Calculate Work: Lifting and Moving a 90 lb Bag of Shingles in 5 seconds

[fuocoso]

can anyone help me out with a simple work problem?

say I lifted a bag of shingles that weighed 90 lbs from the ground to 6 feet in 5 seconds, then took 8 seconds to walk that bag to a scaffold which upon arriving to the scaffold it took me an aditional 3 seconds to put it from my 6 foot shoulder to the 6'6" scaffolding platform.

how much work would be involved?

(note: assume there is no aditional forces work on any part such as friction or wind)

 

[Jonathan]

I think the first part is 108 ft*lb/s and the second is 15 ft*lb/s, but I can't be sure because I get confused when not using metric, and I can't ever remember any conversions except for 2.54cm=1inch.

 

[Integral]

Work is force times distance. Time does not matter. Pounds is a unit of force so 90lb*6.5ft. The only distance that counts is change in elevation.

 

[Jonathan]

YES

That makes much more sense, I was thinking power or something. Foucoso mentioned time, so maybe he meant to ask for the power not work though.

 

[fuocoso]

err, yeah. i meant energy used. i only referred to it as a work problem because that is what i did at work.

 

[Integral]

That is the thing about Physics. It gives definite meaning to many terms which are only loosely defined to the layperson.

unfortunately, the answer to your real question is not as easy as a simple phyisics problem. The human body is a very complex machine, attempting to determine the work done by the body as it moves is very difficult. It will not even be constant from person to person, as it depends on your metaboloism and many other external factors.

If you look, Physically, at the energy involved the answer is the same, you start with a definable amount of potenital, call it 0, and end with a change = mgh where mg=90lbs h = 6.5',

 

[fuocoso]

yeah, i guess i was really tired whenever i wrote the inital post, what i really meant by all of it was, about how much energy would it require to lift the bag of shingles from the ground (saying it is a solid, static weight) to 6' and so on with all of those charecteristics. I guess that's the reason i called it simple work was because i had originally intended it to be a simple problem that i just didn't remember the formulas off hand and figured people in this forum would be able to tell me what i wanted with less effort of looking the formulas up and reaching the extra 12 inches to grab my graphing calc.