Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Let's assume we have two identical electric trains. One has a big electric motor (high power) and the other has small motor (low power). Let us assume the electric motors are of the same brand and the power of the motor is directly proportional to the size. Now, the two trains used their max power to travel between to stations and the powerful train arrives quicker. So can we relate time (the saved time) to the size (mass) of the motor or if we neglect the mass variance of the motors, to the mass of additional coal used at the power station? If this so, we can now measure time in kg's or lb's of coal?

share|improve this question
add comment

2 Answers

In physics, the average power $P$ is defined as the amount of work done per unit time interval, i.e., $P = \frac{\Delta W}{\Delta t}$. So, a greater $P$ implies a smaller $\Delta t$ for the same amount of work. This answers the first part of your question. As regards whether you can measure time in terms of kgs of coal, you could. Assume that the power output of the train engine scales linearly with the mass of the coal. You could define 3 minutes as the time it takes for the train to burn up, say, 100 kgs of coal. Then, any other time interval can be measured in terms of the mass of coal. So, 6 minutes is the time required to burn up 200 kgs of coal. Whereas this may be useful to the driver of the train, it is not a very useful definition to the rest of us because it is not at all clear what the efficiency of the engine is, the mass of the train is, how he accelerates and decelerates, etc.

share|improve this answer
Just want to point out an inconsistency in your example: if it takes the train 3 minutes to burn up 100 kg of coal, it won't burn up 50 kg but 200 kg in 6 minutes. You might want to correct that. –  Wouter Feb 13 '13 at 15:03
Edited, thanks! –  contrariwise Feb 27 '13 at 21:50
add comment

The question you seem to be asking, relating time to mass of coal, is: "Why do we need more energy to do a job fast?"

Power is the energy input per unit time. If two jobs require the same amount of energy, the job that has a higher power input will finish the job in less time. However, ideally, both jobs use the same amount of energy or, said another way, burn the same amount of coal. The more powerful train burns coal faster, but for less time, while the less powerful train burns coal more slowly, but over a longer duration.

The reason that real trains behave this way (i.e., getting there faster takes more energy/coal) is because energy losses tend to increase with increasing rapidity. For the trains, air resistance is a primary source of this loss, since drag increases roughly as the fourth power of velocity. Therefore, if the more powerful train can go twice as fast as the less powerful train, it takes half as long, but requires 16 times more power to move at that speed. Over the whole trip, it will use 8 times as much energy (assuming air resistance is the only loss).

The phenomenon of increased energy cost with decreased time applies to other areas as well. Transferring energy from place to place rapidly has greater energy loss than transferring that energy slowly. Pressurizing a gas rapidly increases entropy more than pressurizing that gas very slowly (adiabatically).

The basic reason for this is that the faster something changes, the farther from equilibrium its components are perturbed. Changes which occur close to equilibrium incur smaller entropic loses than those which strongly perturb equilibrium. How strongly the system is perturbed from equilibrium depends on how rapidly the system can return to equilibrium, which depends on the system's temperature. (Warmer systems can more rapidly return to equilibrium than cooler ones.) The temperature dictates the timescales on which equilibrium is restored. If less time is allowed for a given change to take place, then the system is more strongly perturbed from equilibrium, and irreversible, entropic losses increase. The energy that goes into these entropic losses has to come from somewhere, and that somewhere is burning more coal.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.