Power Consumption Calculation Verification

In summary, the conversation is between a first-year college student seeking help with a calculus problem and other forum members providing guidance and corrections. The topic is about calculating the power and work done when moving a 1kg mass with an acceleration of sin(t). The student presents their solution and asks for verification, while other members point out errors and clarify concepts. The conversation also touches on the importance of correctly understanding and applying calculus concepts, and the use of computer algebra systems for calculations and visualizations.
  • #1
lxman
77
0
Hello all,

I am a first year college student who has just been through Calc I, and I am working to put some concepts together. I am also new to the forum. I am working out a thought experiment, and I have what I believe is a correct solution. I would appreciate it if someone could verify my calculations.

Given a 1kg mass with an acceleration of sin(t), we can then derive that its velocity would be -cos(t) and its position function would be -sin(t). So we have the following:

a(t) = sin(t)
v(t) = -cos(t)
s(t) = -sin(t)

Now, for my purposes, as I will be calculating power used to move this object around, I don't want a negative position, therefore, since the function s is an integral of the function v, I will add a constant of one. Thus I now have:

a(t) = sin(t)
v(t) = -cos(t)
s(t) = -sin(t) + 1

Kinetic energy is 1/2*m*v^2. Therefore I obtain:

J(t) = 1/2 * (-cos(t)) ^ 2

Work is energy expended over time. Therefore I integrate again, and obtain:

W(t) = 1/4 * t + 1/8 * sin(2 * t)

If I understand correctly, I have now obtained how many watts are being expended at time t. The electric company, for instance, does not bill by kW. They bill by kWh. So in order to get the total wattage over time I would need to integrate once again to arrive at:

Ws(t) = 1/8 * t ^ 2 - 1/16 * cos(2 * t)

This, then, should give me the total work done at time t to oscillate my 1kg mass back and forth. Am I deriving this correctly?
 
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  • #2
lxman,

Welcome to the PF!

It has been a LONG time since I took calculus, but I am pretty sure that d/dt sin(t) = cos(t). I am also pretty sure that d/dt cos(t) = -sin(t), so s(t) is right, but for the wrong reason ;-)

http://www.themathpage.com/acalc/sine.htm#sine

The rest of your math looks right to me, but like I said, its been a long time since I took calculus;-)

Fish
 
  • #3
Thank you Fish, for the welcome. I believe I must stand on my reasoning on the point you mention, however.

d/dt s(t) = v(t)
d/dt v(t) = a(t)

Therefore:

d/dt s(t) = -sin(t) = v(t) = -cos(t)
d/dt v(t) = -cos(t) = a(t) = sin(t)

To go the other way you integrate, so the indefinite integral of a with respect to t is v, and the indefinite integral of v with respect to t is s. It's easy to get it turned around.

That's part of the reason I have put this together, to help me get it straight in my head. :wink:
 
  • #4
welcome to pf!

hi lxman! another fishy welcome to pf! :biggrin:
lxman said:
Work is energy expended over time. Therefore I integrate again …

no, work is energy, there's no integration

(work done = ∫ force.distance = m∫ acceleration.distance = m∫ ads = m∫ dv/dt ds = m∫ vdv/ds ds = ∫ d(mv2/2) = ∫ d(energy) :wink:)
 
  • #5
Thank you tiny-tim.

It is at that point that things do indeed become a bit fuzzy for me. Hopefully the fine folks here at PF can help me iron this out. :smile:

To illustrate further:

a(t) = sin(t)

v(t) = -cos(t)

s(t) = -sin(t) + 1

These facts I am fairly confident about.

Proceeding on:

F = ma

I am making my m = 1kg for simplicity's sake. Therefore:

F = a

Therefore:

W = Fd = ad

Distance, as I understand it, should be position integrated with respect to time. Therefore:

d(t) = cos(t) + t

Therefore:

W(t) = (t + cos(t)) * sin(t)

I will stop at this point and ask two questions.

1. Is my reasoning correct to this point?
2. If it is, then what does W(t) represent? Is it the energy (in Joules) at time t, or is it the total energy expended up to time t?
 
  • #6
Work is calculated as Force times distance only when the force is constant. Otherwise, it must be computed as the integral of force along distance. This is a very important difference.
 
