# Kinetic Energy Of Train

### #1

Posted 15 June 2019 - 02:58 AM

Let's train engine of 100 t is at rest on platform & then engine start & increases it's velocity to 4 m/s.

So, energy require for this motion will be

Kinetic energy of engine = 0.5 m v^2

= 0.5 x 100000 x 4^2

= 800000 joule --(A)

1) Now, I consider in steps

Step1:- First train engine increase its velocity from 0 to 1 m/s

So, energy require == 0.5 x 100000 x 1^2

= 50000 joule ----(1)

Step2:- Then after some time, from this inertial velocity of 1 m/s, velocity increases by 1 m/s again

So, train engine increase its velocity from 1 to 2 m/s

So, energy require for this again increase in 1 m/s velocity from previous velocity

== 0.5 x 100000 x 1^2

= 50000 joule ----(2)

Step3:- Then after some time, from this inertial velocity of 2 m/s, velocity increases by 1 m/s again

So, train engine increase its velocity from 2 to 3 m/s

So, energy require for this again increase in 1 m/s velocity == 0.5 x 100000 x 1^2

= 50000 joule ----(3)

Step4:- Then after some time, from this inertial velocity of 3 m/s, velocity increases by 1 m/s again

So, train engine increase its velocity from 3 to 4 m/s

So, energy require for this again increase in 1 m/s velocity == 0.5 x 100000 x 1^2

= 50000 joule ----(3)

So, in steps velocity require to increase from rest to 4 m/s is sum of (1) to (4)

So, total energy require to get velocity 4 m/s by engine by step = 200000 Joules--(

Here, (A) & ( are different

Why

### #2

Posted 15 June 2019 - 06:56 AM

I just come across one problem in enginering.

Let's train engine of 100 t is at rest on platform & then engine start & increases it's velocity to 4 m/s.

So, energy require for this motion will be

Kinetic energy of engine = 0.5 m v^2

= 0.5 x 100000 x 4^2

= 800000 joule --(A)

1) Now, I consider in steps

Step1:- First train engine increase its velocity from 0 to 1 m/s

So, energy require == 0.5 x 100000 x 1^2

= 50000 joule ----(1)

Step2:- Then after some time, from this inertial velocity of 1 m/s, velocity increases by 1 m/s again

So, train engine increase its velocity from 1 to 2 m/s

So, energy require for this again increase in 1 m/s velocity from previous velocity

== 0.5 x 100000 x 1^2

= 50000 joule ----(2)

Step3:- Then after some time, from this inertial velocity of 2 m/s, velocity increases by 1 m/s again

So, train engine increase its velocity from 2 to 3 m/s

So, energy require for this again increase in 1 m/s velocity == 0.5 x 100000 x 1^2

= 50000 joule ----(3)

Step4:- Then after some time, from this inertial velocity of 3 m/s, velocity increases by 1 m/s again

So, train engine increase its velocity from 3 to 4 m/s

So, energy require for this again increase in 1 m/s velocity == 0.5 x 100000 x 1^2

= 50000 joule ----(3)

So, in steps velocity require to increase from rest to 4 m/s is sum of (1) to (4)

So, total energy require to get velocity 4 m/s by engine by step = 200000 Joules--(

Here, (A) & ( are different

Why

Because work done is force x distance. Once the thing is already moving, you have to apply the force through a longer distance to achieve a given change in speed. So you do more work, in other words give it more kinetic energy. This is one reason the acceleration of a train tails off the faster it goes. It is not just air resistance.

**Edited by exchemist, 15 June 2019 - 06:56 AM.**

### #3

Posted 21 June 2019 - 03:45 AM

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Just consider one long spacecraft of 10 t accelerate from velocity 0 to 4 m/s from one space station. In this spacecraft's long cabin,

one small rocket of 10 kg is fixed at the ceiling.

