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Homework Help: Energy Conservation and Kinetic, Wind Energies

  1. Oct 18, 2007 #1
    1. Jane, whose mass is 49.0 kg, needs to swing across a river filled with man-eating crocodiles in order to rescue Tarzan, whose mass is 80.0 kg. However, she must swing into a constant horizontal wind force F on a vine that is initially at an angle of with the vertical. (See Fig. P5.73.) In the figure, D = 50.0 m, F = 126 N, L = 40 m, and theta = 50.0°.

    Figure P5.73

    (a) With what minimum speed must Jane begin her swing in order to just make it to the other side? (Hint: First determine the potential energy that can be associated with the wind force. Because the wind force is constant, use an analogy with the constant gravitational force.)

    (b) Once the rescue is complete, Tarzan and Jane must swing back across the river. With what minimum speed must they begin their swing?

    2. Relevant equations
    W = KE + PE
    KE = 0.5mv^2
    PE = mgh
    W = Fcos(theta)*d
    Ei = Ef

    I first used:
    Ei = Ef
    PE(gravity) + KE(motion) = W(wind)
    mgh + 0.5mv^2 = F(wind)*cos(theta)*d
    49*9.81*(40-40*sin40) + 0.5*49*v^2 = 126*cos(50)*50
    v = (2*((126*cos(50)*50)-(49*9.81*(40-40*sin40)))/49)^0.5
    I got a number of different v:
    using different formula and substitutions but they are ALL incorrect by at least 10%!

    Line of reason:
    Jane needs to move against a force field of wind which exertes 126N, when multiplied by 50m the distance the work needed to go against the wind is 126*50 J multiplied by an angle of 50 degrees? Here is where it gets confusing, if she is moving in an arc wouldn't the energy of the wind change also? W = Fcos(theta)*d
    So she obviously needs a KE equal to W in order to overcome it, also in the diagram I am unsure if there is a height change, but if there is it is negative energy with respect to the y axis? Should I break up the question into an x and y part then say v = ((vx)^2 + (vy)^2)^0.5?
    height difference between starting point and ending point:
    h = L - Lsin40
  2. jcsd
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