  • #7
lxman said:
I believe I must stand on my reasoning on the point you mention, however.
...

To go the other way you integrate, so the indefinite integral of a with respect to t is v, and the indefinite integral of v with respect to t is s. It's easy to get it turned around.
:wink:


Yep, you are 100% correct, I got it backwards. The sad thing is even after you pointed it out, it took me quite a bit to get it straight in my head. I prolly need to add calculus to the list of things I have forgotten.

My mother use to warn me about speaking up, she'd say, "It is better to remain silent and let people assume you are an idiot than open your mouth and remove all doubt."

Good Luck with your studies,

Fish
 
  • #8
Okay, I am continuing to play with these equations with my CAS. I'm using Sage to do plots, etc.

Here's a point where I am stuck. Given the function for position, how do I determine the total distance traveled? To clarify what I mean by 'total distance traveled':

Given s(t) = sin(t)

I thought that to get the total distance traveled you would integrate s(t). This would give me:

d(t) = -cos(t)

So that my distance will start at 0 along with my position, I will use:

d(t) = 1-cos(t)

Then for the total distance traveled at t = pi, I would arrive at 2. This part seems to make sense to me. Here is how I verify it in my mind. I picture myself as a tiny little person walking along the line of the graph. From time 0 to time pi, I have walked from 0, to 1, and back again to 0. Therefore the total distance that I have covered in that time period is 2. That seems well and good to me. As we proceed along past that, though, to 2 * pi, things don't add up to me. I realize at that point we have gone under the x-axis and the negative area under the curve cancels out the positive area under the curve, and our integral at that point equals zero. And from the frame of reference that we have, our position is once again 0.

However, falling back to my point of view of the little guy walking the graph - I have walked from 0 to 1, back to 0, on to -1, and then back to 0 again. So the total distance I have covered is actually 4, not 0.

So, what I am trying to arrive at is, given a position function of s(t) = sin(t), what would be the formula for d(t)?

I apologize if I am a bit long-winded here, but believe it or not, I am phrasing myself as succinctly as I feel I can.
 
  • #9
Fish4Fun, at least you have the courage to make a mistake. I find mistakes to be some of the most educational aspects of my life sometimes, once I survive them.
 
  • #10
Hi lxman :)

If the velocity of your object as a function of time is given by [tex]v(t)[/tex], then its displacement is given by:

[tex]
\int_{t_i}^{t_f} \! v(t) \, \mathrm{d}t
[/tex]

where [tex]t_i[/tex] and [tex]t_f[/tex] is the initial and final time respectively.

The work done by the resultant force on the object is then:
[tex]
W = \int_{t_i}^{t_f} \! F(t) v(t) \, \mathrm{d}t
[/tex]

where [tex]F(t)[/tex] is the resultant force on the object at time [tex]t[/tex].

Is this enough to clear things up?

Edit:
I've reviewed some of your arguments and I think I understand your trouble. When considering work done on a system, it's the displacement of the system under action of a resultant force that's important, not the distance traveled.

Using your example, at some time [tex]t_1[/tex] you will be doing work on the system, and at some other time [tex]t_2[/tex] the system will be doing work "on you".

The chapters on system descriptions and work in any decent mechanics text should make all this a lot more clear :)
 
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  • #11
Well, you could integrate to find the work done, and you might prefer doing that for pedagogical reasons. However, it would be simpler to use the work-energy theorem here:
W(t) = KE(t) - KEt=0
As a check on the final expression you come up with -- note that in this example the KE is initially at it's maximum value, and oscillates between that maximum and zero. Therefore negative work has been done at any time that KE is not at it's maximum. Or put another way, the expression for W(t) should oscillate between zero (initially) and a negative value.
 
  • #12
Yes, thank you milesyoung and Redbelly98. Both of your answers make sense, and I am going to spend some time thinking over how they apply.

However, off the top of my head, I believe that my issue is a point of view one (I am learning that this should properly be phrased 'frame of reference'). Based on this, allow me to restate my concept and what I am trying to derive.

Ultimately, I would like to derive a relationship between frequency and power/energy. I am sure there is a pre-derived formula somewhere which will give me a quick answer, but I want to think this through until I understand the reasoning behind it, and my gut instinct tells me that I have the tools to do it, I just need to learn how to apply them.

