Kinetic energy in frames of reference

[daveed]

i was just reading, and i saw this question, and i don't know how to explain it...

how can you describe why when two things go at a velocity v, and one of them accelerates to 2v, the KE gain w/ respect to the Earth and the other object are different, but the work done is constant?

i just don't know how to explain this... it should not be so difficult, no?

 

[SpaceTiger]

how can you describe why when two things go at a velocity v, and one of them accelerates to 2v, the KE gain w/ respect to the Earth and the other object are different, but the work done is constant?

The work, Fx (assuming constant acceleration), isn't constant. The force is the same in both frames, but the distance isn't.

x=\int v~dt

Moving frame:

v=\int a~dt=at
x=\frac{1}{2}at^2

Stationary frame:

v=v_0+\int a~dt=v_0+at
x=v_0t+\frac{1}{2}at^2

In your example, this leads to work that's three times larger in the stationary frame, just like the change of kinetic energy.