Question

Total power?

Answer 1

I'm trying to calculate the total power(W) that's used on an object.

Answer 2

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?

Answer 3

Total Power = P1 + P2?

Answer 4

@experimentX
@Jemurray3
@TheEric

Answer 5

Basically, how can I find the power if the force + acceleration on not constant over a distance of 2M?

Answer 6

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.

Answer 7

How can I?

Answer 8

If work = 4000J how much power is produced? If the accretion is changing and I think time is also changing...

Answer 9

What I'm struggling with is that time is unknown and changing, acceleration is known and changing...

Answer 10

+ Force is changing as well.

Answer 11

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?

Answer 12

Or if you could simply state the problem, that would be fine.

Answer 13

The force is increasing with time.
Problem:
Work is done on an object over 2M
The force is increasing from 1000N to 3000N

Answer 14

How much time is taken, and how much power?

Answer 15

It starts from rest?

Answer 16

Yes Vi = 0.

Answer 17

iV*

Answer 18

You can find the work easily, but you cannot find the time or power unless you have the mass.

Answer 19

Say the mass was 100Kg

Answer 20

I just want to learn how this is done.

Answer 21

Okay. The force is
\[ F = 1000x + 1000\]
so the work is
\[W = \int F \cdot dx = \int_0^2 (1000x + 1000 )dx=2000 + 2000 = 4000\text{ J} \]

Answer 22

Does that make sense?

Answer 23

It does.

Answer 24

Now how could we figure out the power?

Answer 25

In most cases I would find the total work first.

Answer 26

But what's left know is to figure out the Time & Power

Answer 27

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
\[50 v^2 = 500x^2 + 1000x \]
\[v^2 = 10x^2 + 20x\]
\[ v = \sqrt{10x^2 + 20x} \]

Answer 28

The power being supplied at any particular instant is F*v so the power is
\[P = F\cdot v = (1000x+1000)\sqrt{10x^2+20x} \]

Answer 29

Im going to practice this thank you!

Answer 30

But wait, since the velocity is changing, what to do?
See what's confusing is is the "changing" part.

Answer 31

Lastly the time it would take would be
\[ t = \int dt = \int \frac{dx}{dx/dt} = \int \frac{dx}{v} \]
so
\[t =\int_0^2 \frac{dx}{\sqrt{10x^2 + 20x}}=\frac{1}{\sqrt{10}} \int_0^2 \frac{dx}{\sqrt{x^2+2x}} \]

Answer 32

What is confusing?

Answer 33

Im struggling with calculus, and I use simple algebra to solve this...

Answer 34

This cannot be solved with simple algebra, I'm afraid.

Answer 35

Since things are changing with time.
It makes things a lot complicated.

Answer 36

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.

Answer 37

Meaning that the average power is
\[ \bar{P} = \frac{4000 J}{0.557 s} = 7181 \text{ W}\]

Answer 38

@Jemurray3
What subjects do I have to learn perfectly to understand what you said perfectly?
Integration and Derivatives?

Answer 39

Yes. Integration is an extensive subject though -- you'd be fine with simple antiderivatives and trigonometric substitution.

Answer 40

ok

Answer 41

Thanks!

Answer 42

I have a question though,
Say I have an object that work is being done by it over 10M

Answer 43

I know the force being applied between each meter.

Answer 44

Why not add all the forces and multiply it by the total distance? @Jemurray3

Answer 45

That doesn't make any sense. If you have a changing quantity and you want to sum it up, you need to integrate it.

Answer 46

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?