Assume a truck of mass 1 ton braking from 60mph to 0 on a road surface.

From the reference frame of the road, the following energy is transferred to the brakes: 1 ton * 60mph^2 / 2 = 1800 ton mph.

From the reference frame of the truck, the following energy is transferred to the brakes: 1 ton * 60mph^2 / 2 = 1800 ton mph. The curve is different but that doesn't seem to matter.

But from the reference frame traveling at 30mph we get something different. Here we must consider the braking in 2 segments due to a sign problem: 1 ton * 30mph^2 / 2 + 1 ton * (-30mph)^ 2 / 2 = 900 ton mph.

Back to the first case, if we consider it in two segments we get this: 1 ton * 60mph^2 / 2 - 1 ton * 30mph^2 / 2 + 1 ton * 30mph^2 / 2 = 1 ton * 60mph^2 / 2 = 1800 ton mph.

Yes, I know ton mph is a strange energy unit. The problem is easier to understand without any unit conversions at all in it.

The problem seems to have something to do with picking unnatural reference frames, but defining natural reference frame resists. Normally I am accustomed to solving this problem in momentum only and always get the right result, but that doesn't apply right to friction heating. The explanation of kinetic energy transfer into the fuel doesn't work in this specific scenario.

The term for what I am looking for is the frame invariant calculation. The kinetic energy should indeed be different, but the energy added to the brakes should always be the same.


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The energy transferred to the break is given by the friction force between brakes and tire (or rather the part of the brake attached to the tire), times the relative velocity of tire and breaks, integrated over the total time the break is acting (this is, it is force times distance, as usual).

Clearly, the relative velocity of tires and breaks is the same in all reference frames, and so is the force (which is related to the acceleration of the truck. Thus, the total work transferred to the breaks is the same independent of the reference frame.

Note that energy is conserved in all reference frames if you also take into account the energy of the ground. In particular, if you take the mass of the ground to infinity (which means it is a reference frame), you will find that the change in total kinetic energy is exactly $-mv^2/2$, the original kinetic energy of the truck relative to the ground, which has been converted into heat in the brakes. (If the mass $M$ of the ground is finite, the change in energy is reduced by a factor $M/(m+M)$, which relates to the fact that in that case the change in velocity of the truck is less than its original velocity relative to the ground.)

  • $\begingroup$ If you consider the mass of the ground to be infinite, you can transfer momentum without transferring energy - since $E = \frac12 mv^2 = \frac12 \frac{p^2}{m}$. You seem to be implying the opposite... but I can't quite decipher what you say. $\endgroup$ – Floris Jan 3 '17 at 21:18
  • $\begingroup$ @Floris It's just the same as user2309840's calculation below.: The total change in kinetic energy is the kinetic energy of the truck, independent of the reference frame. (Not a miracle, unsurprisingly.) $\endgroup$ – Norbert Schuch Jan 3 '17 at 22:10
  • $\begingroup$ @Floris On a second thought, what you seem to imply is not right: In the reference frame moving with $v/2$ (with $v$ the speed of the truck w.r.t. the ground), the ground is loosing $mv^2/2$ kinetic energy (which goes into the breaks) if its mass is infinity: Before breaking, its energy is $P^2/2M$ ($P=Mv/2$), and afterwards, $(P-p)^2/2M$ (with $p=mv$, and for $M\gg m$) -- which equals $P^2/2M-Pp/M+O(1/M)$, and the change in energy is $Pp/M\rightarrow pv/2=mv^2/2$. -- And yes, the energy transferred depends on the reference frame (and there is one where it is zero), while the momentum doesn't. $\endgroup$ – Norbert Schuch Jan 4 '17 at 1:34
  • $\begingroup$ Huh - you're right! I hadn't thought of it that way. Thanks. $\endgroup$ – Floris Jan 4 '17 at 4:09
  • $\begingroup$ @Floris Basically, the point is that the leading term in the kinetic energy is $\propto Pp$, so if and only if $P=0$, there is no change in kinetic energy. $\endgroup$ – Norbert Schuch Jan 4 '17 at 16:00

The question seems to be a special case of the following question: A particle of mass $M$ traveling at speed $v$ and a second particle of mass $m$ traveling at speed $v+V$ undergo a totally inelastic collision. Calculate the change in kinetic energy. Show that it is independent of $v$. (In the special case of the question above, $M$ is the mass of the Earth and $m$ the mass of the truck. The velocity $v$ specifies a particular choice of reference frame.)

