AA batteries have about 4 watt-hours (14400 joules, 10620.9 pound-feet).

Ignoring losses, does that mean a AA battery could theoretically lift 10,620.9 pounds, 1 foot high?

You'd need an electric motor, some sort of crane or lift, the right gearing, and a fair amount of time. What's the best efficiency you could hope for? Even with 50% loss, you should still be able to lift a pickup truck, but this feels wrong (impossible).


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Figure 1. Extract from Energiser E91 datasheet.

The datasheet shows that at 100 mA you'll get about 2500 mAh (2.5 Ah) if you discharge to 0.8 V (and it doesn't get hot). Let's say that the average voltage is 1.25 V and we get the total useful energy is about 2.5 × 1.25 = 3.125 Wh = 3.125 × 3600 = 11250 Ws or 11250 J. Your calculation is the correct order of magnitude.

At 100 mA the time to discharge will be 25 hours.

I'm not going to use Imperial units so we'll go with 5000 kg. From $ E = mgh $ we get $h = \frac E {mg} = \frac {11250}{5000 \times 9.81} = 0.23 \ \text m $.

So, again your calculations are about right in theory. The problem is that it will be a 0.15 W motor running for 25 h and I can't think of any gearbox that will be capable of supporting 5000 kg and will run at 0.15 W.

  • $\begingroup$ You could replace the gearbox with a hydraulic system. $\endgroup$
    – D Duck
    May 9 '20 at 15:32
  • $\begingroup$ @DDuck The construction of the hydraulic pump will be really large. Either in regards to the area ratio or the displacement required. And this will also require a large amount of resources to build. $\endgroup$
    – Manu G
    May 9 '20 at 15:42
  • $\begingroup$ How did you get to 0.15 W for the motor? Could you go over or under that, and if so, what would happen? $\endgroup$ May 9 '20 at 16:12
  • 1
    $\begingroup$ I chose 100 mA as my current. $ P = VI = 1.5 \times 0.1 = 0.15 \ \text W $. If you discharge faster the extractable energy decreases as shown by Figure 1. $\endgroup$
    – Transistor
    May 9 '20 at 16:22

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