Why is force "accumulated" more at a higher speed?

Question

I tried to understand why kinetic energy is proportional to the square of velocity. In this endeavor I stumbled upon a book "Emilie du Chatelet: Daring Genius of the Enlightenment" (ISBN 978-0-14-311268-6), where she explains it in one section. The key point I took away is "...a moving body accumulated force, and thus the formula describing this movement must include squaring of the speed." I understand that the statement faster moving bodies "accumulate" even more force implies squaring of the speed, but I don't understand why the initial statement is true. Also, I know that a body with twice the speed will penetrate four times deeper upon collision, but that is a demonstration, not an explanation.

Answer 1

I think Emilie meant this (based on a documentary) :

The maximum height reached by the ball, going upward against Earth's gravity with speed $v$ is given by (assuming it starts at height 0):

$$s=\frac{v^2}{2g}$$

So, if a ball has twice the speed of another ball, it reaches four times the height that the other ball reaches.

Today, we understand this in terms of the work-energy theorem. Gravity has to do work, on the twice-as-fast ball, for four times as long a distance, to convert all of the ball's kinetic energy into potential energy.

Emilie showed the mass times square of the speed of an object was a useful physical quantity, as a "measure of how much motion the object carries", where the "measure of motion" means "how much work it takes to stop the object"

Today, we don't call this quantity "accumulated force". We call it " Kinetic energy".

Answer 2

I intuitively expected the kinetic energy to be the same as momentum. The reason why it cannot be the same lies in the fact that kinetic energy (gained by doing work) is the total force accumulated over some distance, since $$E_k = W = \int F\mathrm d l$$ while momentum is the accumulation of force over time, since from the Newton's second law we get $$\frac{\mathrm d P}{\mathrm d t} = F\to P = \int F\mathrm dt$$ If we set a body in motion from standstill with force $F$ so it gains the speed $v$, it will accumulate a certain amount of force over time and distance $l$. If we use the same force again to double the speed, the force accumulated over this time interval does not depend on $v$. But, since the object was already moving, the total distance travelled in the second "phase" must be the distance the object would travel if it was unaccelerated with speed $v$, plus, the distance the object would travel if it was accelerated by the force $F$ but did not have the initial speed $v$. Thus, the total distance traveled must be necessarily longer and it would also be convenient to introduce a quantity that reflects this increased growth of accumulation of force with respect to distance caused by a dependence on instantaneous speed.