Mess with your minds: A wind powered craft going faster than a tail wind speed.

[electrodacus]

Once you understand the idea of using the propeller to slow down the wnd even if the vehicle is moving faster than the wind, it is not so hard to understand the principle. The calculations are sound and the experiments, especially the one on the treadmill are convincing.
If you add the addtional requirement to have just simple passive wheels the system does not work, but this is not claimed or required.

There is no violation in energy conservation: the vehicle still gets enough energy from the wheels to drive to propeller. The trick is that the wheels relative to the ground are realatively fast, while the propeller relative to the wind is relatively slow.  The power is force times speed. So from the forward force produced by the propeller the faster moving wheels can produce enough power to drive the propeller. Just a little faster than the wind, the propeller sees a very low speed - so it works even with a pretty low propeller efficiency, but it gets increasingly more difficult at higher speeds. The propeller effiency will limit the maximum obtainable speed.

What do you mean by simple passive wheels ?  A wheel on a solid surface is much more efficient at providing traction than a propeller else you will probably drive a propeller based vehicle.
Obviously there is no violation of energy conservation as energy storage is used and that is what everyone seems to ignore.
When you have two isolated mediums like atmospheric air and ground then force at the wheel is not the same with force on the vehicle body. You need to think in therms of power in vs power out not forces.
If you take power from the wheel say 10W for one second you reduce the vehicle kinetic energy by 10Ws (10 Joules if you prefer) and so vehicle speed is also reduced. To get back to the same speed you will need to put back all those 10Ws but a propeller is maybe about 70% efficient so you get just 7W of thrust and if you apply that for 1 second you put back only 7Ws thus vehicle has now lower kinetic energy meaning lower speed as weight has not changed.

[bdunham7]

all available power is used meaning 100% efficient travel.

I don't mean to be insulting, but most of your reasoning is insufficiently clear for me to even point out the errors.  The above statement appears to be an incorrect assertion of some law of physics, but I can't be sure which one.  You need to define 'efficiency' to get started.  If extracting energy from the air molecules is your goal, then if they are still moving at the end of your process you have not extracted all the energy from them and therefore have not achieved 100% efficency--and you never will.  None of these processes are anywhere near 100% efficient and that really has nothing to do with this whole discussion.

If the propeller based vehicle was ruining under water (non compressible material) so a water propeller then vehicle will have no energy storage and exceeding water speed can not be demonstrated even for a sort amount of time as it is the case with air.   

That reasoning is all wet!  It might be impractical to made an underwater version of this, but the device does not depend on a compressible medium.  Please, I implore you, go back and find my previous post on this subject since it explains all this using incompressible mediums (belts and friction instead of air and propellers).

[Kleinstein]

There is no energy storage involved (other than the normal kineric energy).  The small treadmill modes are fully open, no hidden parts. The large one drove for quite some distance faster than the wind, not just a few seconds.

The suggestion with the day with no wind - shows that you don't understand the principle. The final speed of the vehicle would be something like 1.2 times the speed of the wind (because of the limited prop efficiency I doubt it could get much beyound 2 times).So 1.2 times zero wind speed would be zero. There is no more acceleration beyond a speed limit set by the gear ratio, so there is a limit to the speed, but no limit to how long this can be maintained.

One difficulty in the build is to find the right gear ratio / prop angle: the faster the design speed the more breaking force from the wheels and the larger the chance it would not work at all.  The speed dependent prop efficiency and additional friction (e.g. from the wheels) the speed range is limited. So it may not work well at very low or very high wind speed.

One could imagine a purely mechanical analog, and there are even toy like this:  consider a vehicle that uses the wheels to wind up a string and that pull on that string. The vehicle will come to you faster than you pull. Nothing really magic, but essentially the same pinciple, just pulling and not pushing.

[electrodacus]

That reasoning is all wet!  It might be impractical to made an underwater version of this, but the device does not depend on a compressible medium.  Please, I implore you, go back and find my previous post on this subject since it explains all this using incompressible mediums (belts and friction instead of air and propellers).

