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Suppose a car of mass $m$ starts out with an acceleration $a$ in the positive direction (say +ve x-axis), against a constant Resistive force $R$. At the instant it's velocity is $v$, what is power of engine? The answer is $(R+ma)v$ which is intuitive as the engine is working against two forces, but the net Force is $ma - R$, so why isn't power $F_{net} v = (ma-R)v$?

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  • $\begingroup$ $(R+ma)v$ is correct, because the engine has to overcome both resistive forces (e.g., drag of various types) and conservative forces in order to accelerate the car. $\endgroup$ – David White Aug 31 '17 at 20:46
  • $\begingroup$ How is the net force $ma-R$? $\endgroup$ – Steeven Aug 31 '17 at 21:44
  • $\begingroup$ I got my mistake, I considered ma as a separate force instead of net Force. $\endgroup$ – drake01 Sep 1 '17 at 3:12
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The engine does work to overcome friction and increase the kinetic energy of the car.

Let the power developed by the engine be $P=F_{\rm engine}v$.

$Rv$ of that power $P$ is the rate of working against the frictional force.

The rest of the power developed by the engine is the rate of working to accelerate the car (increase its kinetic energy) which has a net force $F_{\rm net}(=ma)$ on it and so is equal to $F_{\rm net}v=mav$

So $P=F_{\rm engine}v= Rv + F_{\rm net}v=Rv+mav$ noting that

$F_{\rm engine}= R + F_{\rm net}\Rightarrow F_{\rm engine}-R = F_{\rm net}$

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The formula for acceleration is

$$ a=\frac{F_{net}}{m} = \frac{\frac{P}{v}-R}{m} $$

which is solved for $P$ as $$P = v\, (R + m a) $$

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  • $\begingroup$ Using words: Power goes into not only accelerating the car but also overcoming the resistive force. This effect is additive. $\endgroup$ – ja72 Sep 1 '17 at 12:15

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