Max speed given power and resistive forces

[Jezza]

Homework Statement

A body of mass M, traveling in a straight horizontal line, is supplied with constant power P and is subjected to a resistance Mkv^2, where v is its speed and k is a constant. Show that the speed of the body cannot exceed a certain value v_m and find an expression for v_m.

Homework Equations

P=\frac{dW}{dt}
F=\frac{dv}{dt}

Where W is work, F is force, t is time.

The Attempt at a Solution

First, find the force applied to the body by the power:

P = \frac{dW}{dt} = \frac{d}{dt}\int{Fdx} = \int{F \frac{dx}{dt}} = \int{Fdv}

We conclude by the fundamental theorem of calculus:

F = \frac{dP}{dv}

The resultant force on the body is therefore F_{res} = \frac{dP}{dv} - Mkv^2 = 0 at v_m.
Now we can write \frac{dP}{dv} = Mkv^2 which gives:

P = \int_{v_0}^{v_m}{Mkv^2dv} = \frac{1}{3}Mk (v_m^3 - v_0^3)

If it's ok to assume that v_0 = 0 (which I don't think it can be - I don't think it should depend on any initial velocity):

v_m^3 = \frac{3P}{Mk}.

The answer is given as v_m^3 = \frac{P}{Mk}.

What am I doing wrong? Also why am I having to set an initial velocity? Thanks for any help :)

 

[Jezza]

OK I've given it some more thought. Obviously because the power is constant and at v_m velocity is also constant, I can write:
W=Fd, which by dividing both sides by time gives P=Fv, giving F=\frac{P}{v}.

So now I can just write:

\frac{P}{v_m}=Mkv_m^2

Which clearly gives the desired answer. That doesn't really explain why my calculus doesn't work though...

 

[haruspex]

First, find the force applied to the body by the power:
$\frac{d}{dt}\int{Fdx} = \int{F \frac{dx}{dt}} = \int{Fdv}$

You can't really do that. The middle term makes no sense - you need one more d in the numerator than in the denominator inside an integral.
In fact, at any instant, P=Fv. No need for an integral.

The resultant force on the body is therefore $\frac{dP}{dv} - Mkv^2 = 0$

This is clearly not right since P is constant. You would deduce that v=0.