# Kinetic energy - why no g divisor ?

## Main Question or Discussion Point

Pound force being equivalent to pound mass at earth surface, we can safely safely say 4000 lbs raised 200 feet has 800,000 ft-lbs energy.

Converting to kinetic energy after falling, however, we can't use 1/2 mv^2 without using g as a divisor. By this I mean if you measure the speed after falling and sub into KE formula you'll get an energy 32 times the PE

Why don't we ever see the KE formula with g ?

Last edited:

Related Other Physics Topics News on Phys.org
Gravity doesn't necessarily affect the kinetic energy of an object. The object can be experiencing 0 gravitational acceleration, for example, and still have a kinetic energy. Physics equations are usually given in the most fundamental form, so it can then be used by a physicist to easily derive the desired equation.

Here's how we derive the equation you're thinking of:
In the specific situation where the object is falling, the velocity "v" that determines the kinetic energy is dependent on g. In fact, the velocity of a falling object is:
$\Large v=gt+v_{i}$ (vi is initial velocity)

So to derive the equation for the kinetic energy of a falling object, we can substitute the velocity of a falling object into the kinetic energy equation to give:
$\Large K_{E}=\frac{1}{2}m(gt+v_{i})^{2}$
and simplifying it, we get:
$\Large K_{E}=\frac{mg^{2}t^{2}+mv_{i}^{2}}{2}+mgtv_{i}$

Of course, another way to formulate an equation for KE is:
$\large ΔU_{E}=mg(h_{i}-h{_f})$
$\large ΔK_{E}=-ΔU_{E}$
$\large ΔK_{E}=-mg(h_{i}-h{_f})$
hi is initial height, hf is final height

Last edited:
AlephZero
Homework Helper
Physics equations use consistent units. The unit of mass consistent with pounds-force, feet and seconds is NOT the pound mass but the slug, which equals 32.2 pounds mass.

The formulas for energy in pounds-force feet are
PE = (mass in slugs) x g x height
KE = (mass in slugs) x v^2/2

Of course (nass in slugs) x g = (mass in pounds), which explains why you got the right answer for the PE but perhaps didn't understand exactly why it was right.

If you do science in SI units, it's less easy to get confused between the kilogram as a mass unit and the Newton as a force unit.

sophiecentaur
Gold Member
Pound force being equivalent to pound mass at earth surface, we can safely safely say 4000 lbs raised 200 feet has 800,000 ft-lbs energy.

Converting to kinetic energy after falling, however, we can't use 1/2 mv^2 without using g as a divisor. By this I mean if you measure the speed after falling and sub into KE formula you'll get an energy 32 times the PE

Why don't we ever see the KE formula with g ?
Because KE depends just on the velocity. How an object reached that velocity (falling, with a rocket or by being thrown) makes no difference. Different formulae can be used to find the velocity an object will reach, depending upon the circumstances. g can often come into those formulae but so can E,H or G - it all depends.

Physics equations use consistent units. The unit of mass consistent with pounds-force, feet and seconds is NOT the pound mass but the slug, which equals 32.2 pounds mass.

The formulas for energy in pounds-force feet are
PE = (mass in slugs) x g x height
KE = (mass in slugs) x v^2/2

Of course (nass in slugs) x g = (mass in pounds), which explains why you got the right answer for the PE but perhaps didn't understand exactly why it was right.

If you do science in SI units, it's less easy to get confused between the kilogram as a mass unit and the Newton as a force unit.
Ok that confirms for me that g does enter as the divisor in KE when using lb mass instead of slugs which I also confirmed when reviewing the derivation of KE and using lb mass as the unit.

The whole point being when using lb interchangeably for mass and force, as in the English Engineering units, I needed to be reminded to divide by g in the simple KE formula

PhanthomJay
Homework Helper
Gold Member
Never use pounds interchangeably for mass and force. The pound-mass term has no standing in Physics. As noted, the slug is the proper unit for mass in the imperial system, and the pound is the proper unit for force. In SI, the kilogram is the proper unit for mass, and the Newton is the proper force unit. On planet Earth, one slug weighs 32 pounds and one kilogram weighs 9.8 Newtons.

sophiecentaur
Gold Member
SI makes life so much easier (and you can multiply and divide by ten in your head!!). It really is time that the rest of the known world got rid of the Units that we British ditched a long time ago. (I was only saying the same thing to a friend of mine last night - over a pint!)

