Energy required to accelerate itself

[greypilgrim]

Hi.

The energy required to accelerate an object of mass $m$ to a speed $v$ is $E_k=\frac{m}{2} v^2$. But how much energy does an object need to accelerate itself to $v$? Say, how much fuel (in energy units) does a car need to accelerate to $v$, if there are no air resistance or friction losses? Is it still $E_k=\frac{m}{2} v^2$? How do you calculate that, as the car's rest frame ist not inertial?

 

[PeroK]

One answer is, of course, that in the car's rest frame its speed by definition remains zero.

 

[greypilgrim]

What I mean of course is speed $v$ with respect to the initial rest frame before the acceleration, or the street's rest frame.

 

[Lukeblackhill]

What I mean of course is speed $v$ with respect to the initial rest frame before the acceleration, or the street's rest frame.

@greypilgrim, as our friend PeroK said, if you consider the car itself, You'll always be talking about a frame at rest. The car applies the same force (and by consequence, the same energy) than any other external object would, for the law doesn't consider sources, but the frame of reference. So, under a fix frame, let us say, the street, it doesn't matter whether the car is generating the velocity by internal processes, or another car collides with it, etc...the Ek=mv²/2 is the value for the kinetic energy.

 

[Ibix]

It does depend on the specific energy of the fuel and the engine efficiency. The mass of the car and remaining fuel after acceleration is lower than before the acceleration. Also there is friction to consider.

I don't think there's an easy answer for a realistic car, since engine efficiency and drag are non-trivial. The Tsiolkovsky rocket equation is the relevant thing for a rocket in vacuum.

 

[greypilgrim]

It does depend on the specific energy of the fuel and the engine efficiency. The mass of the car and remaining fuel after acceleration is lower than before the acceleration. Also there is friction to consider.

I guess the mass difference is negligible for a normal car, or we could even consider an electric car.

I'm asking this question because of the following situation (unrealistic numbers for easy calculation):
Assume two cars with $m=2\text{ kg}$ driving with $v_1=1\text{ m/s}$ with respect to the street. The first car accelerates to $v_2=2\text{ m/s}$, for which he needs the fuel equivalent of
$$\Delta E=\frac{m}{2}(v_2 ^2-v_1 ^2)=3\text{ J .}$$
In the rest frame of the second car, the first car accelerates from $v' _1=0$ to $v' _2=1\text{ m/s}$, for which he needs
$$\Delta E'=\frac{m}{2}((v' _2)^2-(v' _1)^2)=1\text{ J ,}$$
which indicates that it uses a different amount of fuel than in the street frame. But the fuel consumed (or fuel gauge, battery level,...) cannot be frame-dependent, can it?

 

[Ibix]

The car must have pushed against the road to accelerate. What happens to the energy of the road (say it has mass M) in the two frames?

 

[greypilgrim]

Well nothing as far as I can see, it stays obviously at rest in the first frame and at $-v_1=-1\text{ m/s}$ in the frame of the non-accelerated car.

 

[Ibix]

The road can't stay at rest if the car accelerates. That would violate conservation of momentum.

 

[PeroK]

Well nothing as far as I can see, it stays obviously at rest in the first frame and at $-v_1=-1\text{ m/s}$ in the frame of the non-accelerated car.

Alternatively, in a frame moving at $2m/s$ in the same direction, the car decelerates from $1m/s$ to rest. It could do that without the engine. Or, could it?

Change in Kinetic Energy is clearly frame dependent; whereas, fuel consumption is clearly not.

Can you see where the solution lies?

 

[jbriggs444]

Change in Kinetic Energy is clearly frame dependent; whereas, fuel consumption is clearly not.

Change in kinetic energy is frame invariant, once you count the kinetic energy of the road or reaction mass.

 

[Dale]

Hi.

The energy required to accelerate an object of mass $m$ to a speed $v$ is $E_k=\frac{m}{2} v^2$. But how much energy does an object need to accelerate itself to $v$? Say, how much fuel (in energy units) does a car need to accelerate to $v$, if there are no air resistance or friction losses? Is it still $E_k=\frac{m}{2} v^2$? How do you calculate that, as the car's rest frame ist not inertial?

I find that it is often easier to think in terms of power instead of work. $P=f \cdot v$. Let’s assume that the road is perfectly rigid and there is no slipping. As the car accelerates even though the center of mass of the car is moving, the velocity of the material (road and wheel) at the contact point is always $v=0$. So no power is transferred to the car from the road. So the increase in KE is equal to the decrease in internal energy.

Now, in a frame where the road is moving at some nonzero speed then the power from the road will also be nonzero. So the increase in the KE of the car will be equal to the decrease in internal energy plus the work done by the road.