Work and chemical energy "paradox" [duplicate]

Question

This is a mistake I've seen many people make, a few physicists included, but I haven't ever seen a satisfactory explanation for what's going on. Apologies for the lengthy setup.

Setup

Suppose I have a stationary 1 kg weight, and I apply a 1 N force for 1 second. At the end of the second, it will have a velocity of 1 m/s and have moved .5 meters. Its initial kinetic energy is 0 and its final kinetic energy is .5. The force has done .5 Joules of work to the weight (just Force * distance).

Now suppose the weight is initially moving at 10 m/s in the same direction as the force, and I again apply a force of 1 N for 1 second. After the second it will have a velocity of 11 m/s and have moved 10.5 meters. Its initial kinetic energy was 50 and its final kinetic energy is 60.5. The force has done 10.5 Joules of work to the weight.

The work done in either case is not the same, but the result of applying the work to change the velocity is. So Work is a relative quantity depending on your frame of reference, just like kinetic energy. It's not an absolute quantity in the same way that mass or distance or time is (assuming classical mechanics).

Suppose my weight is something with stored chemical potential energy. Maybe a potent chemical or a battery or something like that. The potential energy for it is measured in either Joules or calories or kilowatt hours or all sorts of things, but ultimately its the same unit as Work. Suppose I use it to generate a force to accelerate itself.

"Paradox"

If I'm naive and assume that the chemical energy can be plugged in as work to find the final velocity of a weight, I'd get something like:

$$ v_f^2 = v_i^2 + 2 M W $$

M is 1 kg, and lets say my chemical energy is 0.5 Joules. Let's start with the stationary case, and plug in my chemical energy for W. I end up with a final velocity of 1 m/s, which is right. But with the already-in-motion case, I end up with a final velocity of ~10.05 m/s, which differs from the right answer by quite a bit!

Question

So Work isn't the same as chemical potential energy. The two have the same units but potentially different scales, depending on reference frame, which I think is the answer to this paradox? But then, how do you convert between the two? How do you find the impulse a battery or chemical reaction could produce?

Answer 1

Let's set up an example. Imagine a $10kg$ platform, a spring, and a $1kg$ mass. Suppose the spring is compressed to hold sufficient energy to push the items apart with a relative velocity of $11m/s$. Beginning from rest, this will give the object a forward speed of $10m/s$ and the platform a rearward speed of $1m/s$.

The total kinetic energy after this is $$ KE_{tot} = 0.5(1kg)(10m/s)^2 + 0.5(10kg)(1m/s) $$ $$ KE_{tot} = 55J $$

Because they started at rest ($KE_{tot}=0$), the net energy change is $55J$. So the spring potential energy was also $55J$ when compressed. So far, so good.

Now we imagine beginning this with the platform already moving (in the direction the object will be fired) with a speed of $100m/s$. So the object has an initial KE of $0.5(1kg)(100m/s)^2 = 5000J$. We already know the spring will give it a $\Delta v$ of $10m/s$ for a final speed of $110m/s$. The new KE is therefore $0.5(1kg)(110m/s)^2 = 6050J$.

Wow, somehow a $55J$ spring has provided over $1000J$ of KE to our projectile! But we haven't finished....

The initial KE of the platform is $0.5(10kg)(100m/s)^2 = 50000J$. After the projectile is fired, the platform loses $1m/s$ of forward speed for a final KE of $0.5(10kg)(99m/s)^2 = 49005J$.

So only by actually accounting for the recoil in the platform (and not assuming it to be "infinitely" massive), we find where all the energy is coming from. It's really a transfer of KE from the platform to the projectile.