anonymous
  • anonymous
Total power?
Physics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
I'm trying to calculate the total power(W) that's used on an object.
anonymous
  • anonymous
The object is moving from position A to B The distance between them is 2 meters The force is not constant, 1M F = 1000N and 2M F = 3000N So this is how I calculated the work: \[W _{net} = 3000 \times 1 + 1000 \times 1 = 4000J\] Then tried to calculate the acceleration, since the object is 10Kg at 1M F= 1000 and at 2M = 3000N Then, \[a(at)1 = 100 m/s ^{2}\] and: \[a(at)2 = 300 m/s ^{2}\] So, the time taken: \[ D = \frac{ 1 }{ 2 } at ^{2}\] Time taken for the object to travel at 1 m = 0.141142365 Seconds Time taken fro the object to travel at 2m = 0.01153 Seconds \[P _{1}= \frac{ 1000 }{0.1411} = 7087Watts\] \[P _{2} = \frac{ 3000 }{ 0.01153 } = 260.1Kw\] I understand that I've should have used work integral, but the time is not the same since the acceleration is not the same due to the force not constant, so I think the total power is not the same also. Did I do things right here?
anonymous
  • anonymous
Total Power = P1 + P2?

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anonymous
  • anonymous
@experimentX @Jemurray3 @TheEric
anonymous
  • anonymous
Basically, how can I find the power if the force + acceleration on not constant over a distance of 2M?
experimentX
  • experimentX
power is not additive like work and energy. power is something like velocity ... related to time at some instant. you can find average power like average velocity.
anonymous
  • anonymous
How can I?
anonymous
  • anonymous
If work = 4000J how much power is produced? If the accretion is changing and I think time is also changing...
anonymous
  • anonymous
What I'm struggling with is that time is unknown and changing, acceleration is known and changing...
anonymous
  • anonymous
+ Force is changing as well.
anonymous
  • anonymous
As said above, it's like velocity. \[ \text{velocity = } \frac{\Delta x}{\Delta t} \] \[ \text{power = }\frac{W}{\Delta t} = \frac{\Delta E}{\Delta t} \] where W is work and E is energy. To help with this, I need to know what the force is. Is it steadily increasing the whole time?
anonymous
  • anonymous
Or if you could simply state the problem, that would be fine.
anonymous
  • anonymous
The force is increasing with time. Problem: Work is done on an object over 2M The force is increasing from 1000N to 3000N
anonymous
  • anonymous
How much time is taken, and how much power?
anonymous
  • anonymous
It starts from rest?
anonymous
  • anonymous
Yes Vi = 0.
anonymous
  • anonymous
iV*
anonymous
  • anonymous
You can find the work easily, but you cannot find the time or power unless you have the mass.
anonymous
  • anonymous
Say the mass was 100Kg
anonymous
  • anonymous
I just want to learn how this is done.
anonymous
  • anonymous
Okay. The force is \[ F = 1000x + 1000\] so the work is \[W = \int F \cdot dx = \int_0^2 (1000x + 1000 )dx=2000 + 2000 = 4000\text{ J} \]
anonymous
  • anonymous
Does that make sense?
anonymous
  • anonymous
It does.
anonymous
  • anonymous
Now how could we figure out the power?
anonymous
  • anonymous
In most cases I would find the total work first.
anonymous
  • anonymous
But what's left know is to figure out the Time & Power
anonymous
  • anonymous
Sure. We can't just integrate straightforwardly because the acceleration is a function of x. However, we can find the velocity in this way: The work done as a function of x is \[ W(x) = \int_0^x F(x') dx' = \int_0^x (1000x' + 1000) dx' = 500x^2+1000x\] But since it starts out not moving, the kinetic energy is simply equal to the work done so \[ \frac{1}{2} mv^2 = 500x^2 + 1000x \] if m = 100 kg, then \[50 v^2 = 500x^2 + 1000x \] \[v^2 = 10x^2 + 20x\] \[ v = \sqrt{10x^2 + 20x} \]
anonymous
  • anonymous
The power being supplied at any particular instant is F*v so the power is \[P = F\cdot v = (1000x+1000)\sqrt{10x^2+20x} \]
anonymous
  • anonymous
Im going to practice this thank you!
anonymous
  • anonymous
But wait, since the velocity is changing, what to do? See what's confusing is is the "changing" part.
anonymous
  • anonymous
Lastly the time it would take would be \[ t = \int dt = \int \frac{dx}{dx/dt} = \int \frac{dx}{v} \] so \[t =\int_0^2 \frac{dx}{\sqrt{10x^2 + 20x}}=\frac{1}{\sqrt{10}} \int_0^2 \frac{dx}{\sqrt{x^2+2x}} \]
anonymous
  • anonymous
What is confusing?
anonymous
  • anonymous
Im struggling with calculus, and I use simple algebra to solve this...
anonymous
  • anonymous
This cannot be solved with simple algebra, I'm afraid.
anonymous
  • anonymous
Since things are changing with time. It makes things a lot complicated.
anonymous
  • anonymous
Yep. That integral can be solved via substitution, yielding approximately 1.76 which means that the total time is approximately t = 0.557, by the way.
anonymous
  • anonymous
Meaning that the average power is \[ \bar{P} = \frac{4000 J}{0.557 s} = 7181 \text{ W}\]
anonymous
  • anonymous
@Jemurray3 What subjects do I have to learn perfectly to understand what you said perfectly? Integration and Derivatives?
anonymous
  • anonymous
Yes. Integration is an extensive subject though -- you'd be fine with simple antiderivatives and trigonometric substitution.
anonymous
  • anonymous
ok
anonymous
  • anonymous
Thanks!
anonymous
  • anonymous
I have a question though, Say I have an object that work is being done by it over 10M
anonymous
  • anonymous
I know the force being applied between each meter.
anonymous
  • anonymous
Why not add all the forces and multiply it by the total distance? @Jemurray3
anonymous
  • anonymous
That doesn't make any sense. If you have a changing quantity and you want to sum it up, you need to integrate it.
anonymous
  • anonymous
Okay, all I need is the distance and the initial force? Do I need to know all the forces involved during that change F1/F2/F3 etc... Or just F1?

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