Work done in various frames of reference

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Homework Help Overview

The problem involves calculating the work done on a block situated on a moving train, where a man applies a force to displace the block. The original poster is confused about whether to include the displacement of the train when calculating work done with respect to the ground.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to justify their answer by stating that the displacement of the train should not be included since it is not caused by the applied force. Other participants explore the relationship between forces and displacements in different frames of reference, questioning how to properly account for work done.

Discussion Status

Some participants express agreement with the original poster's reasoning, while others provide alternative perspectives on how to approach the problem. The discussion is ongoing, with various interpretations being explored without a clear consensus.

Contextual Notes

There is a note in the problem statement indicating that work is not absolute, which may influence the participants' reasoning and assumptions about the scenario.

amal
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Homework Statement


The problem, basically, is very simple but, it is confusing me. It says:
"There is block on a moving train, which is being pushed by a man. The man applies force F to displace the block by s wrt Train. The moves S in that period. Find work done on the block by the force wrt ground.


Homework Equations


W=F.s


The Attempt at a Solution


What did is this:
The displacement of the block caused by the force is s. Hence, answer is F.s.
But, the book is including the displacement of train as well. It says that from ground, we see the block displaced bys+S. So, answer is F.(s+S). There is also a note saying that this illustrates that work is never absolute.

I think that my answer is correct because displacement S is not caused by the force and so, we should not consider that.
Please help me.
 
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I believe you're right. Here's how I checked.

b = ball
g = ground
t = train

\vec{F}\bigg|^{b}_{g} = \vec{F}\bigg|^{b}_{t} + \vec{F}\bigg|^{t}_{g}

\vec{s}\bigg|^{b}_{g} = \vec{s}\bigg|^{b}_{t} + \vec{s}\bigg|^{t}_{g}

w\bigg|^{b}_{g} = (\vec{F}\bigg|^{b}_{t} + \vec{F}\bigg|^{t}_{g})(\vec{s}\bigg|^{b}_{t} + \vec{s}\bigg|^{t}_{g})

\vec{F}\bigg|^{t}_{g} = ma_t = m * 0 = 0

w\bigg|^{b}_{g} = \vec{F}\bigg|^{b}_{t} (\vec{s}\bigg|^{b}_{t} + \vec{s}\bigg|^{t}_{g})
 
The work of a force F is W=F.s where s is the displacement, no matter if it is caused by the force or not.

The displacement of a stone thrown up is not caused by the gravity, but the force of gravity does work on the stone, and as the displacement and force are opposite, the work is negative: W=-mgs.

In case of work done by a force on a body on the train: the displacement with respect to the ground is s'=Vt+s and the velocity with respect to the ground is v'=V+v, but the force is the same in both frames of reference.
Imagine that the body is a charged particle, and the force is caused by an electric field E from an external source. The electric force is the same qE in both systems, the displacement is different, and so is the work.

ehild
 
I am clear now. Thank you very much ehild.
 
Last edited:
You are welcome:smile:

ehild
 

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