It will not work under water because there is no compressible medium. And by that I mean it will not exceed the water flow speed as it is the case in air.
But no need to go under water just build a wheel only vehicle like the one in my first diagram if you need physical proof to be convinced.
The wheel only prototype can be as simple as the propeller based prototype that was demonstrated on a treadmill. All you need is a box in the back of the treadmill so that surface is same level as the treadmill and put the back wheels there. You can lubricate the box surface if you want to simulate the lower air density.
No matter what you do the vehicle will not move from let to right.

 

[bdunham7]

Just look at my first diagram.

I just did and I don't see your point.  If you simply couple the two wheels (assume what you have drawn is a drive belt between the two so they go in the same direction at a 1:1 ratio), the whole thing will go from left to right at double your track speed, assuming there is no slippage anywhere.

[electrodacus]

There is no energy storage involved (other than the normal kineric energy).  The small treadmill modes are fully open, no hidden parts. The large one drove for quite some distance faster than the wind, not just a few seconds.

The suggestion with the day with no wind - shows that you don't understand the principle. The final speed of the vehicle would be something like 1.2 times the speed of the wind (because of the limited prop efficiency I doubt it could get much beyound 2 times).So 1.2 times zero wind speed would be zero. There is no more acceleration beyond a speed limit set by the gear ratio, so there is a limit to the speed, but no limit to how long this can be maintained.

One difficulty in the build is to find the right gear ratio / prop angle: the faster the design speed the more breaking force from the wheels and the larger the chance it would not work at all.  The speed dependent prop efficiency and additional friction (e.g. from the wheels) the speed range is limited. So it may not work well at very low or very high wind speed.

One could imagine a purely mechanical analog, and there are even toy like this:  consider a vehicle that uses the wheels to wind up a string and that pull on that string. The vehicle will come to you faster than you pull. Nothing really magic, but essentially the same pinciple, just pulling and not pushing.

If you want to consider pressure differential on each side of the propeller hidden then yes it is somehow hidden as you can not see air. That is the storage device allowing higher than wind speed for a limited amount of time.
If you want an electrical analogy then it is analog to how an inductor stores energy in the magnetic field around.
Yes the large blackbird for with I made the calculations requires just about 6Wh to accelerate from wind speed to 28mph record speed so less energy than you find in a modern smartphone battery.
The propeller is over 5m diameter and has a sweep area of around 20m^2 (larger than the floor area in my living room) and in the back of that there is a pressure gradient meaning quite a few meters of ever decreasing presume so many cubic meters of lightly compressed air more than the 6Wh needed for that record.
If there was no wind in that day he made the record he could have just pushed the vehicle to about  12mph then release and it will have still got to 28mph as above that speed the pressure differential energy is what is used.

If you understand that above wind speed you can no longer use wind energy because all air molecules move in the opposite direction of vehicle and if you understand that a gearbox output can not be larger than the input then you should understand that vehicle can not be powered directly by wind above wind speed in the exact same direction as the wind.
If vehicle is perpendicular to the wind then you can get to any speed you want assuming you have a low enough friction vehicle but that is not possible direct down wind or against the wind.

[electrodacus]

I just did and I don't see your point.  If you simply couple the two wheels (assume what you have drawn is a drive belt between the two so they go in the same direction at a 1:1 ratio), the whole thing will go from left to right at double your track speed, assuming there is no slippage anywhere.

If you use a 1:1 gear ratio and assume ideal case no friction then vehicle will not move in any direction.  In a real case where there is some friction the vehicle will move from right to left.
Not quite sure how you came with the 2x track speed with a 1:1 gear ratio.  I guess you imagined a powered off treadmill dragged under the vehicle as that is what people thing is equivalent when it is not at all.

[Ian.M]

If an ice yacht can tack downwind with a velocity made good (VMG) of over double the wind speed, which is very well documented^*, its obvious that a land vehicle with a  variable angle of attack wind turbine geared to wheels can also beat the wind downwind if its got sufficient blade area and great care is taken to minimize friction.

