find an arguement why the work around any closed rectangular path vanishes for this force (which can be used to argue that the force is conservative)
Added by Darren R.
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This means that the work done by the force depends only on the initial and final positions, not on the path taken. Show more…
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Consider a new force which is defined as a function of two-dimensional coordinates this way: F ≡ Ay i + Bx^4 j Here x and y are standard Cartesian coordinates and A, and B are given real positive constants with appropriate units. Consider the loop that goes around a square path as show in the figure. Each side of the square has a given length C. Part a) What is the Work done by this specific force F for the closed path that goes around the loop in the counter-clockwise direction, starting from the origin (0, 0)? Important: You must explicitly calculate the path integral here. Explain your work. Give your answer in terms of the given parameters. Part b) Based on your calculation from Part (a), answer this question: Is this force F Conservative or Not Conservative? Cite specific evidence and/or examples to support your claim either way. Explain your work completely here to get full credit. Possibly helpful hints: You cannot assume that just because the path is closed the integral is zero. You need to calculate the path integral explicitly here. Don't forget the Definition of Work on a path that goes from point P to point Q is given by: W_F(P → Q) ≡ ∫_P^Q F ∙ dr = ∫_{x_P}^{x_Q} F_x dx + ∫_{y_P}^{y_Q} F_y dy where F_x is the x-component of the force and F_y is the y-component of the force. In order to calculate the Work done by the force around the square loop, you will need to calculate the Work done on each of the four "straight-line" pieces of the path. In other words, you need to break the closed-path integral into four regular 1-D integrals. For each "straight-line" piece you are holding one coordinate fixed at a known value and integrating with respect to the other coordinate.
David M.
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