WEBVTT 00:00:00.046 --> 00:00:02.765 This is College Physics Answers with Shaun Dychko. 00:00:03.406 --> 00:00:05.750 So the box begins in this corner of the rug 00:00:06.062 --> 00:00:10.000 and the options are to move it along the smooth concrete this distance d one 00:00:10.508 --> 00:00:13.843 and then again along distance d two to get to this position, 00:00:14.460 --> 00:00:16.804 or we can move it with -- 00:00:17.180 --> 00:00:20.921 by pushing it with a greater force across the rough rug 00:00:21.703 --> 00:00:24.938 but go the shorter distance which is the hypotenuse 00:00:25.477 --> 00:00:27.335 directly from corner to corner. 00:00:28.648 --> 00:00:31.296 So let's consider the case across the concrete first of all. 00:00:31.375 --> 00:00:34.218 So the total distance traveled would be d one plus d two. 00:00:34.288 --> 00:00:36.953 So that's three meters plus four meters which is seven meters 00:00:37.348 --> 00:00:39.121 and the work done to push along the concrete 00:00:39.123 --> 00:00:41.859 would be the force applied on the concrete which is 14 Newtons, 00:00:42.367 --> 00:00:46.793 times the total distance on the concrete, seven meters, for a total of 280 joules. 00:00:47.961 --> 00:00:50.473 Then for the case where we go across the rug, 00:00:50.773 --> 00:00:55.004 we're going to push with this force, the rug force which is 55 Newtons 00:00:55.569 --> 00:00:58.231 multiplied by this distance across the rug 00:00:58.504 --> 00:01:02.028 which is the square root of d one squared plus d two squared. 00:01:02.687 --> 00:01:05.770 So that's 55 Newtons times square root of three squared plus four squared 00:01:05.795 --> 00:01:07.794 which gives 275 joules. 00:01:08.250 --> 00:01:11.797 So it turns out that there is less work to go across the rug, 00:01:13.497 --> 00:01:17.731 or I guess to put it in terms that work with these options here, 00:01:18.179 --> 00:01:23.359 we should say that across the floor is 5 joules extra compared to the rug. 00:01:23.640 --> 00:01:25.226 So option B is the answer.