The problem is getting past the stagnation point in a steady wind where the velocity is equal to the wind which is going to demand some sort of mechanism so stored energy in the rotor can briefly accelerate the vehicle past the wind speed, then flipping the angle of attack will let the rotor continue to draw power from the relative wind.   

I *think* the vehicle in the video BrianHG linked at the start of this topic gets around the apparent need for variable gearing by utilizing the gustiness of the wind.  At the trailing edge of a gust, the vehicle can be already travelling faster than the wind, so its just a matter of picking the moment to flip the blade pitch control through the dead spot, where the blades are effectively flat to the plane of rotation, to continue accelerating.

Any sort of conveyor belt or wind tunnel modelling of such a vehicle, or anything even vaguely similar, that does not have a variable rotor to wheels gear ratio and doesn't model the gustiness of real wind, is doomed to failure . . . 

* e.g. http://www.nalsa.org/Articles/Cetus/Iceboat%20Sailing%20Performance-Cetus.pdf

[electrodacus]

If an ice yacht can tack downwind with a velocity made good (VMG) of over double the wind speed, which is very well documented^*, its obvious that a land vehicle with a  variable angle of attack wind turbine geared to wheels can also beat the wind downwind if its got sufficient blade area and great care is taken to minimize friction.

The problem is getting past the stagnation point in a steady wind where the velocity is equal to the wind which is going to demand some sort of mechanism so stored energy in the rotor can briefly accelerate the vehicle past the wind speed, then flipping the angle of attack will let the rotor continue to draw power from the relative wind.   

I *think* the vehicle in the video BrianHG linked at the start of this topic gets around the apparent need for variable gearing by utilizing the gustiness of the wind.  At the trailing edge of a gust, the vehicle can be already travelling faster than the wind, so its just a matter of picking the moment to flip the blade pitch control through the dead spot, where the blades are effectively flat to the plane of rotation, to continue accelerating.

Any sort of conveyor belt or wind tunnel modelling of such a vehicle, or anything even vaguely similar, that does not have a variable rotor to wheels gear ratio and doesn't model the gustiness of real wind, is doomed to failure . . . 

* e.g. http://www.nalsa.org/Articles/Cetus/Iceboat%20Sailing%20Performance-Cetus.pdf

None of what is found in that document contradicts what I say. They do not drive down wind faster than the wind powered only by the wind.  They drive in multiple directions as they do a close circuit they do not drive in a straight line in the exact direction as wind.
And guys at NALSA do not understand energy storage or to be more blunt physics in general. Else they will not have allowed blackbird record and say there is no energy storage involved.
I think those that know this is not possible do not see the energy storage so I'm sort of alone in trying to correct this mistake.
I only found out about this due to Derek's (Veritasium) videos but this is more than a decade old vehicle and unfortunately it got enough attention that even questions on some exams are based on this wrong understanding of how this vehicle works.
Just imagine an exam to enter at university will show you a motor connected to a generator (over unity device) sniping forever and powering a light bulb.
This just bothers me more than it should to the point that I have a form of PTSD when I just think about this.
I'm of above average intelligence and if there was something here I will have understood but it seems most people just do not care enough to look more in to the claims made here to understand that the explanations violates the conservation of energy.

[Ian.M]

Nor do the blades of the vehicles rotor travel straight downwind.  Conceptually they could be mounted to a long belt or chain oriented cross wind, and then their movement would be almost indistinguishable from the sail of an ice yacht tacking downwind except at the moment where they go round the end of the belt transitioning from pointing up to pointing down, and the ice yacht gybes with its boom crossing the center line.  Both reverse the blade or sail's angle (as seen from above) with respect to the wind.   The actual case of blades on a rotor is a little different as each blade corkscrews downwind, but if you unroll its path its directly equivalent to the motion of an ice yacht's sail on an indefinitely long single tack.