(So, Energy require to move spacecraft from velocity 0- 4 m/s = 0.5 X 10000 X 4^2 =80000 Joule)

& Part of energy to move rocket of mass 10 kg from velocity 0- 4 m/s with spacecraft = 0.5 X 10 kg X 4^2 =80 Joule -----(1)

Step 2:- Now, small rocket in long spacecraft cabin is released from ceiling of roof & get fired in cabin from velocity 0 - 4 m/s

Now, rocket will require 0.5 X 10 kg X 4^2 =80 Joule of energy again in this new inertial frame of cabin.

Now, Total velocity gain by rocket is 4 + 4 = 8 m/s form space station by consuming 80 + 80 = 160 joule of energy.

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If I want to fire directly the same mass small rocket with velocity 8 m/s.

I will require 0.5 X 10 kg X 8^2 =320 Joule of energy.

Now, both same mass rocket will be at same velocity for space station

but require different energy i.e. 160 joule in steps & 320 joule in no step.

Why?

**Edited by maheshkhati, 21 June 2019 - 03:47 AM.**

### #4

Posted 21 June 2019 - 07:38 AM

I am sorry. I was away from town. So, can't reply in time. To explain my problem, I simplify it again.

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Just consider one long spacecraft of 10 t accelerate from velocity 0 to 4 m/s from one space station. In this spacecraft's long cabin,

one small rocket of 10 kg is fixed at the ceiling.

(So, Energy require to move spacecraft from velocity 0- 4 m/s = 0.5 X 10000 X 4^2 =80000 Joule)

& Part of energy to move rocket of mass 10 kg from velocity 0- 4 m/s with spacecraft = 0.5 X 10 kg X 4^2 =80 Joule -----(1)

Step 2:- Now, small rocket in long spacecraft cabin is released from ceiling of roof & get fired in cabin from velocity 0 - 4 m/s

Now, rocket will require 0.5 X 10 kg X 4^2 =80 Joule of energy again in this new inertial frame of cabin.

Now, Total velocity gain by rocket is 4 + 4 = 8 m/s form space station by consuming 80 + 80 = 160 joule of energy.

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If I want to fire directly the same mass small rocket with velocity 8 m/s.

I will require 0.5 X 10 kg X 8^2 =320 Joule of energy.

Now, both same mass rocket will be at same velocity for space station

but require different energy i.e. 160 joule in steps & 320 joule in no step.

Why?

Not so fast, my friend. First, do you now follow what I am saying in post 2?

Once you have that straight, then we can talk about your rocket scenario.

### #5

Posted 24 June 2019 - 01:37 AM

I except for observer in the 1st frame of space station work done is same in both cases.

but my problem is different. We are actually considering two different inertial frames. In which we can add velocity

but can not add kinetic energy.

So, what is actual consumption of rocket, 160 j or 320 j

### #6

Posted 24 June 2019 - 01:53 AM

Yes,

I except for observer in the 1st frame of space station work done is same in both cases.

but my problem is different. We are actually considering two different inertial frames. In which we can add velocity

but can not add kinetic energy.

So, what is actual consumption of rocket, 160 j or 320 j

You have forgotten that the rocket engine also accelerates a stream of exhaust. The work it does is the sum of the kinetic energy imparted to the exhaust and that imparted to the rocket. If you take that into account, you will find the discrepancy disappears.

### #7

Posted 24 June 2019 - 03:52 AM

exhaust but problem is of displacement calculation of rocket with one inertial frame or two inertial frame.

Which is directly related to relative velocity with frame.

In one frame same force get applied to longer distance (consider acceleration same) & in other frame same force get applied

in shorter distance.

**Edited by maheshkhati, 24 June 2019 - 03:53 AM.**

### #8

Posted 24 June 2019 - 04:38 AM

You are wrong. Accelerated gases create force but in this problem force is not the problem, as it can be remain same by adjusting

exhaust but problem is of displacement calculation of rocket with one inertial frame or two inertial frame.

Which is directly related to relative velocity with frame.

