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Homework Help: Max speed given power and resistive forces

  1. Nov 2, 2015 #1
    1. The problem statement, all variables and given/known data
    A body of mass [itex]M[/itex], travelling in a straight horizontal line, is supplied with constant power [itex]P[/itex] and is subjected to a resistance [itex]Mkv^2[/itex], where [itex]v[/itex] is its speed and [itex]k[/itex] is a constant. Show that the speed of the body cannot exceed a certain value [itex]v_m[/itex] and find an expression for [itex]v_m[/itex].

    2. Relevant equations


    Where [itex]W[/itex] is work, [itex]F[/itex] is force, [itex]t[/itex] is time.

    3. The attempt at a solution

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

    [itex]P = \frac{dW}{dt} = \frac{d}{dt}\int{Fdx} = \int{F \frac{dx}{dt}} = \int{Fdv} [/itex]

    We conclude by the fundamental theorem of calculus:

    [itex]F = \frac{dP}{dv}[/itex]

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

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

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

    [itex]v_m^3 = \frac{3P}{Mk}[/itex].

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

    What am I doing wrong? Also why am I having to set an initial velocity? Thanks for any help :)
  2. jcsd
  3. Nov 2, 2015 #2
    OK I've given it some more thought. Obviously because the power is constant and at [itex]v_m[/itex] velocity is also constant, I can write:
    [itex]W=Fd[/itex], which by dividing both sides by time gives [itex]P=Fv[/itex], giving [itex]F=\frac{P}{v}[/itex].

    So now I can just write:


    Which clearly gives the desired answer. That doesn't really explain why my calculus doesn't work though...
  4. Nov 2, 2015 #3


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