A problem with KE = ½mv². Help.

[Farsight]

I've come across something on the Internet that's causing me a big problem. It's called Energy Misdefined in Physics, by a guy called Gary Novak. I didn't pay too much attention to it at first, but it bothered me a little. It nagged, and I looked at it again and it suddenly clicked. Then the full horror of what it meant came home...

 

Kinetic Energy is defined as KE = ½mv², but a trivial thought experiment says it just has to be K = mv. The underlying problem seems to be with Work = Force x Distance, which ought to be Force x Time.

 

You’re sitting on a 100kg sled on a frozen lake, with a spring gun that fires a puck backwards. The recoil accelerates the sled forwards to a velocity of 0.1 meters per second.

 

Sled and puck have equal action and reaction so you calculate the energy stored in the spring as 2 x ½mv² -> 100 x .01 = 1 Joule.

 

With the assistance of a snowcat crew running alongside with a supply of pucks you fire the spring gun a further nineteen times in quick succession, and the sled is now travelling at 2 meters per second.

 

You recalculate the spring energy as 2 x ½ mv² / 20 -> 100 x 4 / 20 = 20 Joules.

 

Problem.

 

Can anybody point out the flaw in this? Because I sure as hell can't, and it affects everything. Like the dimensions of energy. Which threatens E = MC².

 

Help.

[Farsight]

Sorry, I can't post a link until I've made 15 posts. I'll do it later. Use Google to search on "Energy Misdefined" and you'll find it.

[Tormod]

His page is basically mumbo-jumbo which is more critical than it is scientific. Novak's rants against science (while pretending to use science) are sad. He seems to be lacking even a fundamental understanding of physics (although he seems to be quite well prepared to laugh scientists in their face, which is cool if you want to be considered a crackpot).

 

If - as he claims - 90% of physics has errors in it, how come we can use the very same physics to create superconductors, travel to the moon, build positioning satellites, and create heat pumps that work extremely efficiently?

 

He certainly does not threaten [math]E=mc^2[/math].

 

The example you quote above, for example, seems to forget that a certain amount of energy will be lost to friction as the sled moved forward. The ice will slow the sled down, as will gravity. So you will not get a consistent increase in speed like that.

 

As the sled's speed increases, you need less energy to keep it moving at the same speed, but you still need to add energy or it will stop - just like a car will stop accelerating when you let up the gas pedal, and slow down when the amount of fuel delievered to the engine is reduced. This is basic conservation of energy and is easily graspable.

[UncleAl]

Science is rigorously derived. It contains no internal mistakes or contradictions. Zero. Not a one. Not a single internal problem exists anywhere within its entire fabric. Nada, zip, zilch. That is how science is done. That is the standard that any change or addition must meet.

 

Physics, Resnick, Halliday, Krane, 4th Ed, Vol 1, p 138.

Three text inches. Learn some physics.

 

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

http://www.motionmountain.net

Learn some physics.

 

Whether mathematical prediction exactly matches empirical observation is the giggle. Newton is not off by a factor of two.

[arkain101]

Well, its interesting to see someone agree with me.

 

I have kept quiet on this concept for some time on this forum. But it is in fact true!

 

The potential use of Kenetic Energy is infact = to Mass times velocity squared!

 

Each time I present this to someone who knows the mathamatics they cant answer me properly. Ke=1/2M*V^2. The equation relates to how much work The ONE moving object can inflict on another object it interacts with and not how much work is done in the entire process. So it is only concirned with the work in one side of the system. This is why we have the 1/2 in the equation. The Mass in the equation relates to only the moving object (which is fine) because the equations concirn is aimed at the moving object. The Velocity on the other hand is squared. What does this mean? It means you have two simutanious velocities of the moving object.

 

Now to give an example that proves that the Total Energy useable in a moving object is Ke=M*V^2.

 

In this example we ignore friction and energy losses. We have two reference points. A person who is going to catch an object on a frictionless ice rink. The object is a shaft that has a telescopic action to it. When the shaft compresses inwards it spins a wheel on the far end of the shaft. (to give you better imagery. If we were to hold this shaft and push it in and out the wheel would spin faster and faster on the other end, but it is all one peice of mass).

 

The mass of this entire shaft and all its parts is 10kg. The mass of the person is 100kg. With the equation Ke=1/2(M*V^2) we can figure out how much energy the shaft will transfer to the person at rest when it hits him (say in the chest). The shaft is moving at a velocity of 10m/s.