[bdunham7]

If you use a 1:1 gear ratio and assume ideal case no friction then vehicle will not move in any direction.  In a real case where there is some friction the vehicle will move from right to left.
Not quite sure how you came with the 2x track speed with a 1:1 gear ratio.  I guess you imagined a powered off treadmill dragged under the vehicle as that is what people thing is equivalent when it is not at all.

I'm assuming no slippage, not no friction.  And I did miscalculate, 1:1 won't actually work in practice as the vehicle would try for infinite velocity and something would have to slip or break.  However, with an appropriate gearing between the two wheels, you can get the vehicle to move in either direction at an arbitrary speed.

Let's say V1 is the belt speed, where going from right to left as your arrow is drawn is the positive direction.  Let's say V2 is the speed of the outer circumference of the right wheel (G) and V3 is the speed of the left wheel (M), with a positive value reflecting a clockwise rotation.  The gear ratio R of your drive mechanism is V3/V2, so a value of 0.5 means that M is spinning in the same direction as G but half as fast, and so on.  V4 is the speed of the vehicle from left to right relative to the stationary block on the right and the stationary part of the system containing the moving belt, all presumably attached to some reference like the ground.  What you want is to determine V4 given R and V1.

Now assuming no slippage, the velocity of the vehicle V4 has to be the same as V3, which equals R*V2.  At the same time you should see that V2 has to equal the sum the belt speed and the vehicles speed, so V2 = V1 + V4.

I'm of above average intelligence

Then perhaps you can solve these equations and determine the direct relationship between R, V1 and V4.

EDIT: It is a common misconception to misapply the law of conservation of energy to a system that is not closed.  As an example, regular auto or home air conditioning appears to over unity if you don't consider the appropriate bounds of the system.

[electrodacus]

I'm assuming no slippage, not no friction.  And I did miscalculate, 1:1 won't actually work in practice as the vehicle would try for infinite velocity and something would have to slip or break.  However, with an appropriate gearing between the two wheels, you can get the vehicle to move in either direction at an arbitrary speed.

Well you are wrong about the fact that vehicle can travel in any direction based on a gearbox ratio.  That is not true and the vehicle I show can only travel from right to left it can not advance on the treadmill.
What you are thinking of is maybe something like this

But what that represents is a vehicle traveling against the wind direction at much lower speed than the wind. The wind in that case is the treadmill and the road is the fixed body of the treadmill. So it demonstrates that a vehicle can travel down wind at lower speed than the wind and nobody will disagree with that.  But people including that guy interprets this wrongly and adds the speed of the wind to the speed of the vehicle and consider that to be the speed of the vehicle.

Let's say V1 is the belt speed, where going from right to left as your arrow is drawn is the positive direction.  Let's say V2 is the speed of the outer circumference of the right wheel (G) and V3 is the speed of the left wheel (M), with a positive value reflecting a clockwise rotation.  The gear ratio R of your drive mechanism is V3/V2, so a value of 0.5 means that M is spinning in the same direction as G but half as fast, and so on.  V4 is the speed of the vehicle from left to right relative to the stationary block on the right and the stationary part of the system containing the moving belt, all presumably attached to some reference like the ground.  What you want is to determine V4 given R and V1.

Now assuming no slippage, the velocity of the vehicle V4 has to be the same as V3, which equals R*V2.  At the same time you should see that V2 has to equal the sum the belt speed and the vehicles speed, so V2 = V1 + V4.

Think about in much simple terms as vehicle will not advance from left to right in my diagram (not in this universe).
To get any Power from the generator wheel G you need to break so say you apply a 10W break power then vehicle will just move from right to left pushed by the treadmill against the 10W of breaking power.
Now say you take this 10W all of them ideal case and use to power the M (motor) wheel to rotate in clockwise direction then the vehicle will be in equilibrium so it will not move in any direction.
This equilibrium is only possible in best case theoretical analysis but in reality there are extra losses and so vehicle will always move from right to left thus concision should be no higher speed than treadmill speed is possible.
You are using forces and speed that is true is the same as power but with power you can not make the mistakes of not understanding how forces act against a vehicle that is between two isolated mediums.
So please think in therms of power as it is much easier to understand. 

