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Can energy double?
If an iron ball weighting 10kg is placed on the edge of a table and let it drop pulling a toy car of the same weight from the top of the table, the frictional resistance of the car is 2kg. When the ball is about to reach the ground, the ball and the car both have the kinetic energy (10kg×10m－20)（20 by friction consumption), thus 100( potential energy) = 160(kinetic energy).
During the whole process, the ball travels vertically from the table top to the ground, while the car travels horizontally from one end of the table top to the other. If the car is attached to a generator set on the edge of the table, the colliding force will makes it work. And the same is true with the ball.

Please refer to the following drawing.

so how would you close the loop here? how will this self reset to repeat its process? you're saying the generator will provide enough energy to reset the system?

160－100(reset)＝60（net gained)

is this something you came up with or you found this?

this is like an electro-kinetic-gravity machine type of thing.

deleted

The car is the same weight of the ball? Will the ball really fall?

yes the ball would fall but not at free fall speed. i can be wrong though.

Hmmm... Maybe the ball falls, but at slow speed, so the kinetic energy will be far less than the kinetic energy of the ball falling without having to pull the car.

No, energy won't double.

Experiment：

Set a generator and an iron ball weighing 1kg at one edge of a table.
Connect the ball with a toy car weighing 1kg placed at the opposite end of the table top. The distance between the ball and the car is 10m.
Drop the ball. Pulled by the ball, the car travels from one end of the table top to the other, colliding with the generator to produce electricity.
At the same time, the colliding force of the iron ball makes another generator set on the ground produce electricity, with approximately a half potion of which the ball would be lifted and the car pulled back. Thus the device would circulate on end.
Friction=0.2kg,  g=10 m/s2ã€,
The accelerating speed of the iron ball and the car is:
F=ma, F=G-f=1×10-2＝8N, a=F/m=8/1=8(m/s2)
The end speed of the iron ball and the car is:
Vt2=2as＝2×8×10＝160, Vt =40（m/s）
The kinetic energy of the iron ball and the car is:
Ek=1/2mvt2 ＝1/2×1×160＝80(J)
The energy consumed to set the iron ball back in place.
The energy consumed to set the car back in place is negligible.（At most， E=fs=2×10＝20J）
Ep=GH＝10×10＝100（J）
The energy gained:
E=80×2－100＝60（J）ã€,

i thought this system only had 1 generator and not 2.

anyone going to try this experiment?

Pay  1,  Get 2, That's True.

according to the graph what generator would you recommend to build this in real life?
how would you build this in real life using off the shelf parts?

For example: you modify a generator and connect it to the rear wheel of a bike, when the ball hits the pedal of the bike, the generator works.

Hi Zhang,

How much work should be done to lift the ball up to the same 10m height, after it fell down? Use your 1kg ball weight like in your last calculation.

Thanks,  Gyula

Can energy triple?

It's obvious that a ball that falls pulling a car will fall with much less speed than if it was falling freely, without pulling anything.

With less speed, kinetic energy is less. The kinetic energy will just be "shared" between the ball and the car.

E=Gh=mgh=1×10×10=100J

The energy can't triple, It'll only increase a few energy less than 20J.

Quote from: Zhang Yalin on April 28, 2010, 10:25:38 PM
E=Gh=mgh=1×10×10=100J

Hi,   if you mean the above result (100J) is the answer for my earlier question on "How much work should be done to lift the ball up to the same 10m height",  then your previous energy gain of 60J is not enough to return the ball for a restart.  Now what?

80×2－100＝60.

  So your resulting gain is 60J  and you need 100J to lift up the ball for repeating the process?

a simple test can do done by getting a piece of rope, attach equal weights (one weight should have wheels) to each end of the rope, then wrap the rope around a rotor / axle, then have one end of the rope that goes vertical and the other end horizontally like the graph suggests. just measure the rotor / axle output. if the output of the rotor / axle can pick up the weight of the ball to reset the system then we have perpetual motion.

EDIT -
EDIT -
EDIT - Actually the output should have more than enough strength to also reset the weight with wheels  and not just the ball or it wont work
EDIT -
EDIT -
EDIT -

Oh, No, It can't be changed, or it will not do.

what can't be changed?

The device can't be changed, otherwise, 100=100.

Unfoturnatly if the mass of the car equals the mass of the falling weight
There will be no motion at all. Friction will prevent any movement.
Look over the calculations in the attached file. Note I did not add the
Account of friction that will be present in the pulley guiding the string so
The losses will be higher. I would suggest you to build it and try it out
I think there is a lot to learn from trial and error.

BentOptics

No, you're wrong in the last step.
The ball and the car have the EK(80J) at the end.

the weight going horizontally must have wheels...  :)

Of course, it has wheels.

Sorry everyone I seemed to have dropped the ball here (no pun intended)
I need to go over that last set of equations and see where I messed up.
Attached are the corrected results.

@Zhang Yalin
I do not see any extra energy from this system. There is inertia to overcome in the car when the weight is first released.
BentOptics

You should not add the weight of the ball and the car together. It's not right to do so.
If the power of the car comes from the other sources, how to add it ?

The car is moving when the weight falls due to the tension of the string attached to it.
If there were added forces on the car you would be able to see it.
Such as the string would go limp if the car is able to accelerate from added forces.
The reverse would be true as the string could break if the forces were suddenly forcing the car backwards.
You could make this system and when you release the weight try doing this at the same time drop another weight all by itself no strings attached and I would bet the free weight would hit 10m when the one on the string is only at 5m. Can you show a video of this system perhaps I am looking at it incorrectly.

