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galexander
11-01-2009, 12:10 PM
Consider the problem of powered space flight. The equation giving the kinetic energy obtained by any space vehicle which has been accelerated to a velocity v is of course, K.E. = ½ mv². However observe that to double the velocity of any space vehicle (by increasing v to 2v for example), four times as much energy would be required each time if the laws of the conservation of energy are to be maintained. This results from the relation between velocity and kinetic energy being non-linear. However surely this would go against common sense as the latter would tend to suggest that the formula were as follows, K.E. = mv.

Such an equation would simply state that to increase your velocity by a set amount in the frictionless environment of space the same proportion of energy would be required each time no matter what speed you were travelling at to begin with. The laws of conservation of energy would suggest that the energy supplied to the craft by its power source would at all times equal the kinetic energy obtained, taking into account the various efficiency ratios involved of course. If the kinetic energy did equal ½ mv² then this would result in increasingly larger amounts of fuel being needed to accelerate the craft which in turn would produce a limiting velocity requiring its own separate equation.

Of course we have all been taught since day one that it is actually momentum which equals mv and not kinetic energy. But is it really that straight forward? For example is momentum a force or is it actually a form of energy? Indeed the units it is measured in are rather peculiar, for example kilograms per metre per second. But which do these denote, force or energy?

Consider also that at present the world of physics recognises both the conservation of momentum and the conservation of kinetic energy. However how can both be conserved simultaneously in any given situation if one is equal to mv and the other ½ mv²? One is directly equal to ‘v’ and the other ‘v²’. It doesn’t take a brilliant mathematician to observe that this is clearly impossible.

No, I believe that the current equation for kinetic energy is wrong and should be replaced by the equation for momentum, mv.

But why is the current equation for kinetic energy wrong? This I believe is a result of the equation it is ultimately derived from, Work Done = Force x Distance, which is wrong also. Consider the following two examples:

Two men of exactly the same physical strength push two differing weights mounted on sets of wheels. If the second man pushes his weight for twice the length of time as the first man but with the exact same force surely he would have done twice as much work as a result? However the physics books inform us that this is not actually the case as the accepted relation between work and force is Work Done = Force x Distance.

However the second situation would surely indicate that this accepted relation would never apply. If the weight pushed by the second man is exactly double that of the first’s and is pushed through the exact same distance, each man could not have performed the exact same amount of work while pushing with the same force. On the contrary the second man would have needed to have pushed the heavier weight for a longer duration of time using the exact same force in order to have moved it through the same unit distance (as a result of its acceleration being less from F = ma), and this would have resulted in more work having been done. Again this must be the result of a new relation, Work Done = Force x Time. Our first example would also indicate the same relation.

The equation for kinetic energy is derived from the equation Work Done = Force x Time as follows:

K.E. = W.D. = ∫ F.dx = ∫ m dv/dt . dx = ∫ m dx/dt . dv

= ∫ m v dv = ½ mv²,

However if Work Done is actually equal to Force x Time this would change the existing equation for kinetic energy as follows:

K.E. = W.D. = ∫ F.dt = ∫ m dv/dt . dt = ∫ m dv = mv.


What are the consequences of this?

Well, there are many and I can’t go into them all. However first of all it has to be acknowledged that the current erroneous equation for kinetic energy is used throughout Quantum Physics and Relativity. This fact alone would tend to undermine much of the content of these two schools of thought in the world of physics.

Another stunning consequence is found in the world of astronomy in that meteorites no longer explode on impact. It has always been hotly debated whether the craters on both the Moon and the Earth were volcanic or meteoritic in origin. Certainly some of the low lying craters on Earth are provably volcanic in origin.

Perhaps you can think of other consequence yourselves?

