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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?

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  • $\begingroup$ 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 '18 at 17:33
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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!
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