In one frame same force get applied to longer distance (consider acceleration same) & in other frame same force get applied

in shorter distance.

Well, kinetic energy is obviously frame-dependent. But you still need to take into account the energy expended accelerating the exhaust.

If I'm wrong, then how do you explain it?

**Edited by exchemist, 24 June 2019 - 04:44 AM.**

### #9

Posted 25 June 2019 - 03:37 AM

We are living in different inertial local spaces. So, energy calculation for different inertial spaces are different.

So, energy calculation in spaceship frame is different than in space station frame & can not be compare.

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In 1998, I was discussing relativity with Dr Ram Mohan Rao. I said that we can explain M-Morley experiment finding by using

thermodynamics.

For example, we are in train cabin, moving with some velocity on earth. One photon of light move in the direction of train velocity

& mirror reflect that light perpendicularly to that direction.

Now, if I consider that the Train cabin as closed system then no energy will come into the system & no energy will go out from

the system. So, energy of photon in closed system will remain same after reflection. The velocity of reflected photon can not be

increase or decrease as no energy is added to photon in the closed system from outside.

so, velocity of reflected light remain same in this closed system in all directions.

**Edited by maheshkhati, 25 June 2019 - 03:48 AM.**

### #10

Posted 25 June 2019 - 03:49 AM

I wanted to say if you are doing your calculations like that you are doing them wrong as there is a exponential increase in the amount of energy required to move objects at a velocity you have to take the ΔE you cannot just take E for a train in motion, to accelerator the object from 1 m/s to 2 m/s will take less energy than from 3 m/s to 4 m/s, as velocity is squared in the equation E = 1/2 MV^{2 } to find the actual energy of the movements between energy levels you need to take ΔE which = E_{2} - E_{1 }= 1/2 M_{2}V_{2}^{2} - 1/2 M_{1}V_{1}^{2 }, so maheshakhati you are taking the equation incorrectly in your example. This proves to me you have absolutely no idea what you are talking about with kinetic energies and have no idea how thermodynamics works.(https://www.khanacad.../thermodynamics) and possibly Work and Energy physics(https://www.khanacad...work-and-energy)

Basically, that is not how it works and you calculations are garbage based on further inspection as they don't take in account thermodynamics which you should have learned in physics 101.

**Edited by VictorMedvil, 25 June 2019 - 04:11 AM.**

### #11

Posted 25 June 2019 - 04:46 AM

This is the reason, I said that today's relativity is incomplete. That is reason I write paper http://vixra.org/abs/1903.0178 .

We are living in different inertial local spaces. So, energy calculation for different inertial spaces are different.

So, energy calculation in spaceship frame is different than in space station frame & can not be compare.

------------------------------------------------------------------------------------------------------------------------

In 1998, I was discussing relativity with Dr Ram Mohan Rao. I said that we can explain M-Morley experiment finding by using

thermodynamics.

For example, we are in train cabin, moving with some velocity on earth. One photon of light move in the direction of train velocity

& mirror reflect that light perpendicularly to that direction.

Now, if I consider that the Train cabin as closed system then no energy will come into the system & no energy will go out from

the system. So, energy of photon in closed system will remain same after reflection. The velocity of reflected photon can not be

increase or decrease as no energy is added to photon in the closed system from outside.

so, velocity of reflected light remain same in this closed system in all directions.

But everybody has always recognised that kinetic energy is frame-dependent. That is obvious from the fact that velocity is relative and not absolute.

For instance, read this discussion on physics stack exchange: https://physics.stac...eference-frames

You will see that values of kinetic energy are frame-dependent but that the conservation of kinetic energy, in any given frame, is conserved.

There is more about the frame-dependence of kinetic energy in this Wiki article on kinetic energy: https://en.wikipedia.../Kinetic_energy

This is standard stuff. So put away the Michelson-Morely experiment and any delusions that there is some "incompleteness" in relativity. The only incompleteness is in your own understanding of Newtonian mechanics.

