We know the usual aerodynamic drag equation is given by: $$F_d = -bv^2$$ Where $b$ here is just some constants combination related to the property of the fluid and the material passing through.

My question is, how do we approximate the work done and hence the loss of energy caused by drag to some objects moving in the air? The motivation behind the question is I'm rather curious how much energy is required to move a car from rest to some particular speed with air resistance.

For example, knowing that the average weight of a car is 1 ton. If we wanted to accelerate the car from rest to say 100km/h in vacuum. The KE needed is just about 400 J, which (looking at fuel efficiency page in Wiki for say petrol at 34.8 MJ/L) will only require a minuscule amount of fuel like less than a thousandth of a liter, even if we have engine that has 0.1 efficiency. On the other hand we know that fuel mileage for ordinary car is a lot higher than that. Where all of the energy is gone?

Edit: The link suggested seems to show only the drag force and power at particular speed. I'm interested in knowing the total energy needed to overcome drag force starting from rest. (If I understand it wrong please correct me as well)


First of all, the work needed depends on the way in which the final velocity is reached. If you write the expression for the work done by the drag force you obtain

$$W=-\int F ds = -b \int v^3 dt$$

and it's easy to see that if the object is carried to its final speed in a very short time $W$ can be arbitrarily small because $v$ in the integrand is always less than the final value, and the integration time can be reduced at will.

As a concrete example, let us suppose that the object is accelerated by an engine which has a constant power $P$. We can write

$$\frac{d}{dt} \left(\frac{1}{2} m v^2 \right)=P-bv^2$$

which can be explicitly integrated:

$$\frac{m}{2} \int_0^{v^2} \frac{dv^2}{P-bv^2} = t $$

which gives

$$\log \left( 1-\frac{b v^2}{P} \right) = -\frac{2bt}{m} $$


$$v^2 = \frac{P}{b}\left( 1-e^{-\frac{2bt}{m}} \right) $$

The maximum speed that can be reached is $\sqrt{P/b}$. Now, let us suppose for simplicity that $P$ is large compared with the power of the drag force. In this case we can expand the exponential at the first order obtaining

$$v^2 \simeq \frac{2Pt}{m} $$

The work done by the drag force can now be evaluated using the integral in the first equation,

$$W=-b\int \left( \frac{2Pt}{m} \right)^{3/2} dt =-\frac{4}{5}\sqrt{2} b t \left(\frac{P t}{m} \right)^{3/2}$$

which can be expressed as a function of the speed

$$W = -\frac{b m}{5 P} v^5$$

As expected from the general initial discussion, $W$ can be reduced by increasing the engine power $P$.


Drag force: $$\mathbf{F_{\text{drag}}} = \dfrac{1}{2}\cdot \text{density} \cdot \text{cross sectional area} \cdot \text{drag coefficient} \cdot \text{velocity}^2$$. Now, $$\text{energy lost due to drag} = \mathbf{F_{\text{drag}}} \cdot \text{displacement}$$. It might give the exact or quite straight approximation. :)

  • $\begingroup$ But if the car is accelerating, will the velocity change as well and not be constant? $\endgroup$ – vonTeslacx Apr 18 '15 at 11:29
  • $\begingroup$ that depends on the net unbalanced force acting on the body $\endgroup$ – Abhijeet Apr 18 '15 at 12:13
  • $\begingroup$ Well if it's accelerating, the net force is clearly not 0 right..? $\endgroup$ – vonTeslacx Apr 18 '15 at 12:14
  • $\begingroup$ of course f=ma :) $\endgroup$ – Abhijeet Apr 18 '15 at 12:15

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