Two scenarios...different energy requirement?by maximiliano Tags: energy, requirement, scenariosdifferent 

#1
Apr1112, 07:31 PM

P: 43

Been wondering about this for a while.
Two absolutely identical (in every way) vehicles are traveling 100 miles, at a perfectly constant speed of 75mph, in a vacuum. The starting elevation is exactly 1,000' above sealevel. The finishing point is exactly 1,000' above sealevel. But.... Vehicle 1: encounters hills along the way. Vehicle 2: travels on perfectly flat terrain I'm curious how Mr. Newton's laws apply to these two cars. It seems the energy required would be identical, since one has to do work to climb the hill (resisting the acceleration of gravity), but will catch the "tailwind" of stored potential energy in the form of gravity on the other side of the hill. Thus, making everything exactly equal (in a vacuum anyway). Thoughts?? 



#2
Apr1112, 07:41 PM

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Assuming that none of the hills are too tall for the car to coast over them, then in both cases zero energy is required, assuming no losses. Note the car going over the hills will slow down when going uphill and speed up when going downhill, but it's speed will be back to 75 mph once back at 1000 feet altitude. The car going over the hills will take more time, since it's traveling along a greater distance.




#3
Apr1112, 07:42 PM

P: 276

All the energy you put in pushing the car up the hill will not be lost. At the end destination the car rolling of the hills will move at a greater velocity thus will have greater kinetic energy than the car you didn't had to push so hard.
It all could be written in vie the WorkEnergy theorem i.e. Change in Work = Change in Potential Energy + Change in Kinetic Energy. Cheers. Roman. 



#4
Apr1112, 08:23 PM

P: 43

Two scenarios...different energy requirement?
Okay, slight clarification or refinement to the problem
1) The total distance as would be recorded by the odometer is what I'm looking at...not just the lineal distance. True distance traveled is what we're working with. 2) Although I said "constant speed of precisely 75mph"....that probably creates problems if gravity would, in the given vacuum, cause the vehicle on the hilly course to go faster than 75 on the downside of the hills. Thus, that driver would have to scrub speed with brakes....which screws up the equation. SO......let's assume exactly 75.0000 mph EXCEPT on the way down hills, which is 75 unless gravity would tend to make that vehicle go faster. Simply NO energy is wasted. It's all used for propulsion. I guess the bottom line, in terms of my cogitation, is this If I ride a bike on a course that 10 miles uphill and then I come back down 10 miles......have I done essentially the same work as someone who rode 20 miles on flat terrain? Or, am I trying to create an overlysimple formula to quantify apples in terms of walnuts?? 



#5
Apr1112, 08:30 PM

P: 276

Change in Work = Change in Potential Energy + Change in Kinetic Energy
*Total distance traveled is not important as there are no fricative (tangential) forces. Cheers. Roman. 



#6
Apr1112, 08:54 PM

P: 832

You are asking a problem with unrealistic assumptions, so it's not clear what else you expect us to assume. If both cars go 100 miles at constant 75 miles per hour, then that means the car on hilly road will have to increase the throttle on uphill parts and push the breaks on downhill parts so it will expend more fuel to travel constant 75 miles per hour.




#7
Apr1112, 11:42 PM

P: 43





#8
Apr1112, 11:52 PM

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Max, you appear to be saying that the car on the hilly route has to expend energy to maintain speed up the first hill, but then gets to keep that extra as kinetic energy thereafter. So now it will be traveling faster and may eventually overtake.
Doesn't sound very interesting. Maybe you meant to specify regenerative braking, so it maintains the steady 75mph and at the end has as much stored energy left as when it started. In that case it will take longer because of the extra distance traveled. Still doesn't sound an interesting question. 



#9
Apr1212, 12:03 AM

P: 43

The question is in a nutshell, does someone traveling X distance in Y time, on flat ground expend essentially the same energy as someone traveling the same X distance in the same Y time, but with a hill in the middle, assuming all potential energy gained on the way up is utilized on the way down? If it's "not interesting" to ya.....maybe leave it alone? 



#10
Apr1212, 09:35 AM

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In a frictionless vacuum in a theoretical environment, both routes need the same energy (0).
If you ride a real bike, you'll experience a lot of energy losses, which are larger on the hilly terrain and heat the environment. 



#11
Apr1212, 10:17 AM

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P: 11,340

In an ideal, frictionless, world, you would need NO energy for the journey, other than what would be needed to get the car to its required start speed (Kinetic Energy). You may need to 'borrow' some energy to get up hills but you could get it back on the way down.
The additional criterion of constant speed is a bit of a pain but, even that can be taken care of, as mentioned already, with regenerative braking. So how 'ideal' is this scenario and what practical details do we need to add into the problem? As it stands, the answer is Zero for both cars, I think. 



#12
Apr1212, 11:47 AM

P: 197

No, he has modified his question.
"The question is in a nutshell, does someone traveling X distance in Y time, on flat ground expend essentially the same energy as someone traveling the same X distance in the same Y time, but with a hill in the middle, assuming all potential energy gained on the way up is utilized on the way down?" In this case yes, the energy in both journeys would be 0. (suposing there was no difference in hights at the end of trips) 



#13
Apr1212, 11:54 AM

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[Edit] Also note that the two cars cannot be traveling the same distance according to their odometers. 


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