Tackling the Work-Energy Theorem

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The discussion focuses on understanding the work-energy theorem, emphasizing the relationship between work, force, distance, acceleration, and time. It highlights that distance is influenced by acceleration and time, which leads to the conclusion that work is half the product of mass and velocity squared. The conversation also explores the derivation of kinetic energy and its connection to momentum, questioning the rationale behind the multiplication of acceleration and distance. Additionally, it discusses the integral formulation of work and its equivalence to the change in kinetic energy. Overall, the thread seeks clarity on the conceptual underpinnings of energy, work, and their mathematical representations.
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I'm trying to tackle the work-energy theorum.

It has occurred to me that the distance in work: Force x distance

is determined by the acceleration in the force variable, and time of force as well.

The distance, creates a halving of the final product of the Joule unit, because the distance is half the acceleration. I also see the distance is also a result of the time of force. A longer distance implies more time the force occured. So, acceleration and time result in the distance, but time is sort of embedded in the distance.

I've read, Work = KE: Force x distance = 1/2 mass x velocity ^2.

I see the 1/2ing property in Work, because the distance traveled at an acceleration rate is half, but how is the speed squared explained for the other part of KE representation?

I hope that made sense.
 
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Energy does not make sense unless its units are the same as mass times speed squared. If you are not happy with the "squared", then you need to develop more intuition about energy. I'm not kidding; energy is an abstract quantity, not like volume (you can see), or force (you can feel).

A simple way to derive the result you want is using the formula from elementary physics:\Delta v^2=2ad. This works for a constant acceleration.

A better way that does not require a to be constant is the following (but you need to know calculus): W=\int Fdx=m\int adx=m\int {dv\over dt}dx=m\int {dv\over dx}{dx\over dt}dx=m\int vdv=m\Delta v^2/2
 
\Delta v^2=2ad, seems a bit cumbersome. My textbook explains something similar by comparing momentum to energy, but saying that kinetic energy is equal to momentum squared divided by twice the mass. I just don't get the round about way that shows the corelation.

Since I've asked the question, something occurred to me in looking at the equation, that seems to be a bit more directly representative.

If I think in terms of KE by isolating the the magnitude of acceleration and the magnitude of distance traveled, I see the work acceleration and distance magnitudes corelate exactly.

In terms of there respective magnitudes:

In Work = mass(acceleration)(distance).

In KE = mass (velocity) and either (distance) or (average velocity)

The acceleration magnitude is multiplied by the distance or average velocity magnitude the acceleration produces. Why they are multiplied is my next question. Do you know why?
 
You could use lineal momentum. krab used the chain rule.

W = \int F dx

Newton's 2nd Law

F = \frac{dP}{dt}

Constant Mass

F = m \frac{dv}{dt}

Substituting

W = m \int \frac{dv}{dt} dx

You know

v = \frac{dx}{dt}

so

vdt = dx

Substituing again:

W = m \int \frac{dv}{dt} vdt

W = m \int vdv

W =\frac{m \Delta v^2}{2}

Work done is equal to the change in the kinetic energy

W = \Delta K

For your books way:

\frac{P^2}{2m} = \frac{mv^2}{2} \frac{m}{m}

Same thing.
 
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For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

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