Questions on Kinetic Energy:

(1) “Why is kinetic energy equal to (1/2) mv² ?

The problem here is with d, e, and h and the use of the variable v.

The equation is based on the assumption that you are considering the unique situation of an entity starting at zero speed in your frame of reference, which is, in a lot of applications, invalid. As a matter of course, I don't actually remember ever being taught or told that this particular form of the Kinetic Energy equation was valid only for a speed that starts at your 0 speed and goes to some value from there.

Let's start with equation d and the use of v as a symbol for velocity. Velocity by definition is a vector and has a direction associated with it. Since the energies involved are not direction sensitive, I have changed all the v variables to s, for speed. It is speed that we are working with irregardless of direction. So d becomes:

new d

Average Speed s' is

new e

and for an accelerating object

Which changes the others to:

new f

thus

new g

or

new h

Acceleration

new i

substituting g into h

new j

substituting i into c

So, if the true form of the equation is used, everything below the threshold where quantum interaction rates begin to have a significant effect would produce the proper result irregardless of the reference space and time frame. Long ago, I re-taught myself the new j equation so I wouldn't apply it improperly, all to find that it may have more problems.

To give an example that many can check out, consider the following:

We will be working with a bag of chocolate covered peanuts with a nominal weight (mass) of 1oz. These actually exist and here is a picture of one [Name removed for Copyright/Trademark purposes]:

Setting on the ground where you are, lightly toss this 1oz bag of peanuts to a friend so it gets up to 20mph during flight. If we use:

1

we get that we have to impart: of energy to the bag to toss it.
Now, let's assume we are on an airplane traveling (3) 565mph (very practical) and toss a bag of peanuts to our friend (I have actually done this with chocolate covered peanuts.). According to the standard teachings of today as shown by (2) James W. Brennan, when you deal with an object that started at a non zero speed, you simply take the initial kinetic energy and subtract the final kinetic energy as in

2

And, solving this for our situation we would get Original Kinetic Energy =

Final Kinetic Energy =

For a difference of or 57 times the energy required to throw the 1oz bag on the ground. This would be the same as trying to throw a 3.5lb bag to your friend (3.5lb = 56oz) while you were on the ground. This difference would be very noticeable under most conditions of significant speed. Try it yourself on the ground, throw a 1oz bag of anything and then throw a 3.5 lb bag of something. See if you can tell the difference.

Using the new j equation we would get:

Original Kinetic Energy =

Final Kinetic Energy =

For a difference of . The same as in our original condition, and what I experienced when I did it. Using the new j form of the equation seems to allow you to get the proper answer.

Everything seems to be working, but consider the condition where you and your friend are at the airport and you give him the bag of peanuts. He then boards the plane, which buzzes the airport at 565 mph and he throws the bag of peanuts to you. If you were to try and catch the bag of peanuts, with what force does the bag strike (and terminate) you? (disregarding the obvious windage and gravity effects.)

Using the new j equation, the kinetic energy should be because we gave the bag of peanuts to our friend as he boarded the plane, so its initial speed () was zero when he took it and it is now traveling at 585mph, and we are dealing with a closed system within the accuracy of the numbers we present.

Since this is a closed system and as such we cannot be creating or destroying energy, and we know that the bag had a 565mph kinetic energy of and by our own calculations, our friend added , the total energy added to the system should be the sum of the initial Ke plus the Ke added by our friend, .

These two seemingly logical and valid methods of calculating the Kinetic Energy do not balance with each other. You dissipate an extra of energy to stop the bag. In other words you managed to somehow add that amount of energy to the universe?

Personally, I'm stumped. I know for a fact that that throwing a bag of peanuts on a jet traveling at about 550mph feels like it takes no more energy than throwing the same bag doing 0mph. I also know that K_e is a valid concept, because it is one of the principles that keep an electron in orbit, and on a larger scale, it is obviously present in crash analysis.

Can anyone explain where my logic is wrong?

Thank you for your time

Bruce B Williams

To respond please email bbwilliams_bresnan.net and substitute the at symbol for the underscore

Offered Solutions
None so far



Variables
= Kinetic Energy
= mass
= speed relative to you
= final speed relative to you
= original speed relative to you
= change in distance
= force
= acceleration





Citations

1. http://www.chemistrycoach.com/BohrAssump.htm . Apr 16, 2006 (No longer valid). Contact the author for an html copy of this page.

2. James W. Brennan. Selland College of Applied Technology - Boise State University. Aug. 14, 2007 <http://selland.boisestate.edu/jbrennan/physics/notes/Energy/mechanical_energy.htm>

3. Aerospaceweb.org. Aircraft – Boing 747. Aug 16, 2007. <http://www.aerospaceweb.org/aircraft/jetliner/b747/>

Notes

1. Original Publish Date Aug 16, 2007

2. © 2007 Bruce B Williams

3. No independent math checks as of publish date. If errors they will be corrected at a later date if they do not solve the problem. If this is a math problem then the errors will be noted as the cause of the illogical conditions.

4. Proof read Aug 17, 2007. Changed New Kinetic to Final Kinetic in all instances. Will finish sometime the week of Aug 20.

5. Accepted for use Aug 31, 2007

6. Changed email address Aug 13, 2012