Kinetic Energy and Conservation of Energy, Part 2

See Part 1, where I used conservation of energy of a projectile to determine the formula for kinetic energy, Kmv². The aim was to see if I could derive a more intuitive understanding of this formula — especially why it depends on the square of the velocity. With a constant force and therefore constant acceleration, velocity increases linearly, by the same amount per second, so why does kinetic energy increase at increasingly large rates? (I should specify that I am using the nonrelativistic formula, so we assume speeds much lower than the speed of light.)

Problem: Can I understand the kinetic energy formula, Kmv², in a more intuitive manner?

A couple of connections have occurred to me. One is that ½mv² is ½v² multiplied by a constant (m). The form ½v² reminds me of the integral of a variable with respect to itself:

\displaystyle \int x \, dx = \frac{x^2}{2} + C

Perhaps I can do something similar with velocity:

\displaystyle \int v \, dv = \frac{v^2}{2} + C

In fact, I can multiply this by the constant m:

\displaystyle m \int v\, dv = m\left({\frac{v^2}{2} + C}\right)

and mv is the momentum, p:

\displaystyle \int mv \, dv = \frac{mv^2}{2} + C_1

\displaystyle \int p \, dv = \frac{mv^2}{2} + C_1

\displaystyle \int p \, dv = K + C_1

Since we know the kinetic energy is zero when velocity (and momentum) = 0,

\displaystyle \int p \, dv = K

This is very neat, but I don’t really understand what \int p \, dv , the integral of momentum with respect to velocity, should represent. But the v dv above gives me an idea. I know that v dv = a dx (for movement in the x direction). It is easy to prove this. Start with the following expression, which is of course equal to itself.

\displaystyle \frac{dx \, dv}{dt} = \frac{dx \, dv}{dt}

\displaystyle \frac{dx}{dt}dv = dx\frac{dv}{dt}

\displaystyle v \, dv = dx \, a

I don’t have an intuitive sense of what a dx represents any more than of what v dv represents. But what if I multiply the m back in?

\displaystyle K = m \int v \, dv 

\displaystyle K = m \int a \, dx

\displaystyle K = \int ma \, dx

Of course, F=ma by Newton’s second law.

\displaystyle K = \int F \, dx

This is fantastic. Force times displacement (in the same direction) is equal to work. And by the work-energy principle the work done on an object equals the increase in its kinetic energy. Carrying the whole chain of reasoning:

\displaystyle W = \int F \, dx

\displaystyle W = \int ma \, dx

\displaystyle W = m \int v \, dv

\displaystyle W = \frac{mv^2}{2}  = K    

where the constant is the prior kinetic energy of the object, and where I assumed an initial starting velocity (and kinetic energy) of zero.

I thought about it another way, as well. Intuitively, I might think that energy requirements or transfers depend on force times time Ft, or \int F\, dt . But this is impulse, not work — which we’ve noted earlier is force times displacement Fx, or \int F \, dx . Part of the reason for this discrepancy is that you cannot arbitrarily apply forces to objects. As I explored in the Rocket Propulsion posts, a rocket engine using a steady amount of fuel does provide a constant force to accelerate the spacecraft, but most of the work is done on accelerating the exhaust. The work done on the spacecraft and therefore the spacecraft’s kinetic energy does increase quadratically, but it is only a small fraction of the total kinetic energy. Another example of a steady force is an object falling in a constant gravitational field, as we’ve been exploring in the prior post. But an accelerating object of course falls faster and faster — it drops 1, then 4, then 9, then 16 units of distance from its original height, even as its velocity increases linearly. Therefore, it is losing potential energy at an increasing rate, and so by conservation of energy, the kinetic energy will increase at an increasing rate.

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