BentOptics

If only you would contrast the ball with the car and the free ball 's dropping, then you will get it.

Zhang Yalin
I do not think I will get it.  I guess I can not see outside the proverbial box.

BentOptics

You should ask your teacher, if what I said  is wrong,  I die.

@Zhang Yalin

I understand you believe in your idea, that is fine. However you have to make sure your idea works in practice and gives the results as you imagine in your mind.
This is an 'overunity' forum and if you claim something then you ought to show it in practice you are right. Calculations show you idea is not 'overunity' so if you insist on it is, then you have to prove it is overuntity.  It is not enough to state it on paper with words.

Respectfully, 
Gyula

exactly, a real life working model should answer the question.

i have a feeling this system will not work but then again i can be wrong.

Thank you anyway. I am too poor to do it myself.
You don't know how I suffer, what a mess I am.

this should not be so expensive, just go to a junk yard and you should be able to find necessary parts to build this.

we all suffer, it's part of life. just hang in there, be strong or die trying. soon all things will be alright.  :)

peace.

Yes you can even double the speed of light but the car moves sideways by itself if you change it to water
The cheif prince of meshech

Quote from: FreeEnergy on April 27, 2010, 03:38:09 AM
so how would you close the loop here? how will this self reset to repeat its process? you're saying the generator will provide enough energy to reset the system?

Do water mills that use a gravity wheel to spin a crankshaft work the same way?  Is your thought experiment equivalent to a mill wheel?  Does the mill wheel operating on a stream or dam spillway exhibit OU?

REEDIT:
I'd vote for undecided or "Don't Know" on the poll.  I've never thought of the concept, but the law of conservation of energy may be at work here.  I'm not sure, though.

--Lee

Ya simular to water wheel but leverage the water
by self leveling it the farther it goes the more leverage
the car does not self level
but beings were now off topic drunks kill
victims not due to the crash but due to energy
transfer there so wasted - ph when they hit a car they
are essentialy colliding the bomb
taking the positive energy from the sober people absorbing
it to the drunk the drunk lives sobers up the inocent die
so if you on opiates +ph and get hit by a drunk you neutralize
of course not counting other injuries this is why drunks live
in the wreck sad but true well figured id put that out before i forgot it   

Quote from: Zhang Yalin on April 25, 2010, 11:57:48 PM
Can energy double?
If an iron ball weighting 10kg is placed on the edge of a table and let it drop pulling a toy car of the same weight from the top of the table, the frictional resistance of the car is 2kg. When the ball is about to reach the ground, the ball and the car both have the kinetic energy (10kg×10m－20)（20 by friction consumption), thus 100( potential energy) = 160(kinetic energy).
During the whole process, the ball travels vertically from the table top to the ground, while the car travels horizontally from one end of the table top to the other. If the car is attached to a generator set on the edge of the table, the colliding force will makes it work. And the same is true with the ball.

Some of the kinetic energy in the ball are consumed by the toy car so not all of the initial potential energy in the ball are converted to kinetic energy when it hits the ground. So the equation would be [100N -20N] x 10m = 800Joule, where the consumed energy of the toy car is 200Joule (20N x 10m). 800J + 200J = 1000J (The same as the weight of the ball falling freely for 10 meters). All the energy, no matter how you put it, are accounted for.

Vidar

Quote from: Zhang Yalin on April 25, 2010, 11:57:48 PM
Can energy double?

With respect to this post and posted Reply #30,

Are you simply adding both the horizontal force of the car moving across the table and also the vertical force of the ball dropping?  If you can capture all the energy of both objects moving, I suppose you have a point.

But did anyone think of drag or inertia?  I think either or both would cancel the forward force acting on the objects.  "Conservation of Energy" as a concept comes to my mind, somehow.

--Lee

I think for me the pushing and pulling are two separate energy, then when it hits something it creates another energy effect to that something, therefore i conclude that energy can be trippled!  ;D

And! if there is something in that something then there is a corresponding reaction also to that but it can be lesser or higher effect that depends on that something in that something.  ;D

 .

Quote from: Zhang Yalin on April 27, 2010, 11:32:59 PM
...
The accelerating speed of the iron ball and the car is:
F=ma, F=G-f=1×10-2＝8N, a=F/m=8/1=8(m/s2)
...

Wrong calculus. F=ma with a=g applies only in a free fall. Here the ball is prevented by the pulled car from freely falling, then a is much less than g, even if there was no friction.
Without friction, your equation shows that you are expecting the ball to fall with the same acceleration as it was alone, which is obviously false.

Proof:
Let M be the mass of the ball, m the mass of the car, and a their acceleration.
F=(M+m)*a is the force applied to the whole system when its acceleration is a.
F=M*g is the motor force applied to the system (only the ball can move under g, not the car).
It follows that (M+m)*a=M*g then a = g * M/(M+m).
Therefore a<g (even with no losses)

The balance of energy at any time is given by :
M*g*x(t)-1/2*M*v²(t)-1/2*m*v²(t)-Efriction(v) = 0
where v(t) is the speed of both car and ball and x(t) the distance traveled by the ball from its start point, at time t. Efriction is the losses, generally dependent on the car speed or squared speed.