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From the definition of power:

$$P \ = \ \frac{dW}{dt} \ = \ \frac{dE}{dt}$$

Multiplying both sides by $dt$, we get:

$$dW = dE$$

$1.$ What does this imply?

$2.$ Are there any other (perhaps more rigorous) way of proving this?

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  • $\begingroup$ It implies that you are equating $W$ and $E$ up to a function that has no time dependence. What is $W$ and what is $E$ here? $\endgroup$ – Aaron Stevens Jul 19 '19 at 0:13
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    $\begingroup$ This question scratches the tip of a very large iceberg. Search here for questions on work and energy. $\endgroup$ – garyp Jul 19 '19 at 0:25
  • $\begingroup$ If you ask what $dW = dE$ imeans mathematically: it means $W = E +c$ where $c$ is a constant. $\endgroup$ – Fabian Jul 19 '19 at 2:58
  • $\begingroup$ Work is transferred energy. That's about it: $W=\Delta E$ means that the work $W$ done by a force equals the amount of energy $\Delta E$ that's moved from one (sub)system to another in the interaction. When you write that with differentials, you get $dW=dE$. $\endgroup$ – stafusa Jul 20 '19 at 8:11
  • $\begingroup$ @Fabian I can understand that relationship by either intuition or integrating both sides. But I have one contradictory idea in my mind that $W = E + C$ implies that W and E are both the indefinite integrals of their derivative. As far as I know indefinite integral of a force is energy, and the definite integral is work. How can work, a definite integral, be in the relation of indefinite integral ??? $\endgroup$ – curious Sep 29 '19 at 8:07
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Taking W as work done on the system and E as the change in mechanical energy, the equality implies that when you do work on a body it's mechanical energy will increase by the same amount. The increase in energy could be either in the form of Kinetic energy or Potential energy. (This is strictly on a classical mechanics point of view, and i think you are asking the question from that area.). For a rigourous approach, try differentiating the basic equation for total energy, that is the sum of kinetic and potential energy.

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From thte first law of thermodynamics, considering no heat transfer.

$\Delta U = Q + W$ (internal energy)

$\Delta E = \Delta E_m + Q + W$ (total energy)

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