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^{2}.

Looking at this equation got me thinking, obviously this applies to a system of constant mass, but what if the mass varied? I tried looking it up on the Internet, but either I'm wording it wrong or there isn't much info out there freely available. So I figured I should try to derive an expression on my own.

Now then, consider a box of some mass on a frictionless table in vacuum. Then let's say a piston pushes the box from rest with a force that varies through time i.e. [tex]\overrightarrow{F} = \overrightarrow{F}(t)[/tex]

Now since the mass isn't constant, we'll use a more general equation for Newton's second law of motion: [tex]\overrightarrow{F}(t) = \frac{d(m\overrightarrow{v})}{dt} = \frac{dm}{dt}\overrightarrow{v} + \frac{d\overrightarrow{v}}{dt}m[/tex] where mass and velocity are functions of time.

Then, consider that the work done by a force is [tex]W = \int \overrightarrow{F}\cdot d\overrightarrow{s}[/tex] or in this case: [tex]W_{F}(t) = \int \overrightarrow{F}(t)\cdot d\overrightarrow{s}[/tex]

In this 1D problem, the force acts in the same direction as the displacement, so the angle between the vectors are zero and so we can write the equation thusly:

[tex]W_{F}(t) = \int F(t)ds[/tex] Expanding F(t):

[tex]W_{F}(t) = \int v\frac{dm}{dt}ds + \int m\frac{dv}{dt}ds[/tex] Using the chain rule and cancelling ds:

[tex]W_{F}(t) = \int v\frac{dm}{ds}\frac{ds}{dt}ds + \int m\frac{dv}{ds}\frac{ds}{dt}ds = \int v^{2}dm + \int mvdv = mv^{2} + \frac{1}{2}mv^{2} + C = \frac{3}{2}mv^{2} + C[/tex] Since the box was at rest in the beginning, the constant C = 0.

[tex]W_{F}(t) = \frac{3}{2}mv^{2} \left( = \frac{3}{2}m(t)v(t)^{2}\right)[/tex]

So my question is, is this equation valid classically? Have I made any mistakes in my reasoning?

Thanks in advance.