is the 1/2 from the equation KE=1/2mv^2 arbitrary ?

Question

KE=1/2mv^2 is the equation for kinetic energy. The author of an online physics textbook for scientists/engineers states that the 1/2 constant in the equation is entirely a result of the metric system. That is to say that it is arbitrary. If you were to use another system, say the old English system, the constant would be different.

I believe that the 1/2 is a necessry consequence of the double integration of F=ma from classical mechanics. If by definition f=ma is true, then the 1/2 value for the kinetic must follow regardless of any unit system and is thus not arbitrary.

From my understanding 1/2 in the kinetic energy theorom is dissimilar from say the gravitational constant found in the equation for the universal law of gravitation. The gravitational constant is one of proportionality whose value would change depending on the units used.

Can a chemist or physicist clarify? or anyone else?

Answer 1

No the 1/2 would always stay the same. because this half is figured out by breaking down the distance.No matter what the units are.

So you must know That W = Force * Displacement

Where force is in (N) and displacement is in (m) and W is in (J).

Deriving Kinetic energy equation to clarify question:

F=ma , d=(v1+v2)/2*t , a=(v2-v1)/t

W = (ma)*(v1+v2)/2*t
W= m(v2-v1)/t * (v1+v2)/2*t
t's cancel out.
W= m(v2-v1)*(v1+v2)/2

Since v1=0,


W = .5mv2^2

if we let v2=v then

E=.5mv^2 W>>>>E(replaced).

This expression is the term for kinetic energy. Kinetic energy is the energy of motion when work is done on an object. Kinetic energy is a scalar quantity and its SI unit is (J).

Basically you get .5 because you substitute the (v1+v2)/2*t
formula in displacement. Now if your units are cm (for v1 and v2 its gonna be cm/s not m/s for one of them as units have to be constant.)Now assume if it's cm/s for both you will do the exact same thing you will divide by 2 and the time will get cancelled out and v1=0 and difference of squares would lead you to your Kinetic Energy. What i am saying is that .5 would always be with you whenever you are finding the kinetic energy.

Answer 2

No, the 1/2 in the KE equation is not arbitrary. In fact it can be derived without integration by starting with v^2 = v0^2 - 2adel(x), which is simply one of the SUVAT equations relating velocity and acceleration. [See source.] v0 = initial velocity, v = resulting velocity after accelerating = a over a distance (del(x)); where del(x) = x - x0.

Now multiply v^2 = v0^2 - 2adel(x) by the mass (m) of the moving body (e.g., an airplane); so we have mv^2 = mv0^2 - 2ma del(x). Rearrange the terms and we have ma del(x) = 1/2 mv^2 - 1/2 mv0^2. If the initial velocity is v0 = 0, we have ma del(x) = 1/2 mv^2 = F del(x) = work = kinetic energy; where F = ma = the force generating the work and, thereby, kinetic energy.

And there you have it...no units (no metric system) were assumed to derive the KE equation. Could be SI or English or any other true metric system.

You are also right about G in F = GMm/r^2; it is in fact just a constant of proportionality found through experiment.

Answer 3

You're correct. The 1/2 is part of the formula, not a consequence of the unit system. If you would like a derivation of that equation, try this website:
http://hypertextbook.com/physics/mechanics/energy-kinetic/

If you are using a different unit system, the formula is the same, and only the derived units will change. We commonly use MKS units, so the units of energy are Joules. If we were to measure mass and velocity in CGS units (mass in grams, velocity in centimeters per second), the resulting energy would be in the CGS unit of energy, ergs.

Answer 4

The online book is correct and you are also right about the 1/2 coming from the integration of the force equation (over time to get energy). But you forget that the simplest form of F=ma uses SI units - these use the same scale as metric. As soon as you start putting in ponds of force instead of newtons or miles/hour instead of m/s etc etc you will get constants cropping up all over the place. You'll still get the right answers tho. When you integrate and rearrange to get the KE equation then you'll still get the 1/2 from the integration but you'll be multiplying 1/2 from whatever other constants you have made up from using non - SI units.

Answer 5

You are correct about the 1/2 being the result of integration; it has nothing whatever to do with which units are used, and this author you mentioned has made a serious error.

(I'm not a physicist, but I played one in college.)

Answer 6

How about an old engineer?

I will not attempt to answer your question as it makes little difference as to wether or not the 1/2 is arbitrary. However, the important issue is that it made you think about it rather than just accept it.

Consider where the equation(s) come from and what are their purpose(s)?

Basically, equations are an attempt, by man to record the events around us in such a manner as to be understood by all (mathematics). They are derived from observation and considered fact when they continue to accurately describe our universe.

Remember, however, that these observations are true from our current perspective and IF we change our perspective, they may no longer hold. And that anything derived from our current equation set in the name of advancing theoretical possibilities might not hold true either.

It is very likely that man will never have the physics of the universe down pat. In fact we may never be able to comprehend it all. But we as an intelligent(?) race of beings must be prepared for what is about to become known and act quickly in response, even if it goes against all that we "know to be true".

Good luck and I hope you add to our knowledge.

Answer 7

The relativistic expression for the energy of a moving particle is:

mc^2 / Sqrt(1+(v/c)^2)

Expressed as a series, this becomes:

mc^2 + 1/2 m v^2 + (3 m v^4)/(8 c^2) + ...

Eliminating mc^2 as the rest energy quantity, we have for relativistic kinetic energy:

1/2 m v^2 + (3 m v^4)/(8 c^2) + ...

For low values of v, the first term is used.

Answer 8

The author is *wrong* and you are mostly correct.
http://en.wikipedia.org/wiki/Kinetic_energy#Definition