00:01
So if you examine figure p5 .85, we can see that when the mass sub 3, when the third mass moves downward, that first mass must move upward, and then that second mass must slide a distance to the right across that horizontal tabletop.
00:20
Now here, because they're all connected, each block must always have the same speed as each of the other blocks.
00:26
So if the system starts from rest, we can then use, of course, the work energy theorem, given that there is a frictional force involved.
00:35
So using the work energy theorem, we can say that the negative frictional force times the displacement d, this would be equaling one half multiplied by m sub 1 plus m sub 2 plus m sub 3.
00:51
This would be multiplied by velocity final squared minus velocity initial squared.
00:56
Of course, simply the change in the kinetic energy.
01:00
Plus, and then this would be m sub 1g plus d because m sub 1 is moving upwards now, and then plus m sub 2g multiplied by 0 because m sub 2, the second mass, is moving horizontally.
01:19
Therefore, there isn't any change in height.
01:21
And then plus m sub 3g, and it's moving down that same displacement d.
01:29
And so we can then say that knowing that v initial is zero we can eliminate that term and we can then solve for the final velocity...