Thus I have the following scenario to examine the situation:

I have a 1kg mass, which I am moving by hand from -1 to 1. With my magic arm, I am able to move this mass just perfectly so that its motion over time describes a sine wave (My magic arm moves it in a perfect one dimension, of course, over a straight line, but plotted over time the acceleration and deceleration that I provide plot as a sine wave). Therefore, in 2 * pi seconds, I am able to complete one complete circuit and start over.

So the mass's kinetic energy would reduce to zero for an instant as I hit -1 and 1, finished decelerating (negatively accelerating, actually) and started accelerating again in the opposite direction. At the instant that I cross the point of origin, also, my displacement returns to zero, and my net displacement is also zero.

That part I understand. Here's what I am trying to figure out.

I once read that engineers are some of the laziest people in the world, because they always figure out the most efficient/easiest way to do anything. Not being a physical type of guy, all of this mass-moving to me means one thing, WORK (such an ugly four letter word :rolleyes:). So, if I move this thing through x cycles (with my magic arm, of course, which also, btw, magically eliminates the effects of air resistance and gravity), after I am finished, how much work have I done? EXACTLY :bugeye:

Then, having come to that result, move the same mass the same distance, from -1 to 1, back and forth, but at a higher frequency. Now compare how much work I have done with what I arrived at previously and come up with a formula for when the amplitude does not change, but the frequency changes, what is the total amount of work that I will have to do to run my magic arm for time x.

As I stated earlier, I am sure that someone has already worked this out and they can drop down an equation which will provide a solution. However, I wish to go through this process myself, as it teaches me how to think and how to analyze situations so that when I come into an unconventional situation that doesn't quite fit, I can figure out what is going on.

I hope I am not coming across as pedantic or argumentative. If I am, I apologize.
 
  • #13
I see what you're getting at, but the thing is, you're talking about work - which is the amount of energy transferred to a system by a force acting on the system as it undergoes some displacement.

Regardless of the frequency, the net work done on your object will always be 0 at the end of n number of cycles. This is because your object isn't displaced under action of a force.

It might not seem intuitive, but that's how it works :)
 
  • #14
OK, thank you for the answer. Now here comes a question (and it seems awfully silly to me, but I have to ask it anyway).

Based on this, in theory, I should then be able to design a system which oscillates my 1kg mass back and forth, and the only energy I should have to input to the system should be for friction losses, power conversion losses, etc. But the theoretical system, the mass oscillating back and forth, should continue to oscillate infinitely with no energy input?
 
  • #15
Yes. It's called a spring :)
 
  • #16
OK, adding a spring makes perfect sense. Then my spring would modulate the energy and continue to try to return the mass to point zero.

But in doing so, this modifies my original concept. Would you have suggestions on what tools to analyze my system in the framework that I am seeing it? In short, I guess that I am translating all of my motion into linear motion, so in the frame of reference of 'there is no spring and I have to keep pushing this thing to get it to move,' how would I derive an equation to determine how much energy I am expending over time?
 
  • #17
As you force your object to follow a sinusoidal trajectory, you are, in effect, a spring. You're performing the action of a simple harmonic oscillator, see here:

http://en.wikipedia.org/wiki/Harmonic_oscillator

Either way you cut it, you're doing zero net work.

Edit:
We're starting to move into differential equation territory which might needlessly complicate things. I think you're better off going back to basics, like the concept of work and the Work-Energy Theorem as Redbelly98 suggested.
 
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  • #18
I guess that I will have to hold what you say in the back of my mind and examine it over time. Perhaps it will make sense to me later. I can say at this point that it definitely does not, simply for the fact that if I push this mass back and forth, I am expending energy. I cannot continue the process forever, and I do not believe the only losses in the system are air resistance, conversion losses, etc. Unlike a spring, when the mass moves away from me and then back, calories are not put back into my muscles to revitalize me for the next push. I am working in a total energy output situation.

