Power = Force*Speed = Force*Distance/Time = Work/Time It doesn't matter whether the force is producing acceleration or just overcoming friction or gravity. Brent On 4/20/2020 3:46 PM, Brad Klee wrote:
Arg, wrong again! Yes that is the work integral, so needs another 1/delta t. Thanks for the correction.
However, the mistaken formula doesn't change my argument, which is that P = m*(delta x)^2 / (delta t)^3 is not the correct formula because actually, P is proportional to a Force F, which usually isn't proportional to (delta x) / (delta t)^2.
The math works out nice in this block problem, but we don't really need another block problem. It isn't physically realistic anyways. What is a "frictionless surface"? And what about drag? Is the block in a vacuum? Why does the block matter? Etc.
In the problem about runners, the resistive force combines a percentage of body weight and drag force. Drag is significant especially for sprinters like Usain Bolt, who are running very fast. The drag proportionality is v^2.
We need runners b/c physical fitness helps to improve bodily health, and to improve chances of surviving viral sickness!
An even better runner problem is to look closely at what happens during the acceleration phase of a 100m dash. Not only are forces imbalanced, the resistive force increases as a function of time and /or velocity. Yet Usain Bolt can still exceed 2 kilowatts of human body power.
https://www.popsci.com/science/article/2013-07/physics-record-breaking-run/
Amazingly, some of the beetles reflect mostly green, left-circular polarized light. If that is not enough they are stronger on average than the strongest humans, most likely more powerful, and easy to find in Arkansas (if you are willing to break open a dung pat):
https://www.livescience.com/8145-super-bug-world-strongest-insect-revealed.h... https://www.inaturalist.org/observations/41877497
I don't know if anyone has directly measured the beetle power (I haven't read about it), but it should be done! Too bad about the march purge. The labs are shut down right now anyways...
Cheers,
Brad
--Brad
On Mon, Apr 20, 2020 at 1:21 PM William R Somsky <wrsomsky@gmail.com> wrote:
Way too unspecified.
Brad, you seem to be considering "steady-state" motion at a constant velocity, where the power counteracts some resistive force. And your integral is wrong. P = F v. The time integral of this is E, energy. Check the units.
Bill, you would seem to I be considering accelerating an object, where the power is used to the increase the kinetic energy if the object. It's all going to be dependent on the initial velocity before the power/force is applied.
Let me take a stab at an example elementary physics problem:
You have a block of mass m on a frictionless surface, initially at rest at x = 0. Starting at t = 0, you are going to apply a force to the block in the positive x direction in such a way that the rate of change of the kinetic energy of the block is a constant, ie, dK/dt = P, where P is the power "being applied"/required/however you want to term it in words.
Now: a) What is the kinetic energy of the block as a function of time? b) What is the velocity of the block as a function of time? c) What is the position of the block as a function of time? d) How does the position of the block at a time 2t compare with the position at time t?
a) The block starts at rest so K(t=0)=0, and K is increasing at a constant rate dK/dt=P, hence: K = P t ((for t>=0))
b) The kinetic energy of the moving block is K=(1/2) m v^2, so v = Sqrt[2K/m] = Sqrt[2 P t / m]
c) The block starts at x=0 so x(t=0)=0, and v=dx/dt, hence x = Int v dt = Int Sqrt[2 P t / m] dt = ( Sqrt[2 P / m] (2/3) t^(3/2) )
d) x(2t)/x(t) = ... = 2^(3/2), so the block goes Sqrt[8] times as far.
On Mon, Apr 20, 2020, 10:28 Brad Klee <bradklee@gmail.com> wrote:
No, forces are applied, power is an intrinsic property, so you might rather say "Given two movers with equal average power, one that goes twice as long also goes eight times as far"; however, this is also incorrect. The integral equation P = Int F*v*dt, says that power depends on resistive force. This is somewhat confusing when thinking of, say, a sprinter or a marathon runner because it isn't clear what the resistive force is. It is more complicated than you might guess:
http://www.georgeron.com/2017/09/kipchoge-running-power-marathon.html
If you want a high school problem to work out, here is one I wrote earlier this month:
--Brad
On Mon, Apr 20, 2020 at 12:06 PM Bill Gosper <billgosper@gmail.com> wrote:
In[395]:= UnitConvert[Quantity[42, "Watts"], "Kilograms" "Meters"^2/"Seconds"^3]
Out[395]= Quantity[42, ("Kilograms" ("Meters")^2)/("Seconds")^3]
This seems to say: If you apply a given amount of power (to a freely moving object) for twice as long, you go eight times as far. —rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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