Ke=1/2(10*10^2)= 1/2(10*100) = 1/2 (1000) = 500joules. This is how much energy is in the moving shaft.

But here is the catch for the Total Potential Usage of Kinetic Energy.

In this frictionless example.

The shaft hits the man in the chest and delivers 500 joules. However, the shaft does not hit the man and bounce off right away. Because it has a telescopic device that spins a wheel containg 5kg (which makes up half the mass of the shaft that is 10kg) as it his the man it begins decelerating as the wheel spins. The force is transfered over a longer time than compared to a fixed 10kh object. The wheel is spinning rather quickly by the time the shaft reaches its final travel point, now fully compressed it sends the last bit of its momentum into the man and sends him moving. While the wheel was spinning up as the shaft compressed on the man, the total mass of the shaft and wheel is still acting upon the man as it transfers momentum. It transfers it over a slightly longer time and makes the force less, but the transfer of momentum remains the same.

Now the man goes drifting backwards (at whatever velocity this would be Im not going to calculate that part). The man has taken the full energy of a 10kg object moving at 10m/s. But before the shaft-wheel falls to the ground, a person catches it. The person that catces it holds it in both hands as he watches the 100kg man drift away, but to his suprise the wheel on the end of the shaft is spinning!!.

If you calculate this with all the right equations you will find the total energy spent in the system is exactly equal to Ke=M*V^2. Using only the Mass of the object concirned to be in motion.

 

 

Lets take a a look at how to write the equation Ke=1/2(M*V^2) expanded and what each character and functino is representing.

Ke=(M*V) (M*V) / (2*M)

 

The first (M*V) is the moving shaft-wheel

 

The second (M*V) is an equal portion of the mass of the man to the mass of the shaft-wheel.

 

The interaction force can only be as great as the lesser mass. Hence the reason for only using it.

 

Then we devide by the lesser mass value (10kg) (multiplied by two to relate to both objects), and there we have it same thing as 1/2M*V^2 (I was only showing the obvious derivitivation between the two equations).

 

So Low and behold the total potential use of Kinetic Energy of a moving body is M*V^2 of the moving object.

 

This leads all the way back to the theory of duality relativity I presented here in the past. MOTION requires two reference frames. Motion is relative to the observer. Each can consider itself moving, and an equation can be written to prove in the grand frame, they can both be considered moving, and the force is only equal to the lesser mass. The larger mass in a way donates the equal mass to create the energy in the system, but the leftover mass dictates the trade off in momentum.

[Erasmus00]

Kinetic Energy is defined as KE = ½mv², but a trivial thought experiment says it just has to be K = mv. The underlying problem seems to be with Work = Force x Distance, which ought to be Force x Time.

 

You are confusing impulse/momentum with work/energy.

 

Sled and puck have equal action and reaction so you calculate the energy stored in the spring as 2 x ½mv² -> 100 x .01 = 1 Joule.

 

This is inappropriate, as I've explained on another thread. The sled and the puck have different masses, and the same momentum so different energies.

 

You recalculate the spring energy as 2 x ½ mv² / 20 -> 100 x 4 / 20 = 20 Joules.

 

Now this is very wrong. Each puck travels slower than the previous, as it is fired from a moving sled. Your oversimplified math does not take this into account.

-Will

[IDMclean]

I would like to take a moment to point out that K=1/2mv^2 is a antiquated equation, the proper form is:

 

For a better explination of the whole concept here you go:

http://hyperphysics.phy-astr.gsu.edu/HBASE/ke.html#ke

 

I do have a question though, What is the difference between mv^2, mv and ma? As I understood it a = v^2; Force, F = ML/S^2 and Energy, E = ML^2/S^2.

[kmarinas86]

I do have a question though, What is the difference between mv^2, mv and ma? As I understood it a = v^2; Force, F = ML/S^2 and Energy, E = ML^2/S^2.

 

a = acceleration = m/s^2

v^2 = velocity^2 =m^2/s^2]

 

a≠v^2 ("a=v^2" is not correct)

 

Energy = kg m^2/s^2 (You're correct here) = Force * Distance

Momentum = kg m/s = Force * Time

Force = kg m/s^2 (You're correct here)

[ronthepon]

I just wanted to put up a very very basic proof of the kinetic energy expression. I am not putting up the full derivation because I am not interested in typing it here without learning Latex.

