a) $W=920 J$
Transcript for this Giancoli solutionIn part A, we are going to be pushing this crate at constant speed horizontally. So here's the ground, and the crate is on the ground. There is a friction force we're told going back. There's a friction force of, 230 newtons in the 6th Edition. And so we're going to be applying a force of equal magnitude in the other direction, since the speed is constant. We know that these forces have to be the same size. So the applied force is going to be also 230 newtons. The work that has to be done in this case is going to be the applied force times the distance over which it is applied, that's 230 Newtons times 4.0 meters, and this gives # joules. That's the 6th edition answer for Part A.
Part B, we're going to lift it up in this case, so we're going to be applying a force of equal magnitude to the gravity we're going to lift it up at constant speed. So we have gravity times the height that's 1300 newtons, because that's how much it weighs times 4 Meters because that's how high it will be lifted. And this gives us 5200 Joules. In the 6th Edition, you change 230 newtons to 180 Newtons, as the size of the friction force and the box is pushed 6 meters, giving a 6th Edition answer of 1.1 times 10 to the 3 joules. And the box weighs only 900 Newtons instead of 1300. And it again is raised 6 Meters and our 5th Edition answer in the 5th Edition we have # times # to the # joules is the answer to part B.