I have recently learned about three types of energy. Kinetic, elastic and gravitational potential energy. I have also leaned about Work done on a particle.

I would like to know if the Work done on a system is equivalent to the total energy in a system?

I ask this because when we determine the work done by a force compressing or extending a spring, we do this by finding $$W = \int \frac{\lambda x}{l} \thinspace dx= \frac{\lambda x^2}{2l}$$ and then we define this result as the Elastic Potential Energy.

Does this mean that total energy is equivalent to work?

  • $\begingroup$ In the absence of heating, the work done on the system is equal to the change in energy of the system. $\endgroup$
    – march
    May 14, 2018 at 17:33

1 Answer 1


The short answer is "no", although it will depend on what you call "total energy".

The point is that work done equals the variation of kinetic energy:

$$W=\Delta E_k$$

(I'm not considering heating, just mechanics).

But, there are two types of forces. We can divide the work in two parts:

work done by conservative forces and work done by non-conservative forces.

The first one can be written as $W_c=-\Delta E_p$.

So then you have

$$ \Delta E_k = W_c + W_{nc} = -\Delta E_p +W_{nc} $$

so, if you rearrange it, you have

$$\Delta E_k + \Delta E_p = W_{nc}$$

So total mechanical energy variation equals the work done by non-conservative forces. $\Delta E_m=W_{nc}$.

  • If there isn't any non-conservative force on the system, the energy will be conserved ($E_m$ won't vary$.
  • When you calculate that integra, you are calculating the work done by a conservative forces. Potential energy is "minus the work done by a conservative force".
  • But you must be careful to account all forces in the system. There can be many conservative forces (ellastic, gravitational, electrostatic...), and there can also be non-conservatives too!

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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