Physics Professors...I'm confued about Kinetic Energy numbers?

[arizonaguide]

To clarify further...in the definition of KE "Foot-Pound" is measured as the Energy required to raise a 1LB weight a vertical distance of 1FT. This makes no mention of Velocity units (only "distance")...yet velocity (including time units) is one of the variables in the KE formula.

(1/2Mass x Velocity^2 = FtLb.)

My "problem" Simply,

1.) if Kinetic Energy of the projectile is measured in FtLb,

and 2.) one of the KE formula's multipliers is "Velocity"...

then 3.) where the hell is the complete Velocity Unit (to include time) in the result?

Why do we throw away the TIME factor of velocity in the KE result?
Shouldn't the KE result be in FtLb/second rather than just FtLb? (as a full measure of the "capacity to do work")

If we're going to debate "stopping power" then lets talk about "power" being perhaps more accurate than "energy". I believe the KE formula itself is flawed in that it drops the unit of TIME from the velocity multiplier.
Someone REALLY smart PLEASE explain this to me.

[lksteve]

T13.) where the hell is the complete Velocity Unit (to include time) in the result?.Kinetic energy is defined by an object's motion. Time is not a factor. Here's another way of looking at your question...you make a cake, comprised of sugar, flour, eggs, milk and whatever else...the discreet ingredients are lost in the mix and are indistinguishable when the cake comes out of the oven...mass and velocity determine kinetic energy...time as a discreet ingredient of velocity and gets lost in the result.

Unit definitions are also at play here...in physics class, when you determine kinetic energy, the answer you end up with will most likely be in joules...here's a conversion table.

http://www.uwsp.edu/CNR/wcee/keep/Mod1/Whatis/energyresourcetables.htm

A foot-pound is 1.356 joules. When you calculate in one system and convert to another, each discreet element gets lost.

[arizonaguide]

Steve, I agree with you mostly, but

we didn't lose the discrete distance measurement part of velocity, just the time part.

We keep the units of mass (weight) in the form of the LB, and we keep only HALF of the velocity units in the form of FT...and we just drop the Seconds unit of time.

If a foot-pound is a measurement of KINETIC ENERGY (or joules) and is the "capacity" for a moving object to do work...then how can moving 1LB a distance of 1FT in 1Second be the same capacity to do work as moving the same 1LB by 1FT in several hours. That would seem to me to be a different "capacity" to do work.

PLUS, mathematically...you can't just "drop" units of measure from a formula.
Something doesn't add up...but I'm sure it's a lack of understanding on MY part.

I need to fully understand the difference between "energy" (kinetic, and other), "momentum", and "work" and power". The answer HAS to be in there somewhere.

[Buffalobob]

Foot pounds is an old system of energy where

E = 1/2M*V*V/g

where g = gravity = 32.2 ft/sec *sec


If you wish to be further confused it involves two concepts using the same name -- pound as a unit of weight and pounds as a unit of force.



This old system will also give you fits when you try to calculate momentum.

[BrainStorm]

Mass and weight are often confused as we mistakenly use them interchangeably. If you recall ideas such as "how much will I weigh on the moon," what you are being reminded of is that your mass is unchanged but your weight is different because the force of gravity is different on the moon than on earth.

The equation for kinetic energy references "M" for mass. As mentioned earlier, converting weight to mass requires taking into consideration the force of gravity. So where you are when making that calculation makes a difference. The force of gravity differs slightly depending on where you are on earth as well. Gravity can be thought of as an artifact of the concentration of mass. So granite mountain ranges might be expected to have a slightly higher gravity than other places.

[lksteve]

I need to fully understand the difference between "energy" (kinetic, and other), "momentum", and "work" and power". The answer HAS to be in there somewhere.
It's been 40 years since Freshman Physics...it's been 25 years since I took geophysics...for me, the concepts have been reduced to abstractions since I don't use them in my profession...as far as reloading or other applications go, what used to pass for understanding has faded to acceptance. If you really need to understand these concepts, it might not hurt to enroll in an elementary mechanics class. I got rid of my last college physics text some time back.

Remember that the value for kinetic energy changes with velocity. As the velocity changes, the capacity to move weight changes. 1 pound moving at 1 foot per second will have a different kinetic energy value than one pound moving at an inch per second.

Hell, inertial moment is really gonna give you fits.

[Maytime]

If you decompose the units of kinetic energy, we can see how time is accounted for in this handy dandy pdf I whipped up.

Let me know if you have any questions!

[Chris Cram]

In another form...

