Why is the kinetic energy equation multiplied by ½?

[Dale]

I thought the whole point of the thread was ...

Well, the whole point of the thread is still unclear since the @rajen0201 never bothered to clarify what it was that he wanted in an explanation.

But I disagree that the 1/2 is a convention except insofar as the definition of any word is a convention. If I do 3 J of work I gain 3 J of KE, not 6 J. The definition of work and the conservation of energy require the 1/2.

 

[jbriggs444]

Well, the whole point of the thread is still unclear since the @rajen0201 never bothered to clarify what it was that he wanted in an explanation.

But I disagree that the 1/2 is a convention except insofar as the definition of any word is a convention. If I do 3 J of work I gain 3 J of KE, not 6 J. The definition of work and the conservation of energy require the 1/2.

"except insofar as the definition of any word is a convention". Yes.

It would be silly. [Full stop]. But we could have decided to measure work in Newton-meters and kinetic energy in half-joules. Then we could do 3 Newton-meters of work and gain 6 half-joules of kinetic energy. $W=F\cdot s$ and $E=mv^2$. It would then be a mathematical fact that the two units differ by a factor of two ($\Delta E = 2W$) That we choose not to do so is a matter of convention. A convention adopted for good and obvious reasons.

 

[Dale]

we could have decided to measure work in Newton-meters and kinetic energy in half-joules.

Changing the units doesn’t change the rules of integration nor does it get rid of the dimensionless constants of integration. It just hides them. That factor of 1/2 still arises in the integration and it is still present in the formula, it is just hidden by the units and the resulting conversion factors.

 

[Mister T]

All you have to do is define $W=2 F\cdot s$.

 

[Dale]

All you have to do is define $W=2 F\cdot s$.

Yes. Which is why I said “except insofar as the definition of any word is a convention”

 

[sophiecentaur]

All you have to do is define $W=2 F\cdot s$.

But that would be totally arbitrary and would require much more than 75 posts to discuss and justify. The 'half' in the KE formula would be there, even on the planet Zog and whatever units they built their Physics on. It would be just as loony to try to fit 19e or π/93 into our formulae.

 

[A.T.]

...just as loony to try to fit 19e or π/93 into our formulae.

Since you mention π, the common convention there went exactly the other way, and we ended up with many '2π' in the formulas. I personally would prefer to have no factor 2 in the circle circumference, and instead a factor 1/2 in the circle area, because that can be derived from the triangle area as well.

 

[PeroK]

Since you mention π, the common convention there went exactly the other way, and we ended up with many '2π' in the formulas. I personally would prefer to have no factor 2 in the circle circumference, and instead a factor 1/2 in the circle area, because that can be derived from the triangle area as well.

You're alluding to the Tau Manifesto?

https://tauday.com/tau-manifesto

 

[jack action]

All you have to do is define $W=2 F\cdot s$.

Wouldn't it leads to the question «Why is the work equation multiplied by 2?»

To which the probable answer would be: «To eliminate the ½ in the kinetic energy equation.»

 

[Dale]

This thread went from off topic to further off topic. Time to put it out of its misery