1D (net) work done by (net) force on a variable mass system.

[da_nang]

So I was sitting on the train last weekend, reading through my physics book on mechanical work and its relation to kinetic energy. One example would be that a box on a frictionless table being pushed and they would conclude that W = ΔK = ½mΔv².

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. \overrightarrow{F} = \overrightarrow{F}(t)

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

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

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:

W_{F}(t) = \int F(t)ds Expanding F(t):

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

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 Since the box was at rest in the beginning, the constant C = 0.

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

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

Thanks in advance.

 

[Ken G]

The integrals are not valid-- you did the integral over dm as if v was constant, and the integral over vdv as if m was constant. Where it gets interesting is if you imagine an object moving with constant velocity that starts out with negligible mass an gradually accumulates to mass m, maintained at v by a force. You then can take just your first integral to show that the work done by the force is mv², even though the final kinetic energy is 1/2 that. The remaining 1/2 is thermalized internally when you stick on mass, i.e., it goes into heat.