1 J = 1 N * m = [(kg * m)/(s^2)] * m = kg * [m/s]^2

K = n [ mass * speed^2]

So from the first equation, N is being replace with lb, and m is being replace by ft…

The energy of the object in motion is converted…
in effect, to heat on impact… or to a hell of shock wave.
So velocity (speed) becomes meaningless?

[BrainStorm]

If you decompose the units of kinetic energy, we can see how time is accounted for in this handy dandy pdf I whipped up.

Let me know if you have any questions!

Nice work for just "whipped up."

What environment did you use? I typically use MiTeX, but always looking to add to my tool kit.

[arizonaguide]

Maytime, thank you.
I will sit down and attempt to get a handle on that.
Thank you for the time. (no pun intended).

:)

I did have college physics a lifetime ago...and now it's probably just enough knowlege (left in memory) to be dangerous to myself! ;) I think I will take a community college class just to settle my ongoing "stopping power" debate once and for ALL!

And, everyone please notice that this "stopping power" debate didn't turn into the opinionated "insult exchange" that typically happens on that topic...thanks again to the "quiet professionals" on this forum. Thanks to ALL for that!
MUCH appreciated. Any more clarifications/explainations/simplifications are WELCOME!

Oh yeah, for what it's worth I'm talking about small arms at sea level on the surface of the earth.
I'm not too worried about the difference in LB (weight/force) if it's on the moon, and all my projectiles are small arms so I don't have to adjust my trajectory for the gravitational pull of the moon either. ;) We can assume sea level small arms calculations.

[Buffalobob]

it is more correct

Had a little trouble with freshman English did you, Maytime? :D

[incarcerated]

To clarify further...in the definition of KE "Foot-Pound" is measured as the Energy required to raise a 1LB weight a vertical distance of 1FT. This makes no mention of Velocity units (only "distance")...yet velocity (including time units) is one of the variables in the KE formula.

(1/2Mass x Velocity^2 = FtLb.)

My "problem" Simply,

1.) if Kinetic Energy of the projectile is measured in FtLb,

and 2.) one of the KE formula's multipliers is "Velocity"...

then 3.) where the hell is the complete Velocity Unit (to include time) in the result?

Why do we throw away the TIME factor of velocity in the KE result?
Shouldn't the KE result be in FtLb/second rather than just FtLb? (as a full measure of the "capacity to do work")

If we're going to debate "stopping power" then lets talk about "power" being perhaps more accurate than "energy". I believe the KE formula itself is flawed in that it drops the unit of TIME from the velocity multiplier.
Someone REALLY smart PLEASE explain this to me.

KE is the energy of a projectile in flight. That projectile’s ability to do work in the target is a different commodity: how that given KE is applied to a (living) target. Too many variables here (Time being one of them; geometry, target composition, projectile size and construction being others) for a handy, simplistic formula to hold sway: real world results are the preferred yardstick.

For myself, I try not to get carried away with the KE formula when considering ’stopping power,’ or a cartridge’s ability to perform in the real world. A few decades ago, we tended to confuse KE and Lethality, treating them as being synonymous. This notion led to a brief love affair with the 9mm Parabellum, which looks real good on paper when one is looking at KE numbers. Real world results did not align with the projections on paper, and the love affair ended badly in an FBI shootout in Florida on 11 April 1986.

With regards to Time, I suspect that this is what you’re looking for:
http://en.wikipedia.org/wiki/Impulse_(physics)
Put crudely, the Impulse deals with how quickly you lift that pound a distance of one foot.


DISCLAIMER: I do not fit this description:

Someone REALLY smart PLEASE explain this to me.

Where the heck is Team Sergeant?

[Maytime]

Had a little trouble with freshman English did you, Maytime?

Haha Buffalobob you got me. I'm definitely not the best wordsmith in the world.

What environment did you use?

I used MS Word 2007. It comes with a nifty PDF publisher.

arizonaguide, you're quite welcome! The beauty of US Customary units over Imperial is that they are more like metric in that you don't have to mess with slugs, which can be a pain in the ass when designing thermal/fluid systems, but that's another can of spam.

[Dark Matter]

The unit question has been well answered, but I thought the following was interesting in regards to an actual understanding of the difference between mass and weight.

Perhaps a good thought experiment is this:

You are on the ISS. Being in orbit, you (and everything around you) is experiencing no gravity*. There is a large object, say an anvil, floating on one side of the room. You would like to move it to the other side of the room. Will this object be easy to move?

Or another way to think about it. The anvil is floating toward you. Will it be easy to stop?