Never use pounds interchangeably for mass and force. The pound-mass term has no standing in Physics. As noted, the slug is the proper unit for mass in the imperial system, and the pound is the proper unit for force. In SI, the kilogram is the proper unit for mass, and the Newton is the proper force unit. On planet Earth, one slug weighs 32 pounds and one kilogram weighs 9.8 Newtons.
Unfortunately, it does have standing in engineering. Should I have posted this question in the engineering section ? I didn't realize this board was so segregated.

AlephZero
Homework Helper
Unfortunately, it does have standing in engineering.
Only in the USA. The UK doesn't use what you called "English unts" for engineering any more. In fact using them for any commercial purpose in the UK has been illegal for about 30 years now, except for a few things like speed limits in miles per hour, and beer sold in pints.

The term "English Units" seems have been invented in the USA, and found its way into some textbooks. They should call them "American Units" IMO.

sophiecentaur
Gold Member
I'm sure the Americans couldn't bring themselves to use what we call "Imperial Units". They spent enough time in shrugging off all that stuff of Empire (Macdonalds and Microsoft excluded, of course).
Strange thing about the 'slug', though. I was taught the last knockings of Imperial Units whilst at School in the late 50s and then some cgs on the way through Univeristy. But I was never taught, formally, about the slug (I'm sure that would have stuck in my memory because all the rest did). We used Pounds for mass and Poundals for the unit of Force. One Poundal was the force needed to accelerate a Pound Mass by 1ft/sec/sec. Then we used One Pound Force, which was 32 Poundals.

PhanthomJay
Homework Helper
Gold Member
In structural engineering in the USA, the mass concept of slugs and the mass concept of pound-mass, are seldom used. That's because we deal mostly with forces and weights. When required to use mass, as in seismic or frequency calculations, we just divide the weight by g, per m = W/g, and the mass comes out in slugs and we don't give it a name. Well, sometimes, when forced to. With a chuckle.

SI is all but dead in the States. 40 years ago, we were mandated to convert to SI. The weather guys started to give temperature readings in degrees C as well as degrees F. The banks proudly displayed the outdoor temps in both units on their digital signboards. Distance signs on highways showed both kilometer and mile units. The kids were given their height in centimeters, although their weight was still measured in pounds because no on dare weigh them in kilos which is a mass unit, and Newtons, well...those are cake covered figs. And Shell Gas started to pump out their gas in liters instead of gallons, an experiment that lasted a month or 2 before they quickly had to change back to gallons because the public was so confused they stopped going there. Topo maps were converted to meters, and highway plans were developed in meters also, which caused huge and costly errors in construction. Now 40 years later, no more bank displays in Celsius, the km signs are long gone, the kids don't know what a centimeter is, and Shell gas is still pumping in gallons...US gallons, that is. The engineering Codes are still written with both systems, but it only serves to double the size and cost of the book, since we don't pay attention to the SI part. Like it or not, the 'american' units are here to stay for a long long long time.

sophiecentaur
Gold Member
I wonder what the actual cost of NOT changing to SI use has been and will be. The foot pound system is such a bodge and there must have been so many cockups resulting from the clash of units - not least in space!
The UK still hangs on to Miles for road distances but sells fuel in litres - making mpg a total nonsense, except on the dashboard display in cars, which will give you whichever combination you want.

PhanthomJay
Homework Helper
Gold Member
As I recall, we crashed landed a ship on mars a few years ago because someone forgot to program the computer to convert inches to millimeters or something like that. Wonder what that cost, zillions? But on the other hand, I can easily calculate the max allowed point load at the free end in pounds for a 6 x 6 x 1/4 inch 10 feet long tube cantilevered using
ASTM grade 50 Corten with a 50 ksi yield strength. Now ask me to size a beam in mm using newtons and mega pascals, it would take me hours and I'm not cheap so who's going to pay me and I would never be sur if I sized it properly without a good feel for the numbers and th US mills would laugh me off and send be to China for the order and my calculator would burn out counting the ridiculous number of zeros befire or after the decimal point. I don't know how English engineers converted without a bunch of errors and a lot of cursing.

sophiecentaur
Gold Member
Now ask me to size a beam in mm using newtons and mega pascals, it would take me hours and I'm not cheap so who's going to pay me and I would never be sur if I sized it properly without a good feel for the numbers and th US mills would laugh me off and send be to China for the order and my calculator would burn out counting the ridiculous number of zeros befire or after the decimal point.
Yebbut you're the sort of guy who can be bothered to be on PF. You clearly have the right stuff. You would find it a bit of a pain but you'd get there in the end. I made the transition years ago and 'it just happened'. So could you. Then just think of the benefits for future generations.

