Calories expended to lift 48800 pounds 12 feet

[davidsp]

Hello;

 

I'm carrying 8800 pounds of soil up to my garage roof. It is 12 feet above the ground. I have to make 200 trips with pails of soil and I weigh 200 pounds - so my weight in the equation is an accumulated 40,000 pounds, thus the total weight lifted 12 feet is 48,800 pounds.

 

Let's assume complete efficiency of the calories burned in lifting the weight (unless someone knows a reasonable efficiency percentage for the human body in an activity like this - I'm climbing a ladder).

 

I suppose some consideration should be given to acceleration and deceleration, but l don't think that will significantly alter the end product, so I'm happy to ignore that as well.

 

Let's not calculate all of the extraneous activities like filling the pails, walking to the ladder with them, etc. My wife takes them when I get to the top and dumps them where they need to be.

 

It seems to me that the most efficient way to calculate this is to determine the amount of energy (joules? ergs? dynes?) required to raise this weight that height, then convert that to calories. Perhaps there's a significant adjustment in terms of the efficiency of the body's conversion of pure calories into work, but I don't know how significant that is.

 

At the end of the day I want to know how many pounds I'll lose if I don't change anything else - which I believe would require dividing the number of calories expended by 3500 (calories stored in a pound of fat). And, of course, I also want to know how much cheesecake I can eat if I want to stay the same weight after all this work.

 

David

[ecoli]

calories is just a measurement of energy so converting from joules to calories is trivial.

 

Estimating calories burned is a messy business, particularly because so much of it depends on the individuals basal metabolic rate. We burn most of our calories just by keeping our bodies at 37 degrees.

 

However you can simply calculate the amount of potential energy due to gravity it takes to lift the mass of sand up the distance. The equation is U = mgh; where m = mass in kilograms carried, g = acceleration due to gravity (9.8 m/s^2) and h is height in meters.

[swansont]

One also needs to know that the dietary Calorie is actually 1000 calories from physics (kilocalorie; that's why the C is normally capitalized). For a rough approximation, you can take the body's efficiency to be about 25%. It's basically a combustion engine. Since Joule is 4.18 calories, if you figure the energy in kiloJoules it will be the approximate number of Calories burned.

 

So, roughly, you have ~21,800 kg raised 4 meters. That's ~870 kJ, meaning you burn about 870 Calories, give or take.