1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    [itex]P=\frac{dW}{dt}[/itex]
    [itex]F=\frac{dv}{dt}[/itex]

    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:

    [itex]\frac{P}{v_m}=Mkv_m^2[/itex]

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

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Max speed given power and resistive forces
Loading...