Ah, and there, I think, is my main conceptual difference. With the spring and mass harmonic oscillator, energy is conserved. As the mass stretches the spring, the mass loses energy, but the spring gains energy, until the point where the energy of the spring is greater than the energy of the mass, at which point the energy transfer begins to flow the other way. So the total energy of the system remains constant. Whereas, when I replace the spring (with me), as the mass and I are not sharing energy, the energy is not conserved. Or, to view it another way, imagine having a mass in space with rocket engines on opposite ends. One fires to move the object in one direction until midpoint in the cycle, at which point the first shuts off and the second begins firing to apply negative acceleration and reverse the process. Once again, there is no conservation of energy, as the engines do not regain fuel as they are being pushed in the opposite direction by the one doing the pushing. Therefore all of the motion requires an expenditure of energy.

This does present an interesting idea to me, though. Getting philosophical for a moment, there would be some things in life which would be totally non-productive. If all I am doing is oscillating something back and forth, then I am basically accomplishing nothing. And I can definitely see that. :smile:
 
  • #19
I understand your point of view, but let's be clear about the meaning of work in physics - a transfer of energy from one system to another.

In the case of your example with the mass following a sinusoidal trajectory:
No energy is being transferred to the mass system in the course of a number of cycles because you're doing no net work on the system. This can be verified using any number of methods, the Work-Energy Theorem being one of the easiest.

This should be clear in the sense that work has a well defined meaning in physics.

Your body system does fully recover the energy when it's doing negative work on the mass system, i.e. the energy is being transferred from the mass system to your body system - it just mostly gets converted to heat in the process. So yes - your body needs to convert more of its chemical energy to mechanical work to accelerate the mass system.

Calculating the conversion rate of chemical energy for your body is no simple task. When your body system is doing negative work on the mass system, most of the energy gets converted to heat, but some is converted to a range of different types of potential energy which is readily available for doing positive work on the mass system.

I hope this helps in some way. To go further, you probably need to make some assumptions with regards to energy conversion in your body system.

Cheers
 
  • #20
Yes, your explanation helps tremendously. Thank you.

So, let me adjust things a bit. You are right, calculating this from a biological perspective would greatly raise the complexity. So, let's go with my second example.

A mass with variable thrust rocket engines on opposite ends, in space (no gravity, no air resistance), is being pushed back and forth. The plot of the magnitude of the thrust over time of one engine forms the 'bottom' half of the sine wave, and the thrust of the other engine plots to the 'top' half of the sine wave.

Now let me try to reason through my definitions.

KE = 1/2 * m * v ^ 2 (Measured in Joules, kg * m ^ 2 / s ^ 2)
W = F * d (Measured in Joules also)

So that tells me (and seems to be what I am understanding from the kinetic energy-work principle) that KE = W.

Is this correct?
 
  • #21
Okay, I believe I have come up with a set of equations which explain what I have been trying to describe sufficiently. I would appreciate you folks perusing my work and letting me know if I have made any mistakes.

Here are the particulars:

m = 1 kg
a(t) = -cos(t)
v(t) = -sin(t)
s(t) = cos(t) - 1
KE(t) = 1/2 * sin(t) ^ 2
KE_tot(t) = 1/4 * t - 1/8 * sin(2 * t)
P(t) = 1/8 * (2 * t - sin(2 * t)) / t
P_tot(t) = 1/4 * t - 1/8 * integral(sin(2 * t)/t)

Of course, I obtained v(t) and s(t) through integration from a(t), as a(t) is non-constant. KE = 1/2 * m * v ^ 2, my mass is 1 kg (for simplicity's sake). therefore, KE(t) = 1/2 * v ^ 2 = 1/2 * sin(t) ^ 2. As KE is the square of velocity, the kinetic energy is never negative - a critical point here. W = KE. KE_tot is the integration of KE over time, as KE is non-constant also. P = W / t, so integrate W(t) and divide by time. And then P_tot is total power input to the system.

I am fairly certain all of these calculations are correct. In order to demonstrate further, I present two graphs (click on the thumbnails to see a larger image):

http://www.freeimagehosting.net/uploads/th.8d08cfa446.png

This first graph presents the system from t = 0 to t = 2 * pi.
The red trace is acceleration.
The green trace is velocity.
The blue trace is position.
The purple trace is instantaneous kinetic energy.
The black trace is the total kinetic energy over time.
The orange trace is the average power input to the system (watts).

