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I understand this topic well enough to get all the task done because they aren't very creative. But for my exam I think I should have this clear.

During the acceleration the force from the engine is of course bigger than air resistance and friction. This force, can we find it? And then the entire force the engine applies for the acceleration. Not just the stub you after subtracting for friction and air resistance.

$W = F \times s$
$F = m \times a$

We have all the work done by forces at work, and the stretch of road is easy to calculate. If I now do this $\frac{F}{m} =a $ will that output be the correct acceleration? And this force that we found, is that a sum force? Because if that's the force sum on the car I can't find the engines power output which is what I want. And what about the friction force at work, I think we can't find it when we just have the change in kinetic energy. Is that right? Primarily I would like to know if the change in kinetic energy can be tied somehow to the engines output during the acceleration.

The book has this nice equation too: $P = F \times v $
But that's just constant speed.

Since we know the time maybe this can be used: $P = \frac{W}{T}$
That just seems a little too easy.

Edit: I got a B on the exam, which means I'll be at again this fall. Not due to this question. Haha.

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3 Answers

Making the question as simple and concise as possible is a very good way to address it; the way it stands, I can't understand your question. So in the hope that it helps, I'm going to present a simpler question, partly resembling yours, and address the simpler question: a car is moving at a constant velocity on a horizontal road, what are the different forces involved and what's their work and power?

An easy way of thinking about the problem, at least as a starting point, is that there are two external forces acting horizontally on the car: the air resistance and the friction between the tires and the road; interestingly enough, the friction with the road is the driving force which cancels the air resistance, hence the net force becomes zero and the car maintains its velocity. Now how about the works? The work (and power) due to the friction with the road is positive (for it to be possible we should assume the tires slide on the road) and the work (and power) due to the air resistance is negative; the two cancel each other and the net work (and power) is zero, hence the kinetic energy (and the velocity) is constant.

Now what's the role of the engine? In the engine, chemical energy gets converted to mechanical energy and this mechanical energy gets transfered (in the rotational form) to the tires. If one is interested in the dynamics of the tires then, the engine is crucial to be considered, but if one is only interested in the dynamics of the whole car, the interaction between the engine and the wheels is only internal and according to the 3rd law can be neglected when considering the whole car as the system. So as said earlier, the friction with the road (as the driving force) and the air resistance (as the resistive force) are the only two forces to be considered for the dynamics of the whole car.

Try to adapt similar arguments to the case of an accelerating car on a horizontal road.

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Lets say a car changes speed from 10 m/s to 20m/s. That's an increase in kinetic energy of 60kj. Let's assume weight of car 1200kg, and that the acceleration takes 8 seconds. delta v / delta t, gives 1.25m/s^2 as acceleration. Using road formula i get 120 meters. 60kj/12 = 500N. That would be the force that actually accelerates the car. But this isn't general, because I had to make up a few facts. If I would have chosen a different time the force would have been much greater. But one more detail, now!. 1200kg*1.25 = 1500N. 1500-500 = 1000N. Is the friction 1000N i this case? –  Algific May 22 '12 at 7:10
    
What I meant was that 1500 is the total pull forward by the car and 1000N is the friction. –  Algific May 22 '12 at 7:24
    
the difference between the initial and final kinetic energy is the work made by the Friction force $ \int F_{friction}.dr =E_{final}-E_{initial} $ –  Jose Javier Garcia Sep 20 '12 at 18:06
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No, you cannot do that unless you account for change in internal energy (temperature changes, dissipation to atmosphere ...) Ref: Section 13 Chapter 3 on Fictional Work, Physics by Resnick Halliday & Krane

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Use this

$$ a= \frac{P_{\rm engine}}{m\,v} - \frac{F_{\rm resist}}{m} $$

and

$$ x = \int \frac{v}{a}\,{\rm d} v \\ t = \int \frac{1}{a}\,{\rm d} v $$

which applies for any speed $v$ and for whatever power $P_{\rm engine} = P(v)$ as a function of speed and for whatever resistance force $F_{\rm resist} = F(v)$ also a function of speed.

Example 1

Engine of constant torque, with $P_{\rm engine} = F_e v$ and constant resistance $F_{\rm resist} = F_d$

$$ a = \frac{F_e-F_d}{m} \\ x-x_1 = \frac{m (v^2-v_1^2)}{2 (F_e-F_d)} \\ t-t_1 = \frac{m (v-v_1)}{F_e-F_d} $$

Example 2

Engine of constant power $P_{\rm engine} = P$ with no resistance $F_{\rm resist} = 0$

$$ a = \frac{P}{m v} \\ x-x_1 = \frac{m}{P} \left(\frac{v^3}{3} - \frac{v_1^3}{3}\right) \\ t-t_1 = \frac{m}{P} \left(\frac{v^2}{2} - \frac{v_1^2}{2}\right) \\ $$

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