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I pretty much had what the solutions had, but we disagreed with one thing, a minus sensitive sign.

$\sum W = \Delta K$

$\int_0^{-s} - ks^2 ds = \frac{ks^3}{3} = \frac{1}{2}m(v^2-4^2)$

Rearranging, I get $v^2 = 4^2 - \frac{2k(-0.2)^3}{3m} = 19.2$

So v = 4.38m/s

But the key did $v^2 = 4^2 - \frac{2k(0.2)^3}{3m} = 12.8$ where they let $\int_0^{s} - ks^2 ds = \frac{ks^3}{3}$

EDIT

Apparently I am stupid and I made a question that didn't even need answering...

I had this on paper and I typed up the wrong thing, that's why I was confused.

$\sum W = -\int_{-s}^{0} ks^2 ds = \frac{1}{2}m(v^2-4^2)$

Sorry everyone

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You've already taken into account that the force is against the direction of displacement in the minus sign inside the integral. Your $ds$ itself is (explicitly) negative, so no need to integrate to $-s$. –  Manishearth Mar 5 '12 at 3:50
    
If the spring was linear, F = -kx, I still would've needed to use take care of the additional minus sign. –  Hawk Mar 5 '12 at 3:51
1  
Usually in such problems I find it better to qualitatively analyse if velocity increases or decreases first and then explicitly tack on the sign onto $\Delta W$ –  Manishearth Mar 5 '12 at 3:52
    
Oh I see what you mean. I guesss it would make sense for the velocity to drop since it had more kinetic energy in the beginning. But this confuses with what I thought I knew before. –  Hawk Mar 5 '12 at 3:56
    
Its a common confusion. I have to go now, but I'll try to later post an answer explaining the multitude of negatives that can crop up here. –  Manishearth Mar 5 '12 at 4:07

1 Answer 1

up vote 1 down vote accepted

There's a multitude of negatives here:

Firstly, the force would be more correctly written as $-ks^2\hat{s}$, which allows us to analyse it like a normal spring. $F=ks^2$ on its own becomes unidirectional, which can lead to confusions. I think this is your main issue here.

Now, $W=\int \vec{F}\cdot d\vec{s}$ try not to get confused with work done by internal/external force. Here, let's look at work done by the external(spring) force. Now, since work is a path integral, our $d\vec{s}$ and thus our $\vec{s}$ must be in the direction of movement. If not, we get an extra negative. For convenience, we'll try to avoid that extra negative and take $\vec{s}$ as positive-right. Now, since the final displacement is also right, our upper limit becomes positive.

So, the only negative involved is the one in $-ks^2\hat{s}$, which bubbles out of the dot product.

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Doesn't make sense still. $W = \int \vec{F} \cdot d\vec{s} = \int -ks^2 \hat{s} \cdot d\vec{s} $. Our final displacement is to the left, not to the right. So I don't understand why you would still say the upper limit is positive. –  Hawk Mar 6 '12 at 2:49
    
@jak I said that work is a line integral. We integrate it in the direction that we are moving. As such, we move to the left, so $\hat{s},\vec{s},\mathrm{d}s,s$ are all positive-left. –  Manishearth Mar 6 '12 at 2:53
    
@jak or, you can define $ds'=-ds$ and use it in the integral, where s is now positive-right. –  Manishearth Mar 6 '12 at 3:41
    
Oh okay I see what you mean on the line integral stuff now. That argument also make sense (and consistent with a linear spring too). Not sure what your ds' meant. But how exactly are you suppose to know that when you tackle the problem the 1st time without the solutions if you were to use the method I chose from before? –  Hawk Mar 6 '12 at 6:20
    
Oh you know what? I actually screwed my LaTeX in the beginning of the problem, that's why I was confused. I actually did this right the 1st time lol... I'll vote you up. I edited my post. This was so stupid of me... –  Hawk Mar 6 '12 at 6:27

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