[bdunham7]

Well you are wrong about the fact that vehicle can travel in any direction based on a gearbox ratio.  That is not true and the vehicle I show can only travel from right to left it can not advance on the treadmill.
What you are thinking of is maybe something like this

I'm not thinking about random videos, I'm applying basic principles that I thoroughly understand to your example.  Now even though I have that understanding, I still could make a mistake, so check my work.  I explained where I think your main error lies, and that is in assuming that conservation of energy is somehow relevant here.  It isn't.  If you assume a belt moving at a given speed, as you have, then it would be possible to extract any arbitrary amount of energy from that subject to actual physical limitations in practice.  Stop thinking in terms of energy when you can solve the whole thing with elementary mechanics.

So please, at least solve those basic equations and show me if there is any part of the solution that involves any error like infinite force, division by zero or the like.  Even if you can't solve the equations, you should be able to see that for R=1/2, the vehicle will move from left to right at V1.

[electrodacus]

I'm not thinking about random videos, I'm applying basic principles that I thoroughly understand to your example.  Now even though I have that understanding, I still could make a mistake, so check my work.  I explained where I think your main error lies, and that is in assuming that conservation of energy is somehow relevant here.  It isn't.  If you assume a belt moving at a given speed, as you have, then it would be possible to extract any arbitrary amount of energy from that subject to actual physical limitations in practice.  Stop thinking in terms of energy when you can solve the whole thing with elementary mechanics.

So please, at least solve those basic equations and show me if there is any part of the solution that involves any error like infinite force, division by zero or the like.  Even if you can't solve the equations, you should be able to see that for R=1/2, the vehicle will move from left to right at V1.

Conservation of energy is relevant everywhere.  Nobody has proven that conservation of energy law can be broken tho many have claimed that including here with blackbird.

Yes you can extract energy if you lock the M wheel (but that will be like bolting the vehicle to the ground) and you can use that energy to power a light bulb or anything you like but you can not take that energy and trying to move from left to right.
Nobody has shown an example where the vehicle in my diagram works as it is impossible. Not quite sure why that is not obvious.

At some point I asked a question in a similar manner but was unanswered.
Say you have two identical vehicles (say they are electric vehicles but with a gearbox) and so they have exactly same power motors but one is in first gear and the other in 4th gear and they are back to back connected trough a rope and so question is witch one of the two vehicle will win? 

[bdunham7]

Conservation of energy is relevant everywhere.  Nobody has proven that conservation of energy law can be broken tho many have claimed that including here with blackbird.

OK, so tell me in your own words:  What is the law of conservation of energy?

Nobody has shown an example where the vehicle in my diagram works as it is impossible. Not quite sure why that is not obvious.

I showed you and provided the equations that would apply and even supplied the solution for the case R=1/2.  What error do you find in that explanation?

Say you have two identical vehicles (say they are electric vehicles but with a gearbox) and so they have exactly same power motors but one is in first gear and the other in 4th gear and they are back to back connected trough a rope and so question is witch one of the two vehicle will win?

A perfect example of where your thinking is going wrong.  Power is completely irrelevant.  Torque--which does not mean power--will determine which way the vehicles go, subject to the limitations of the strength of the rope and the friction of the tires with the ground.  If the motors have the same torque, then the one with the greater torque multiplication--the one in first gear--will be the one pulling harder.

[Kleinstein]

But what that represents is a vehicle traveling against the wind direction at much lower speed than the wind. The wind in that case is the treadmill and the road is the fixed body of the treadmill. So it demonstrates that a vehicle can travel down wind at lower speed than the wind and nobody will disagree with that.  But people including that guy interprets this wrongly and adds the speed of the wind to the speed of the vehicle and consider that to be the speed of the vehicle.

A similar vehicle would go faster than the band, if the gear ratio is negative, so invert the direction of rotation.

