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

1. Oct 3, 2011

### 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?