Also British and American engineers both tend to have a problem learning 'foreign' languages (unlike other European and Far Eastern Engineers who put the Brits to shame). Same thing at work here in that they are in a big enough minority not to need to make the effort. I can only say that it's worth it in the end.

K^2
The whole point being when using lb interchangeably for mass and force, as in the English Engineering units, I needed to be reminded to divide by g in the simple KE formula
Not really. The units aren't units of g. It's the unit conversion factor, actually. Let me show you exactly where it comes from.

Lets start with fact that I can write g=9.8N/kg. These are consistent units that give me the correct answer. Similarly, g = 1lb(force)/lb(mass) = 32 lb/slug. Again, that's exactly the same value with different units. So say you have an object with mass m=x lb(mass) traveling at v=y ft/s. You then compute the kinetic energy according the the formula E=(1/2) mv² = (1/2) xy² lb(mass) ft²/s². But these aren't the units you want. You want E in lb(force) ft. What do you do? As usual, you multiply by 1. Except, in this case, 1=g/g. Fair?

E = (1/2) gxy²/g lb(mass) ft²/s²

Except now I insert two different definitions of g from above.

E = (1/2) (1 lb(force)/lb(mass) xy² / (32 lb(force)/slug) lb(mass) ft²/s².

If I simplify the above, I get this.

E = (1/2) xy²/(32 lb(mass)/slug) lb(mass) ft²/s² = (1/2) xy²/32 slug ft²/s² = (1/2) xy²/32 lb(force) ft.

These are finally the units you want with the factor of 32 popping up. But that 32 doesn't have units of ft/s² attached to it, so it's not equal to g. It has units of lb(mass)/slug attached to it, which makes it a conversion factor, which only happens to have same numerical value as g in these particular units because of how pounds and slugs are defined.

Always, and I mean always carry units through your computation. If units don't match, insert factors that make them match, keeping in mind that the net result should always be multiplication by 1. If you start dropping units and just punching numbers into calculator, you'll run into confusion like this.

Alternatively, of course, you can always pre-convert all your numbers and stick to a specific set of units. But it's not always 100% practical. Sometimes you do want to use a formula with mixed units, and in these cases, just make sure you don't forget to carry units through.

By the way, Google Calculator is really good with this. You can enter a formula with mixed units, making sure you specify them, and it will spit an answer in any units you like. A word of warning, however. It does interpret pounds as unit of mass by default, so be careful when working with these.

Not really. The units aren't units of g. It's the unit conversion factor, actually. Let me show you exactly where it comes from.

Lets start with fact that I can write g=9.8N/kg. These are consistent units that give me the correct answer. Similarly, g = 1lb(force)/lb(mass) = 32 lb/slug. Again, that's exactly the same value with different units. So say you have an object with mass m=x lb(mass) traveling at v=y ft/s. You then compute the kinetic energy according the the formula E=(1/2) mv² = (1/2) xy² lb(mass) ft²/s². But these aren't the units you want. You want E in lb(force) ft. What do you do? As usual, you multiply by 1. Except, in this case, 1=g/g. Fair?

E = (1/2) gxy²/g lb(mass) ft²/s²

Except now I insert two different definitions of g from above.

E = (1/2) (1 lb(force)/lb(mass) xy² / (32 lb(force)/slug) lb(mass) ft²/s².

If I simplify the above, I get this.

E = (1/2) xy²/(32 lb(mass)/slug) lb(mass) ft²/s² = (1/2) xy²/32 slug ft²/s² = (1/2) xy²/32 lb(force) ft.

These are finally the units you want with the factor of 32 popping up. But that 32 doesn't have units of ft/s² attached to it, so it's not equal to g. It has units of lb(mass)/slug attached to it, which makes it a conversion factor, which only happens to have same numerical value as g in these particular units because of how pounds and slugs are defined.

Always, and I mean always carry units through your computation. If units don't match, insert factors that make them match, keeping in mind that the net result should always be multiplication by 1. If you start dropping units and just punching numbers into calculator, you'll run into confusion like this.

Alternatively, of course, you can always pre-convert all your numbers and stick to a specific set of units. But it's not always 100% practical. Sometimes you do want to use a formula with mixed units, and in these cases, just make sure you don't forget to carry units through.

By the way, Google Calculator is really good with this. You can enter a formula with mixed units, making sure you specify them, and it will spit an answer in any units you like. A word of warning, however. It does interpret pounds as unit of mass by default, so be careful when working with these.

yes you are right and I actually went through the exercise after I posted in the same manner you did.

Also found some formula lists on the internet that acknowledge needing to divide by "g" (number,not units) when using Lb mass and force.