 

Now, let a body be accelerated from rest to a constant speed 'x'.

 

Now,

work done to speed the body up = Change in kinetic energy = The total kinetic energy assuming the body was initially at rest. - (Code RED)

 

So say the body was accelerated by 'a' and has mass 'm'

 

so force applied = ma

 

And work done = x(ma) - (Eq. 1)

 

now,

 

v^2 = 2ax (Newton's third eq.) where v is the final velocity and u, the initial velocity is zero.

 

so, x = v²/2a

 

substitute the value for x in eq. 1

 

We get

 

Work done = (ma)v²/2a

 

Or, work done = mv²/2

 

Using code RED, as I defined it above,

 

Total kinetic energy = ½mv²

 

A very very simple derivation method of the expression. I'd like to say that this does hold true in any other circumstance, including variable acceleration.

Though there is another method of proving in that circumstance, I'll just say:

 

The kinetic energy of a body depends on it's mass and velocity, not on it's history, how it was brought to that speed.

 

Hope I'll make an impact in this thread...:eek:

[IDMclean]

Thanks, I always get confused with the whole PEMDAS. I was thinking of the E=mc^2 equation, thinking of c as v. wait. Isn't c = 299792458v?

 

:eek: :)

 

so if, a != v^2, then doesn't that mean that c^2 != a? So E=mc^2 is not with light in acceleration?

[kmarinas86]

Thanks, I always get confused with the whole PEMDAS. I was thinking of the E=mc^2 equation, thinking of c as v. wait. Isn't c = 299792458v?

 

no, unless v=1 meter per second.

 

c/299792458 = 1 meter per second

 

so if, a != v^2, then doesn't that mean that c^2 != a? So E=mc^2 is not with light in acceleration?

 

yes. yes.

 

E is total energy not kinetic energy.

[IDMclean]

That's right, I've been bussy with other things lately, have to brush up again.

 

e = mc^2 = ML^2/S^2. So where does the other L come from? c = 299792458 m/s right or is that wrong, wait, huh? c^2 is = (299792458m/s)^2 = 89875517873681764 m/s, :) ?

 

Alright I haven't done this for far to long, I think...

[arkain101]

Any thoughts on this? I dont know if people bothered to read my rather long post.

There is,

[math]\normalsize Ke=\frac12(M*V^2)[/math]

and

[math]\normalsize Ke=M*V^2[/math]

 

(in this example we also assume the part of the telescoping shaft that impacts the person first is almost massless to make all things easier).

 

In this example we ignore friction and energy losses. We have two reference points. A person who is going to catch an object on a frictionless ice rink. The object is a shaft that has a telescopic action to it. When the shaft compresses inwards it spins a wheel on the far end of the shaft. (to give you better imagery. If we were to hold this shaft and push it in and out the wheel would spin faster and faster on the other end, but it is all one peice of mass).

 

The mass of this entire shaft and all its parts is 10kg. The mass of the person is 100kg. With the equation [math]\normalsize Ke=\frac12(M*V^2)[/math] we can figure out how much energy the shaft will transfer to the person at rest when it hits him (say in the chest). The shaft is moving at a velocity of 10m/s.

[math]\normalsize Ke=\frac12(10*10^2)= \frac12(10*100) = \frac12(1000) = 500[/math]joules. This is how much energy is in the moving shaft.

But here is the catch for the Total Potential Usage of Kinetic Energy.

In this frictionless example.

The shaft hits the man in the chest and delivers 500 joules. However, the shaft does not hit the man and bounce off right away. Because it has a telescopic device that spins a wheel containg 5kg (which makes up half the mass of the shaft that is 10kg) as it his the man it begins decelerating as the wheel spins. The force is transfered over a longer time than compared to a fixed 10kh object. The wheel is spinning rather quickly by the time the shaft reaches its final travel point, now fully compressed it sends the last bit of its momentum into the man and sends him moving. While the wheel was spinning up as the shaft compressed on the man, the total mass of the shaft and wheel is still acting upon the man as it transfers momentum. It transfers it over a slightly longer time and makes the force less, but the transfer of momentum remains the same.

Now the man goes drifting backwards (at whatever velocity this would be Im not going to calculate that part). The man has taken the full energy of a 10kg object moving at 10m/s. But before the shaft-wheel falls to the ground, a person catches it. The person that catces it holds it in both hands as he watches the 100kg man drift away, but to his suprise the wheel on the end of the shaft is spinning!!.

