Kinetic energy as seen from a different frame

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Homework Help Overview

The problem involves analyzing the increase in kinetic energy of a mass under the influence of a constant force from two different reference frames. The original poster describes a scenario where a mass is acted upon by a force while initially at rest, and then considers the perspective of an observer moving relative to the mass.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the transformation of kinetic energy equations between reference frames, questioning the validity of the original poster's approach and the equations used.
  • Some participants suggest clarifying the relationship between force, distance, and kinetic energy in different frames.
  • There are inquiries about how to properly account for the displacement of the mass as perceived by the moving observer.
  • Participants express confusion regarding the mixing of quantities from different frames and suggest using notation to differentiate them.

Discussion Status

The discussion is ongoing, with participants providing hints and guidance rather than direct solutions. There is recognition of the need to clarify the transformation of variables between frames, and some participants are exploring how to quantify the work done in different reference frames.

Contextual Notes

There are indications that the original poster is struggling with the problem and has requested direct solutions, which is not in line with the forum's guidelines. The conversation reflects a mix of attempts to guide understanding while adhering to the forum's learning-first approach.

Amitayas Banerjee
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Homework Statement


A mass m is initially at rest. A constant force $F$ (directed to the right) acts
on it over a distance d. The increase in kinetic energy is therefore $Fd$.
Consider the situation from the point of view of someone moving to the left
at speed $V$ . Show explicitly that this person measures an increase in kinetic
energy equal to force times distance.

Homework Equations



KE=\frac{m(V_f^2+2VV_f)}{2}
3. The Attempt at a Solution
Let $V$ be the velocity of the person and $V_f$ be the final velocity of the object
So, change in
KE=\frac{m(V_f^2+2VV_f)}{2}
 
Last edited:
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Your method appears to be to write down an answer without any explanation or calculation.

You need to transform the data you have to the second frame.

Note also that your answer makes no sense to me.
 
Amitayas Banerjee said:
energy
PeroK said:
Your method appears to be to write down an answer without any explanation or calculation.

You need to transform the data you have to the second frame.

Note also that your answer makes no sense to me.
I have not given the answer. I was just attempting to solve the question.
 
Amitayas Banerjee said:
I have not given the answer. I was just attempting to solve the question.
Where are you getting your equation from?
 
Amitayas Banerjee said:
A mass m is initially at rest. A constant force $F$ (directed to the right) acts
on it over a distance d. The increase in kinetic energy is therefore $Fd$.
Consider the situation from the point of view of someone moving to the left
at speed $V$ . Show explicitly that this person measures an increase in kinetic
energy equal to force times distance.

Homework Equations



KE=m(V2f+2VVf)2KE=m(Vf2+2VVf)2​

KE=\frac{m(V_f^2+2VV_f)}{2}
3. The Attempt at a Solution
Let $V$ be the velocity of the person and $V_f$ be the final velocity of the object
So, change in
KE=m(V2f+2VVf)2​

Imagine the moving observer to be at rest ...then the body will have two types of displacements say
ds1 and ds2 in a time interval dt then calculate the velocity as observed by static observer and then calculate accelerations.
naturally he acceleration caused by change in relative velocity will be zero whereas the displacement due to force will have constant acceleration.

by decoupling the two motion...one due to observer and another due to actual force one can calculate the change in KE and it will be actual work done b the force.
the cross term will not appear in such analysis.
 
PeroK said:
Where are you getting your equation from?
I just took the difference of the final and initial kinetic energies as seen from that inertial frame
 
Amitayas Banerjee said:
I just took the difference of the final and initial kinetic energies as seen from that inertial frame
Okay, but you are mixing quantities from the two different frames there.

I suggest you use a prime for quantities in the second frame.

Your equation should be for ##\Delta KE'##: the change in KE measured in the second frame.
 
PeroK said:
Okay, but you are mixing quantities from the two different frames there.

I suggest you use a prime for quantities in the second frame.

Your equation should be for ##\Delta KE'##: the change in KE measured in the second frame.
Sir, I am unable to do it, would you please do it for me?
 
Amitayas Banerjee said:

The Attempt at a Solution


Let $V$ be the velocity of the person and $V_f$ be the final velocity of the object
So, change in
KE=\frac{m(V_f^2+2VV_f)}{2}
Good so far, but now you need to figure out how to transform F and d into this frame. Clearly F does not change, but what about d? How far does the person think the force moved?
 
  • #10
Amitayas Banerjee said:
Sir, I am unable to do it, would you please do it for me?
This is not how this forum works. We can give you hints and guidance, but you will have to do the work yourself.
 
  • #11
Amitayas Banerjee said:
Sir, I am unable to do it, would you please do it for me?
I can't do that, but I can give you a relevant equation. In any frame:

##v_f^2 - v_i^2 = 2as##
 
  • #12
haruspex said:
Good so far, but now you need to figure out how to transform F and d into this frame. Clearly F does not change, but what about d? How far does the person think the force moved?
d depends on the reference frame(inertial), but f does not...so my Q is how are the work done same?
 
  • #13
Amitayas Banerjee said:
so my Q is how are the work done same?
Are they though?
 
  • #14
Amitayas Banerjee said:
d depends on the reference frame(inertial), but f does not...
yes, but you need to quantify that. How far does the mocing observer see the force as having advanced?
 
  • #15
Orodruin said:
Are they though?
It depends which two works Amityas is comparing. Maybe it is the KE and Fd in the observer's frame.
 
  • #16
I have tried my best to solve this question...I am unable to proceed further...please provide me with the soln.
 
  • #17
Amitayas Banerjee said:
I have tried my best to solve this question...I am unable to proceed further...please provide me with the soln.
We don't do that here. We can help you get to the solution, but we will not do your work for you.

I will close this thread. If you write an actual attempt at a solution, send me a PM and I can reopen the thread.
 

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