EireEngineer
11-01-2009, 04:33 PM
While it may have been hotly debated in the past that craters on the moon are volcanic, it hasnt been in quite sometime. Since you can clearly see the ejecta fans around them it is fairly clear that they are impact craters. Not to mention the lack of evidence for any substantial vulcanism at any time in the moons history. In addition, whether or not a meteorite "explodes" on impact or simply dives its way into the ground has far more to do with its composition then with its kinetic energy at impact.

I think you might benefit from a kinematics course at your local community college.

BlueAngel
11-01-2009, 11:19 PM
You might want to check with Gale, Eire.

Apparently, he knows how to access the superscript capability and, as proven by his post, he has mastered it.

I might add that all of the members and viewers of the forum are so grateful that Gale has masterd the superscript capability because, without it, he would not be able to provide us with all of those equations and, if he couldn't do that, we would have no clue as to what was talking about.

:rolleyes:

galexander
11-02-2009, 11:45 AM
While it may have been hotly debated in the past that craters on the moon are volcanic, it hasnt been in quite sometime. Since you can clearly see the ejecta fans around them it is fairly clear that they are impact craters. Not to mention the lack of evidence for any substantial vulcanism at any time in the moons history. In addition, whether or not a meteorite "explodes" on impact or simply dives its way into the ground has far more to do with its composition then with its kinetic energy at impact.

I think you might benefit from a kinematics course at your local community college.

There are several different explanations for the ejecta fans on the Moon. My favourite is as follows. The Moon's crust has just solidified and is still thin but as the rest of the Moon solidifies and shrinks as a result, pockets of gas are forced towards the surface causing the solidified surface to bulge into domes. These domes then crack under the pressure causing a sudden release of gas followed by a complete collapse of the dome. The result of this collapse is ejecta fans.

Such unexplained domes, though far smaller in size can be seen on the Moon to this day.

If you believe I have my kinematics wrong then please inform me where my logic is faulty.

EireEngineer
11-02-2009, 12:24 PM
Its a cute theory, but no, it doesnt work that way. And wouldnt we still see these domes forming?

galexander
11-03-2009, 12:08 PM
Its a cute theory, but no, it doesnt work that way. And wouldnt we still see these domes forming?

No, the Moon is now almost entirely dormant.

http://www.phys.ncku.edu.tw/~astrolab/mirrors/apod/image/9809/copernicus_apollo17.jpg

Take a look at this image of the Copernicus Crater which famously has extensive 'rays' connected to it which span a large part of the Moon's globe. Does it look as if it were formed by a meteorite?

Take a closer look.

See how the inner parts of its walls consist of a series of collapsed terraces?

See also the central mountain peaks on the crater floor? How were these formed?

See also how flat and level the crater floor is. Why is the floor not curved and bowl shaped at the bottom?

Still convinced it was formed by a meteorite?

In my opinion the terracing is evidence of a domed feature collapsing and this explains the central mountain peaks which 'Bounced back' against the collapse.

EireEngineer
11-05-2009, 10:14 AM
No, the Moon is now almost entirely dormant.

http://www.phys.ncku.edu.tw/%7Eastrolab/mirrors/apod/image/9809/copernicus_apollo17.jpg

Take a look at this image of the Copernicus Crater which famously has extensive 'rays' connected to it which span a large part of the Moon's globe. Does it look as if it were formed by a meteorite?

Take a closer look.

See how the inner parts of its walls consist of a series of collapsed terraces?

See also the central mountain peaks on the crater floor? How were these formed?

See also how flat and level the crater floor is. Why is the floor not curved and bowl shaped at the bottom?

Still convinced it was formed by a meteorite?

In my opinion the terracing is evidence of a domed feature collapsing and this explains the central mountain peaks which 'Bounced back' against the collapse.
You seriously dont know how the peaks in the middle were formed? lol They are prime evidence that this is an impact crater. Its called a central uplift.
As for the collapsed terraces you often to see the ejecta form just such a pattern, especially on Earths moon and other terrestrial bodies in the solar system.
At least try to learn some of the science before spouting off such nonsense.

galexander
11-05-2009, 12:16 PM
You seriously dont know how the peaks in the middle were formed? lol They are prime evidence that this is an impact crater. Its called a central uplift.
As for the collapsed terraces you often to see the ejecta form just such a pattern, especially on Earths moon and other terrestrial bodies in the solar system.
At least try to learn some of the science before spouting off such nonsense.

