How Does Observer Motion Affect Calculated Work in Relativity?

[Dale]

Remember that he is not standing at one place. He is pushing the cart & applying force Fg on ground & move on platform ground. So, Work done by Fg = Fg. dx & dx is not zero

This is a slightly different scenario from what I had in mind, but your analysis is not correct.

The displacement at the contact patch for $F_g$ is always 0 (in the ground frame) under the usual assumptions that the ground is rigid and his feet are not slipping. So in the ground frame the work done by $F_g$ is $F_g\cdot 0$. It is not the displacement of the center of mass which is important, it is the displacement of the contact patch. That is 0 for $F_g$

force is applied by old man on cart is Fx, force applied by old man on ground = -Fx & force applied by ground on old man = Fx
Resultant force = Fx-Fx+Fx =Fx=force applied by old man on cart.

This is a very common error for beginning physics students. The force applied by the man on the ground and the force applied by the ground on the man act on two different objects. You never add them. They cannot be added to get a resultant force because they don't act on the same object. This is a common misunderstanding of Newton's 3 rd law, but all of the subsequent reasoning is wrong.

Ask one question, Who is losing energy?
Answer:- In this total event only old man is losing energy. Means' he is only doing work.

That is true in the ground frame. In other frames the ground could be losing energy, or even the cart could lose energy.

 

[ravi#]

Honorable Dale if you are right then
In frame of train observer, person in resting room on platform, slipping back on his seat is also doing work because his body is applying force Fx on seat in back ward direction & there is displacement V.dt.
Can you say that train rider observered that this slipping man in rest room is loosing his energy & look tired because he is doing work Fx. (V.dt) with relative to it.
Honorable Dale, there are infinite number of balance forces present on platform frame. They are not doing any work in any inertial frame & loose energy in any frame. Other wise whole system dies down automatically.

Only one unbalance force is present on platform , that is given below
Just consider only cart, Force applied on it in horizontal direction is force applied by old man & that unbalance force create acceleration, displacement on platform.
This force is only responsible for change in state of motion of body(cart) on platform.
So, I think in this complete event the only force applied by old man on cart has to be consider . That force is only doing work & making old man tired.

 

[A.T.]

there are infinite number of balance forces present on platform frame. They are not doing any work in any inertial frame

Balanced forces can do work.

 

[ravi#]

Can you say that
Slipping man in rest room is loosing his energy with relative to train rider & look tired because he is doing work Fx. (V.dt) with relative to it. where Fx is force applied by him on the back of chair.

 

[ravi#]

This can be done in following way also
Honorable A.T. you are true, cart apply the same force on old man but in this event there are two actors
One is old man:- on this old man, two forces are acting in horizontal direction -Fx is force by cart & Fx is the force by ground.
so, resultant force in X-direction = Fx-Fx = 0
On Cart :- There is only one force is acting i.e. Force applied by old man
so, resultant force in X-direction = Fx
As this force is unbalance, this do work on cart & displace it in the direction of force.
One thing must remember that old man is doing work. No one other is doing work on old man.
(It is not necessary that old man move on platform, he can pull the cart by rope also.)

 

[A.T.]

Slipping man in rest room is loosing his energy with relative to train rider

Is that rest room on the train?

where Fx is force applied by him on the back of chair.

Why is there a horizontal force? Is the train accelerating?

 

[Dale]

Honorable Dale if you are right then
In frame of train observer, person in resting room on platform, slipping back on his seat is also doing work because his body is applying force Fx on seat in back ward direction & there is displacement V.dt.

Yes. That is how work is defined.

Can you say that train rider observered that this slipping man in rest room is loosing his energy & look tired because he is doing work Fx. (V.dt) with relative to it.

Draw a free body diagram of the man and identify all of the forces acting. For each force calculate the displacement. Then determine the NET work.

Honorable Dale, there are infinite number of balance forces present on platform frame. They are not doing any work in any inertial frame & loose energy in any frame. Other wise whole system dies down automatically.

You should actually calculate it sometime and see. Your intuition is wrong here. This is why it is important to do homework problems.

Just consider only cart, Force applied on it in horizontal direction is force applied by old man & that unbalance force create acceleration, displacement on platform.
This force is only responsible for change in state of motion of body(cart) on platform.

Yes. For the cart the force from the man is the only force responsible for the change in the state of motion. You can see that by drawing the free body diagram.

So what is the change in the KE of the cart? In the ground frame the initial KE is $0.5 m 0^2 = 0$, and the final KE is $0.5 m (\Delta v)^2$ so the change in KE is $0.5 m (\Delta v)^2 = F \Delta x$

In the train frame the initial KE of the cart is $0.5 m v^2$ and the final KE is $0.5 m (v+\Delta v)^2$ so the change in KE is $0.5 m (v+\Delta v)^2-0.5 m v^2 = 0.5 m (\Delta v)^2 + m v \Delta v = F \Delta x + F v \Delta t = F (\Delta x + v \Delta t)$

As you said, the force of the man on the cart is the only force responsible for changing that state of motion. Therefore the force of the man is the one that did work $F(\Delta x + v\Delta t)$, which is the total displacement in the train frame as I described in my previous post.

Your suggested method for calculating the "work" violates the conservation of energy because the cart gains more KE than the "work" done on it.

That force is only doing work & making old man tired.

If you want to determine the "tiredness" of the man then you obviously need to consider the forces acting on the man rather than the ones acting on the cart. It is obviously insufficient to consider only the forces acting on the cart when making conclusions about the man.

 

[Dale]

Since the OP is clearly not interested in learning, and since this thread has become repetitive, and since this thread is about a mistake in Newtonian physics rather than relativity, it is now closed.