$2400 \textrm{ J}$

The work output is less than for $W_{ABCD}$ since the area enclosed by the PV loop is less.

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This is College Physics Answers with Shaun Dychko. For a cycle beginning at point <i>A</i> going to <i>B</i> and then it crosses diagonal to <i>D</i> and then back to <i>A</i>. The net work done will be the area enclosed by the curve. So the area enclosed by the full cycle so we have find the area of that triangle, in other words, and so it’s going to be one half base times height and so we have one half a base which is between 2.6 and 1. So we are finding this distance here we will call that the base. So 2.6 times 10 to the 6 Newtons per square meter minus 1 times 10 to the 6 Newtons per square meter. And then we multiply that by the height which is from here to here and that is 4 millimeters and then minus 1 millimeter written as times 10 to the minus 3 and then that's cubic meters, by the way, not millimeters. So 4 times 10 to the minus 3 cubic meters minus 1 times 10 to the minus 3 cubic meters and all this works out to 2400 Joules and the work output in this cycle going from <i>ABDA</i> is less than the total work for <i>ABCDA</i>. Since the area enclosed by this loop is less and so if the cycle had gone from <i>A B</i> and then down to <i>C</i>, we would have had a cycle with this green loop and this is a trapezoid, which would have a larger amount of enclosed area compared to the enclosed area by the yellow cycle.