ch08-p120 - x direction ( d → = +(3.0 m)i ^ ), and it is...

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(c) The two integrations that need to be performed are each of the form ´ 2 x dx so that we are adding two equivalent terms, where each equals x 2 (evaluated at x = 4, minus its value at x = 1). Thus, the work done is 2(4 2 – 1 2 ) = 30 J. (d) This is another conservative force field, as can be easily verified by calculating that the net work done here is zero. (e) The forces in (b) and (d) are conservative. 120. (a) The table shows that the force is +(3.0 N)i ^ while the displacement is in the +
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Unformatted text preview: x direction ( d → = +(3.0 m)i ^ ), and it is –(3.0 N)i ^ while the displacement is in the – x direction. Using Eq. 7-8 for each part of the trip, and adding the results, we find the work done is 18 J. This is not a conservative force field; if it had been, then the net work done would have been zero (since it returned to where it started). (b) This, however, is a conservative force field, as can be easily verified by calculating that the net work done here is zero....
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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