# Work and chemical energy “paradox” [duplicate]

This question already has an answer here:

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.

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?

## marked as duplicate by knzhou, Sebastian Riese, Jon Custer, ZeroTheHero, Qmechanic♦ classical-mechanics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 26 '18 at 12:06

• The work is the work. It has nothing to do with the frame dependence of the kinetic energy. Your paradox simply stems from your assumption that the energy in a chemical can be converted into kinetic energy in the greater than 100% efficient and reference frame independent way that you propose by handwaving. Every rocket scientist can tell you that that's not the case and the detailed calculations are usually compiled in the first chapter of an aerospace textbook on rocket design. – CuriousOne May 29 '15 at 4:36
• I did point it out. You have to take both momentum conservation and energy conservation into account, which leads to the rocket equations. – CuriousOne May 29 '15 at 5:06
• That's exactly your problem. You are violating momentum conservation without even thinking twice about it. :-) – CuriousOne May 29 '15 at 5:18
• Momentum is always conserved in a properly set up physics problem. :-) – CuriousOne May 29 '15 at 5:39
• Possible duplicate of Where does the extra kinetic energy of the rocket come from? – knzhou Sep 24 '18 at 13:30

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.

• That's called "momentum conservation". :-) – CuriousOne May 29 '15 at 8:49
• Okay, so the trick here is that while the Work done on something by a force is not frame independent, because forces come in complementary pairs, the total Work done by a complementary pair of forces is frame independent. That is, in your spring example, the sum of the changes in the KE of both platform and 1kg mass add to 55J regardless of frame. So you could replace the spring here with anything, and as long as you account for the other half of the force used to accelerate the weight you should add to the total potential energy of the reaction. – Jay Lemmon May 29 '15 at 18:07
• What about the second half of my question now? If I just give you the fact that 55J of Work was done (and the directions), how do you determine the impulse delivered? Or is it not well defined by Work alone? – Jay Lemmon May 29 '15 at 18:08