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Today my friend told me something that blew my mind completely.

He said:

The energy necessary to stop a train is equal to the energy in a pack of cookies.

How is that possible? Is he right? I'm done understanding energy if he's right...

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    $\begingroup$ If it seems hard to understand, consider how far a trained cyclist can ride after only eating a pack of cookies... (many, many kilometers) now pack that small constant expenditure of energy over a long period of time into a single short expenditure of energy on a larger object. $\endgroup$ – Michael Sep 16 '15 at 17:23
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    $\begingroup$ It's probably important to distinguish the difference between caloric energy from digestion (not the most efficient process), and the energy of chemical/nuclear bonds (as efficient as you can get). If you split all the chemical bonds in those cookies, or better yet split the atoms themselves, I'm willing to bet the effect would be far more devastating than you'd expect. However when you digest something, you'll rarely explode (I hope). $\endgroup$ – thanby Sep 17 '15 at 6:28
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    $\begingroup$ How certain are you that your friend was talking about chemical energy? $\endgroup$ – R.. Sep 17 '15 at 8:27
  • $\begingroup$ This question is wrong, because it implies that stopping a train requires different amount of energy than making it move. Real world train has energy stored in compressed air in brake cylinders so stopping it requires merely releasing it - that's not the point and it only distracts from the point. $\endgroup$ – Agent_L Sep 17 '15 at 10:01
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    $\begingroup$ @Agent_L: How does any of that make the question "wrong"? You're bringing up something the OP didn't even mention then claiming "that's not the point and it only distracts from the point" ... well, yes, and you raised it not the OP! $\endgroup$ – Lightness Races in Orbit Sep 17 '15 at 11:46
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In the UK a packet of biscuits would typically be 200g and contain about a thousand Calories or 4.2MJ. By contain I mean that if the biscuits were burned in oxygen the energy released would be about 4.2MJ.

If a train has a mass $m$ and is moving at a speed $v$ then its kinetic energy is:

$$ T = \tfrac{1}{2}mv^2 $$

Equating this with the energy in the biscuits we find:

$$ v = \sqrt{\frac{8.4 \times 10^6 \text{J}}{m}} $$

Googling suggests the weight of a train would be 100 to 1,000 tonnes depending on the type of train. Using the lower figure we get $v \approx 9$ m/sec while the higher weight gives $v \approx 3$ m/sec.

So the two energies are actually comparable (if it's a slow train :-).

But it's important to be clear what we mean when comparing the energies. What we mean is that if we put a packet of biscuits into the burner of a 100% efficient steam train then the energy released as the biscuits were burned would accelerate the train from standstill to the velocity calculated above.

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    $\begingroup$ In reality steam trains are actually pretty inefficient beasts. But given that the latent heat of water steam is about 2MJ/kg, we're talking about boiling around 2kg of water here, which when fully expanded would be about 3 cubic metres. Full analysis is more complicated and would require specifications of the steam conditions prior to expansion, but that's certainly enough to move a (frictionless) train. Giving the biscuits to a horse or a very strong man and having him pull the train is probably slightly more efficient. $\endgroup$ – Level River St Sep 16 '15 at 12:11
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    $\begingroup$ Steam engines have single digit percent efficiency efficiency ratings. $\endgroup$ – Joshua Sep 16 '15 at 15:16
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    $\begingroup$ @steveverrill: It would be even more efficient to use a very strong and stupid man, and keep the biscuits just out of his reach. $\endgroup$ – Beta Sep 16 '15 at 15:17
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    $\begingroup$ @Joshua: Some steam engines were pretty efficient, but steam locomotives were optimized for power per unit size rather than efficiency. The earliest steam engines got much (or in some cases all) of their power from condensing steam (so as to reduce its pressure below atmospheric), but one can't get more than an atmosphere of pressure differential doing that, and condensing a lot of steam quickly requires a very large apparatus. Locomotives vent steam rather than condense it, which means they forego a lot of the energy contained therein, but one can easily vent a huge amount of steam... $\endgroup$ – supercat Sep 16 '15 at 15:35
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    $\begingroup$ What do you mean "100% efficient steam train?" Often, when we talk about the efficiency of an heat engine, we are talking about its work output relative to a Carnot-cycle engine. That is not the same as the total amount of heat released by burning the fuel. $\endgroup$ – Solomon Slow Sep 16 '15 at 16:00
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I believe he is wrong, but not about the number.

From @JohnRennie answer we see the energy of the cookies is roughly equal to the train's kinetic energy, so we need such amount of energy to accelerate the train to this speed. However,

Energy to stop train...

Energy , or work required to stop a train generally doesn't equal to the train's kinetic energy $E_k$, and can be extremely little. Edit: as @kojiro and @Asher pointed out in the comments, according to Work-Energy Theorem, the work done on the train equals $-E_k$. But I think it doesn't invalidate the reasoning: the ground lose no energy when doing work on the train.

