ws1_work_flux_soln

ws1_work_flux_soln - Math 1920 Workshop 1 Work and Flux...

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Math 1920 Workshop 1: Work and Flux Solutions Problem 1: Work a) We know that ~u · ~v = k ~u kk ~v k cos θ . Hence if the angle θ is acute the cosine will be positive and so the dot product would be positive too, because magnitudes are always positive. In reverse if the angle is obtuse then the cosine and thus the dot product would be negative. So we should look at the angles for each segment. From A to D, the path vector points straight down, while the force points upward and to the right; the angle between them is obtuse, so the dot product (work done) is negative. From D to C, the path and the force are perpendicular, so their dot product, and therefore the work done, is zero. Thus the work done along this path from A to C is negative. b) ~ F is not conservative. Consider if we traveled from A to C via point B instead– the work done along this path is positive (why?), so the work cannot be path–independent. Problem 2: Flux through a Pipe a) Assuming that the fluid velocity is the same over the entire cross-section, the volumetric flow rate is then 5 m/s × 3 m 2 = 15 m 3 /s. Why? Consider a column of fluid of area A = 3 m 2 and width v = 5 m, as shown in the part (a) of figure below. This is the amount of liquid that will flow through A in 1 second. The volume of this liquid is vA = 15 m 3 , and the flow rate is vA m 3 /s. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
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