I was using what's known as rhetoric. FOOL!! Check the dictionary under 'r'.

If you're a devout believer in NASA inspired quack science then bully for you.

Wasn't it NASA who 'finally proved' that the Moon's craters were meteoritic as a result of the Apollo landings? And don't 30% of the US public disbelieve that man ever set foot on the Moon?

Don't be so smug EireEngineer.

EireEngineer
11-06-2009, 01:03 PM
I was using what's known as rhetoric. FOOL!! Check the dictionary under 'r'.

If you're a devout believer in NASA inspired quack science then bully for you.

Wasn't it NASA who 'finally proved' that the Moon's craters were meteoritic as a result of the Apollo landings? And don't 30% of the US public disbelieve that man ever set foot on the Moon?

Don't be so smug EireEngineer.
Actually, I believe that it was research into terrestrial impact craters that explained the structure of those on the moon. Look at Crater Lake with its prominent central uplift. It wasnt NASA, it was geology.

galexander
11-07-2009, 12:23 PM
Actually, I believe that it was research into terrestrial impact craters that explained the structure of those on the moon. Look at Crater Lake with its prominent central uplift. It wasnt NASA, it was geology.

Yes and take a look at the world famous 'Meteor Crater' in Arizona. It is said to have been formed by a nickel-iron meteorite 50,000 years ago but if you look at photographs of the crater taken over-head its more square than round!

I mean its as square as your TV!

http://faculty.fortlewis.edu/tyler_c/classes/206/apod_az_crater.jpg

EireEngineer
11-07-2009, 05:32 PM
Yes and take a look at the world famous 'Meteor Crater' in Arizona. It is said to have been formed by a nickel-iron meteorite 50,000 years ago but if you look at photographs of the crater taken over-head its more square than round!

I mean its as square as your TV!

http://faculty.fortlewis.edu/tyler_c/classes/206/apod_az_crater.jpg
Only because you are looking at it from an angle. Its anything but square. You cant possibly be serious.

galexander
11-08-2009, 12:08 PM
Only because you are looking at it from an angle. Its anything but square. You cant possibly be serious.

Get real EireEngineer.

Looking from directly over-head you get exactly the same impression. Unfortunately I couldn't immediately find such an image on the internet.

Anyway you can see it quite clearly from an angle as well.

EireEngineer
11-08-2009, 02:21 PM
Get real EireEngineer.

Looking from directly over-head you get exactly the same impression. Unfortunately I couldn't immediately find such an image on the internet.

Anyway you can see it quite clearly from an angle as well.
Then you must truely suck at using google. It took me all of five seconds to find a overhead view, and i will concede the point that it is a very rounded square. However, you have to realize that the impact is not the only force at work there. The sides of the crater have erroded and collapsed over time, giving it a slightly more square shape.

galexander
11-09-2009, 11:41 AM
Then you must truely suck at using google. It took me all of five seconds to find a overhead view, and i will concede the point that it is a very rounded square. However, you have to realize that the impact is not the only force at work there. The sides of the crater have erroded and collapsed over time, giving it a slightly more square shape.

The next question is if its sides have eroded then why is the crater still so deep and its sides so precipitately steep?

And what is it about the soil/rock in the region that resistance to erosion is found in four directions?

Still an enigma for me.

JazzRoc
11-24-2009, 03:33 AM
if Work Done is actually equal to Force x Time

IT ISN'T

galexander
11-25-2009, 12:01 PM
if Work Done is actually equal to Force x Time

IT ISN'T

Please explain your logic JazzRoc.

