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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