I'm attempting to solve a simple equation for work done when pulling a spring with a mass attached to it from point A to point B. Point A is the length of the spring with a mass attached to it, and point B is a fixed length after I've done an unknown amount of work. I've already figured out the spring constant.
Work = force x compression distance, and since the force applied is changing over time and I'm trying to find total work I have to take the integral from A to B with respect to distance; that's the work done if the mass wasn't on the spring, I understand that much. But how does the mass already pulling on the spring come into this, when work is involved? Do I ignore that force since it was what pulled down the string from relaxed to point A to begin with, or do I have to account for that when integrating somehow?
...I'm not quite sure why I'm asking B8, but there's gotta be someone here who's a science-related major who finds this easy to conceptualize <_<
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"Duel audio? You mean like rap battles?" - termina_rzident KrahenProphet and Kana are on opposite ends of the Awesome Spectrum.
If I'm following you correctly, the spring is being pulled on by an external force causing a reaction force in the spring. If you're looking for the work done by the spring force, you do not need to account for the external force.
And remember that work is a vector, so what your sign.
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[witty phrase including the guru champ SuperNiceDog] If there is one thing I know, it is that I know nothing.
Yeah, I set up my integral exactly like that and got an answer. I'm pretty sure that answer is right if I disregard the mass. The only thing I'm thinking I got wrong in my calculations is that since I have been considering my system as the spring and the mass suspending from it so I could solve for the external force, would the mass modify the force you have to apply to displace it to the final x value? Or is the mass a nonfactor when calculating work in this situation?
I'm thinking about how it would take less force to displace the spring with the mass on and more force if the mass were removed, and it's making me second guess myself >_>
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"Duel audio? You mean like rap battles?" - termina_rzident KrahenProphet and Kana are on opposite ends of the Awesome Spectrum.
The point is that the mass doesn't matter. If a spring goes from point A to point B, there is a set amount of work done, which is represented by that integral, no matter how that work got done. There could be a mass pulling it down, you could be manually pulling it, whatever.
-- _foolmo_ 'but that statement is something only an Aspergers patient would say' - UltimaterializerX
Oh I see your confusion. If there is already a mass there, you can manually displace it easier than if there wasn't a mass there, but the work done is the same. It just depends whether how much work is being done by gravity, and how much by your own hand.
But the problem is only asking about the work done so yeah.
-- _foolmo_ 'You are obviously intelligent and insightful' - Sir Chris about me
frankftw posted... If I'm following you correctly, the spring is being pulled on by an external force causing a reaction force in the spring. If you're looking for the work done by the spring force, you do not need to account for the external force.
And remember that work is a vector, so what your sign.
Ah yeah, that's exactly what I was trying to figure out. And I got a negative sign as my x values are positive; I'm guessing that's the resistance in the spring, whereas inverting the sign is the work done by the hand that's pulling the spring? I'm probably phrasing this wrong.
I'm just trying to understand this problem completely since I'm doing a screencast of it as homework and he grades understanding as much as getting the right answer.
And thanks a ton guys, my physics teacher's a genius but doesn't teach all that well >__>
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"Duel audio? You mean like rap battles?" - termina_rzident KrahenProphet and Kana are on opposite ends of the Awesome Spectrum.
if the mass is part of your system, then no work is done.
Work is done by systems. Work is done when a system applies a force over a distance.
Work cannot be done inside a system. Any forces that act within systems cause changes in internal energy.
Lets go back to the spring and mass system. What if you wanted to know the work done on a mass, that is hanging from a spring?
Your work integral will still be Int[initial position, final position](Sum of forces in x direction)(dx).
Now, instead of only using the spring force, you would add the gravitational force to your sum of forces. If you defined up as the positive direction, your sum of forces would look like kx-(mass of block)g
In Summary, when you want to find the work done on a mass, find the forces that act on that mass, sum them together, and integrate them over the distance traveled in the direction that the forces act.
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"What if you just eat vegetarians?"-neonreaper "^ You are what you eat. ^_~"-Koiji