00:01
You and three friends stand at the corners of a square whose size are 8 meters long.
00:06
In this problem, you will be taking your physics book and pushing it around the square.
00:11
The book has a mass of 1 .5 kilograms and the coefficient of kinetic friction between the book and the floor is 0 .25.
00:22
In the first part of this problem, the book slides from you to beth and then from beth to carlos.
00:30
And the problem wants you to find the work done by friction during this movement of the book.
00:38
So here, recall that the formula for kinetic friction is fk equals mu fn, where fn is the normal force.
00:54
However, since the book is only moving horizontally, normal force will be equal to fg, which is equal to m g so if we plug that back into the kinetic force equation we get fk is equal to new m g but here we're not trying to find force we're trying to find the work done by the friction so work is equal to negative force of friction times the distance that the book travels here the negative sign is because the direction of the friction force is opposite of the direction of movement so if we plug everything into the work equation we have negative mu times m times g times d where d is again distance of travel so for part a we have negative 0 .25 for mu 1 .5 kilograms for the mass of the book 9 .8 meters per second squared for g and 16 meters for d since it moved from you to beth and then beth to carlos.
02:35
This will give us about negative 58 .8 jules and that would be the work done by the friction force as it travels from you to beth and then from beth to carlos.
02:57
So now in part b we have the book sliding from you to carlos along the diagonal of the square.
03:08
Let's go ahead and copy this diagram and i'll put it on the next page so that we have more space.
03:29
All right now we want to look at the book's movement along the diagonal of the square from you to carlos.
03:40
Again, the work equation will be negative mu m g times the distance that it travels.
03:52
So here we have 0 .25 for the mu 1 .5 kilograms for the mass, 9 .8 meters per second square for g.
04:11
And here d will be the diagonal, which from trigonometry, remember, is 8 times rad 2 meters.
04:29
So if we crunch all the numbers in a calculator, we'll end up with negative 41 .6 joules for the work done in part b by the friction force.
04:53
In part c, you slide the book to kim and then kim slides it back to you.
05:00
So again, i'm going to copy the diagram so that we have more space...