Power Consumption Calculation Verification

[Redbelly98]

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.

 

[lxman]

Then would it be reasonable to say that I have done work, but no net work?

 

[lxman]

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.

 

[milesyoung]

On average, you're doing no work.

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

Graph P(t). 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 P(t) 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.

 

[milesyoung]

So where is this negative work coming from? All of the work I am showing here is positive.

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.

 

[lxman]

I believe you gentlemen are corrupting me. This is starting to make sense.

I am reworking my equations (again) and I will post a graph here momentarily that I believe wraps everything into a nice package.

 

[lxman]

OK,

From scratch, here are my equations. I will not go into the derivation details, if anyone sees a difficulty with any of my equations, I will address that in a subsequent post.

m = 1kg
a(t) = -cos(t) red
v(t) = -sin(t) green
s(t) = cos(t) - 1 blue
KE(t) = 1/2 * m * v(t) ^ 2 brown
W_tot(t) = 1/2 - 1/2 * cos(t) ^ 2 black
P(t) = sin(t) * cos(t) purple

http://www.freeimagehosting.net/uploads/th.290bc3ee4a.png

What I see happening here is the following:

v, s, KE, W_tot and P all begin at zero, as they should. a begins at -1, which I believe is stretching things a bit, but to make everything else work, I left it that way. Now, KE represents the instantaneous energy of the object, and W_tot represents the net work done over time, which in my system KE and W_tot happen to be equivalent. W_tot oscillates regularly between 0 and a positive value. What illustrates the negative force here (and the positive one as well, is P. P is the derivative of W_tot with respect to time, so P is the instantaneous power at time t. Therefore P illustrates where the negative energy is coming from.

How does this sound?

 

[milesyoung]

Looks good but:

KE represents the instantaneous energy of the object

I don't know what you mean by this. KE is simply the kinetic energy of the point mass system.

W_tot represents the net work done over time

W_tot is the net work done on the point mass system at some instant in time.

which in my system KE and W_tot happen to be equivalent

They're equivalent in general. That's the Work-Energy Thorem.

Edit:
Note that we're talking about a change in kinetic energy for this to hold.

In this case, the change in kinetic energy is equal to KE since the kinetic energy of the point mass is zero initially.

 

[lxman]

Points taken, milesyoung. Thank you.

So, what this shows is that in order for my point mass to continue its oscillations I have to affect it with alternating positive and negative power. The sum of this power over time will be zero, of course. But nevertheless it does have to be there.

Therefore, back to my earlier postulation (a 1kg mass in deep space with a variable thrust rocket engine on each end), in order to calculate the total fuel that I would require to operate the system for time t, I would have to integrate the absolute value of P over time t. That mass of fuel would then be substituting for the spring.

Is this a reasonable conclusion?

 

[milesyoung]

Note that doing positive work on the point mass system means transferring energy to it, and doing negative work means transferring energy from it.

If you make the assumption that you're unable to recover the energy you receive from the point mass system as you do negative work on it, in the sense that you cannot use this energy for doing positive work on the point mass system, then the net positive work done on the point mass system should be what you're after.

This is just off the top of my head and I'll leave it to you to prove me right or wrong. I really need some sleep :)

Hope you find your answers. Cheers.

Edit:
If you decide to look up something relating to system descriptions and work, beware that some might define work the other way around, i.e. positive work done on a system means a transfer of energy out of the system. This should be clear from their definition of work though.

 

[lxman]

Yeah, that works.

Thanks much for taking the time.