# Work as an integral of mass over velocity?

As I've understood it, the area under $F$-$s$-graph is the work done, so then :$$W(s)=\int{F(s)ds}$$ I am also given this equation ($W_k$ is kinetic energy, which is equal the work done to set the object in motion): $$W_k=\frac{1}{2}mv^2$$ Does this mean that work is also the area under a $m$-$v$-graph, like so: $$W(v)=\int{m(v)dv}$$ Could anyone explain why this is, or isn't true?

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Hint: Did you check that the units (or more correctly, the physical dimensions) are the same on the left-hand side and the right-hand side of your last formula (v2)? – Qmechanic Mar 24 '13 at 14:24
I don't quite understand. The dimensions? – Daniel Beecham Mar 24 '13 at 15:22
And that's only the first test! More fundamentally, how should one understand that mass $m(v)$ could depend on speed in your last formula (v2), especially seen in the light that we are just doing basic non-relativistic Newtonian mechanics? – Qmechanic Mar 24 '13 at 18:05

What is the net force? Newton:

$F_{net}=ma=m\frac{dv}{dt}$

So the net work done to accelerate a particle from $v_0=0$ to final velocity $v_f$ is

$W_{net}=\int F_{net}ds=\int m\frac{dv}{dt}ds=m \int_{v_0}^{v_f} v dv=\frac{1}{2}mv_f^2$

where in the last step I used $ds/dt=v$ and $m$ constant. If you had some crazy system where $m=m(v)$, then that mass could be a function of velocity. Maybe some effective mass due to interactions...

So what you wrote is $not$ correct since (as Qmechanic said), the units don't match; $mdv$ has units of $\text{kg ms}^{-1}$, and units of work are $\text{J=kg m}^2s^{-2}$.

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