Deriving the Work Energy Theorem – Don’t need to use differentials as fractions

When I first saw this derivation my teacher allowed differentials to cancel each other out. I found this strange since differential notation is based on a limiting process. How could they become simple fractions? I learned later that this is possible due to the chain-rule.

Here I will derive the Work Energy theorem in 1-D without canceling any differential fractions.

Using Newton’s Second Law: F=ma we replace the acceleration with the time derivative of velocity.

    \[ F=m\frac{d v}{dt} \]

Work is defined as the integral of force times distance integrated over its trajectory. The reason for this is that this integral can be evaluated.

    \[ W=\int F dx \]

    \[ W=m \int \frac{d v}{dt}dx \]

Due to chain rule we have the following:

    \[ \frac{d v}{dt}=\frac{d v}{dx} \frac{d x}{dt} \]

    \[ W=m \int \frac{d v}{dx} \frac{d x}{dt}dx \]

One of the terms is simply the velocity.

    \[ W=m \int \frac{d v}{dx}vdx \]

This term below looks very interesting. It seems like someone took the derivative of a velocity function f(v) with respect to position x using the chain rule.

    \[\frac{d(f(v))}{dx}=\frac{d v}{dx}v\]

f(v) is the square of the velocity.

    \[ f(v)=\frac{v^2}{2}\]

Now subbing into the work integral.

    \[ W=m \int \frac{d(f(v))}{dx}dx \]

We can see that the integral undoes the derivative since they are both respect to position. We should clarify that the integral is taken from the initial position to the final which correspond to the initial and final velocity.

    \[ W = m \frac{v^2}{2} \Bigg|_{v_i}^{v_f} \]

    \[ W = \frac{1}{2}m{v_f}^2-\frac{1}{2}m{v_i}^2 \]

Define Kinetic Energy as the following KE = \frac{1}{2}m{v}^2

    \[ W = \Delta KE \]

Done! Let me know if I made a mistake.

 

 

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