Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum  7Likes
Apr 8th 2018, 09:25 AM

#21  Physics Team
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Posts: 2,285

Originally Posted by OlegGor Do you admit that the 2nd seconds in these two processes are completely identical?
Do you admit that the Karlson and the gravitational Force are one and the same thing  each one does the same thing with the stone (during the 2nd second)? 
Already asked and answered, so yes.
Originally Posted by OlegGor Do you admit that Karlson’s Energy (Resource) will not be spent on this Inertial Displacement of the stone (9.8m, during the 2nd second), since the stone moves this 9.8m displacement, during the 2nd second, also by inertia, as during the 1st second? 
No.
The problem is you use the term "inertial displacement" without defining it adequately. You are arguing that Karlson need not expend any energy to cause the stone to move the first 9.8m during the 2nd second, but you ignore the fact that the stone is accelerating to a higher velocity during that first 9.8m, and accelerating it to that higher velocity requires energy.
Last edited by ChipB; Apr 8th 2018 at 09:29 AM.

 
Apr 8th 2018, 09:49 AM

#22  Junior Member
Join Date: Apr 2018
Posts: 16

Originally Posted by topsquark Yes. Please read the responses. If nothing changes the motion due to gravity and the motion due to Karl are the same for all time, t.
Dan 
I have read your answers and I asked ChipB these questions .
(Only because of that I repeated my 3 questions for ChipB)
It's important to understand that I say about only the 2nd second! Only.
So I take your answer as Yes.
Originally Posted by topsquark Sort of. Again, so long as the initial velocity and acceleration are the same in both cases the motion will be the same.
Dan 
"so long as"  the answer is allways during the 2nd second.
So I take your answer as Yes.
Originally Posted by topsquark
That depends on what an "Energy Resource" is. Sorry, I'm cranky, tired, achy, feverish and I'm not going to read your article to point out all the flaws. We can do that a lot more simply in the thread. So please define what an energy resource is.
Dan 
I consider that in the most general sence:
Work is any activity that requires effort;
Energy is an ability (resource) to do Work.
It is important to understand that when any Work is being done, Energy (resource to do Work) is always being spent on this Work. Always.
So I will ask my question once again, 
Do you admit that Karlson’s Energy (any resources of Karlson) will not be spent on this Inertial Displacement of the stone (9.8m, during the 2nd second), since the stone moves this 9.8m displacement, during the 2nd second, also by inertia, as during the 1st second?

 
Apr 8th 2018, 10:16 AM

#23  Forum Admin
Join Date: Apr 2008 Location: On the dance floor, baby!
Posts: 2,355

Originally Posted by OlegGor I have read your answers and I asked ChipB these questions .
(Only because of that I repeated my 3 questions for ChipB)
It's important to understand that I say about only the 2nd second! Only.
So I take your answer as Yes.
"so long as"  the answer is allways during the 2nd second.
So I take your answer as Yes.
I consider that in the most general sence:
Work is any activity that requires effort;
Energy is an ability (resource) to do Work.
It is important to understand that when any Work is being done, Energy (resource to do Work) is always being spent on this Work. Always.
So I will ask my question once again, 
Do you admit that Karlson’s Energy (any resources of Karlson) will not be spent on this Inertial Displacement of the stone (9.8m, during the 2nd second), since the stone moves this 9.8m displacement, during the 2nd second, also by inertia, as during the 1st second? 
Any impulse that deals with the effects of inertia is taken care of when the object starts to move under the applied force. You don't need to keep adding it, otherwise you are "overcounting."
In both cases $\displaystyle W = F \cdot d = (ma) \cdot \left ( v_0 t + (1/2) at^2 \right )$
What part of these equations do you think is wrong?
Dan
Addendum: And why are you so hot on t = 2 s? The effect you are trying to explain has to do with the time interval. Your effect would start taking place at just over 0.0000000000000000000000000000001 seconds, not for just t = 2 s.
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Last edited by topsquark; Apr 8th 2018 at 10:19 AM.

