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It has nothing to do with SI units.SI units are foreign to me so I'll need a tutorial as how to handle equations.
Odd thing I am trying to get a handle on, is, from physics help forum, when you see a general formula i.e. PE = mgh, one is to assume that mass will be entered in SI units. I am going to make a WAG here and speculate that the slide rule generation brought about the change whereas the kg is the base unit of mass rather than the gram. I you were ever graced with having to use a slide rule, the reason may be obvious. Please see attachment. I support the metric system because of its accurate repeatability. Going from memory, I think a gram is equal to the mass of a cubic centimetre of H2O at a specified temperature. A value arrived at using naturally occurring events. Good Idea. Now since the standard (SI) unit of mass is the kilogram, there are a number of locations that store an object with an exact mass of 1kg. Weights and measures yada, yada, yada. Unfortunately the concern lately is that these holy grails of W&M are changing in mass. Degradation of material used for Kg reference. So, button, button, who has the button. Who holdeth the rightful cosmic truths and can certify the correct methodology of formula massage and certify a correct result? Using SI may very well produce a correct result but physics etiquette, or ease is somewhat foreign to me. Personally, I'm hooked on engineering notation as opposed to scientific notation. Engineering notation blends so well into my familiar, comfy world of kilo, Mega, Terra, milli, micro, nano, etc. Scientific notation seems to reduce a value where the most significant digit is placed in the one's column and the exponent lands willie nillie wherever. I can live with it but just another bottleneck to deal with.It'S easy as pie: k stands for a factor of 1000. This has, by the way, nothing to do with SI or imperial units.
Therefore 1 kW = 1000 W and consequently 1 kWs = 1000 Ws.
It's totally the same as with e.g. resistors: 1 kΩ = 1000 Ω. No special magic of any kind involved.
Please see table and footnotes in post #7.It has nothing to do with SI units.
K means 1000, so you divided when you should have multiplied.
Cool beans. I do think you are correct about kilo abbreviation. kWHr not KWHr. You have me wondering about hour now. I prefer Hr but the community rarely yields to my personal preference. Hz I'm pretty sure is convention.Hello,
BTW, the prefix is a small k for kilo or 1000.
View attachment 55322
In the attached PDF you will find more info on the SIsystem.
The complete document can be found here:
https://usma.org/wpcontent/uploads/2015/06/Practical_Guide_to_the_SI.pdf
Bertus
The reason I did another division of 1000 is bc, as I see it, 3,600,000 joules = 1kWHr. That factor of 1000 is skewed when working in SI units. If m is in the equation as kilograms (SI) then you would intuitively divide by 3,600 instead of 3,600,000. Looking for the definitive answer as to how m should be represented in PE = mgh. It is not a judgement call. Going on gut instinct, and working the problem backwards, 8.17kWHr seems a reasonable amount of energy to raise 2,200 lb 9 feet high, whereas 0.00817kWHr seems like a ridiculous value of effort in elevating the mass.You need to multiply KWs by 1000, not divide by 1000 to get Ws.
So the the answer is 2270 Ws (if the rest of you calculations are correct).
That may very well be the correct answer. 29,400 / 3,600 = 8.17kWHr. I can more easily visualize using a constant flow of 1kW for 8 hours (something like an eight amp, 115V motor plus some mechanical advantage [gear heads] for 8 hours and lifting mass the predetermined height. Can ya FEEL me? ;~)1 tonne 3 metres 1 second?
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E = mgh = 1000 * 9.8 * 3 = 29400 Joule = 29400 Watt seconds
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Power to do this in 1 second 29400 Watt = 29.4 kWatt
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The reason I did another division of 1000 is bc, as I see it, 3,600,000 joules = 1kWHr. That factor of 1000 is skewed when working in SI units. If m is in the equation as kilograms (SI) then you would intuitively divide by 3,600 instead of 3,600,000. Looking for the definitive answer as to how m should be represented in PE = mgh. It is not a judgement call. Going on gut instinct, and working the problem backwards, 8.17kWHr seems a reasonable amount of energy to raise 2,200 lb 9 feet high, whereas 0.00817kWHr seems like a ridiculous value of effort in elevating the mass.
I guess there was a time when I had a feel for what a degree Fahrenheit was and knew a stone, a pound and the other pound and had a feel for them too.
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I can say that having moved to SI, or just to mks, and metrics in general, I'd never look back.It's good to be basically rid of many of the strange constant factors.
I have found it so much easier in use and one quickly enough builds up a feel for how big or small they are, so that hey too gain that natural feeling.
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For those who have more of a feel for horse power or kWh, ok, it's just a constant factor away.
I assume that your 1000kg is = to 1000 grams and is not = to a Mg (a tonne)29.4kJ = 0.00817 kWhr to raise a metric ton three meters against 9800 Newtons of gravitational force is correct. The takeaway is that 1kWhr = 3.6MJ is a buttload (technical physics term) of energy that could life 1000kg 367 meters.