Constant performance vs friction and drag

[bXD01Q99]

Hi, I wanted to make an equation with Newtonian physics, but failed. Here's the problem:

                                    Consider an object with that
              ^                     has [B]m[/B] mass, and a built in
    drag &    |                     magic motor that has [B]P[/B]
   friction   |   magic force       constant performance. The
 <----------[OBJ]------------->     motor is magically propelling
==============|=================    the object in a direction,
              | gravity             parallel with the ground,
              v                     which has [B]u[/B] friction.
                                    Gravity is [B]g[/b].
What is the velocity and the traveled distance after [B]t[/B] time?

I've done this so far:

Symptoms: the object starts moving with high acceleration, velocity gain speed gets slower over time, and the velocity converges to (and reaches?) its maximum speed. Then, all the additional performace goes, through friction, into heat energy.

The engine gives [tex]P \times \Delta t[/tex] joules of energy every [tex]\Delta t[/tex] time, and friction takes away [tex]F_f \times \Delta s[/tex]. Energy at a given time can be written as:

[tex]E = P t - F_f s[/tex]
[tex]\frac{m v^2}{2} = P t - \mu m g s[/tex]
[tex]\frac{m v^2}{2} = P t - \mu m g v_{avg} t[/tex] ?

[tex]s[/tex] is dependant on [tex]t[/tex] and [tex]v[/tex], but [tex]v[/tex] depends on the [tex]E[/tex], which depends on [tex]s[/tex], so I don't know how to continue. I need to get [tex]s[/tex] and its derivative, [tex]v[/tex], for any given [tex]t[/tex] time.

So I've tried a temporaty iterational method:
Start with [tex]E = 0[/tex], [tex]v = 0[/tex], then, iterate with very small [tex]\Delta t[/tex] time intervals, and do this:
1. to [tex]E[/tex], add [tex]( P - \mu m g v ) \times \Delta t[/tex]
2. set [tex]v[/tex] to [tex]\sqrt{ 2 E / m }[/tex]

With this algorithm, I've drawn this velocity-time graph:
img182.imageshack.us/my.php?image=47155991qs7.png or [ATTACHMENT]
Iteration count: Blue=5, Green=25, Red=625.
It looks like some sort of logarithmic or root curve.

Well, that's it. If anyone can give me equation instead of an iterating algorithm, and/or also consider slope, drag, and 2D plane vector directions, I'll really appreciate it. Thanks in advance.

 

[Matthaeus_]

Please use LaTeX.

Friction force dissipates a power:
[tex]\displaystyle P_f = \frac{F\cdot \mathrm{d}s}{\mathrm{d}t} = \mu m g v[/tex]

Drag force:
[tex]\displaystyle P_d = c_1v^2 + c_2 v^3[/tex]
For some drag coefficients [tex]c_1[/tex] and [tex]c_2[/tex]

Therefore the limit speed is the real solution of:

[tex]\displaystyle c_2 v^3 + c_1v^2 + \mu m g v - P = 0[/tex]

Assuming that the speed is high enough you can approximate this by:

[tex]\displaystyle v \approx \sqrt[3]{\frac{P}{c_2}}[/tex]

 

[bXD01Q99]

Thanks.

I've gotten acquanted with this LaTeX thingy...

You've looked into the case when the object reaches it's maximum speed: [tex]\displaystyle P_f = \frac{F\cdot \mathrm{d}s}{\mathrm{d}t} = \mu m g v[/tex]. I can get the maximum speed of the object. But I knew that. This formula can be used in a special case of [tex]\frac{m v^2}{2} = P t - \mu m g v_{avg} t[/tex], when v is a constant, so that [tex]v_{avg} = v[/tex]. This is not usable for the curved part of the graph. However, I am interested in the curve.

You've also considered drag, uhm, drag is approximately proportional to [tex]v^2[/tex] or even [tex]v^3[/tex], depending on [tex]v[/tex]?
[tex]\displaystyle v \approx \sqrt[3]{\frac{P}{c_2}}[/tex]
This is a simplification, which I'm not really in to, because there are no guarantees (I'd rather chose the 3rd degree solution formula). Generalization is needed.

OK, to ease things, let's not consider drag and slopes yet, just friction. There is a flat ground, and the object can move in any direction parallel to the ground (forward, left, back-right, ...). It can change its direction of the magic force any time it wants, say, every 10-20 milliseconds. I need an equation that can determine the new location and the speed of the object after it has traveled in that given direction for a given time.

 

[Matthaeus_]

Then you'll have to solve the differential equation:

[tex]\displaystyle F_{motor} - \mu m g = m \ddot{x}\qquad \text{(Newton's II Law)}[/tex]

[tex]\displaystyle \ddot{x} + \mu g - a_0 = 0[/tex]

Where [tex]a_0[/tex] is [tex]F/m[/tex].

In this case, it is clear that the acceleration [tex]\ddot{x}[/tex] of the body will be zero if and only if [tex]\mu g = a_0[/tex].The drag force of a fluid comes in its most general form as:

[tex]\displaystyle \vec{F_{drag}} = -(c_1 v + c_2 v^2)\hat{v} [/tex]