I'm not a chemist, and I am not exactly sure how energy is stored in ATP and fat.
Here's my thoughts.
If you lift a 1kg object 1 m, the work done is
W=f*d=mgh=9.8 joules
1 calorie = 4.18 joules, so
1 can of mountain dew = 170 calories, but i recall that these are actually kilocalories (right?),
so 1 can of mountain dew = 170*1000 = 170,000 real calories, or 710600 joules/dew.
9.8 joules/lift,
710600/9.8 = 72,510 lifts per dew.
So I would have to lift 1 kg 1 meter in the air 72,510 times to burn off the calories from 1 can of dew?
That seems a little off. Am I missing something? I wonder how inefficiency figures into this.
On to biology, which I am less familiar with:
ATP stores energy for our muscles. At what point is fat tapped for energy, to be transferred to ATP?
It seems like the core of a fat-burning work-out is to deplete the supplies of ATP such that fat reserves are used.
Can some more knowledgeable person set me straight on this subject?
Here's my thoughts.
If you lift a 1kg object 1 m, the work done is
W=f*d=mgh=9.8 joules
1 calorie = 4.18 joules, so
1 can of mountain dew = 170 calories, but i recall that these are actually kilocalories (right?),
so 1 can of mountain dew = 170*1000 = 170,000 real calories, or 710600 joules/dew.
9.8 joules/lift,
710600/9.8 = 72,510 lifts per dew.
So I would have to lift 1 kg 1 meter in the air 72,510 times to burn off the calories from 1 can of dew?
That seems a little off. Am I missing something? I wonder how inefficiency figures into this.
On to biology, which I am less familiar with:
ATP stores energy for our muscles. At what point is fat tapped for energy, to be transferred to ATP?
It seems like the core of a fat-burning work-out is to deplete the supplies of ATP such that fat reserves are used.
Can some more knowledgeable person set me straight on this subject?