For example, we can push an obstacle into its way: ( Overlooking from the sky )

                           (rock)
[  train  ][  train  ]>
                         | (rock)
                         |
                         |
              push this thing to the north

Theoretically we need zero energy if the ground is friction-less. All the train's kinetic energy becomes heat.

More practically we just brake the train. Braking a fast bicycle doesn't need so much energy from the rider's hands, right?

In conclusion, we cannot calculate the "energy" needed from the train's mass and speed.

What we can estimate is the impulse needed to stop the train. The train's momentum decreases from $mv$ to $0$, so it must be given impulse $J = mv$ in the reverse direction.

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    $\begingroup$ If the ground is frictionless, how do you stop a train? $\endgroup$ – kojiro Sep 16 '15 at 11:40
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    $\begingroup$ @kojiro e.g. the ground under the bar is frictionless, but the ground under the train and the rock has friction. This is just an theoretical example showing that the energy needed is not the train's kinetic energy. $\endgroup$ – jingyu9575 Sep 16 '15 at 12:08
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    $\begingroup$ @jingyu9575 Kojiro's point is that the total energy to stop the train doesn't change. If the train stops due to friction, then the ground has done work on it, which requires energy equal to the initial kinetic energy of the train, regardless of source $\endgroup$ – Asher Sep 16 '15 at 15:08
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    $\begingroup$ @jingyu9575 both the ground and the train will heat up, scratches and cracks will appear... The energy still goes somewhere. $\endgroup$ – Davidmh Sep 16 '15 at 20:28
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    $\begingroup$ @kojiro If the ground is frictionless, how did you start the train? ;) $\endgroup$ – jpmc26 Sep 17 '15 at 2:55
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The energy needed to stop a train is the energy needed to open the air brake valve and let the air out of the air brake system (at least with US trains). It's hard to guestimate the amount of energy required to do this, but I'd guess to turn even a moderately stiff lever would require significantly less than one kilogram-meter == 9.8 joules.

The kinetic energy of the train, of course, is converted to heat by the brake shoes rubbing against the wheels. That doesn't factor into the equation.

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  • $\begingroup$ OK, I'm a bit confused. Is not the Westinghouse-style air brake system (or the conceptually equivalent vacuum brake system) pretty much standard for trains worldwide? $\endgroup$ – Hot Licks Sep 16 '15 at 23:18
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    $\begingroup$ I didn't downvote you, but this seems pretty clearly to miss the point of the question (maybe on purpose). The OP is asking about the kinetic energy of the entire train, not how much work needs to be applied to the controls to trigger braking. $\endgroup$ – zwol Sep 17 '15 at 1:10
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    $\begingroup$ @zwol - The question says "the energy necessary to stop a train". I don't see how one can read that to mean the entire kinetic energy of the train. $\endgroup$ – Hot Licks Sep 17 '15 at 1:14
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    $\begingroup$ You have to read it that way for the comparison to be meaningful and interesting. $\endgroup$ – zwol Sep 17 '15 at 14:08
  • $\begingroup$ @zwol - I think it's more interesting when you consider this as sort of fluttering about on the edge of the "butterfly effect" -- a very small amount of energy can control what happens with a much larger amount of energy. $\endgroup$ – Hot Licks Sep 17 '15 at 20:26
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In his excellent answer John Rennie gives the numbers. If this sounds incredible, A more intuitive approach that indicates that's roughly in the correct ballpark is to take a look at strongman pulling contests.

https://www.youtube.com/watch?v=hP00VmKx_No shows you a dude pulling a 150 tonne train. He's not going particularly fast, but still, he's moving at, maybe, say 0.5 m/s? How many cookies should he eat more than if he didn't move that train?

Obviously, that's impossible to say, but it's clearly nothing in the league of 10 packs. It's probably more than a single pack. Looking at it this way gives you some intuitive feeling of how much energy there is in a pack of cookies compared to the kinetic energy of a moving train.

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  • $\begingroup$ But keep in mind that kinetic energy is proportional to the square of velocity. Most people would consider a "moving train" to be going at least 30 mph/50 kph. If one is allowed to assume a vanishingly small velocity then the energy is likewise vanishingly small. $\endgroup$ – Hot Licks Sep 17 '15 at 20:31
  • $\begingroup$ @HotLicks True. But I'm only looking at an order of 10 estimation. This answer doesn't try to quantify how much energy there is in a pack of cookies or in a moving train, but just that it's plausible the numbers are in the same ballpark. $\endgroup$ – Martijn Sep 18 '15 at 7:43
  • $\begingroup$ Given those criteria I'd assert that the question is essentially "not a question" (as interpreted by you and others), since it's a tautology so long as the energy levels exceed those where quantum mechanics would rule. $\endgroup$ – Hot Licks Sep 18 '15 at 12:46

protected by Qmechanic Sep 16 '15 at 11:04

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