WEBVTT 00:00:00.000 --> 00:00:02.860 This is Giancoli Answers with Mr. Dychko. 00:00:04.120 --> 00:00:06.020 Here's a free-body diagram of the crate 00:00:06.020 --> 00:00:08.020 and we can see there's an applied force to the right 00:00:08.490 --> 00:00:10.090 and there's friction force to the left 00:00:10.250 --> 00:00:12.650 and since the crate is sliding at constant speed, 00:00:12.860 --> 00:00:15.140 that means the applied force equals the friction force 00:00:15.610 --> 00:00:17.960 and so then we need to turn to the vertical direction in order to 00:00:17.960 --> 00:00:19.250 figure out what the friction force is 00:00:19.380 --> 00:00:22.840 because friction equals the coefficient of friction times the normal force. 00:00:23.250 --> 00:00:26.040 And the normal force is gonna equal gravity 00:00:26.040 --> 00:00:28.360 because there's no vertical acceleration of this box 00:00:28.610 --> 00:00:30.500 and so the up force has to equal the down force 00:00:31.010 --> 00:00:34.170 and we can substitute in m g for F N. 00:00:34.410 --> 00:00:37.700 So the friction force is μmg, in other words. 00:00:38.080 --> 00:00:41.970 And since the crate is moving at constant speed, 00:00:41.970 --> 00:00:43.440 the applied force equals that friction force 00:00:43.440 --> 00:00:45.610 so the applied force is μmg. 00:00:45.850 --> 00:00:49.540 And this applied force is in the same direction, or parallel, 00:00:49.720 --> 00:00:52.080 to the displacement—it's going to be moving to the right here— 00:00:52.340 --> 00:00:54.210 and so that means the work done 00:00:54.220 --> 00:00:56.760 is the applied force multiplied by the displacement. 00:00:57.540 --> 00:01:01.120 And, applied force is μmg 00:01:01.760 --> 00:01:03.960 and so we have, 0.5 times 00:01:03.960 --> 00:01:06.520 46 kilograms times 9.8 newtons per kilogram 00:01:06.520 --> 00:01:07.890 times 10.3 meters 00:01:08.170 --> 00:01:11.340 and we get 2300 joules with two significant figures.