Kinetic energy as seen from a different frame

[Amitayas Banerjee]

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
$$KE=\frac{m(V_f^2+2VV_f)}{2}$$

 

[PeroK]

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]

energy

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.

 

[PeroK]

I have not given the answer. I was just attempting to solve the question.

Where are you getting your equation from?

 

[drvrm]

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.

 

[Amitayas Banerjee]

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

 

[PeroK]

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.

 

[Amitayas Banerjee]

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?

 

[haruspex]

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?

 

[Orodruin]

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.

 

[PeroK]

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:

$v_f^2 - v_i^2 = 2as$

 

[Amitayas Banerjee]

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?

 

[Orodruin]

so my Q is how are the work done same?

Are they though?

 

[haruspex]

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?

 

[haruspex]

Are they though?

It depends which two works Amityas is comparing. Maybe it is the KE and Fd in the observer's frame.

 

[Amitayas Banerjee]

I have tried my best to solve this question...I am unable to proceed further...please provide me with the soln.

 

[DrClaude]

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.