x$ and the sping constant was $k$ then the work done by the spring on the particle is $\frac 12 k x^2$.
Please note that I have not used the term (elastic potential energy) as in the context of what is happening to the particle (the system) it is irrelevant as to the origin of the external force.

Reverse the process with a moving particle hitting an unextended spring and compressing it, then the force (external) is in the opposite direction to that of its displacement and so the work done on the particle by the spring is negative.
In this case the work done by the external force is $-\frac 12 k x^2$.

Let me explain why you cannot use the work - kinetic energy theorem for a spring.

Imagine an uncompressed spring (the system) fixed at one end.
You apply a force (external) to the spring and whilst you are compressing the spring you are doing work on the spring.
You compressing the spring and the spring has no kinetic energy so the work - kinetic energy theorem does not hold.
It does not hold because in doing the spring another form of energy is stored in the spring - elastic potential energy.
So in such a case you should be thinking of the work- energy theorem which states the the work done on a system by external forces is equal to the change in energy of the system.

Using the spring being compressed by a moving particle as an example of a system you could say that no work is done on the system and so the total energy of the system does not change.
The force on the spring due to the mass and the force on the mass due to the spring are internal forces so cannot be counted in the work done by external forces.
The external force exerted on the spring at the fixed end of the spring does not move and so does no work.

As there is no change in the total energy of the system (spring and particle),

$\rm (kinetic \:energy)_{\rm initial}+(elastic \:potential\: energy=0)_{\rm initial}=\rm (kinetic \:energy=0)_{\rm final}+(elastic \:potential\: energy)_{\rm final}$.

  • $\begingroup$ Even after editing, the question promotes misleading or even wrong statements. $\endgroup$
    – nasu
    Commented Mar 31, 2019 at 13:51
  • $\begingroup$ 1. A key point which is omitted is that the change in kinetic energy equals the NET work. 2. The W-E theorem works for individual particles as well for systems of particles. Most mechanics books treat the case of systems of particles a little bit after the case of point particles. 3. There are no cases when the W-E does not work. It works very well for the compression of a spring or any other process. You just have to apply it properly. $\endgroup$
    – nasu
    Commented Mar 31, 2019 at 14:04
  • $\begingroup$ @Nasu please note that I have differentiated between the work - kinetic energy theorem and the work - energy theorem. The second works for all situations the first does not. I did this because of the title of this question. $\endgroup$
    – Farcher
    Commented Mar 31, 2019 at 14:08
  • $\begingroup$ The term "Work-energy" theorem refers to the connection between work and KINETIC energy. There is no separate theorem called "work-energy" theorem and having a different meaning. courses.lumenlearning.com/boundless-physics/chapter/… $\endgroup$
    – nasu
    Commented Mar 31, 2019 at 14:13
  • $\begingroup$ You cannot compress a spring without moving the parts of the spring and even the center of mass of the spring. If you compress an ideal spring for a while and then remove the compressing force the spring will oscillate so it does have kinetic energy. $\endgroup$
    – nasu
    Commented Mar 31, 2019 at 14:26

In simple words, the W-E theorem states that the net work done by forces on a body is equal to the change in kinetic energy of the body. Kinetic energy includes the sum of rotational and kinetic energies. That means simply summing up the work done by forces on the body: it is equal to the change in $KE$ of the body. Remember here that work done by force is important, not the force itself. It may be that force is present but it's work is zero. Now for the sign convention. Using a fixed sign convention for a question will give the same answer. But the generally adopted convention is that if displacement is in the same direction as force, work is positive, while in the opposite direction, it is taken to be negative. A point at which confusion occurs is that one should be careful about work done by the body and work done on the body. The sign convention is for work done by the body. For example. In the compression of a spring, we apply a force to compress it. The displacement of the spring is in the same direction as when we apply force, so the work done by us is positive. Or we say work is done by us (work done on us is negative). But for spring, it applies a force outwards and instead gets displaced inwards. So the work done by spring is negative. Or you say work is done on the spring. (Work done in spring is positive) I hope it clarifies your doubts.

  • $\begingroup$ Sorry , added by mistake $\endgroup$
    – Tojra
    Commented Mar 31, 2019 at 9:28

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