# Power in different reference frames

I would like to reopen the question asked in this post because I am not quite satisfied with the accepted answer.

Imagine observer A stationary (in world reference frame) and observer B moving with a constant velocity v. They observe a car with a mass m moving with the same velocity v relative to the world. Coordinate axes of the frames are all parallel so problem is one-dimensional. Let's assume that a car has a battery with finite amount of energy. At time $$t_1$$ car starts to accelerate with constant acceleration a until car drains all energy from the battery.

Clearly the conservation of momentum does not hold due to acceleration of a car and what interests me is this:

Both observers agree on amount of energy stored in the battery. If I do the calculation from both reference frames, similar to this, I get that the power that accelerates the car is:

$$mv_A(t)a_A(t) = m (v +at)a = mav + ma^2t$$

$$mv_B(t)a_B(t) = m (at)a = ma^2t$$

Integrating and subtracting gives: $$\Delta W = W_A - W_B = m\ v\ a\ t$$

This implies that the work done by battery is not the same in those 2 reference frames which seems to be a contradiction. I get that amount of kineric energy is not invariant under changing reference frames, but they should agree on amount of work done. So, what is the catch in this setup?

• "but they should agree on amount of work done" ...why? The very answer you link is based on the friction force doing different amounts of work in the different frames! Oct 3 '19 at 12:50
• It would help if you specifically said why the answer in the question you link to is not sufficient for you Oct 3 '19 at 13:12
• So, they both agree on the amount of energy stored in the battery. All that energy is then converted to the kinetic energy of a car. By the conservation of energy in each reference frame individually, they should have gained the same amount of kinetic energy which is not true in my calculations since $\Delta W$ is not zero Oct 3 '19 at 13:16
• @LukaMandić I could be wrong, but I think the key is, in the absence of dissipative influences (air friction, rolling resistance, etc.) no work is done by the battery when the car is moving at constant velocity, whatever that might be, in either frame. I am trying to develop an answer along that line of reasoning. Oct 3 '19 at 15:03

Clearly the conservation of momentum does not hold due to acceleration of a car

I don't see why you say this. Momentum conservation should hold in all cases.

I get that amount of kineric energy is not invariant under changing reference frames, but they should agree on amount of work done. So, what is the catch in this setup?

You haven't considered the change in KE of whatever the car is driving on. In the frame where the world/surface is at rest, a teeny change in velocity means a teeny change in energy. We generally ignore this change.

But since KE scales as the square of the speed, then as the speed increases, the incremental change in speed represents a larger and larger change in KE.

For something like the earth, the velocity change that represents an increase in a few million joules is not measurable, but still exists.

So the battery (via the car) is doing work on both the "ground" and the car, changing KE in some manner. The sum of both KE changes is the same as the battery energy, minus losses.

If you were to do this with a smaller reaction mass (say a spaceship instead of a planet), the calculations are easier to see.

I was wondering the exact same thing and I didn't find the answer in the other post satisfactory either.

The key insight for me was that every acceleration should be viewed as a collision. You can't accelerate only one object (Newtons third law.)

I found this answer to be more thorough.