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Old Apr 6th 2018, 11:23 AM   #1
Pmb
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Without reading it I can tell you that you made a mistake somewhere. I know the principles extremely well and know where and how they came from. The cost to my soul being an ungodly number of study time.

If you knew/know how to derive these things you'd see that I speak the truth. When we derive such things its done very very carefully and in all generality. I can read what you wrote and show my work as well. However I'm not feeling well today so maybe this weekend. Is that a good time for you?

Pete
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Old Apr 6th 2018, 02:57 PM   #2
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the displacement caused by initial velocity and inertia is irrelevant when calculating work
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Old Apr 6th 2018, 05:16 PM   #3
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Originally Posted by Pmb View Post
Without reading it I can tell you that you made a mistake somewhere. I know the principles extremely well and know where and how they came from. The cost to my soul being an ungodly number of study time.

If you knew/know how to derive these things you'd see that I speak the truth. When we derive such things its done very very carefully and in all generality. I can read what you wrote and show my work as well. However I'm not feeling well today so maybe this weekend. Is that a good time for you?

Pete
I am Oleg Gorokhov.
Thank you for answering.

I don't know if I сan put pics and tabs in the text. If I can not, then please read these 3 links on Medium carefully:

https://goo.gl/aETJdQ (3 min read)
https://goo.gl/YffRQL (5 min read)
https://goo.gl/1q9HaU (32 min read)

Yes, this weekend is a good time for me.

Last edited by OlegGor; Apr 9th 2018 at 03:36 AM.
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Old Apr 6th 2018, 05:26 PM   #4
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Originally Posted by nightmare3399 View Post
the displacement caused by initial velocity and inertia is irrelevant when calculating work
Thank you for answering.

"inertia is irrelevant when calculating work"

Exactly!
I just say the same thing.
But now, according to this “current” formula,

W(E)=F*D ---- where, D=D1+D2

the Inertial Displacement (D2) erroneously increases the Work (W) of the Force (F) and, accordingly, erroneously increases the Energy (E) spent on this Work.

I don't know if I сan put pics and tabs in the text. If I can not, then please read these 3 links on Medium carefully:

https://goo.gl/aETJdQ (3 min read)
https://goo.gl/YffRQL (5 min read)
https://goo.gl/1q9HaU (32 min read)

Last edited by OlegGor; Apr 6th 2018 at 10:36 PM.
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Old Apr 6th 2018, 05:51 PM   #5
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Work done to accelerate an object by applying a force does in fact get larger the longer distance that it is applied for. This means the change in kinetic energy from when the force started to be applied also gets larger. Hence the speed of the object gets larger. This is true whether or not the object already had a speed when the force was applied.

What is incorrect about this? I don't know where your objection comes from.

Again, the formula for the work done on an object is a definition...there is no Physical content to this equation. Work is measured in N m = J. It's just a unit.

The physical content comes from equating the net work done on an object with its change in kinetic energy. I don't know what else to say. There is nothing wrong with all this. It has been successfully tested time and time again over the last few centuries.

-Dan
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Last edited by topsquark; Apr 6th 2018 at 05:54 PM.
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Old Apr 6th 2018, 07:33 PM   #6
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Originally Posted by OlegGor View Post
Hello.
I found a logical mistake at the very beginning of physics..
Ha ha .... I like your attitude OlegGor ....It's clear from your post that you have not studied the subject to degree level , yet you are confident you've discovered a big flaw at the heart of physics ...

Never loose that know it all confidence , it will take you far ....

I'll let others address your main question , I'll just point out that a jet engine also needs an atmosphere (containing oxygen) to operate , it sucks in the air , burns the oxygen with fuel which exits from the back at high speed.
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Old Apr 6th 2018, 08:27 PM   #7
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Originally Posted by oz93666 View Post
Ha ha .... I like your attitude OlegGor ....It's clear from your post that you have not studied the subject to degree level , yet you are confident you've discovered a big flaw at the heart of physics ...

Never loose that know it all confidence , it will take you far ....

I'll let others address your main question , I'll just point out that a jet engine also needs an atmosphere (containing oxygen) to operate , it sucks in the air , burns the oxygen with fuel which exits from the back at high speed.
Thank you for answering.

" I'll just point out that a jet engine also needs an atmosphere (containing oxygen) to operate , it sucks in the air , burns the oxygen with fuel which exits from the back at high speed."

You totally get this point wrong. I say about the completely different thing.

I don't know if I сan put pics and tabs in the text. If I can not, then please read these 3 links on Medium carefully:

https://goo.gl/aETJdQ (3 min read)
https://goo.gl/YffRQL (5 min read)
https://goo.gl/1q9HaU (32 min read)

Last edited by OlegGor; Apr 6th 2018 at 10:36 PM.
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Old Apr 6th 2018, 09:12 PM   #8
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Now science ERRONEOUSLY THINKS that the Energy of the gravitational Force (Karlsson’s Energy) is spent on this Inertial Displacement of the stone (9.8m, during the 2nd second).
No it doesn't. Look at the motion equation:
$\displaystyle d = v_0t + \frac{1}{2} at^2$

If what you say is correct then this equation would have to be wrong as well. But it is not because the v_0t term is included to take care of the problem that you have just mentioned. There is no problem with the work-energy theorem.

Something tells me that you are a stubborn person (so am I) and I suspect my reasoning will not convince you. So let's put it on the balance scales...Physics is an experimental science. Do an experiment on the W-K theorem. I'm sure any High School or College will have some decent experiment for you to work at.

-Dan
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Old Apr 6th 2018, 10:24 PM   #9
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Originally Posted by topsquark View Post
No it doesn't. Look at the motion equation:
$\displaystyle d = v_0t + \frac{1}{2} at^2$

If what you say is correct then this equation would have to be wrong as well. But it is not because the v_0t term is included to take care of the problem that you have just mentioned. There is no problem with the work-energy theorem.

Something tells me that you are a stubborn person (so am I) and I suspect my reasoning will not convince you. So let's put it on the balance scales...Physics is an experimental science. Do an experiment on the W-K theorem. I'm sure any High School or College will have some decent experiment for you to work at.

-Dan
Thank you for answering.

I don't disagree with this formula, d=v0t+at2/2
I absolutelly agree with it.

I see that you totally get this point wrong. I say about the completely different thing.
Please before you write your comment, read these 3 links on Medium carefully:

https://goo.gl/aETJdQ (3 min read)
https://goo.gl/YffRQL (5 min read)
https://goo.gl/1q9HaU (32 min read)

Last edited by OlegGor; Apr 6th 2018 at 10:36 PM.
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Old Apr 7th 2018, 10:58 AM   #10
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It seems to me that all you have doe is decide to create a new type of measurement that has units kg-m/s, and you call it "work." But you don't explain what advantage there is in using this new unit? Does it help us understand the motion of objects in a way that classical mechanics does not? In classical mechanics the definitions of work and energy are useful in helping us calculate the velocity of an object that is subjected to a net force operating over a distance. But your approach doesn't do that, so your approach seems to be of little value. Does it make predictions of an object's motion that are different than Newtonian mechanics would predict, and more importantly - can you demonstrate that your results are more accurate?

Also, the concept that a force acting for a period of time (rather than distance) should contribute to "work" falls apart when you consider that multiple forces working on an objects that cancel each other out would lead to the object having infinite "energy." The 20 Kg object resting on a table is constantly subjected the force of gravity, so over time gains an infinite amount of "energy" from gravity. How is that concept useful? How much "energy" does a 2 billion year old rock sitting on the ground have?
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Last edited by ChipB; Apr 7th 2018 at 11:02 AM.
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