JazzRoc
11-25-2009, 02:56 PM
Please explain your logic JazzRoc.
Work done is NOT equivalent to FORCE times TIME. I reckon it's FORCE times DISTANCE.

A rock rests on a flat plain on the surface of a gravitational body - say the Earth. It exerts a force on the plain - and the plain exerts a force on it, and nothing goes anywhere.

No work is being done on anything. But time is still passing....... and the force still pressing......

Work is only done when a force moves a mass some distance (lifting the rock off the plain against the gravitational acceleration). Or when a force accelerates a specified mass for a period of time. Acceleration involves a length dimension.

Therefore "Work" involves a length dimension. Distance.

I don't think that much in physics theory these days. I hope you understand what I write here. I knew you weren't correct here but to my astonishment I initially found it difficult to enunciate why. :)

galexander
11-26-2009, 12:20 PM
Work done is NOT equivalent to FORCE times TIME. I reckon it's FORCE times DISTANCE.

A rock rests on a flat plain on the surface of a gravitational body - say the Earth. It exerts a force on the plain - and the plain exerts a force on it, and nothing goes anywhere.

No work is being done on anything. But time is still passing....... and the force still pressing......

Work is only done when a force moves a mass some distance (lifting the rock off the plain against the gravitational acceleration). Or when a force accelerates a specified mass for a period of time. Acceleration involves a length dimension.

Therefore "Work" involves a length dimension. Distance.

I don't think that much in physics theory these days. I hope you understand what I write here. I knew you weren't correct here but to my astonishment I initially found it difficult to enunciate why. :)

Okay JazzRoc if work is not being done by the solidity of the Earth's surface against the solid rock then what about the following example?

Try holding a heavy sack of potatoes at arms length with just the one arm held horizontally. Hard WORK isn't it? And yet according to your logic no work is being done because the sack is stationary! :)

JazzRoc
11-26-2009, 01:18 PM
Okay JazzRoc if work is not being done by the solidity of the Earth's surface against the solid rock then what about the following example?

Try holding a heavy sack of potatoes at arms length with just the one arm held horizontally. Hard WORK isn't it? And yet according to your logic no work is being done because the sack is stationary! :)
Ah. I see where you're coming from.

Of course it's work for a human arm. A constant stream of work is being done maintaining muscular tension. That you might call STRAIN. But muscles work by electrochemical pressure, and their electrical "circuits" are very lossy. A constant stream of ATP (the body's gasoline) is required to maintain each muscular fibre's position. It gets "burnt" to lactic acid which is flushed to the kidneys.

But consider a rock on a plain on the Moon.

An old BIG rock which might have whacked down in an aftermath of an early asteroid strike and would have sat there for a million years. Straining. Consuming nothing. You need to understand the difference between plasticity and elasticity as well, by the sound of it.

Has it done a virtually infinite amount of work?

Or the Earth? The whole of it is in strain searching to fall to the centre. An almost infinite amount of work?

No. Work is done raising up a weight up using a pulley. Or the same amount of work done when it is released and falls back to where it was lifted from. Technically described as accelerating a mass over a period of time, and thus involving the dimension of length. In free fall the acceleration will be the local gravitational constant for the gravitational body. The product of the particular mass of that body, the acceleration, and the distance travelled will determine the work done.

In deep space, the acceleration will be determined by the mass divided by the force. The work done (say by a rocket engine) would then be the mass times times acceleration times distance the rocket engine works over.

Get your head round it. Pick up a schoolbook on mechanics and read and understand it.

Again I had to re-edit my work: I really must pick up a physics book again.

galexander
11-28-2009, 12:13 PM
Ah. I see where you're coming from.

Of course it's work for a human arm. A constant stream of work is being done maintaining muscular tension. That you might call STRAIN. But muscles work by electrochemical pressure, and their electrical "circuits" are very lossy. A constant stream of ATP (the body's gasoline) is required to maintain each muscular fibre's position. It gets "burnt" to lactic acid which is flushed to the kidneys.