 
Apr 8th 2018, 10:16 AM

#24  Junior Member
Join Date: Apr 2018
Posts: 16

Originally Posted by ChipB Already asked and answered, so yes. 
So your answer for 1 and 2 questions is Yes.
Good.
Originally Posted by ChipB No.
The problem is you use the term "inertial displacement" without defining it adequately. You are arguing that Karlson need not expend any energy to cause the stone to move the first 9.8m during the 2nd second, but you ignore the fact that the stone is accelerating to a higher velocity during that first 9.8m, and accelerating it to that higher velocity requires energy. 
"inertial displacement" is the displacement that the stone moves by inertia through (without spending any resources)
For example, the stone will move through a displacement of 9.8m during the 1st second only by inertia (i.e. without spending any resources). It is important to understand that no Energy (no resources) will be spent in this process.
"first 9.8m during the 2nd secon"
Perhaps here is a disunderstanding.
This 9.8m displacement is not the first one.
The stone does not move first the 9.8m displacement by inertia (during the 2nd second) and then 4.9m displacement (under the pushing of the gravitational Force).
The stone is being under the pushing of the gravitational Force during the whole 2nd second also and as the stone is being moved by inertia also during the whole 2nd second. These 2 movement happen simultaneously during the whole 2nd second.
Last edited by OlegGor; Apr 8th 2018 at 11:17 AM.

 
Apr 8th 2018, 10:22 AM

#25  Forum Admin
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Originally Posted by OlegGor So your answer for 1 and 2 question is Yes.
Good.
"inertial displacement" is the displacement that the stone moves by inertia through (without spending any resources)
For example, the stone will move through a displacement of 9.8m during the 1st second only by inertia (i.e. without spending any resources). It is important to understand that no Energy, that is, no resources will not be spent in this process.
"first 9.8m during the 2nd secon"
Perhaps here is a disunderstanding.
This 9.8m displacement is not the first one.
The stone does not move first the 9.8m displacement by inertia (during the 2nd second) and then 4.9m displacement (under the pushing of the gravitational Force).
The stone is being under the pushing of the gravitational Force during the whole 2nd second also and as the stone is being moved by inertia also during the whole 2nd second. These 2 movement happen simultaneously during the whole 2nd second. 
I'm going to take a stab at a new direction. Are you aware that inertia is not a force, right? It is for this reason that W = Fd doesn't need to make any explicit reference to it.
Dan
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Apr 8th 2018, 10:46 AM

#26  Junior Member
Join Date: Apr 2018
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Originally Posted by topsquark Any impulse that deals with the effects of inertia is taken care of when the object starts to move under the applied force. You don't need to keep adding it, otherwise you are "overcounting."
In both cases $\displaystyle W = F \cdot d = (ma) \cdot \left ( v_0 t + (1/2) at^2 \right )$
What part of these equations do you think is wrong? 
As an user "nightmare3399@ has already said in the comments, 
"inertia is irrelevant when calculating work"
That part of displacment (9.8m of 14.7m) that the stone is moving by inertia through can not increase the Work. (But this happens exactly so, accoding to the "current" formula).
For example, the stone will move through a displacement of 9.8m during the 1st second only by inertia (i.e. without spending any resources). It is important to understand that no Energy (no resources) will be spent in this process.
It is very important to understand that in this “old” formula for Work and Energy,
W(E)=F*D (where, D=D1+D2)
in the calculation of Work and Energy, the Summary Displacement (D) was erroneously taken into account, in which 2 completely different displacements were summarized:
 the NonInertial Displacement (D1), which the object moves as a result of the Work (Pushing) of the gravitational Force; and
 the Inertial Displacement (D2), which the object moves as a result of motion by inertia.
We may not “give credit” to the gravitational Force for the Inertial Displacement (D2), in other words, Inertial Displacement (D2) must not increase the Work (W) of the Force (F).
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) expended on this Work.
Thus, the using of Inertial Displacement (D2) in the calculation of the Work and the Energy (that is, the calculation of the Joule) is absolutely erroneous!
The answer is the v*t part (it is the D2) into the D (v*t+(1/2)at2) can not increase the Work.
Only the (1/2)at2 displacement is the result of Work of Force and can increase the Work of the Force.
Last edited by OlegGor; Apr 8th 2018 at 11:17 AM.

 
Apr 8th 2018, 10:52 AM

#27  Junior Member
Join Date: Apr 2018
Posts: 16

Originally Posted by topsquark I'm going to take a stab at a new direction. Are you aware that inertia is not a force, right? It is for this reason that W = Fd doesn't need to make any explicit reference to it.
Dan 
But inside the D (in F*D) "sits" the inertial displacement.
It is very important to understand that in this “old” formula for Work and Energy,
W(E)=F*D (where, D=D1+D2)
in the calculation of Work and Energy, the Summary Displacement (D) was erroneously taken into account, in which 2 completely different displacements were summarized:
 the NonInertial Displacement (D1), which the object moves as a result of the Work (Pushing) of the gravitational Force; and
 the Inertial Displacement (D2), which the object moves as a result of motion by inertia.
We may not “give credit” to the gravitational Force for the Inertial Displacement (D2), in other words, Inertial Displacement (D2) must not increase the Work (W) of the Force (F).
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) expended on this Work.
Thus, the using of Inertial Displacement (D2) in the calculation of the Work and the Energy (that is, the calculation of the Joule) is absolutely erroneous!