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P.S. You introduce a needless complication by dragging light into it. But the explanation of that is that the frequency of the light, as seen by an observer inside the train, is different from that seen by an observer outside. And the energy of a photon is given by E=hν, i.e. depends on its frequency.

So the energy of a photon varies according to the relative motion between emitter and receiver, even though its speed is the same for both.

**Edited by exchemist, 25 June 2019 - 05:05 AM.**

### #12

Posted 26 June 2019 - 02:37 AM

Only problem is rocket fired in space ship....I am sorry. I was away from town. So, can't reply in time. To explain my problem, I simplify it again.

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Just consider one long spacecraft of 10 t accelerate from velocity 0 to 4 m/s from one space station. In this spacecraft's long cabin,

one small rocket of 10 kg is fixed at the ceiling.

(So, Energy require to move spacecraft from velocity 0- 4 m/s = 0.5 X 10000 X 4^2 =80000 Joule)

& Part of energy to move rocket of mass 10 kg from velocity 0- 4 m/s with spacecraft = 0.5 X 10 kg X 4^2 =80 Joule -----(1)

Step 2:- Now, small rocket in long spacecraft cabin is released from ceiling of roof & get fired in cabin from velocity 0 - 4 m/s

Now, rocket will require 0.5 X 10 kg X 4^2 =80 Joule of energy again in this new inertial frame of cabin.

Now, Total velocity gain by rocket is 4 + 4 = 8 m/s form space station by consuming 80 + 80 = 160 joule of energy.

----------------------------------------------------------------------------------------------------------

If I want to fire directly the same mass small rocket with velocity 8 m/s.

I will require 0.5 X 10 kg X 8^2 =320 Joule of energy.

Now, both same mass rocket will be at same velocity for space station

but require different energy i.e. 160 joule in steps & 320 joule in no step.

Why?

One thing you are missing is rocket when get fired in space ship cabin. It move to much away from space station.

So, Space ship is completely independent inertial frame without any relation with space station.

So, rocket will get velocity 4 m/s with relative to space ship by consuming 0.5 X 10 kg X 4^2 =80 Joule of energy only.

For example, if I fire the bullet in sky with velocity V, I will require 0.5 x bullet mass x V^2 energy.

I do not need to add or subtract earth's velocity around Sun or even consider that velocity.

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2nd point:- For photon velocity in rail cabin..

This is not only applicable to only photon but also to any perfect elastic ball.

If in rail cabin, elastic ball is get reflected in any direction then also it will have same velocity to conserve its energy.

&

To observe the frequency of photon in rail cabin, we have to do experiment in rail cabin.

### #13

Posted 26 June 2019 - 02:40 AM

Only problem is rocket fired in space ship....

One thing you are missing is rocket when get fired in space ship cabin. It move to much away from space station.

So, Space ship is completely independent inertial frame without any relation with space station.

So, rocket will get velocity 4 m/s with relative to space ship by consuming 0.5 X 10 kg X 4^2 =80 Joule of energy only.

For example, if I fire the bullet in sky with velocity V, I will require 0.5 x bullet mass x V^2 energy.

I do not need to add or subtract earth's velocity around Sun or even consider that velocity.

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2nd point:- For photon velocity in rail cabin..

This is not only applicable to only photon but also to any perfect elastic ball.

If in rail cabin, elastic ball is get reflected in any direction then also it will have same velocity to conserve its energy.

&

To observe the frequency of photon in rail cabin, we have to do experiment in rail cabin.

Photons all have the same velocity you don't seem to be understanding some very basic physics principals, as previously stated. Use E=hv not the energy of a massive object like you have been.

**Edited by VictorMedvil, 26 June 2019 - 02:41 AM.**

### #14

Posted 27 June 2019 - 01:25 AM

Now, our space ship problem which is more important:-

In any inertial frame of reference physics remain same. So,

The rocket of 10 kg will get velocity 4 m/s with relative to space ship from 0 m/s by consuming 0.5 X 10 kg X 4^2 =80 Joule

of energy only.

