Work-Energy Theorem in Inertial Reference Frames

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The discussion revolves around the work-energy theorem as applied to two observers in different inertial reference frames (IRFs). Observer A, on the ground, and Observer B, on a moving train, analyze the work done on a particle acted upon by a constant force. The key confusion lies in how the work done differs between the two frames due to their relative motion, as the distance moved by the particle varies for each observer. While both observers agree on the force applied, the work calculated is different because it depends on the distance the particle travels in each frame. Ultimately, the work-energy theorem holds true in both frames, but the values of work and kinetic energy changes differ due to the relative motion of the observers.
Quelsita
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OK, I'm working on a question regarding IRFs, but I seem to be a little confused.

Question:
Observer A is on the ground and Observer B in on a train moving with uniform velocity v wrt the ground. Each observes that a particle of mass m, initially at rest wrt the train, is acted upon by a constant force F applied to it in the forward direction for a time t.
a)What is the work done on the particle by F in referece frame A and B? Are they equal?
b)What are the changes in kinetic energy observed by A and B?
c)Does the Work-energy theorem hold in reference frames of observers A and B?

So, as I understand it, the particle is on the train as it is initially at rest wrt the train (i.e. it is moving with the train). The Force applied in the positive direction indicates that the train has slowed or stopped causing it to jerk forward.

Here is where I get confused, how is the work different for each reference frame? If the train is moving in the +x direction, then the particle is also moving in the +x direction, then does A see the train move in the +x and B see the train move in -x direction?

How does this change kinetic energy?

Thanks!
 
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Quelsita said:
So, as I understand it, the particle is on the train as it is initially at rest wrt the train (i.e. it is moving with the train). The Force applied in the positive direction indicates that the train has slowed or stopped causing it to jerk forward.

No, I don't think that's the situation here. Imagine a food cart resting in the aisle of the train, stationary as seen by observer B on the train, but moving (in the +x direction) as seen by observer A on the ground. A train attendant comes along and pushes the cart down the aisle. Imagine that we can ignore friction, so if the attendant pushes with a constant force, the cart accelerates (in both reference frames).
 
I see your point, that does make sense.

So, the particle is moving wrt to both reference frames, but I'm still stuck on how the different reference frames affect work...
 
Hi Quelsita! :smile:
Quelsita said:
…The Force applied in the positive direction indicates that the train has slowed or stopped causing it to jerk forward.

No … where do you get that from? :confused:
Here is where I get confused, how is the work different for each reference frame? If the train is moving in the +x direction, then the particle is also moving in the +x direction, then does A see the train move in the +x and B see the train move in -x direction

Hint: Force = rate of change of momentum, which you can fairly easily prove is the same for both frames.

But the work done = force x distance moved by the point of application of the force (the particle), which is different for both frames.

How far does the particle move in time t in frame A? and how far in frame B? :smile:
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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