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Math Help - Work done moving against a force

  1. #1
    Mar 2007

    Work done moving against a force

    Currently stuck on the following question:

    A cyclist traveling on a straight road at 20km/h is subject to wind resistance proportional to the square of the wind speed. When a wind of 40km/h develops blowing in the direction perpendicular to the road, the cyclist must increase power output to maintain constant speed. By what factor must the cyclist increase power to maintain constant speed in the cross wind?

    The answer is root 5, but I don't know how to attempt the question. I did manage to work out that the resultant velocity of the cyclist is 44.7km/h at an angle of 63 degrees to the direction of travel, but I don't know what to do after that.

    Help would be hugely appreciated.
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  2. #2
    Oct 2008
    First, lets find power output of the cyclist when there is no wind:
    <br />
P_0=Const.Vc^2.Vc<br />
    With the wind blowing we find that the speed of wind which the cyclist feels has value V equal to
    <br />
V=Sqrt[Vc^2+Vw^2]<br />
    and makes an angle a with the vector of cyclist'sdisplacement:  cos(a)=Vc/V
    Power output with the wind present is thus
    <br />
P_w=Const.V^2.cos(a).Vc<br />
    Let me explain a little: the force of the wind is proportional to V^2 but work done is  W=\vec F.\vec r =F.r.cos(a). where \vec r is the vector of displacement. Power is therefore: P_w=dW/dt=F.(dr/dt).cos(a) = F.Vc.cos(a)
    You are sked to compute P_w/P_0. If You do the calculations you'll get required answer Sqrt[5].
    Good luck
    Attached Thumbnails Attached Thumbnails Work done moving against a force-cyclist.bmp  
    Last edited by katastrofa_nadfioletu; October 27th 2008 at 12:50 AM. Reason: latex error :(
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