The sytem with the moving  ground on one side or the wind blowing has plenty of energy to go around. So it is not a good idea to argue with conservation of energy. Even a standing vehicle would not conserve the enery and standing in the wind is definitely possible.  It is more about arguing with a balance / imbalance of forces.

[electrodacus]

A similar vehicle would go faster than the band, if the gear ratio is negative, so invert the direction of rotation.

The sytem with the moving  ground on one side or the wind blowing has plenty of energy to go around. So it is not a good idea to argue with conservation of energy. Even a standing vehicle would not conserve the enery and standing in the wind is definitely possible.  It is more about arguing with a balance / imbalance of forces.

Yes please build such a vehicle and you will be surprised when it will not work as you expect.
I think most people can not simulate correctly what will happen in their head and need to see a real example. I can not explain how else I'm unable to describe how vehicle works to others.
We are talking only about particular case where vehicle moves in exact same direction as the wind any other vehicle orientation except this and the against the wind can allow a vehicle to exceed wind speed.

What is wrong in my simple explanation where all I say is the following.
-In my diagram ideal case vehicle can not move from left to right because any breaking power at wheel G will need to be fully available to wheel M just so that vehicle remains stationary.

I can not think of a simpler description.

[bdunham7]

What is wrong in my simple explanation where all I say is the following.
-In my diagram ideal case vehicle can not move from left to right because any breaking power at wheel G will need to be fully available to wheel M just so that vehicle remains stationary.

What is wrong is that force is not power.  It is possible (and completely ordinary) to continuously exert a force without expending any power.  Braking force is what is important, if the components are stationary, then the braking power is zero.  If they are moving, braking power is negative, but that's another issue altogether. 

[electrodacus]

A perfect example of where your thinking is going wrong.  Power is completely irrelevant.  Torque--which does not mean power--will determine which way the vehicles go, subject to the limitations of the strength of the rope and the friction of the tires with the ground.  If the motors have the same torque, then the one with the greater torque multiplication--the one in first gear--will be the one pulling harder.

You are ignoring a very important factor.  Hope I will be able to explain.
You are thinking on a vehicle with both generator wheels and motor wheels on the same medium instead of two isolated mediums.
Also even in a single medium you can not generate more power than you need to put back in just to maintain speed in ideal case.

What you are saying is having a vehicle at a fixed speed with no energy source you can have a generator on the front wheels that will transfer the produced power with any sort of gear ratio you want to the motor wheels at the back and somehow the vehicle speed will increase.
You may not think that what I mentioned above is the same thing with my drawing because in example above generator and motors are on same mediums and in my diagram generator and motor are in two isolated mediums.
The difference between the two is only the frame of reference as the vehicle is stationary at start while the road if you want is moving and equivalent to wind speed is zero.  When changing reference frames you need to consider the implications else your interpretation of what happens will be wrong.

[electrodacus]

What is wrong is that force is not power.  It is possible (and completely ordinary) to continuously exert a force without expending any power.  Braking force is what is important, if the components are stationary, then the braking power is zero.  If they are moving, braking power is negative, but that's another issue altogether.

Force is contained in power. Conservation of energy will not mention force as energy is power over time so force in isolation is irrelevant.
It is like saying that you have a power supply that is 10A at 24V so 240W max and that if it needs to move a vehicle will provide you something extra compared to a 20A at 12V power supply so same 240W

Braking power is what is important. A breaking force alone will say nothing (provide no useful information).


 

[bdunham7]

with no energy source

When changing reference frames

You're making a complete hash of elementary level physics.  I based what I said on your diagram, not any 'single medium' or anything else.

There's no reason to change the reference frame at any point here and doing so will only confuse you.  The appropriate reference frame is the one where the platform on the left is fixed, the mechanism holding the belt is fixed, the belt moves a per your arrow from right to left and the car does whatever it is going to do.

