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Kinetic Energy Of Train


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#18 VictorMedvil

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Posted 06 July 2019 - 03:29 PM

Just consider that one space ship A is moving in space, far away from astrological bodies. It's engine is shut down then definitely
it is in inertial state of motion. This inertial state is consider as state 1.
Another space ship B is moving in parallel direction with A space ship with constant velocity 4 m/s with relative to A & its engine
is also shut down. This inertial state is consider as state 2.
Another space ship C is moving in parallel direction with A space ship with constant velocity 8 m/s with relative to A & its engine
is also shut down. This inertial state is consider as state 3.
Now, as physics is same in all inertial reference frame, So if rocket of 10 kg is released & fired from space ship A with
velocity 4 m/s in same direction. It will consume 0.5x m x v^2 = 80 joule of energy.
Similarly, rocket of 10 kg is fired from spaceship B with velocity 4 m/s. It will consume 0.5x m x v^2 = 80 joule of energy.
But, rocket is released & fired from space ship A with velocity 8 m/s then it will consume 0.5x m x v^2 = 320 joule of energy.
This physics clearly shows that 10 kg mass rocket require 80 joule of energy to change the state of motion from state 1 to 2
& 2 to 3. So, total energy require to change the state of motion from state 1 to 3 in steps is 80 + 80 =160 joule. ---(1)
but to change the state of motion 1 to 3 directly, rocket will consume 0.5x m x v^2 = 320 joule of energy. -----(2)
(1) & (2) are different.
why?

 

Because you still didn't do the equation correctly in one of the states due to the fact it is not 80 + 80 but rather 80 + 240 you forgot to add the increase in energy requirements as it is V and not V while still taking E and not ΔE, as I said you have no idea what you are talking about and are high on that crack again. Do the equation correctly and you will get the correct answer! Stop solving it as a linear and not exponential. You cannot just take (1/2)MVAB2 + (1/2) MVAB, you must take   (1/2)MVAC- (1/2) MVABΔEBC

 

image-qgfhfe.png


Edited by VictorMedvil, 06 July 2019 - 03:58 PM.


#19 maheshkhati

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Posted 11 July 2019 - 02:23 AM

You are true but these are the calculations done by you in inertial frame of space ship A only.
here, we are actually interested in energy consumption of rocket.
If all work is done in space ship A inertial frame then work done = 320 joule. you are ture.
but work done in two stages then 1st work done will happen in inertial frame of shape ship A & then work done happen
in inertial frame space ship B. In both inertial frames, initial velocity is 0 & final velocity is 4 m/s so work done= 80 joule.
--------------------------------------------------------------------------------------------------------------
Problem is very simple. Bullet is fired in sky with velocity v then actual energy consumption by bullet to gain energy is
0.5 m v^2 or {0.5 m (v+vo)^2-0.5 m vo^2} where vo is velocity of earth around Sun in the direction of bullet.
Both will give different results.
If answer is 0.5 m v^2 then my calculations are true.

Edited by maheshkhati, 11 July 2019 - 02:34 AM.


#20 VictorMedvil

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Posted 11 July 2019 - 12:41 PM

You are true but these are the calculations done by you in inertial frame of space ship A only.
here, we are actually interested in energy consumption of rocket.
If all work is done in space ship A inertial frame then work done = 320 joule. you are ture.
but work done in two stages then 1st work done will happen in inertial frame of shape ship A & then work done happen
in inertial frame space ship B. In both inertial frames, initial velocity is 0 & final velocity is 4 m/s so work done= 80 joule.
--------------------------------------------------------------------------------------------------------------
Problem is very simple. Bullet is fired in sky with velocity v then actual energy consumption by bullet to gain energy is
0.5 m v^2 or {0.5 m (v+vo)^2-0.5 m vo^2} where vo is velocity of earth around Sun in the direction of bullet.
Both will give different results.
If answer is 0.5 m v^2 then my calculations are true.

 

Yes, but your are missing the change between the frames you gotta take the difference between the two, then lets say you want the BC area of the curve you gotta take the integral or the derivative.


Edited by VictorMedvil, 11 July 2019 - 12:42 PM.


#21 maheshkhati

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Posted 12 July 2019 - 01:44 AM

That is the problem of present relativity. We get energy consumption different for different inertial frames but
actual energy consumption with relative to only one reference frame is true.
For example;- old man pulling box with velocity v may be tired in platform frame but having not doing any work for observer
in train moving with velocity v parallel to him. Which is wrong observation.
So, I say that some are the special frame with relative to that we do work when we change the state of motion.

Edited by maheshkhati, 12 July 2019 - 01:48 AM.


#22 exchemist

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Posted 12 July 2019 - 09:39 AM

That is the problem of present relativity. We get energy consumption different for different inertial frames but
actual energy consumption with relative to only one reference frame is true.
For example;- old man pulling box with velocity v may be tired in platform frame but having not doing any work for observer
in train moving with velocity v parallel to him. Which is wrong observation.
So, I say that some are the special frame with relative to that we do work when we change the state of motion.