But consider a rock on a plain on the Moon.

An old BIG rock which might have whacked down in an aftermath of an early asteroid strike and would have sat there for a million years. Straining. Consuming nothing. You need to understand the difference between plasticity and elasticity as well, by the sound of it.

Has it done a virtually infinite amount of work?

Or the Earth? The whole of it is in strain searching to fall to the centre. An almost infinite amount of work?

No. Work is done raising up a weight up using a pulley. Or the same amount of work done when it is released and falls back to where it was lifted from. Technically described as accelerating a mass over a period of time, and thus involving the dimension of length. In free fall the acceleration will be the local gravitational constant for the gravitational body. The product of the particular mass of that body, the acceleration, and the distance travelled will determine the work done.

In deep space, the acceleration will be determined by the mass divided by the force. The work done (say by a rocket engine) would then be the mass times times acceleration times distance the rocket engine works over.

Get your head round it. Pick up a schoolbook on mechanics and read and understand it.

Again I had to re-edit my work: I really must pick up a physics book again.

Why does a gravitating body ONLY do work on an object when it is moving? The gravitational field is there all the time.

Forget about nerves and ATP, consider the following example. I am on a steep hill in my car and I take the handbrake off. Instead of rolling back down the hill I have my clutch part way out and the accelerator pedal slightly depressed. The revs of the engine combined with clutch control stops the vehicle rolling back down the hill. The car engine is therefore doing work but yet the car is not moving. How is this possible?

Going back to the example of the rock sat on the surface of the Moon, on a quantum level you could indeed consider that the gravitational field was doing work on the stationary rock. Between atoms there is mainly space and what stops the atoms of the rock and the lunar surface combining in the gravitational field is an electrostatic repulsion in the other direction. This electrostatic repulsion is continuous and therefore the electric charges in the atom are continually doing work.

galexander
11-28-2009, 12:49 PM
Get your head round it. Pick up a schoolbook on mechanics and read and understand it.

Again I had to re-edit my work: I really must pick up a physics book again.

You're the one JazzRoc that needs to do the reading. I suggest you carefully re-read my original post, specifically the example of the two men pulling the two different weights.

Read carefully, understand it and get your head round it.

JazzRoc
11-29-2009, 06:16 AM
Why does a gravitating body ONLY do work on an object when it is moving? The gravitational field is there all the time.
Because a stationary object in the gravitational field will FALL unless it is restrained by a force. Then it won't. (Say.) That force is unmoving and therefore no work is being done.

Forget about nerves and ATP, consider the following example. I am on a steep hill in my car and I take the handbrake off. Instead of rolling back down the hill I have my clutch part way out and the accelerator pedal slightly depressed. The revs of the engine combined with clutch control stops the vehicle rolling back down the hill. The car engine is therefore doing work but yet the car is not moving. How is this possible?
I searched in vain for a good analogy for muscular effort the other night. This is it. When a human lifts a weight, his muscular function is exactly analogous to a slipping clutch. Humans do work to create a force. Machines don't.

Going back to the example of the rock sat on the surface of the Moon, on a quantum level you could indeed consider that the gravitational field was doing work on the stationary rock.
That's a FAIL. It's just exerting a force upon it.

Between atoms there is mainly space and what stops the atoms of the rock and the lunar surface combining in the gravitational field is an electrostatic repulsion in the other direction.
ALL gravitational fields pass through ALL materials without exception. You would get NOTHING in a physics paper for this.

This electrostatic repulsion is continuous and therefore the electric charges in the atom are continually doing work.
The electrostatic repulsion is FRICTIONLESS. Quantum rules. No work can ever be done upon them, (let's not consider electrochemical energies right now) except in the following manner:
ANY work done in the readjustment of an electron orbit will either require a photon input or produce a photon as output.
No atomic nuclei are ever exposed outside of nuclear fission or fusion.
All we EVER "feel" are the repulsions of such electron orbits.
EVEN when a tank is penetrated by an anti-tank round (possibly one of the hardest man-made impacts), NO nucleus of an atom of the metal of the round gets anywhere near contacting any nucleus of an atom of the metal of the tank.

galexander
11-29-2009, 01:47 PM
Because a stationary object in the gravitational field will FALL unless it is restrained by a force. Then it won't. (Say.) That force is unmoving and therefore no work is being done.