If you calculate this with all the right equations you will find the total energy spent in the system is exactly equal to [math]\normalsize Ke=M*V^2[/math]. Using only the Mass of the object concerned to be in motion.

[ronthepon]

Any thoughts on this? I dont know if people bothered to read my rather long post.

There is,

[math]\normalsize Ke=\frac12(M*V^2)[/math]

and

[math]\normalsize Ke=M*V^2[/math]

(in this example we also assume the part of the telescoping shaft that impacts the person first is almost massless to make all things easier).

In this example we ignore friction and energy losses. We have two reference points. A person who is going to catch an object on a frictionless ice rink. The object is a shaft that has a telescopic action to it. When the shaft compresses inwards it spins a wheel on the far end of the shaft. (to give you better imagery. If we were to hold this shaft and push it in and out the wheel would spin faster and faster on the other end, but it is all one peice of mass).

The mass of this entire shaft and all its parts is 10kg. The mass of the person is 100kg. With the equation [math]\normalsize Ke=\frac12(M*V^2)[/math] we can figure out how much energy the shaft will transfer to the person at rest when it hits him (say in the chest). The shaft is moving at a velocity of 10m/s.

[math]\normalsize Ke=\frac12(10*10^2)= \frac12(10*100) = \frac12(1000) = 500[/math]joules. This is how much energy is in the moving shaft.

But here is the catch for the Total Potential Usage of Kinetic Energy.

In this frictionless example.

The shaft hits the man in the chest and delivers 500 joules. However, the shaft does not hit the man and bounce off right away. Because it has a telescopic device that spins a wheel containg 5kg (which makes up half the mass of the shaft that is 10kg) as it his the man it begins decelerating as the wheel spins. The force is transfered over a longer time than compared to a fixed 10kh object. The wheel is spinning rather quickly by the time the shaft reaches its final travel point, now fully compressed it sends the last bit of its momentum into the man and sends him moving. While the wheel was spinning up as the shaft compressed on the man, the total mass of the shaft and wheel is still acting upon the man as it transfers momentum. It transfers it over a slightly longer time and makes the force less, but the transfer of momentum remains the same.

Now the man goes drifting backwards (at whatever velocity this would be Im not going to calculate that part). The man has taken the full energy of a 10kg object moving at 10m/s. But before the shaft-wheel falls to the ground, a person catches it. The person that catces it holds it in both hands as he watches the 100kg man drift away, but to his suprise the wheel on the end of the shaft is spinning!!.

If you calculate this with all the right equations you will find the total energy spent in the system is exactly equal to [math]\normalsize Ke=M*V^2[/math]. Using only the Mass of the object concerned to be in motion.

 

You asked for thoughts and that's what I'm giving. So please forgive a kid rapping on your knuckles.

 

but the transfer of momentum remains the same.

That is absolututely incorrect. Not only will the transfer of momentum be lesser in the shaft-having-wheel case, but also will the energy transferred be less than 500J in both cases.

 

I mean that if the shaft transfers 500J to the man, then the man and the shaft will still have as much energy as it had before. Thus, the shaft will stop moving and the man only will start moving. This is wrong.

Such a case will only take place when the man and the shaft have equal weights, and the collision is perfectly elastic.

 

The shaft hits the man in the chest and delivers 500 joules. However, the shaft does not hit the man and bounce off right away. Because it has a telescopic device that spins a wheel containg 5kg (which makes up half the mass of the shaft that is 10kg) as it his the man it begins decelerating as the wheel spins.

I'm not sure if that will be worth noticing. The collision time difference will be of the order of nanoseconds.

Further, we can consider this as a inelastic collision with the lost energy not dissipated, but sent into the spinning wheel

 

You know, even the big guys need some revision at times;) ...

 

PS: As I said earlier, please don't take offence. :)

[Racoon]

Ronthepon is Artificial Intelligence! :)

 

RUN!

[ronthepon]

Ronthepon is Artificial Intelligence! :)

 

RUN!

No, kiddo, I am not AI. I come from te planet Zorg. Every time you look at my avtar, some of your brains come to me.

[Racoon]

No, kiddo, I am not AI. I come from te planet Zorg. Every time you look at my avtar, some of your brains come to me.

 

ahhh, that would explain it Galaxy Traveler. :)

Alas, I have no Brains to extract...:)

 

What does KE= mean to Zorgians?