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For example consider that I am driving a car with a mass of $200$ kg moving at $10$ m/s relative to an outside observer, with an assumed constant opposing friction of $50$ N, for simplicity (it won't make much of a difference in my main point). I then press the pedal to exert a force of $100$ N for 2 seconds. After using $F=ma$ and $s = ut+(1/2)at^2$, the distance traveled within these $2$ seconds can be calculated as $20.5$ m. So the work done would be:

$$W = F d$$

$$W = 100 \times20.5 = 2050 J$$

If I applied the same force for the same time if the car was not moving at first, it would result in a much smaller value for work done because the distance covered would be less. To be exact it would be:

$$W = 100 \times 0.5 = 50 J$$

In both cases I'm pressing the pedal the same way, for the same duration, so the amount of chemical energy converted from fuel is the same. But then how is there such a big difference in the change in kinetic energy depending on the speed? The first result is 41 times the second result. That would mean that if the kinetic energy calculated were equal to the chemical energy ignited from the fuel, going from 10m/s to 10.5m/s would require 41 times more fuel than going from 0 to 0.5m/s. This is considering there is no energy loss, but even if there were, the energy loss wouldn't even be close to the same degree as calculated, and the values would still be proportional. There is no increase in potential energy as well. But the law of conservation of energy has to be true. To elaborate further, if W =Fd were converted to a function of force and time it would become:

$$W = Fut + F^2t^2/2m$$

Where u is initial velocity, which is frame dependent. This shows that according to the equation, W will change with the initial velocity and the frame of reference, is a quadratic function of force and time, and strangely, inversely related to the mass. The contradiction here is that this implies the same amount of fuel will appear to have different energies depending on the frame

I would also like to point out that the equation for kinetic energy is derived from the equation for work done. So then does the equation for work done and kinetic energy have problems/limitations? Should work done be rather represented as a function of force and time and modified further?

Update: This might be like answering my own question here, but in the equation for W that I have derived, only the first half is velocity and frame dependent. If that part is removed then W becomes independent of frame, so one of the contradictions goes away.

So it would be:

$$W = F^2t^2/2m $$

This equation reduces to W=Fd at 0 initial velocity and if you derive kinetic energy from this formula, it still gives K.E. = 1/2mv^2, which probably means that the equation for K.E. is correct. Is this an appropriate solution?

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  • $\begingroup$ Please be mindful that every edit to a post bumps it to the top in the queue of active posts. If you see yourself revising a post over a period of time, try to collect several minor edits you want to make and make them in one larger substantial edit instead of individual small edits. $\endgroup$
    – ACuriousMind
    Commented Dec 22, 2020 at 18:32

3 Answers 3

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You have hit upon the apparent paradox of the Oberth effect, and it is discussed in some detail here and here.

Tl;dr: No, the formula $W = F\cdot s$ is correct for a constant force.


What this really amounts to is the fact that kinetic energy is a non-linear function of the velocity. Let us look at your problem from the viewpoint of energy.

If the car initially has velocity $v_0 = 10$m/s, and you push with a net force of $50$N for $2$s, it's final velocity is $v = v_0+at = 10.5$m/s.

If the initial velocity is zero then $v_0 = 0$m/s, and the final velocity is $0.5$m/s.

So the change in velocity is the same in both scenarios, but the change in kinetic energy is hugely different!

In the first scenario, the gain in kinetic energy is: $$ \frac{1}{2}m(v^2-v_0^2) = 100\cdot(10.5^2-10^2 ) = 1025\ \mathrm{J} $$ In the second scenario however: $$ \frac{1}{2}m(v^2-v_0^2) = 100(0.5^2) = 25\ \mathrm{J} $$

Which is perfectly consistent with your calculation (half the work you do goes to friction, the other half to increasing the kinetic energy).

So why is the kinetic energy a quadratic function of velocity? This may at first seem counter-intuitive (see the "paradox" section of the Wikipedia article on the Oberth effect). But the crucial thing is that energy is relative to frames of reference. This is true for potential energy (what your potential energy is depends on what you define as zero, at least in non-relativistic physics). But it is equally true for kinetic energy: the kinetic energy depends on what frame you choose, e.g. which velocity you are moving. This is consistent, because energy is not absolute, and you can only measure energy differences. Thus, you have to be careful which frame you work in (the same is true for gravitational energy).

So it is consistent with frame invariance to have a kinetic energy which is a quadratic function of velocity. Is it intuitive? Well, that depends on your intuition.


One final comment: real-life spoils this a little bit, because of the difference between static and dynamic friction. If you start an object from zero velocity with respect to the ground, you first have to overcome static friction, which tends to be larger than dynamic friction. Thus, it will be somewhat harder to get an object moving initially than you may expect.

