Kinetic Energy and Observer Velocity

## Question:

If there is an equivalence between inertial reference frames, can you conclude that delta KE (as in the work-energy theorem) are the same across all inertial frames? It's hard to find a discussion of this on the internet - I don't know much physics, but I tried to prove it by substituting v+u for v in m(vf^2 - vi^2)/2, and u doesn't seem to cancel out.

What am I doing wrong? Does net work change with the observer's velocity?

Interesting question. Kinetic energy is different as measured in different reference frames. Clearly so since it is a function of velocity and velocity is frame dependent. We do need to check that the change in KE is also reference frame dependent as you suggest by measuring it in two frames with relative velocity u. If my algebra(see below) is correct, always a chancy proposition, the change in KE differs by a term m*u*(vf-vi).
(vf+u)^2-(vi+u)^2=
vf^2+2*vfu+u^2-(vi^2+2viu+u^2)=
vf^2-vi^2+2*u*(vf-vi)
Multiplying by m/2 we get
m/2*(vf^2-vi^2)+m*u*(vf-vi)

This tells us that the change in KE also depends on the relative velocity of the measuring frames. Observers in two reference frames moving relative to each other then would conclude that different amounts of work were done by a force acting on an unconstrained particle. The real question buried in yours is, "Is this fact surprising?". Well on the face of it does seem a bit odd that they could not agree on the work done. Each should agree that the work is the force times the distance traveled during the application of that force. Let's calculate the work done on that basis for each reference frame.

The force applied is mass times acceleration and acceleration is the difference in velocities divided by the time for the change. Neglecting relativistic effects, both observers will agree on the change in velocity and the time for the change. The u's do cancel in this calculation. So both agree on the force.

Will they both agree on the distance over which that force is applied. Each observer will calculate the distance as the average velocity times the time in motion. One observer will find that to be (vi+vf)/2*t. The other will find it to be (vi+u+vf+u)/2*t. The difference is u*t which seems intuitively correct to me.

The difference in work then will be the force (m*a) times the distance difference(*u*t). Multiplying this out and substituting for a the quantity (vf-vi)/t we get
m*u*(vf-vi)
Yup... The same as we found calculating the work from the change in KE. I guess the answer to your questions are, no, nothing and yes, respectively.

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JDJ