A bit of an underachiever, but full of potential (energy, that is)

I have just been given God’s bathroom scale!

Yes, this is truly a devine bathroom scale. Jesus himself has weighed himself with it every morning, sucking in his gut and peeking through the holes in his feet at the numbers. Why, hark! Angels can be heard singing with every spin of the numbered dial.

So, anyway, that afternoon I lug my trusty Brunswick bowling ball up the Empire State building.

If I’m remembering my (somewhat remedial) high school physics, I have just charged said bowling ball with alot of Potential Energy. This energy is powerful enough to crack the sidewalk if released (by dropping the ball off the side), so I know this energy exists (you still with me?).

If I weigh this bowling ball charged with potential energy on God’s Devine Bathroom Scale (keep in mind its perfect in every way) would it weigh MORE than it does on the ground (where theres NO potential energy)?

Aren’t you sorry I found this message board?


Mass would not change. Gravity’s pull would not. Weight is gravity’s pull. It would hit with massive momentum, but if it didn’t bounce, and stayed on the scale, it would quickly flicker to very high, to owling ball weight.

"No job’s too small, we bomb them all."
-Ace Wrecking Company

from high, TO bowling ball weight.

"No job’s too small, we bomb them all."
-Ace Wrecking Company

Actually, technically it would weigh slightly less, because it is farther from the surface of the Earth. If you were thinking that E=mc[sup]2[/sup] would give you a little bit of extra mass, I’m afraid that wouldn’t apply here, because potential energy isn’t something that is truly internal to the object. It is more of a relation between the object and the field it is in.

Okay, I think I get the idea.

Potential Energy is just a convienient abstract measure, its like a price tag that tells you what it COULD be worth under certain circumstances/conditions (?).

But since I’ve got this here dead horse, and this here stick…

Would a depleated 9-volt (or whatever) battery weigh less than a charged one (have I mentioned this is a DEVINE bathroom scale?)?


In a word, no.

The energy of a battery comes from the chemical reaction inside. A 9-volt, or any small cell, is sealed so that the products of the reaction cannot escape. Not even its number of electrons changes; for every one that leaves the - terminal, one arrives at the + terminal. So nothing in it leaves, and nothing from outside enters, therefore it weighs exactly the same.

Now if it were a car battery, that’d be a different story. When you charge one of those, the reaction inside those generates hydrogen gas, which is allowed to escape. (This is why you need to be careful not to make sparks while jump-starting your car.) So a depleted car battery would weigh ever so slightly less than a fully-charged one.

Laugh hard; it’s a long way to the bank.

That makes sense in an elegant sort of way. Athough I suspect it weighs slightly less (probably point 00009 of a kazillionth) because of friction or heat loss or something (I seem to remember hearing some obscure law of thermodynamics sez so) but I’m not gonna’ press the point.

This whole message board makes me feel dumb. From now on I’m sticking to the “how’s that drinking bird toy work” questions.


I hate to quibble, especially with AuraSeer, but chemical reactions do, in fact, involve very tiny changes in mass. Even a sealed, insulated system (no heat loss, no reaction products escaping, etc.) would be subject to this.

We are trained to believe that chemical reactions do not involve mass changes because to a very good approximation this is true, and extremely useful, but it is an approximation.

If you could weigh, using Inky’s scale, two otherwise identical batteries, one charged and one depleted, they would differ exactly by the mass equivalence of the energy difference. The mass-energy equivalence Einstein described is universal, although it is (almost always) negligible in chemical reactions.

“Finally, consider Kottke’s voice which sounds like geese farts on a muggy day.”
Leo Kottke
6- And 12-String Guitar

And it’s divine, Inky!

“Finally, consider Kottke’s voice which sounds like geese farts on a muggy day.”
Leo Kottke
6- And 12-String Guitar

Actually, the extra potential energy of the bowling ball at the top of the Twin Towers can be weighed!!

Here’s how you do it: Put the divine scale on the sidewalk. Go up to the observation deck of the skyscraper. Lean over the rail, and then place the ball on the scale. In about five seconds, you should see the result in the scale.



I can’t leave a dead horse be…

Since we’re already here, and we’ve already got the building, the elevator, a bowling ball, and we’ve got God’s bathroom scale lets do an exparament.

We get in the elevator, I put the bowling ball on God’s Bathroom Scale (It’s devine you know, ask for it by name!) and push the button for the top floor…

As we accelerate and gain momentum the bowling ball goes from standard bowling ball weight to HEAVIER than bowling ball weight, back to standard bowling ball weight, then suddenly LIGHTER as the elevator slows to a stop.

We get to the top, spit off the ledge for a few minutes, and get back in the elevator.

As we decend, the bowling ball briefly gets LIGHTER, then again becomes standard bowling ball weight and then as we slow to a stop it gets HEAVIER.

Now, if I take the sum stotal of all the high and low weight, does it all break even? Is the bowling ball slightly slimer for the journey?

Adam “Inky” Greene

Er, rather I ment to say “Slimmer”.

It might be slimier if I set it in the puddle of spit…


pluto, I was under the impression that the energy change of a chemical reaction did not involve a mass change. But I won’t argue, because Chemistry was definitely not my best subject. (Of all my science classes, Chem was the only one which I disliked.)

Aura, Cecil actually dealt with that question (and a Moderator will be along soon to post the link, I hope).

Any reaction with a net energy change does change the mass. Even rubbing your hands along the object in question would. The point Cecil made was that that mass change is insignificantly small, though extant, in non-nuclear reactions, giving rise to the idea that E=mc^2 applies only to the nuclear kind.

To answer you question about whether or not the ball is any skinnier as a result of the lift trip, or to phrase it differently, does gravity do work? The answer is no. Gravity is a conservative force, and thus the total work require to move some mass in a closed loop in a graviational field is always zero. Friction is not a conservative force, and thus the path you take determines the work that needs to be exerted.

:::Fleeing thread, Fear of UD on face:::

Such remarks will not sit well with the holey one.

LOL!! I’m being a little bit more careful these days, so I wouldn’t worry too much. :slight_smile:

The ball won’t get “skinnier”, in that it won’t lose any mass. It’d just have slightly less weight.

Well, there would need to be some way for energy to escape, in order to get a mass change. Heat loss is the easiest way to lose energy.

The ball will have lost a teensy bit of weight.

Of course in your description, it is a little bit dubious to refer to the ball as increasing in weight during acceleration. Certainly it will act as if it is heavier, but that is not really related to gravity. That is related to inertia. So I wouldn’t really go for describing that jump on the scale as a real increase in weight.