Oh yeah, for what it's worth I'm talking about small arms at sea level on the surface of the earth.
I'm not too worried about the difference in LB (weight/force) if it's on the moon, and all my projectiles are small arms so I don't have to adjust my trajectory for the gravitational pull of the moon either. ;) We can assume sea level small arms calculations.

So in regards to the above, the projectile will have the same energy on the moon as it does on the earth. That is the point being made in distinguishing mass from weight. If we use metric units (because I think they are generally more clear and easy to work with), one unit of energy (Joule) is kg*m/s^2, as mentioned earlier. The unit of mass, kg, does not change (mass is conserved). The acceleration obtained from the chemical propellent (m/s^2) is also the same.

A way to think about the difference between the two is this. Think of mass as some quantity that every object has. Think of gravity (or any other accelerations) as an arrow (vector) from the object in the proper direction. Weight is a combination of the quantity mass, and the length or size of the arrow.

Taking this further, if the size of the arrow is zero (like in our thought experiment), the object still has mass. So in trying to push the object you are in effect making the arrow point towards yourself. So the object has weight relative to you as long as you push on it.

This representation can be very powerful for looking at mechanics problems with clarity. Drawing it out is called a Free Body Diagram, which I'm sure is explained better on the internet with pictures than it is by me.

Cheers.

*Not strictly true, since objects in orbit are actually in a state of freefall inside the earth's gravitational field.

[Mississippi10]

Most of the confusion comes from the fact that we're used to using "pound" as both a unit of force and unit of weight, and that we seldom stop to think about the fact that gravity isn't the same everywhere. If you take a pound weight up 1000 feet, it has 1000 foot-pounds of potential energy. If you then drop it, in the absence of air resistance, it will start to fall, and every second it will be falling 32 feet per second faster than it was a second before. In slightly less than eight seconds, it will be falling a little over 250 feet per second, and it will have fallen 1000 feet. If you plug 252 ft/sec and 1 pound into the proper formula, you'll get 1000 foot pounds, only now the potential energy it had a thousand feet up has been changed into kinetic energy by the force of gravity.

[nukem]

I didn't go through and check but I'm sure if you wright it out as a conversion problem you will find that the time unit cancels somewhere along the line.

[Maytime]

Most of the confusion comes from the fact that we're used to using "pound" as both a unit of force and unit of weight, and that we seldom stop to think about the fact that gravity isn't the same everywhere. If you take a pound weight up 1000 feet, it has 1000 foot-pounds of potential energy. If you then drop it, in the absence of air resistance, it will start to fall, and every second it will be falling 32 feet per second faster than it was a second before. In slightly less than eight seconds, it will be falling a little over 250 feet per second, and it will have fallen 1000 feet. If you plug 252 ft/sec and 1 pound into the proper formula, you'll get 1000 foot pounds, only now the potential energy it had a thousand feet up has been changed into kinetic energy by the force of gravity.

I hate to be a stickler and I mean no disrespect, but I feel I need to clarify a few things in your post. Please don't take it the wrong way, you're not wrong per se, but I need to add a few things:

1) Weight is force with a specific vector, i.e. it has magnitude and direction, which points downward to the center of the Earth at all times. Force is a vector that may point any which way at any given time.

2) On or near Earth's surface, gravity is nominally the same. If a space shuttle wasn't in orbit, i.e. it stood still, it would still feel about 90% of Earth's gravity and start to move towards the center of the Earth at about 29ft/(s^2). Keep in mind a typical space shuttle orbit is about 250 miles above sea level.

3) Your example with the falling object is fundamentally correct. Mechanical energy is conserved at all times. In a closed system, Total Energy = Potential Energy + Kinetic Energy = mass*gravity*height + 0.5*mass*(velocity^2). A 1lbf object at rest at 1000ft AGL will have the same energy if it was released to free fall to 0ft, because 100% of its potential energy was converted to kinetic energy. Total Energy on its way down does not change since as potential energy is getting smaller, kinetic energy is getting larger and they add up to total energy at all times (assuming no friction, etc).

Again, please don't take this as I'm saying you're wrong, because you're not, it's just that since we've opened this can of worms, I feel like things should be defined as clearly as possible.

[Mississippi10]

Of course, you're absolutely right, Maytime. We all live in a world where it takes the same effort (as far as we can discriminate) to lift a given object on top of Pikes Peak as it does in Death Valley, and, if we drop it, it always falls straight towards the center of the earth. The fact that other results are possible other places in the universe, and those results have to be correctly predicted by our formulas, causes complication, and confusion.

[Maytime]

causes complication, and confusion.

Amen to that!