Can an Electric Car Run Forever on a Slope with Rollover Power Generators?

In summary: So you can't say the textbook: gpe at top of slop = kinetic energy at the bottom.As that's not the case here.In summary, the car will always generate more power going down the slope than it needs to get back up.
  • #36
spikenigma said:
I don't understand why the slowdown is even an issue, because between each ramp the car accellerates back up to terminal velocity - which is what prompted the entire scenario.

If we assume mathematically that each ramp slows the car down by x % of it's velocity, does that % - even an arbitrary one - make the car unable to return to the top of the hill?

I was thinking about this on my way back from work.

It takes energy to accelerate. We haven't factored in any energy for acceleration from gravity back up to terminal velocity.

In the scenario before we only had two losses:
Air resistance
The pads

Now we have three:
Air resistance
The pads
Acceleration

If we say each pad causes a 10% loss in velocity. Then the energy needed to accelerate back up to the terminal velocity is:
0.5m(v1^2-v2^2).
=250*(165^2-149.5^2)
=1.2MJ for each ramp.

So in reality the activation of each ramp that stores up 1.65MJ of energy costs us 1.65 (to store the energy in the device) + 1.2 MJ to get back to where we were before we hit the ramp. Meaning a total loss of 1.2 MJ every time we activate a ramp.

The problem is we can't get any real data, and we can't assume everything (becuase it's likely to be wrong and losses missed out etc). So in this case we have to assume that the laws of physics are correct and that the energy is like a bucket filled with water. (you only have a finite amount)

If we could do a practical experiment we wouldn't need to assume anything. The experiment would clearly show the premise in the OP is unsound.What would happen if you put too many of these pads is that the car would slow down enough and come to rest on one of the pads becuase it didn't have the energy to activate it. If you put in just enough then you would have a scenario where the car comes to rest exactly at the bottom.
 
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  • #37
spikenigma said:
Hi,

(Firstly, This is not a perpetual motion discussion. Nor am I suggesting anything of the sort. I'm just wondering where the energy losses enough to stall the system creep in.)

I find your statement in your first post to be disingenuous. You have wasted enough of everybody's time. Thread closed.
 
  • #38
A note on conservation of energy after the lock: CoE isn't just a physical law, it is a mathematical/logical necessity and therefore a useful tool for solving problems. All that you accomplish by trying to solve a problem without COE is to first derive COE, then use it to solve the problem! That's just a waste of time. Consider the simple example of calculating the speed at which a dropped rock hits the ground. The two approches are:

1. Using COE, you can set the potential energy at the beginning equal to the kinetic energy at the end and solve for V.
2. Combine f=ma, V=at and d=Vt, and solve for V. Along the way, you'll discover you're doing most of the work to derive KE from PE!

Why bother re-inventing the wheel? There would need to be a pretty compelling reason and I'm not seeing one here. More importantly, the more complicated the problem gets, the more difficult the derivation gets. Therein lies the essence of most perpetual motion claims: the more complicated a problem gets, the more likely the claimant is to make a mistake that implies perpetual motion is possible (interestingly, they never make a mistake that implies it isn't!). In this problem, the mistake was simple: increase the number of "pads" by a factor of four and you must decrease the amount of energy each collects by a factor of four. This is easy enough to see by simplifying the problem to isolate the "pads":

Consider a situation where there are no losses except for the extraction of energy by the "pads". The car rolls freely from the top of the hill to each pad, where the pad stops the car and recovers all the energy. Increasing the number of "pads" by a factor of four simply divides the distance the car travels between "pads" by 4, making the COE equation look like:

PE = EPad1 + EPad2 + EPad3 + EPad4
mgh = mgh/4 + mgh/4 + mgh/4 + mgh/4
 

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