This is all very silly.

 

Whenever there is a change in kinetic energy, as a result of doing work, it will be the same regardless of reference frame. It is only the values before and after that will be different between frames.

 

So if someone at the roadside throws a cricket ball at the same speed and direction as a passing car, The car occupant will see the ball having a k.e. before it is thrown and none after, while the thrower sees it have no k.e. before and some after. But the difference in energy before and after is identical in both cases, as demanded by energy conservation.  

 

None of this has anything at all to do with SR or GR of course. It is school-level Newtonian mechanics.    


Edited by exchemist, 12 July 2019 - 09:39 AM.


#23 maheshkhati

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Posted 15 July 2019 - 03:12 AM

Your event is wrong, ball can not do any work in motion in space as cart do against friction.
--------------------------------------------------------------------------------
Let, consider man on Platform pulling the cart again friction with constant velocity V.
Now, he cross the poll A & then Poll B in the event.
Then for man on platform work done between A & B polls = F. V . dt
But, for observer in train with velocity v
Work done = F. 0 . dt = 0
As relative velocity is zero.
So, work done & energy consumption are different in this event in two inertial frames.
(Sorry due to official work load, I can not give reply in time)

Edited by maheshkhati, 15 July 2019 - 03:21 AM.


#24 maheshkhati

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Posted 15 July 2019 - 03:31 AM

This is all very silly.
 
Whenever there is a change in kinetic energy, as a result of doing work, it will be the same regardless of reference frame. It is only the values before and after that will be different between frames.
 
So if someone at the roadside throws a cricket ball at the same speed and direction as a passing car, The car occupant will see the ball having a k.e. before it is thrown and none after, while the thrower sees it have no k.e. before and some after. But the difference in energy before and after is identical in both cases, as demanded by energy conservation.  
 
None of this has anything at all to do with SR or GR of course. It is school-level Newtonian mechanics.

This is also the wrong. Just consider the velocity of ball is V then
Energy consumption for man on platform dE= 0.5m V^2 - 0 = 0.5 m V^2
Now, consider that second observer is in train moving with velocity u in direction opposite to ball velocity then
Energy consumption for observer in train dE=0.5 m (V+u)^2 - 0.5mu^2
Both are different.
-------------------------------------------------------------
In your case also it is not identical, in one case it is +ve & in other case it is -ve

Edited by maheshkhati, 15 July 2019 - 03:46 AM.


#25 VictorMedvil

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Posted 19 July 2019 - 02:03 AM

I am telling you, you have no idea what you are talking about this is all literal word salad with misused equations and concepts that have been proven to work flawlessly being questioned, please just stop.



#26 maheshkhati

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Posted 19 July 2019 - 02:16 AM

Just consider on road side well, on one pulley, one bucket of weight Fv is tied to rope & other end of rope is pulled by man on ground
& move horizontally with velocity ux then
Energy consume by man =Fh . ux . dt where Fh=horizontal force applied by man, ux=horizontal velocity of man
Work done = Fy . uy . dt where Fh=vertical force applied to bucket, uy= vertical velocity of bucket
So, as Fh=Fy & ux=uy in earth frame
on earth frame
Fh . ux . dt = Fy . uy . dt
Energy consume by man = Work done on bucket
----------------------------------------------------------
Now, for observer in car moving with velocity ux on ground
Energy consume by man =Fh . ux . dt = 0 as ux=0 (relative horizontal velocity is zero)
& Work done = Fy . uy . dt ...... is not zero as uy is not zero
So, in this inertial frame moving with velocity ux.
Fh . ux . dt... is not equal to..... Fy . uy . dt as ux=0
Energy consume by man is zero ... is not equal to..... Work done on bucket.
--------------------------------------------------------------

Edited by maheshkhati, 19 July 2019 - 02:19 AM.


#27 maheshkhati

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Posted 19 July 2019 - 02:23 AM

I am telling you, you have no idea what you are talking about this is all literal word salad with misused equations and concepts that have been proven to work flawlessly being questioned, please just stop.

Please tell me
In post 23, 24 & 26, where have I done the wrong?
In every post, I use kinematics correctly.

Edited by maheshkhati, 19 July 2019 - 02:27 AM.


#28 VictorMedvil

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Posted 19 July 2019 - 02:33 AM

Please tell me
In post 23, 24 & 26, where have I done the wrong?
In every post, I use kinematics correctly.

 

Here is your problem as exchemist said these are not relativistic formula these are Newtonian meaning you cannot take them as frames of reference, they are taken as a system together with one generalized solution such that there are no frames of reference the entire system is taken into account with one solution, taking one of these equations from a frame of reference is nonsense. When I spoke about frames earlier I meant the different energy states and not as a relativistic frame of reference only in relativity are there such things these are Newtonian.