If you hold the car in a fixed position in this reference frame, which can be done without the expenditure of any power whatsoever by simply mechanically locking wheel M, you can then generate any arbitrarily large amount of power with wheel G, subject to physical limitations.  In this case, and in any case where the car remains fixed or moves at a constant speed, the force applied to wheel G will match that applied to or by wheel M.  This results simply from F=MA, where if A = 0 (car not accelerating)  then the total F must also equal 0.

What you do with this power is up to you--you can make toast, smelt aluminum or you can apply some of that power to wheel M and move the car.  Now once the car is moving there will be a different set of calculations you will need to do to define its steady state--there will be a continuous expenditure of energy at wheel M because there is a force over distance, but you can generate that from wheel G without violating conservation of energy because the energy is coming from the belt.  As long as the ratio R is less than one, wheel G will be rotating at a faster rate then wheel M, and since the forces must be equal in a steady state, that means there is always more power available from wheel G than is needed by wheel M.  The maximum speed you can attain, which is correlated to the ratio R by those equations you still haven't solved, will be determined by the overall efficiency of the generator/motor combination.

[bdunham7]

Force is contained in power. Conservation of energy will not mention force as energy is power over time so force in isolation is irrelevant.
Braking power is what is important. A breaking force alone will say nothing (provide no useful information).

That's unparsable gibberish to me....

[electrodacus]

That's unparsable gibberish to me....

OK let me give you an example. You have an ideal vehicle driving at constant speed and say has 500Ws of stored kinetic energy.
Then I say to you that I apply a 10N breaking force for 1 second so then I ask. What is the vehicle current kinetic energy?
You will not be able to answer that question.
But if I say something like I apply a breaking force of 10Ws for 1 second you will be able to say that vehicle has now just 490Ws of kinetic energy so lower speed since weight is unlikely to have changed.

Also in keeping with this example if you want to get back to the same speed you where at before the 10Ws for one second breaking you will need to apply the same in propulsion so if you stored those 10Ws in a battery or capacitor instead of wasting as heat then you will need all that energy to get back to where you where and there is no way to use those 10Ws to increase the vehicle speed (meaning increase kinetic energy).

[bdunham7]

You will not be able to answer that question.

Only because you haven't told me the mass of the vehicle and it would be required to solve that problem.

But if I say something like I apply a breaking force of 10Ws for 1 second you will be able to say that vehicle has now just 490Ws of kinetic energy so lower speed since weight is unlikely to have changed.

Aside from the fact that 10W is not a 'force', OK.  Say a force such that 10W is dissipated. So what?

Also in keeping with this example if you want to get back to the same speed you where at before the 10Ws for one second breaking you will need to apply the same in propulsion so if you stored those 10Ws in a battery or capacitor instead of wasting as heat then you will need all that energy to get back to where you where and there is no way to use those 10Ws to increase the vehicle speed (meaning increase kinetic energy).

What does all that have to do with the example we are talking about? 

One thing about physics is that often different sets of rules apply to the same problem and the solutions have to work in all of them, otherwise you have made an error.  You can try to solve an energy equation here if you like, but you should also be able to come up with a solution in simple mechanics, Newton's laws and so forth, and if the methods don't yield the same result, then one of them hasn't been done right or doesn't apply.  The simplest trick exam question involves firing a bullet into a pendulum and then determining the resulting motion of the pendulum afterwards.  In that case, the common mistake is to use conservation of energy instead of conservation of momentum to try and solve the problem.  That doesn't work because (coherent) kinetic energy is not conserved in that case. You're making a similar mistake here--this is not a closed system and you cannot use conservation of energy laws to solve it.

[iMo]

OK let me give you an example. You have an ideal vehicle driving at constant speed and say has 500Ws of stored kinetic energy.
Then I say to you that I apply a 10N breaking force for 1 second so then I ask. What is the vehicle current kinetic energy?
You will not be able to answer that question..

Knowing the mass "m" (ie. 1000kg) of the vehicle you can, imho.
KE=p^2/(2m)
KE=10Ns*10Ns/(2m)=100/2000=0.05Ws
500-0.05=499.95Ws