Simple question about energy used.

[dennis]

Not in the context of this thread it isn't.

[The Natural Philosopher]

remember if 'g' varies so will the 'weight' of a 1kg mass..

So as long as you have an accurate scale, it doesn't really matter :-)

[Tim Streater]

Scale as in "spring", or a proper balance?

[Ghostrecon]

I salute you - at first I thought that they would not bite but ....

[geoff]

even if its just to say that he won't

[dennis]

I haven't replied to ARW at all.

[Neil]

Still completely wrong.

m g h

Depends if 120Kg is the weight or the mass, but let's assume it's the weight and dvide it by 9.81 to get the mass which would be

12.23Kg

Then multiply by G gets you to120 (yes I know) then x 5 meters equals

600 and thats Joules because both metres and kilograms are SI units.

What that is in watts depends on how long it takes you to lift it.

[Neil]

Only if 120Kg is the mass not the weight. If 120Kg is the weight then it's 600 Joules and any calculation is independent of gravity.

[John Rumm]

The number is ok, just three orders of magnitude out ;-)

kg would be mass not weight....

Work done is force x distance, so 1200N x 5m = 6kJ near enough.

Indeed, but that is power, not energy required.

[John Rumm]

Depends on if you want to know the mass or the weight in the true sense... a spring will let you work out the second, the balance will factor local gravity out of the equation for you ;-)

[Tim Streater]

Well, quite :-)

[Tim Streater]

And I'm not sure why he thinks dividing a mass by an acceleration gets him weight.

[Neil]

I would bet that the OP was refering to weight not mass. Who in everyday usage (apart from a physicist) would ask about lifting a mass unless they didn't understand what one was.

[Neil]

It doesn't and I didn't say it did. Dividing the weight by the acceleration does. Back to school for you.

W = m g where W = weight (N) m = mass (kg) g = acceleration of gravity (m/s2)

so

mass = weight / acceleration.

[Neil]

When you say 120Kg are you talking about the weight or the mass?

120Kg is what the weighing scale would tell you it weighed, assuming graivity was normal round your house :-)

[John Williamson]

And with normal spring or load cell type scales, that would also be the mass of the object as convention is that at 1g (9.81 m/s2), as is normal(ish) at the Earth's surface, scale weight ( kgw) is equal to mass ( kgm). That's what I was taught in school, at least.

120kgm gives 1177(120*g) Newtons force on the scale, aka 120kgw. On the moon, where g= 1.63 m/s2 (approx.), then a spring scale calibrated for Earth would read 20kgw. All approximately.

Tciao for Now!

John.

[ARWadsworth]

Are you related to dennis?

[Neil]

And when you go back to school take me with you because i hadn't realised that spring balance machines were normalised for earths graivty and so weight and massmeasured by them on the earth's surface is essentially the same.

[Neil]

turns out i was wrong too but less so.

6 Kjoules, give or take.

[Ghostrecon]

oh I believe you I really do (uncrosses fingers)

shame you reply to graham had nothing to do with grahams post :-)