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  • $\begingroup$ Thanks for the answer, it makes a lot of sense. Also I am aware of the friction effect, I assumed a constant friction for simplicity. I have one problem though, isn’t the equation for kinetic energy derived from W = Fd? So problems in the first can lead to problems in the other? I haven’t done the exact calculations yet, but it seems like energy can seem to be conserved even if there is a problem. But only considering there is a problem, otherwise its ok. $\endgroup$
    – Neelim
    Commented Dec 20, 2020 at 14:38
  • $\begingroup$ I'm not entirely sure I understand what you mean, but whether you consider the definition of work or of energy more fundamental is in some sense a matter of taste. You are free to modify the definition by including the relative velocity, but you should not expect that this will lead to all the nice results associated with energy. Ultimately, we find that $(1/2)mv^2$ is a useful quantity for a host of reasons, so it is what we define to be the kinetic energy. $\endgroup$
    – G.Lang
    Commented Dec 20, 2020 at 16:28
  • $\begingroup$ I am wondering if the equation for W and therefore also kinetic energy accurately represents the energy content in physical reality. E.g. if the car moved using fuel, the same amount of chemical energy would create such a big difference in kinetic energy. Unlike the answer you linked to, there is no change in potential energy to account for this difference. Then doesn’t that imply that there could be a problem in the definitions? If the calculations are self contained, the difference will cancel out. But if you were to convert one form of energy to another, that could create problems $\endgroup$
    – Neelim
    Commented Dec 20, 2020 at 16:49
  • $\begingroup$ The definition is flexible, but the problem is I could arbitrarily write kinetic energy = (1/2)mv^3 for example, but that won’t necessarily represent the energy content accurately in physical reality. It seems to me like W=Fd only applies to conservative forces, but not all forces are conservative. So I’m wondering if this is not the correct universal equation for W and K.E. and if not, what would be the correct equations $\endgroup$
    – Neelim
    Commented Dec 20, 2020 at 16:54
  • $\begingroup$ In classical physics, $W = Fd$ (for constant F) is the correct equation. Whether or not the force is conservative only makes a difference to the total energy balance (hence why the kinetic energy increase above is not equal to the work done). An alternative way to view this is in terms of power, $P = F \cdot v$. From this expression you can see that how much energy you get into the system per unit time depends on the velocity. That's the idea behind the Oberth effect. $\endgroup$
    – G.Lang
    Commented Dec 20, 2020 at 19:50
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So doesn't the equation need to account for initial velocity as well?

No, the difference that you posted is in fact correct. We want the same force to give different values for work in the different cases.

The purpose of work is to quantify a transfer of energy from one system to another. Recall that kinetic energy is $KE=\frac{1}{2}mv^2$. So if you start from a higher initial velocity then the change in KE is larger.

Since the change in KE is larger it is necessary that the work also be larger. The issue that you found is therefore not an accident nor a mistake. It is a necessary feature of work for it to serve as a measure of the transfer of energy, which is the whole reason we are interested in the quantity.

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  • $\begingroup$ Thanks for the answer, there is one problem though, isn’t the equation for kinetic energy derived from the equation for work? It seems to me like W=Fd only works for conservative forces. If that is the case, how can it be applied to all forces? $\endgroup$
    – Neelim
    Commented Dec 20, 2020 at 14:42
  • $\begingroup$ @Neelim W=F.d is valid for non-conservative forces also. The equation for KE is properly derived from the time symmetry of the Lagrangian using Noether’s theorem. Obviously, this is an advanced topic so it is not presented that way to beginning students. $\endgroup$
    – Dale
    Commented Dec 20, 2020 at 17:16
  • $\begingroup$ does time symmetry here mean that the order of events do not matter? in that case it would not actually be true. If you apply a force on an object for some duration, and then apply an equal force in the opposite direction for the same duration, there will be displacement in the direction of the first force even though the net force over time is 0. So order of events can change the outcome even if there is symmetry in the magnitude, direction and duration of the force. I am not sure if this is what you meant by time symmetry, but I wanted to point this out in case its relevant $\endgroup$
    – Neelim
    Commented Dec 20, 2020 at 17:25
  • $\begingroup$ @Neelim no, it is time translation symmetry. This means that the laws of physics are the same today as they were yesterday. As I mentioned, this is an advanced topic but the bottom line is that KE = 1/2 mv^2 is well founded. And because it is, the formula for work must have the feature that you identified. It is not a mistake, it is necessary. $\endgroup$
    – Dale
    Commented Dec 20, 2020 at 17:46
  • $\begingroup$ en.wikipedia.org/wiki/… Even if the laws don't change with time, it wouldn't matter if what is considered the law itself is wrong. Not saying that's necessarily the case here. Also we don't even know all the laws of physics yet and if there are an infinite number of laws, it won't even be possible. $\endgroup$
    – Neelim
    Commented Dec 20, 2020 at 18:45
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Several answers have simply been affirming that $K := \frac{1}{2}mv^2$ is the appropriate definition and alluded to the Oberth effect (which this isn't quite), but I don't think they've addressed the crux of your issue, namely the apparent violation of energy conservation.

