Examples from Principles of Physics (by Walker,Resnick,Halliday) will say it better:

Let us throw a tomato upward. . .as the tomato rises, the work $\mathbf{W_g}$ done on the tomato by the gravitational force is negative as the force transfers energy from the $E_K$ of the tomato. We can now finish the story by saying that this energy is transferred by gravitational force to the gravitational potential energy of the tomato-earth system. The tomato slows down, stops & then begins to fall back down. During the fall, the transfer is reversed: the work $\mathbf{W_g}$ done on tomato by the gravitational force is now positive - that force transfers energy from the gravitational potential energy of the tomato-Earth system to the $E_K$ of the tomato.

If we abruptly shove a block to send it moving rightward, the spring force acts leftward & thus does negative work on the block, transferring energy from the $E_K$ of the block to the elastic potential energy of the spring-block system. . . .

And many,many other examples. Now, why is the potential energy attributed to the system and the kinetic energy to the object itself? What is the logic/cause?

  • 1
    $\begingroup$ We also call sometime potential energy of an object. Infact wiki gives defination of potential energy like this,''In physics, potential energy is the energy that an object has due to its position in a force field or that a system has due to the configuration of its parts.''en.m.wikipedia.org/wiki/Potential_energy $\endgroup$
    – Paul
    Commented Jan 28, 2015 at 5:19

4 Answers 4


Here's an example of why this is done using a gravitational context.

Kinetic energy

The kinetic energy of a system of particles is defined to be $K_\text{tot} = \sum_i K_i$, where the sum runs over all particles in the system. This is just a definition, and choosing to use such a definition allows one to speak of the kinetic energy $K_i$ of the $i$th particle, for example, since it's just one of the terms in a sum.

Okay, that was boring.

Potential energy

Gravitational potential energy $U_\text{g} = -\frac{Gm_im_j}{d_{i,j}}$ is defined for pairs of particles.$^\text{1}$ Mathematically you can see that in the subscripts of the masses $m_i$ and the distance between the two particles $d_{i,j}$. Why? Because to determine potential energy from scratch, you must consider how two objects interact, which in classical mechanics is where forces come in. And forces themselves come in pairs.

Now, the fact that they come in pairs means one must be careful when determining the potential energy in some given situation, otherwise you might over-count. Suppose, for example, there are 4 particles with masses $m_i$, and the distance between the $i$th and $j$th objects is $d_\text{i,j}$. What, then, is the potential energy of the masses and distances are all given? The answer is $$U_\text{g}=-G\left( \frac{m_1m_2}{d_{1,2}} + \frac{m_1m_3}{d_{1,3}} + \frac{m_1m_4}{d_{1,4}} + \frac{m_2m_3}{d_{2,3}} + \frac{m_2m_4}{d_{2,4}} + \frac{m_3m_4}{d_{3,4}} \right)$$ First, note that since there are six terms you can't make a one-to-one associating of particles with potential energy. Second, if someone tried to associate potential energy with a single particle, then that person might mistakenly include too many terms in their sum:

$$U_\text{g}\ne-G\left( \underbrace{\frac{m_1m_2}{d_{1,2}}}_\text{a term for particle 1} + \underbrace{\frac{m_2m_1}{d_{2,1}}}_\text{a term for particle 2} + \cdots \right)$$ This incorrect expression would over-count the number of terms in $U_\text{g}$.

To avoid such over-counting, and to make our language more inline with the mathematics, we tend to speak of the potential energy of the system.

$^\text{1}$The expression $-\frac{Gm_1m_2}{d_{1,2}}$ is a more general form of the more-common $U=mgh$, but that's a separate issue.


Let me put it this way,all forces that we know arise from interactions between two particles/systems. This is true either for fundamental forces such as gravitation, electromagnetism as well as derived forces as friction,viscosity,etc.

The potential or the potential energy as a concept is introduced to take advantage of the conservative character of most forces say gravitation or the spring force and so on.By introducing the potential as $$\textbf{F} = -\nabla \phi$$ we get a vector function from a scalar function which is more easier to deal with.

With this being said,the final reason that the potential arises as a system dependent function is because the underlying force itself is system dependent.It does not make sense to talk force of a single particle.We always talk of a force of a particle/system of particles on a particle/system of particles.

The force as an entity does not stand on its own.It is from a system on a system.

  • $\begingroup$ Wouldn't you say that the gravitational force is being exerted on an object neare earth is not really arising from the interaction between earth and object, but rather between interaction between object and the curved space-time ? Isn't this a big difference ? $\endgroup$
    – nerdy
    Commented Feb 13, 2015 at 17:21

In case of tomato falling down, kinetic energy is gained both by the tomato and the earth. However since $E_K = \dfrac{\mathtt{P^2}}{2M}$ (Same $\mathtt{P}$ for both) the heavy mass of earth shows very small change of kinetic energy. The kinetic energy of the system would therefore include both, the earth and the tomato. The $E_K$ of earth is neglected and therefore we speak of $E_K$ of tomato.

  • $\begingroup$ Sir,what let you to assaign same PE to both tomato and earth? $\endgroup$
    – user36790
    Commented Jan 28, 2015 at 13:32
  • $\begingroup$ @SAKhan you don't explain the things clearly. Indeed, potential energy is due to the interaction between the two bodies and is stored in the field between the two bodies. However, about the kinetic energy we can say that the Earth moves toward the tomato or vice-versa. Usually, we work in the center-of-mass (C-o-M) frame, defined as the frame in which the total linear momentum of the objects is zero, $mv_1 = -Mv_2$. So, if $m/M << 1$, one gets $|v_2| << v_1$, s.t. the big mass is practically at rest. So, we can take it as C-o-M frame. This is the issue, we work in fact in the C-o-M. $\endgroup$
    – Sofia
    Commented Jan 28, 2015 at 14:03
  • $\begingroup$ They should learn to do things on their own. Also I try not to introduce things for which I am not certain they know and which would move them away from what they are actually looking for. $\endgroup$
    – SAKhan
    Commented Jan 29, 2015 at 19:00

A field, like the gravitational or electric fields, can be treated as a store of energy, but you must have something to equate to the energy.

Potential is a measure of space and is purely relative. It is in effect, the moment of field. eg $g = \nabla\phi$ or $E = \nabla V$.

If you apply a charge (say a mass in gravity or a charge in electricity), it requires a measure of work to move it against the fields to get it to a place, and this energy is in effect, stored in the field.

Moving a tomato up ten feet gives it 'energy of location', which the field attempts to extract from it. But since we measure moment of field from a relative position, we have no way of knowing the true moment of field, so it is possible to have a negative potential, and the tomato, for falling 11 feet, releases more energy than it took to raise 10 feet. But to put it back to where it came from, will produce 'zero energy'.


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