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Problem 47 Hard Difficulty

(a) Show that a constant force field does zero work on a particle that moves once uniformly around the circle $ x^2 + y^2 = 1 $.

(b) Is this also true for a force field $ \textbf{F(x)} = k\textbf{x} $, where $ k $ is a constant and $ \textbf{x} = \langle x, y \rangle $?


a) $W=0$
(b) Yes


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Video Transcript

So So we have particle moves uniformed around this uncle. So I Paramount tries it. We get this and we need a force. If we do computer work done by a force field, the first problem is a constant force field. So that's the F O. C. That would come here. That f thought pr is the simplest computers T from zero to two pi No, no, a partial. Sorry. The force should be a vector. So sorry about that. BC one seat there, say and see, once you two are constant is a top product f end our prime, which is minus science minus C one society pussy to co sign T. And, uh, yeah, actually, I write the way. So the site he and I drove here this nephew, of course, sci fi and the Coast ninety and either to society and regardless of hac Wannsee to is this copy zero because cause and here inside he's about is a are period, a functional period to pie. That means your problem to pipe fucking zero, you're going to get the same number and therefore the differences here and that concludes the first part. And that's also solves a seven part. So in this case, f equals K X for X is X Y S o f basically host k x k y, and we won't compute f off are So X is replaced by co sign wise, replaced by side. So is f our prime. That's the same. We want to compute the first Hamud private negative scientist. So if you pay close I nt sci fi and the second time is cosign These are plus Keiko seventy a society which keeps us zero. And therefore if this zero then this inter crow No, that they are or simply be zero. Plus you're just integrating zero. So this is also true for for this case.