I searched in vain for a good analogy for muscular effort the other night. This is it. When a human lifts a weight, his muscular function is exactly analogous to a slipping clutch. Humans do work to create a force. Machines don't.


That's a FAIL. It's just exerting a force upon it.


ALL gravitational fields pass through ALL materials without exception. You would get NOTHING in a physics paper for this.


The electrostatic repulsion is FRICTIONLESS. Quantum rules. No work can ever be done upon them, (let's not consider electrochemical energies right now) except in the following manner:
ANY work done in the readjustment of an electron orbit will either require a photon input or produce a photon as output.
No atomic nuclei are ever exposed outside of nuclear fission or fusion.
All we EVER "feel" are the repulsions of such electron orbits.
EVEN when a tank is penetrated by an anti-tank round (possibly one of the hardest man-made impacts), NO nucleus of an atom of the metal of the round gets anywhere near contacting any nucleus of an atom of the metal of the tank.

Well how about the following example. Two teams contest in a tug of war. While both sides take the strain pulling either end of the rope and the red rag tied at the centre of the rope is not moving, you claim no work is being done. According to work done = force x distance, work is only ever done when the red rag is moving. Bravo, bravo!

I could go on and on, there are so many other examples I just couldn't think of them all!

And you claim I have a fail in physics?

RussyB
11-29-2009, 11:53 PM
Interesting post but how it can be brought into public?

JazzRoc
11-30-2009, 01:34 AM
Well how about the following example. Two teams contest in a tug of war. While both sides take the strain pulling either end of the rope and the red rag tied at the centre of the rope is not moving, you claim no work is being done. According to work done = force x distance, work is only ever done when the red rag is moving. Bravo, bravo! I could go on and on, there are so many other examples I just couldn't think of them all! And you claim I have a fail in physics?
That again is an example of musculature emulating a slipping clutch, and has nothing to do with simple mechanics. Do you KNOW how muscles work?
Engineering solved three hundred years ago the problem of how to avoid human beings doing WORK while in a motionless state. They used a RATCHET. As in a vehicular handbrake lever. The pawl locks the mechanism which remains under tension.
Because you wilfully will not understand the science of mechanics you believe the millions of engineers who preceded you have all missed something essential.
Any year ten science book will show you otherwise.
Waste your own time attempting to prove it wrong if you must. You won't.
The principle of how work is done is one of the most basic principles in Physics. It's less than 1% of all the physics theory there is, yet underlies all else.
If you cannot understand this clearly you're in a world full of pain, full of misconception after misconception. Yes, FAIL.

albie
11-30-2009, 06:03 AM
The tug of war is surely TWO works cancelling each other out. Not NO work, but they do seem the same.

galexander
11-30-2009, 11:31 AM
That again is an example of musculature emulating a slipping clutch, and has nothing to do with simple mechanics. Do you KNOW how muscles work?
Engineering solved three hundred years ago the problem of how to avoid human beings doing WORK while in a motionless state. They used a RATCHET. As in a vehicular handbrake lever. The pawl locks the mechanism which remains under tension.
Because you wilfully will not understand the science of mechanics you believe the millions of engineers who preceded you have all missed something essential.
Any year ten science book will show you otherwise.
Waste your own time attempting to prove it wrong if you must. You won't.
The principle of how work is done is one of the most basic principles in Physics. It's less than 1% of all the physics theory there is, yet underlies all else.
If you cannot understand this clearly you're in a world full of pain, full of misconception after misconception. Yes, FAIL.