Edited by VictorMedvil, 19 July 2019 - 02:39 AM.


#29 maheshkhati

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Posted 19 July 2019 - 02:48 AM

I have not consider any newtonian meaning or consider any frame of reference for entire system.
I just consider relative velocity which is frame dependent & it is different for different inertial frames.
In one inertial frame, ux is absent & in other it may present.
What is wrong in it?
One event can be observer by multiple observer.

Edited by maheshkhati, 19 July 2019 - 02:59 AM.


#30 exchemist

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Posted 19 July 2019 - 11:29 AM

I have not consider any newtonian meaning or consider any frame of reference for entire system.
I just consider relative velocity which is frame dependent & it is different for different inertial frames.
In one inertial frame, ux is absent & in other it may present.
What is wrong in it?
One event can be observer by multiple observer.

There is no point is discussing any of this further unless you are prepared to read, and learn, for yourself.

 

I suggest you start here: https://en.wikipedia.../Kinetic_energy and read in particular the section on frame of reference, plus this paragraph:

 

QUOTE

 

The kinetic energy of any entity depends on the reference frame in which it is measured. However the total energy of an isolated system, i.e. one in which energy can neither enter nor leave, does not change over time in the reference frame in which it is measured. Thus, the chemical energy converted to kinetic energy by a rocket engine is divided differently between the rocket ship and its exhaust stream depending upon the chosen reference frame. This is called the Oberth effect. But the total energy of the system, including kinetic energy, fuel chemical energy, heat, etc., is conserved over time, regardless of the choice of reference frame. Different observers moving with different reference frames would however disagree on the value of this conserved energy.

The kinetic energy of such systems depends on the choice of reference frame: the reference frame that gives the minimum value of that energy is the center of momentum frame, i.e. the reference frame in which the total momentum of the system is zero. 

 

UNQUOTE

 

If you will not do this, we must conclude that you have delusions that you have you made a great discovery, which you are determined not to disturb.

 

Furthermore, if you want to have a discussion you will have to be a lot clearer in how you express yourself. Post 26 in particular is total gibberish. My fear, however is that you are a person who is unable to think clearly. This may explain both your inability to express yourself and your misconceptions about kinetic energy. After all you do have a track record on this forum of posting gibberish. In fact you've been at it for four years. So probably we are wasting our time and should just ignore you. 


Edited by exchemist, 19 July 2019 - 11:33 AM.


#31 maheshkhati

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Posted 22 July 2019 - 04:28 AM

After week, I will answer your post, I am going away from city urgently for inspection.
I only say that you are mixing the correct things incorrectly.

#32 exchemist

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Posted 22 July 2019 - 08:56 AM

After week, I will answer your post, I am going away from city urgently for inspection.
I only say that you are mixing the correct things incorrectly.

Don't bother. I'm going away too and I am not going to devote much time to you on my return, given your track record here. There's only so much one can  get out of sparring with internet cranks.  



#33 maheshkhati

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Posted 02 August 2019 - 02:09 AM

The kinetic energy of such systems depends on the choice of reference frame: the reference frame that gives the minimum value
of that energy is the center of momentum frame, i.e. the reference frame in which the total momentum of the system is zero.
This is your quote:- This quote is completely wrong when event happen, force get applied then total moment in the system
do not remain zero. In every inertial frame of reference, when work happens & moment changes in that system from one value to other.
(so, when there is force, summation of moment is not zero in any system)
in your post 22, when cricket ball throw by person, in no frame you get zero moment.
Because F= dP/dt where P is moment
Or dP= F dt
---------------------------------------------------------------------------------------------------------
This F remains same in every reference frame. So, when you consider any event & force,
P will not remain same & zero.
For example:- Your above example,
Man throw the ball with velocity V
Work done = ds . F = ds . dp/dt = ds . ( mv-0)/dt
So, in this event moment of ball changes from 0 to mv
This observation of change in moment value is not depend on frame. It remain constant in every frame.

For example, for man on platform
F= (mv-0)/dt =m v/dt ------(1)
For frame 2 , man in train with velocity u in direction opposite to v
F={m(v+u)-mu}/dt = m v/dt -----(2)
Both (1) & (2) are same.
-----------------------------------------------------------------------------------------------
So, problem is not of moment as change in moment remain same in every inertial frame.
But problem is of ds i.e. displacement.
That remain different.
Please, explain wrong things in post 23, 24.
-------------------------------------------------------------------------------------------------------
I think we have to stop here because we both can not change each other view but
my view is very clear. We are actually living in different inertial space frames like rail cabin,
earth space etc where we get actual true energy consumption. For other inertial observer we get wrong energy consumption. As in
post 23, 24 & 26 ....1st chapter of http://vixra.org/abs/1903.0178
This thought if extended then it can solve the problem of dark energy as given in 2nd chapter of paper http://vixra.org/abs/1903.0178

Edited by maheshkhati, 02 August 2019 - 02:13 AM.