Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
1  
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
add comment

1 Answer

up vote 1 down vote accepted

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 $\text{J=kg m}^2s^{-2}$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.