Okay JazzRoc if you're so sure of yourself then explain away my original example in my first post. You haven't done this yet.

Indeed this is my best example:

Two men of exactly the same physical strength push two differing weights mounted on sets of wheels. If the weight pushed by the second man is exactly double that of the first’s and is pushed through the exact same distance, each man could not have performed the exact same amount of work while pushing with the exact same force. On the contrary the second man would have needed to have pushed the heavier weight for a longer duration of time using the exact same force in order to have moved it through the same unit distance (as a result of its acceleration being less from F = ma), and this would have resulted in more work having been done. This contradicts the accepted relation, Work Done = Force x Distance.

Take it away JazzRoc.

JazzRoc
12-01-2009, 02:04 AM
Two men of exactly the same physical strength push two differing weights mounted on sets of wheels. If the weight pushed by the second man is exactly double that of the first’s and is pushed through the exact same distance, each man could not have performed the exact same amount of work while pushing with the exact same force. On the contrary the second man would have needed to have pushed the heavier weight for a longer duration of time using the exact same force in order to have moved it through the same unit distance (as a result of its acceleration being less from F = ma), and this would have resulted in more work having been done. This contradicts the accepted relation, Work Done = Force x Distance.
That's from its definition.
You've missed out the essential point: PER UNIT OF MASS
Work Done = Force x MASS x Distance
W1 = F*M*D
W2 = 2*F*M*D
Therefore the work done in the second case is twice as much.

You'd best set yourself to re-evaluating where you stand.

galexander
12-01-2009, 11:58 AM
That's from its definition.
You've missed out the essential point: PER UNIT OF MASS
Work Done = Force x MASS x Distance
W1 = F*M*D
W2 = 2*F*M*D
Therefore the work done in the second case is twice as much.

You'd best set yourself to re-evaluating where you stand.

In my physics book, and yes I do actually own one which I have read again and again in the relevant sections, work is always quoted as equal to force multiplied by distance.

W.D. = F x d

I have never heard of this equation W.D. = F x m x d and believe you have just made it up in a vain effort to win the argument.

Nice try but no.......

I suggest you read a physics book yourself carefully.

And yes you do score a FAIL.

JazzRoc
12-01-2009, 01:39 PM
In my physics book, and yes I do actually own one which I have read again and again in the relevant sections, work is always quoted as equal to force multiplied by distance. W.D. = F x d I have never heard of this equation W.D. = F x m x d and believe you have just made it up in a vain effort to win the argument. Nice try but no.......I suggest you read a physics book yourself carefully. And yes you do score a FAIL.
Text books quoting FORCE are doing so in terms of MASS x ACCELERATION being FORCE. They are talking about LIFTING WEIGHTS, and as you know, the WEIGHT of a body is actually its mass times its acceleration due to gravity.
Hence the true equation is as I have said Work done = Distance x Mass x Acceleration (frequently G). In space (free of fields) this still applies. The acceleration is whatever is imparted to the mass over the particular distance.
It's something that most people intuit, but apparently you don't. This "intuition" has allowed most physics book writers to abbreviate what they mean to the point where it's possible to misunderstand. It really should be obvious to you that pushing an object twice as heavy as another object should involve twice as much work. More obvious, surely, than suddenly imagining you had discovered a new branch of physics.
In your options, the second requires twice as much work as the first.
And you're still as wrong as you were before. You aren't reading what I'm writing.
You may think what you like. If you're in a permanent mess because you can't see it any other way, this will be quite to my taste. I like seeing hubris. :)

JazzRoc
12-01-2009, 01:55 PM
Text books quoting FORCE are doing so in terms of MASS x ACCELERATION being FORCE. They are talking about LIFTING WEIGHTS, and as you know, the WEIGHT of a body is actually its mass times its acceleration due to gravity.
Hence the true equation is as I have said Work done = Distance x Mass x Acceleration (frequently G). In space (free of fields) this still applies. The acceleration is whatever is imparted to the mass over the particular distance.
It's something that most people intuit, but apparently you don't. This "intuition" has allowed most physics book writers to abbreviate what they mean to the point where it's possible to misunderstand them.
It really should be obvious to you that pushing or lifting an object twice as heavy as another object should involve twice as much work. More obvious, surely, than suddenly imagining you had discovered a new branch of physics.
In your options, the second requires twice as much work as the first.
And you're still as wrong as you were before. You aren't reading what I'm writing.
You may think what you like. If you're in a permanent mess because you can't see it any other way, this will be quite to my taste.