The issue you've laid out is that, since you have pressed the pedal at the same level for the same time in both scenarios of differing initial velocity, the engine is exerting the same uniform force, and expending the same chemical energy in doing so, yet the kinetic energy increases by much more when the initial velocity is larger. Where did the additional energy come from in the latter scenario, then? Could it be that our formulation of kinetic and chemical energies isn't describing all of the energy present?

As in most cases where one is driven to ask "is our understanding of physics and energy utterly wrong?" when reviewing simple classical systems, the answer is, of course, "No". The mistake in your conception of the problem is that the engine does not expend the same chemical energy per unit time simply because it is imparting the same force. To see why, we simply need to consider the basic picture of how an internal combustion engine works:

enter image description here

Upon each rotation of the crankshaft, the piston in the engine moves up, compressing fuel and air before it is ignited to push the piston back down, accelerating the crankshaft through another turn. Clearly, the expenditure of chemical energy occurs during the ignition of the fuel-air mixture. The expenditure of chemical energy per unit time while maintaining a constant torque on the crankshaft (or constant translational force on the car from friction with the ground), then, is proportional to the ignitions per second, which is exactly the rotations per second of the crankshaft. But the rotational rate of the crankshaft is proportional (depending on the gear) to the speed of the car-- the faster the car is going relative the ground, the faster its wheels turn, and the faster the crankshaft turns. Putting these together, the chemical energy expended per unit time (while maintaining a constant force) is proportional to the speed of the car.

Lo and behold, the kinetic energy per unit time gained by the car according to $K = \frac{1}{2} mv^2$ is exactly

$$\frac{d}{dt} K = m\vec{v} \cdot \frac{d \vec{v}}{dt} = m\vec{a} \cdot \vec{v} = \vec{F}_\text{net} \cdot \vec{v},$$

or the power gained in kinetic energy is $P = Fv$ for one-dimensional motion, also proportional to the speed of the car when the force is fixed.

Indeed, notice that the average speeds in your two scenarios are $0.25$ m/s and $10.25$ m/s, which have a ratio of $41$, so the chemical energies expended should also have a ratio of $41$, as you've found the gained kinetic energies do.

(Image Credit)

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  • $\begingroup$ en.wikipedia.org/wiki/… According to this article the optimum speed would be close to 70 km/h, compared to 36 km/h here, but that would result in a decrease in fuel efficiency of less than 40%, certainly not close to the 41 fold margin according to the equations. $\endgroup$
    – Neelim
    Commented Dec 21, 2020 at 0:21
  • $\begingroup$ so the fuel consumption would actually decrease at higher speeds, not increase, which contradicts with your argument $\endgroup$
    – Neelim
    Commented Dec 21, 2020 at 0:27
  • $\begingroup$ @Neelim Notice you're using different notions of "fuel consumption". Fuel efficiency as discussed in that article is fuel (or chemical energy) expended per unit of distance-- my argument has been about fuel expended per unit of time (relevant to your scenario because you're comparing equal time segments). The speed minimizing fuel/distance does not minimize fuel/time. Indeed, fuel/distance is (fuel/time)/speed. My argument indicates Fuel/time $\propto$ speed in a fixed gear and force provided by the engine, so that these parameters determine the fuel/distance fuel economy. $\endgroup$
    – jawheele
    Commented Dec 21, 2020 at 2:10
  • $\begingroup$ @Neelim Of course, both my analysis and your posed problem assumed speed-independent friction, so they aren't good models at all for understanding fuel economy. $\endgroup$
    – jawheele
    Commented Dec 21, 2020 at 2:11
  • $\begingroup$ This is a valid point, I was thinking the of this possibility. In this case the fuel consumed would be almost proportional to the distance. But this may be the result of friction, which wouldn't give the accurate picture like you said. That is why I assumed a constant value for friction, but still that would not be accurate. But it still holds that without friction, there would be no force needed to maintain any velocity, and also acceleration i.e. increase in velocity is proportional to force $\endgroup$
    – Neelim
    Commented Dec 21, 2020 at 2:14

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