Note - the black line is the total kinetic energy over time, which is steadily increasing. The orange line is the average power input to the system over time in watts. It asymptotes to about 1/4 watt.

My second graph presents the same information over the span of t = 0 to t = 10 * pi, just to show that the system is stable over time, with the proper power input.

http://www.freeimagehosting.net/uploads/th.4a15db929d.png

I therefore conclude that this system cannot continue to oscillate on its own. It requires a constant regular input of energy. The energy is not lost. It is being converted to motion (kinetic energy) primarily, but it must continually be added to or the mass would simply drift off in one direction or the other.

I have made a case, but I am not unwilling to listen to reason if someone can logically show me where my errors are. In fact, I would be indebted to such an individual if they would be willing to do so.

Thank you.
 
  • #22
I see a problem.
lxman said:
... KE_tot is the integration of KE over time, ...
I fail to see any physical significance in this KE_tot quantitiy. It does not have the units of energy, but rather energy*time, so it is not equivalent to any energy in any form.

The purple trace is instantaneous kinetic energy.
The black trace is the total kinetic energy over time.
What do you mean by "total kinetic energy over time"? There is only one mass in the system, so it's kinetic energy is the total kinetic energy.
 
  • #23
Thank you Redbelly98. I will do some further research on KE then. The bottom line, though, it is an instantaneous value and totaling it over time is meaningless. Am I paraphrasing you correctly?
 
  • #24
lxman said:
Thank you Redbelly98. I will do some further research on KE then. The bottom line, though, it is an instantaneous value and totaling it over time is meaningless. Am I paraphrasing you correctly?
Yes, correct.
 
  • #25
Gordianus said:
Work is calculated as Force times distance only when the force is constant. Otherwise, it must be computed as the integral of force along distance. This is a very important difference.

Would someone be willing to explain to me how this would be accomplished? I understand that to get work, you integrate force with respect to distance, and that means W(t)=integral(F(t), dx). I don't know how to go about doing that particular integral mathematically, though. Would that mean that you would take the integral of F with respect to the distance you have traveled to that point?
 
  • #26
milesyoung said:
Hi lxman :)

If the velocity of your object as a function of time is given by [tex]v(t)[/tex], then its displacement is given by:

[tex]
\int_{t_i}^{t_f} \! v(t) \, \mathrm{d}t
[/tex]

where [tex]t_i[/tex] and [tex]t_f[/tex] is the initial and final time respectively.

The work done by the resultant force on the object is then:
[tex]
W = \int_{t_i}^{t_f} \! F(t) v(t) \, \mathrm{d}t
[/tex]

where [tex]F(t)[/tex] is the resultant force on the object at time [tex]t[/tex].

Is this enough to clear things up?

Edit:
I've reviewed some of your arguments and I think I understand your trouble. When considering work done on a system, it's the displacement of the system under action of a resultant force that's important, not the distance traveled.

Using your example, at some time [tex]t_1[/tex] you will be doing work on the system, and at some other time [tex]t_2[/tex] the system will be doing work "on you".

The chapters on system descriptions and work in any decent mechanics text should make all this a lot more clear :)

Well, that's a hoot! In googling around for the answer to my previously posted question, google eventually led me back to this post from earlier in the thread, which I read, but obviously didn't understand. Now it makes sense. Sorry milesyoung, for being so thick-headed. OK, picking up from here (I'm not quite done in yet):

As my mass is 1kg, in this case, F(t) = a(t), so F(t) = -cos(t). v(t) = -sin(t). Therefore W(t) = integral(-cos(t) * -sin(t), t). So W(t) = -1/2 * cos(t) ^ 2. A graph of this function:

http://www.freeimagehosting.net/uploads/th.5969474e7a.png

Now as I understand from my reading work is a scalar, not a vector. It can be positive or negative, but this is all negative, so the sum of the work is not zero over time.

To take this one step further, power is W/t. So my power function would be -1/2 * cos(t) ^ 2 / t. A graph of this:

http://www.freeimagehosting.net/uploads/th.efcc91be1d.png

Can power be integrated over time? Either way, the graph shows that the oscillations approach zero over time, but the power never quite reaches zero until infinity. So, and I am not being smart here, I am genuinely asking, how am I to interpret these results in the light of accomplishing zero work over time? Am I looking in the completely wrong direction, perhaps? Newton's 3rd law does state that for every action there is an equal and opposite reaction, and I am only considering the force on the mass. Since the force the mass would be exerting would be equal, would they cancel each other out? But if that applied to work, then it would be impossible to ever get any work done, as any force we applied would be met with an equal force back, and we could never move anything.

