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System S1 moves at constant speed V with respect to S0 in one dimension:

               S1                                      x
S0                               x

A particle with mass m, moves at speed v with respect to S1

then at a speed of $v+V$ respect to S0 (if we can neglect relativity effects)

Kinetic energy

$Es1=\frac{1}{2} m v^2$ (respect to S1)

$Es0=\frac{1}{2} m (v+V)^2$ (respect to S0)

Then an external influence change its speed

(from S1) = $v+dv$

(from S0) = $(v+dv)+V $

New Kinetic energy

$Es1=\frac{1}{2} m (v+dv)^2$

$Es0=\frac{1}{2} m (v+dv+V)^2$

Gain of energy from S1

$\frac{1}{2} m (v+dv)^2 - (\frac{1}{2} m v^2)= \frac{1}{2} m (dv^2+ 2 v dv)$

Gain of energy from S0

$\frac{1}{2} m (v+V+dv)^2 - (\frac{1}{2} m (v+V)^2) = \frac{1}{2} m (dv^2+2(v+V)dv )$

Difference between energy gains = $dv*m*V$

Where did that energy came from?

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In that 2nd equation you wrote, shouldn't it be $Es1=\frac{1}{2}m(v+V)^2$? I mean, only a few words before that you stated that it travels at $v+V$ with respect to s1. – Alan Rominger Oct 17 '11 at 18:29
shouldn't the next to last equation be $\frac{1}{2}m(v+V+dv)^2-\frac{1}{2}m(v+V)^2$? – Alan Rominger Oct 17 '11 at 18:44
@Zassounotsukushi yes, a typo, difference remains the same, thanks – HDE Oct 17 '11 at 18:56
The kinetic energy is a velocity-dependent thing. Even without force it can be different in different reference frames. – Vladimir Kalitvianski Oct 17 '11 at 21:45

Let me go ahead and answer the question. I think I have enough information now. We've calculated two expressions for the change in energy, specific to the reference frame. I'm going to use the assumption $dv\ll v$ and $dv \ll V$, because I'm just allergic to heavy algebra like that.

$$\Delta E_{s0}=\frac{1}{2}m(v+V+dv)^2-\frac{1}{2}m(v+V)^2$$ $$\Delta E_{s0} \approx \frac{m}{2}((v+V)^2+2(v+V)dv-(v+V)^2) = m dv (v+V)$$

$$\Delta E_{s1}=m(\frac{dv^2}{2}+v dv) \approx m v dv$$

So there we go, yes! The energy change is different depending on the reference frame. But what reference frame was the force exerted from? Imagine that a spaceship with near infinite mass exerted the force to speed up the object and was in the s1 reference frame. I will denote the kinetic energy of the spaceship as $E'$. The change in kinetic energy of the spaceship according to s0 is nothing since with infinite mass the spaceship velocity changes virtually none, and started at nothing.

Conservation of momentum (valid in both s1 and s2)

$$\infty \Delta v' = -m dv$$

Spaceship energy change in s1:

$$\Delta E'_{s1} = \frac{1}{2} \infty ( 0^2 - (0-\frac{m dv}{\infty})^2) = 0$$

So, I'm pretty sure I'm going to make someone upset with my notation in this answer, but allow me to continue. Now, write the equation for s0.

$$\Delta E'_{s1} = \frac{1}{2} \infty ( (-V)^2 - (-V-\frac{m dv}{\infty})^2) \approx \frac{1}{2} \infty(V^2-V^2 - 2 V \frac{m dv}{\infty})$$

$$\Delta E'_{s0} = - V m dv$$

So there you go. The reason that s0 and s1 disagreed about the change in energy in the object of mass $m$ was because the change in kinetic energy of whatever pushed that object was neglected. Both reference frames agree that the total change in kinetic energy of all objects combined, or $\Delta E+\Delta E'$ is equal to $m v dv$, and this is the amount of energy the spaceship had to spend in order to deliver that impulse to the object with mass $m$. That quantity would change given a different speed of the spaceship.

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Here's a knee-jerk answer (I probably should think more before committing to paper): The work done to increase the speed is greater measured in s0 compared to what it is as measured in s1. Whatever force is acting on the particle acts over a longer distance as measured in s0 (but over the same duration of time) than in s1.

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This is woefully incomplete--- you absolutely need to take into account back-reaction on the pushing body and conservation of momentum. – Ron Maimon Oct 25 '11 at 20:27

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