# Work done in various frames of reference

## 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 by$s+S$. 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.

Related Introductory Physics Homework Help News on Phys.org
I believe you're right. Here's how I checked.

b = ball
g = ground
t = train

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

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

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

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

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

ehild
Homework Helper
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:
ehild
Homework Helper
You are welcome ehild