Still scratchin' my head. Thanks for indulging me.
 
  • #27
You're getting there! :smile:
lxman said:
As my mass is 1kg, in this case, F(t) = a(t), so F(t) = -cos(t). v(t) = -sin(t). Therefore W(t) = integral(-cos(t) * -sin(t), t). So W(t) = -1/2 * cos(t) ^ 2.
Very close, except you're forgetting the integration constant:

W(t) = -(1/2) * cos2(t) + C

To find that constant, note that the integration limits are from 0 to some later time. So at t=0, the work done must be zero. You can figure out what C must be to make that true.

To take this one step further, power is W/t.
That would be true if force and velocity were constant, but doesn't work here.

It's best (and correct) to think of power is the rate of doing work, so it's dW/dt.
 
  • #28
Let's say you're accelerating some point mass as in your example. The instantaneous power being delivered to your system [itex]P(t)[/itex] is then given by:

[tex]
P(t) = F(t) v(t) = \cos(t) \sin(t)
[/tex]

This function is periodic and the area under its curve is the amount of energy that has been transferred in some time frame, i.e. the amount of work you have done on the system in that time frame. It should be clear that the average of [itex]P(t)[/itex] over a period is 0, so you're doing no net work on the system.

This is all there is to

[tex]
W = \int_{t_i}^{t_f} \! P(t) \, \mathrm{d}t = \int_{t_i}^{t_f} \! F(t) v(t) \, \mathrm{d}t
[/tex]

You're calculating the area under the [itex]P(t)[/itex] curve.

If you include the integration constant as Redbelly98 showed you, you're evaluating:

[tex]
W = \int_0^{t_f} \! P(t) \, \mathrm{d}t
[/tex]

You should find that this function is 0 at the period intervals.
 
  • #29
Thank you gentlemen for being patient with me.

So I understand that P(t) (the instantaneous power at time t) should indeed be sin(t) * cos(t).

That makes sense. Now if I integrate power with respect to time, I arrive at W(t) = -1/2 * cos(t) ^ 2. That gives me an initial value of W(0) = -.5, so I add an integration constant of .5 and arrive at W(t) = (-1/2 * cos(t) ^ 2) + 1/2, which gives me W(0) = 0. Now I have:

http://www.freeimagehosting.net/uploads/th.760343bbdf.png

How do I interpret this?
 
  • #30
That's correct:

[tex]
W(t_f) = \int_0^{t_f} \! P(t) \, \mathrm{d}t = \int_0^{t_f} \! \cos(t) \sin(t) \, \mathrm{d}t = \left[-\frac{1}{2}\cos(t)^2^]\right]_0^{t_f} = \frac{1}{2} - \frac{1}{2}\cos(t_f)^2
[/tex]

[itex]W(t_f)[/itex] is the amount of work you have done on the point mass system at time [itex]t_f[/itex].

As you can see from it's graph, it's 0 at the period intervals. When you have moved the point mass from it's initial position, through the sinusoidal trajectory and back to it's initial position, you have done no net work on the point mass system.
 
  • #31
lxman said:
http://www.freeimagehosting.net/uploads/760343bbdf.png

How do I interpret this?
The net work done on the mass is initially zero, as it must be. Moreover, it is zero at t=π, 2π, etc., as well. That makes sense, since the particle has the same speed (and therefore kinetic energy) at those times that it had initially.

At other times the work done is positive. That makes sense, because at those other times the speed (and therefore kinetic energy) of the mass is greater than it was initially, and there must be a net positive work done in order for that to be true. Doing positive work increases the kinetic energy, and negative work decreases the kinetic energy.
 
  • #32
Then would it be reasonable to say that I have done work, but no net work?
 
  • #33
Redbelly98 said:
The net work done on the mass is initially zero, as it must be. Moreover, it is zero at t=π, 2π, etc., as well. That makes sense, since the particle has the same speed (and therefore kinetic energy) at those times that it had initially.