I like seeing hubris in others. :)

galexander
12-02-2009, 11:28 AM
Text books quoting FORCE are doing so in terms of MASS x ACCELERATION being FORCE. They are talking about LIFTING WEIGHTS, and as you know, the WEIGHT of a body is actually its mass times its acceleration due to gravity.
Hence the true equation is as I have said Work done = Distance x Mass x Acceleration (frequently G). In space (free of fields) this still applies. The acceleration is whatever is imparted to the mass over the particular distance.
It's something that most people intuit, but apparently you don't. This "intuition" has allowed most physics book writers to abbreviate what they mean to the point where it's possible to misunderstand them.
It really should be obvious to you that pushing or lifting an object twice as heavy as another object should involve twice as much work. More obvious, surely, than suddenly imagining you had discovered a new branch of physics.
In your options, the second requires twice as much work as the first.
And you're still as wrong as you were before. You aren't reading what I'm writing.
You may think what you like. If you're in a permanent mess because you can't see it any other way, this will be quite to my taste.

I like seeing hubris in others. :)

You're just trying to cloud the issue now.

Also you have just contradicted what you previously claimed. Earlier you said it was W.D. = Force x Mass x Distance and now its W.D. = Distance x Mass x Acceleration.

Make your mind up JazzRoc.

Anyway your last equation still doesn't satisfactorily explain my example. The mass is fixed as is the force pushing the weights. And I don't dispute that the acceleration is different for each weight.

Looks like I've got you on the run JazzRoc.

JazzRoc
12-03-2009, 02:18 AM
You're just trying to cloud the issue now.
Also you have just contradicted what you previously claimed. Earlier you said it was W.D. = Force x Mass x Distance and now its W.D. = Distance x Mass x Acceleration.
Make your mind up JazzRoc. Anyway your last equation still doesn't satisfactorily explain my example. The mass is fixed as is the force pushing the weights. And I don't dispute that the acceleration is different for each weight. Looks like I've got you on the run JazzRoc.
"W.D. = Force x Mass x Distance and W.D. = Distance x Mass x Acceleration"
Well. The latter is correct.

I had to dredge back some fifty years. You should try that sometime.

When a force acts on a body and causes it to move against a resistance the force is said to do work.
If the force does not vary in magnitude and is in the direction of motion, the amount of work done is equal to the product of the force and the distance moved, or -
Work = force x distance
If a force of one pound acts through a distance of one foot, the amount of work done is -
1 foot x 1 lb = 1 foot-pound
and one foot-pound is called the unit of work. If a weight of 1 lb is lifted through a vertical distance of 1 foot, 1 foot-pound of work is done. If a weight of, say, 7 lbs is lifted through a vertical distance of 9 feet, the work spent in lifting is -
9 x 7 = 63 foot-pounds

Now the "pound" is the force a pound mass exerts in the earth's gravitational field.
The "pound" force is the MASS of that pound times the ACCELERATION due to gravity.
hence W.D. = Distance x Mass x Acceleration

My stickiness with this subject is to do with Metric and Imperial units. Imperial (American) I find much easier to deal with: foot-pounds instead of G x newton-metres.

But the important point is that weight = mass x G, which is where confusion may spring.

Also work calculations done outside a gravity field have to take into account the acceleration involved from one speed to another to arrive at resolution: they no longer involve a return to a rest state.

Good luck. :)