 
Apr 8th 2018, 04:16 PM

#28  Physics Team
Join Date: Apr 2009 Location: Boston's North Shore
Posts: 1,526

Wow. You must have little face of the millions of physicists since Newton who know you're wrong. I explained to you why too. So instead of telling you again I'll now show you. See: WorkEnergy Theorem 
 
Apr 8th 2018, 07:30 PM

#29  Junior Member
Join Date: Apr 2018
Posts: 16

Originally Posted by Pmb Wow. You must have little face of the millions of physicists since Newton who know you're wrong. I explained to you why too. So instead of telling you again I'll now show you. See: WorkEnergy Theorem 
topsquark liked your comment.
So sweet.
So you both (Pmb and topsquark) have one (such a pitiful) argument from you both.
Do not drag Newton into your pitiful stuff.
I have said in the main text that
"In my opinion, basic errors in science occurred because people have not fully understood all the three Newton’s discoveries: Newton’s 1st, 2nd and 3rd Laws."
And I explained it.
So what is your logical argument on my logical ones?
Before you write your comments please do your work and read (at least here) all my texts once again but now carefully.
Also https://goo.gl/aETJdQ (3 min read) https://goo.gl/YffRQL (5 min read) https://goo.gl/1q9HaU (32 min read)
As the user "nightmare3399@ has already said in the comments, 
"inertia is irrelevant when calculating work"
I would say that the inertial displacement must be irrelevant when calculating Work and Energy.
Because when the stone moves by inertia (the inertial displacement)  no Energy, no resources are being spent in this process.
But you somehow do not understand it.
Last edited by OlegGor; Apr 8th 2018 at 08:01 PM.

 
Apr 8th 2018, 08:35 PM

#30  Forum Admin
Join Date: Apr 2008 Location: On the dance floor, baby!
Posts: 2,355

I got some sleep so I'm a little less cranky now. I finally got around to reading the long document.
If I can wade through your 30 minute essay then I ask that you carefully read what I am writing here.
You aren't trying to discuss work, which is why we are having a problem. The concept you are referring to is called "impulse" and is defined as the average force applied to an object over a given time that the force acts. The impulse given to an object is equal to its change in momentum. (For example, I = Ft = mat, whereas you wrote W = mat.) If you change your discussion and replace the word work with the word impulse you pretty much have the base of a good argument. Conceptually this works.
Of course you have started redefining other concepts in terms of the impulse as being work so the rest of it gets skewed.
I would like to conclude this post with two comments.
1) We can define anything we want as long as it gives us something useful. We can always define work as Fd. We can always define kinetic energy as (1/2)mv^2. The Physics doesn't come in until we note that the work done by a force is equal to the change in kinetic energy of that object...which extends to the Law of Conservation of Energy. (At the current level of discussion. Work is a more general concept. See 2). ) What you are saying is that this principle is wrong. Okay, I'll play with it for a while, but you have to do me a favor and design an experiment that proves your statements about "colossal clean energy." If "standard" Physics doesn't meet expectations then you have some room for change. But it's getting to an equation for conservation of energy that is the whole goal.
2) The definition for the work done is far more abstract than you seem to think. The definition of work being done is that the total energy of an object changes. For example if we have a weight lifter that is lifting and holding a barbel up into the air W = Fd just doesn't cut it. The lifter is obviously putting forth some kind of effort. But we know that work has to be done in holding it up... it's the change in the chemical potential energy within the lifter that provides the force to keep the barbel up. Hence work is being done even if the weight isn't moving. Your discussion doesn't seem to be capable of deriving this result.
You are a creative person and have obviously spent much time on this question. I congratulate you on this... we need more people like that. For the record I am not disregarding your views out of hand because they don't match up with what I've learned. I am disregarding them in the face of you having no physical proof of your statements whereas current Physics has made and confirmed experimental results since the Renaissance era.
Dan
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Last edited by topsquark; Apr 8th 2018 at 08:47 PM.

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