At other times the work done is positive. That makes sense, because at those other times the speed (and therefore kinetic energy) of the mass is greater than it was initially, and there must be a net positive work done in order for that to be true. Doing positive work increases the kinetic energy, and negative work decreases the kinetic energy.

Hmm, this confuses me a bit. You say doing negative work decreases the kinetic energy. I would agree with this, as a consequence of Newton's first law of motion. But the plot I am presenting here is (theoretically) a plot of work, not kinetic energy. Although in my hypothetical system, W and KE are coming out equivalently to this point. So where is this negative work coming from? All of the work I am showing here is positive. :confused:
 
  • #34
On average, you're doing no work.

Let me just go back to your example with your hand accelerating the point mass.

Graph [itex]P(t)[/itex]. When it's positive, you're transferring energy from your body system to the point mass system. When it's negative, you're transferring energy from your point mass system to your body system. Notice how this is in equal amounts (remember we're talking areas under the [itex]P(t)[/itex] curve - area = energy).

Now, it's clear that you're not, on average, doing any work on the point mass system, so why do you start to feel tired after a while?

This is because you're not, as you stated yourself, recovering all the energy in your muscles, but your body system is. It's recovering all of the energy - mostly as internal energy.

You start to get tired because you're running out of chemical energy. It's getting converted to kinetic energy in your point mass system as you do positive work on it, and this kinetic energy gets converted to internal energy in your body system as you do negative work on the point mass system. The amount of chemical energy in your body system is thus decreasing.
 
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  • #35
lxman said:
So where is this negative work coming from? All of the work I am showing here is positive. :confused:

No no.. your W(t) function is the net work done. You're still doing negative work at some point in time, otherwise your net work would never decrease.
 
<h2>1. How is power consumption calculated?</h2><p>Power consumption is calculated by multiplying the voltage (V) by the current (I) in a circuit. This can be represented by the equation P = VI, where P is power in watts (W), V is voltage in volts (V), and I is current in amperes (A).</p><h2>2. What factors affect power consumption?</h2><p>Several factors can affect power consumption, including the number and type of devices connected to a circuit, the length and thickness of wires, and the efficiency of the power supply. Temperature and environmental conditions can also impact power consumption.</p><h2>3. How can power consumption be verified?</h2><p>Power consumption can be verified through measurements using a multimeter or power meter. This involves connecting the device to be measured in series with the meter and recording the voltage and current readings. These values can then be used to calculate power consumption using the equation P = VI.</p><h2>4. What is the importance of verifying power consumption?</h2><p>Verifying power consumption is important for several reasons. It can help identify any potential issues with a circuit, ensure that devices are operating within their specified power limits, and help with energy efficiency and cost-saving measures.</p><h2>5. Are there any limitations to power consumption calculations?</h2><p>Yes, there are limitations to power consumption calculations. These calculations assume ideal conditions and may not account for factors such as power fluctuations, power supply inefficiencies, or variations in device usage. It is important to regularly verify and monitor power consumption to ensure accurate measurements.</p>

1. How is power consumption calculated?

Power consumption is calculated by multiplying the voltage (V) by the current (I) in a circuit. This can be represented by the equation P = VI, where P is power in watts (W), V is voltage in volts (V), and I is current in amperes (A).

2. What factors affect power consumption?

Several factors can affect power consumption, including the number and type of devices connected to a circuit, the length and thickness of wires, and the efficiency of the power supply. Temperature and environmental conditions can also impact power consumption.

3. How can power consumption be verified?

Power consumption can be verified through measurements using a multimeter or power meter. This involves connecting the device to be measured in series with the meter and recording the voltage and current readings. These values can then be used to calculate power consumption using the equation P = VI.

4. What is the importance of verifying power consumption?

Verifying power consumption is important for several reasons. It can help identify any potential issues with a circuit, ensure that devices are operating within their specified power limits, and help with energy efficiency and cost-saving measures.

5. Are there any limitations to power consumption calculations?

Yes, there are limitations to power consumption calculations. These calculations assume ideal conditions and may not account for factors such as power fluctuations, power supply inefficiencies, or variations in device usage. It is important to regularly verify and monitor power consumption to ensure accurate measurements.

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