This is simple.

For example, if I fire the bullet in sky with velocity V, I will require 0.5 x bullet mass x V^2 energy.

I do not need to add or subtract earth's velocity around Sun or even consider that velocity.

As earth frame is consider as inertial frame.

### #15

Posted 28 June 2019 - 02:09 AM

In any inertial frame,I am sorry. I was away from town. So, can't reply in time. To explain my problem, I simplify it again.

---------------------------------------------------------------------------------------------------------

Just consider one long spacecraft of 10 t accelerate from velocity 0 to 4 m/s from one space station. In this spacecraft's long cabin,

one small rocket of 10 kg is fixed at the ceiling.

(So, Energy require to move spacecraft from velocity 0- 4 m/s = 0.5 X 10000 X 4^2 =80000 Joule)

& Part of energy to move rocket of mass 10 kg from velocity 0- 4 m/s with spacecraft = 0.5 X 10 kg X 4^2 =80 Joule -----(1)

Step 2:- Now, small rocket in long spacecraft cabin is released from ceiling of roof & get fired in cabin from velocity 0 - 4 m/s

Now, rocket will require 0.5 X 10 kg X 4^2 =80 Joule of energy again in this new inertial frame of cabin.

Now, Total velocity gain by rocket is 4 + 4 = 8 m/s form space station by consuming 80 + 80 = 160 joule of energy.

----------------------------------------------------------------------------------------------------------

If I want to fire directly the same mass small rocket with velocity 8 m/s.

I will require 0.5 X 10 kg X 8^2 =320 Joule of energy.

Now, both same mass rocket will be at same velocity for space station

but require different energy i.e. 160 joule in steps & 320 joule in no step.

Why?

Energy consumed dE = F x ds

If V is final velocity & u is initial velocity of object then

F = m . a = m . (v-u)/dt

& ds = (v+u)/2 x dt

So, dE = 1/2 x m x (v^2 - u^2)

here, u=0

So, dE = 1/2 m v^2

So, For any inertial frame, energy require to move object with velocity v from rest is

dE = 1/2 m v^2

The rocket of 10 kg will get velocity 4 m/s with relative to inertial frame of space ship from rest by

consuming 0.5 X 10 kg X 4^2 =80 Joule of energy only.

This is simple.

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Here, we are interested in consumption of energy require by rocket directly in two different inertial frame.

In Inertial space ship:-

Can we have to consider velocity related to space station even space ship is moved to much distance away moving with constant

velocity or space station may be at other side of planet or not to previous state.

According to me state of motion of space station do not have any effect on above calculation in inertial ship cabin.

as it happen in space ship inertial frame.

**Edited by maheshkhati, 28 June 2019 - 02:22 AM.**

### #16

Posted 01 July 2019 - 03:47 AM

Kinetic energy or work done is frame dependent but actual energy consumption for doing work can be calculated

only in one frame of reference.

& Other are just observation in other frame & may be wrong.

For example:- One old man is pulling the box on platform in x-direction with very small velocity Vx

by applying the force.

Then observer on Platform will find the exact work done by old man.

1)Now, For Observer in train moving with velocity Vx parallel to old man movement will find that old man is tired but

not doing any work

because relative displacement is zero.

This is wrong calculation because old man is tired means he has definitely doing the work.

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I am on earth do the work then energy consume by me can be properly calculated if I fixed frame

of reference on earth

& if I do work in some rail cabin then I have to fixed frame of reference in that cabin.

I called this as local inertial spaces created due to electromagnetic flux of nearby substance.

http://vixra.org/abs/1903.0178

This will easily solve the problem of rocket energy consumption.

**Edited by maheshkhati, 01 July 2019 - 03:49 AM.**

### #17

Posted 04 July 2019 - 02:01 AM

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?

**Edited by maheshkhati, 04 July 2019 - 02:05 AM.**