Imagine a moving object at constant speed (like a car). This object is, then, accelerated for a brief moment. In different reference frames (at rest and moving along with the object), the variation of the car's kinetic energy is not the same.

My question is the following: Suppose I have a battery that's holding a certain amount of energy. Connecting it to the motor, I discharge it completely to power up the wheels and increase the car's speed. If I used the battery's energy to increase the car's, how can I explain the kinetic energy difference between the frames now? I exchanged the same amount of the battery's energy in both reference frames (or not?, this is the important part), so why did the kinetic energy of the car increase comparatively more in the rest frame?


1 Answer 1


A closed system can not speed itself up, that's the momentum conservation law which is also the key to your problem.

As far as I can see you are implicitly supposing the following three equalities to hold $$E_i+A=E_f\\p_i=p_f\\m_i=m_f$$ where subscripts $i$ and $f$ stays for the initial (before acceleration) for the final (after acceleration) states respectively and $A$ is the energy stored in your battery. However, all three can not exist simultaneously: if $p_i=p_f$ and $m_i=m_f$ then from $E=\frac{p^2}{2m}$ one obtains $E_i=E_f$.

Thus, in order to increase the energy of the system you need to relax one of these conditions. Let us suppose that $m_i=m_f$ but $p_i\neq p_f$, i.e. the momentum is not conserved. For example, our 'car' interacts with the road via friction force $F_{fr}$. Suppose that the speed of the car was increased with the constant acceleration $a$ from the initial value $v$ to the final value $v+\Delta v$. The work done by the friction force on the car is $$W_{fr}=F_{fr}\int_0^{\Delta v/a}v(t)dt=F_{fr}\int_0^{\Delta v/a}(v+at)dt=F_{fr}\frac{v\Delta v+\Delta v^2}{a}$$

Since the acceleration is caused only by that friction force we also have $ma=F_{fr}$ and therefore

$$W_{fr}=m(\Delta v^2+v\Delta v)$$

Now consider the reference frame moving with a constant speed $v$. In this reference frame the car have traveled less, and less is the work done by the friction force.

$$W'_{fr}=F_{fr}\int_0^{\Delta v/a}v'(t)dt=F_{fr}\int_0^{\Delta v/a}(at)dt=F_{fr}\frac{\Delta v^2}{a}=\frac{m\Delta v^2}{2}$$

You can see that the difference between these two works $W_{fr}-W'_{fr}=mv\Delta v$ is exactly the difference in variations of kinetic energy calculated in these two frames.

Similarly, one can keep condition $p_i=p_f$ but relax condition $m_i=m_f$ thus considering the case of a jet engine. If done accurately, calculations in this case also yield perfect conservation of energy.

  • $\begingroup$ Then, to solve the problem of the car running off batteries, I need to take into account other factors, such as energy dissipation? Thinking about it another way: if I took the road into account (acceleration of the floor), could the calculations work out? $\endgroup$ Commented Sep 24, 2014 at 22:37
  • $\begingroup$ @André Pereira I'm not sure I got your question right but still I'll try to answer. The work done by the friction force actually goes to the Earth kinetic energy accelerating it a little bit :) The amount of kinetic energy that Earth will acquire depends on the reference frame. Surely this difference exactly matches the difference for the kinetic energy of the car. $\endgroup$ Commented Sep 25, 2014 at 0:29

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