Conservation of momentum asserts that the final velocity $v_f$ is given by $$ Mv + m(v+V) = (M+m) v_f $$ or equivalently $$ v_f = \frac{M v + m(v+V)}{M+m}. $$ The total change in kinetic energy is then $$ \frac{1}{2} (M+m) v_f^2 - \frac{1}{2} M v^2 - \frac{1}{2} m (v+V)^2= - \frac{ m M V^2}{2 (M+m)} $$ which is independent of the velocity $v$. In the limit $M \to \infty$, the change in energy simply reduces to $-\frac{1}{2} m V^2$.

In the statement of the problem, the issue with the calculation in the reference frame with $v = 30$ mph is that it does not take into account the change in kinetic energy of the Earth. Also, as has been stated in the comments, the change in kinetic energy of the truck is actually zero in this case.

  • $\begingroup$ I think you have a mistake in the second line. Why are you saying $mV=Mv$? If you fix that it will still work out in the end. $\endgroup$ – octonion Jan 2 '17 at 2:55
  • $\begingroup$ @octonion fixed $\endgroup$ – user2309840 Jan 2 '17 at 3:03

In Newtonian mechanics the (apparent) kinetic energy of a body depends on the reference frame used to measure it, as you have noticed. That paradox is resolved by relativistic mechanics.

In Newton's time there was no clear concept of "energy" at all - at best there was a confused notion about the "quantity of motion" of an object, which didn't distinguish clearly between momentum and kinetic energy. Newton's important contribution was to clearly define momentum and note that it is conserved. Newton never used the concept of "energy" at all in his publications.

The principle of "conservation of energy" can only be used in a straightforward way in Newtonian mechanics in the limited sense of "conservation of mechanical energy". This clearly does not apply to your scenario, because mechanical energy is being converted into heat in the truck brakes - and that concept was not clearly understood by physicists until 200 years after Newton!

In fact some of the numerical examples in your question are misleading, because you ignored the sign of the kinetic energy change. In a frame fixed to the road, the KE of the truck decreases by 1800 ton mph, but in an inertial frame moving at the initial speed of the truck (note: that is not the same as the non-inertial frame "fixed to the decelerating truck" in your question) the KE of the truck increases from 0 to 1800. In the third example, the KE of the truck changes from +450 to 0 and back to +450, so the total change is zero, not 900.

You can get consistent answers in any inertial reference frame by considering the mechanical work done by the truck brakes, since all of that work is converted into heat. You can measure that work as $$(\text{friction force in the brake}) \times (\text{relative motion between the parts of the brake}),$$ and the relative motion is the same in any reference frame - it depends only on the total angle through which the wheels turn.

  • $\begingroup$ I don't have any non-inertial frames in the question. The "truck" frame is inertial with the truck's initial condition. $\endgroup$ – Joshua Jan 2 '17 at 0:16
  • 5
    $\begingroup$ You could add that if you consider the change in kinetic energy of the earth, you get the same kinetic energy decrease in each frame. $\endgroup$ – Brian Moths Jan 2 '17 at 0:18
  • $\begingroup$ @Joshua I interpreted "the reference frame of the truck" to mean "the reference frame fixed to the truck at all times." But in my "day job" I hardly ever work in inertial reference frames anyway, so that might affect how I interpret the words! $\endgroup$ – alephzero Jan 2 '17 at 2:31

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