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$W_8=20J$

Giancoli 6th Edition, Chapter 6, Problem 6

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Chapter 6, Problem 6 is solved.

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Transcript for this Giancoli solution

We are going to have eight books stacked on top of each other; the question is what is the total work done to create that stack? So here is one of the books. It has a thickness; we call that 'h' for height. The height of that book is 'h' then we have another book that we are going to stack on top of it, equal height and equal mass. Then we are going to lift it up and put it on top of the blue box. Now we only lift it up this high, that height 'h', so the thickness of the book is how high we lift it and we ignore this horizontal acceleration that's needed to get it there. So we assume that everything is negligible horizontally speaking and the only force that we need to think about is the one going up against the gravity to get it up this high. So the work done is the weight of that green book times the height 'h'. I'm going to give you a kind of a bit of a sophisticated answer to this question; we are going to derive the general amount of work done for creating a stack of any number of books. The simpler approach would be to just answer the specific question which would be this one book put on top it had the work done 'mg' times 'h', and then you take the next book put it on top and you add 'mg' times '2h' and so on. But we are going to make a table to show this and then see the pattern that results there.

And then we're going to use the sum of an arithmetic series to derive an interesting equation. So, the number stacked is going to go in this column and then the work done we will put in this column and then we will see a pattern. If you had one book stacked, and the way I'm counting here is this is the total number of books in the stack and so maybe I should even just say the size of the stack, stack size. Number of books in the stack. If there was one book in the stack, well, there is nothing to lift and the work done would be zero. If there are two books in the stack, as illustrated here, this is the case with two books in the stack, the work done is going to be 'mg' that's the weight of the book that was lifted times 'h', the height that it was lifted. We are going to need some more room so let's move this down here. If you have three books in the stack, that's going to be 'mgh' plus 'mg2h' and this is a little cleaner if you read it as 'mg' factored out times 'h' plus '2h' am choosing not collect the 'h' and '2h' into '3h'. I'm going to leave it like this because it will be a little easier to see the arithmetic series that we get in a few minutes. So the case where the stack size of three is if you had...let me illustrate that up here. Let's get rid of that little bit. Here is a third book that we are going to put on top so making a stack size of three books here. This book is lifted up '1h' and then another 'h'. So that explains where this '2h' comes from here. That just explain that by showing that this red book had top be lifted a height 2 times of the thickness of one of the books. Stack size of 4 would be 'mg' times 'h' plus '2h' plus '3h', a stack size of five 'mg' times 'h' plus '2h' plus '3h' plus '4h'. You can see the pattern, can't you? You can see that this coefficient of 'h' is always increasing by 1. And the way we'll make use of that is to recognize that this is what happens in an arithmetic series. Now, telling you deriving the formula for an arithmetic series is a little bit beyond the scope of this video. That could be another video, if you'd like to know more about this formula just let me know on the comments for this video. So the sum of an arithmetic series of 'n' terms is 'n' over 2 times the first term plus the last term and we know this is an arithmetic series because there is a common difference so you are always adding the same amount to each subsequent term. There is an amount 'h' added, so to get from the first term to the second term you add 'h', to get from here to here, add 'h'. That is the sign of an arithmetic series when you have this being true that you can add the same number to get to each of the next terms. Okay.

So the series we are actually going to be dealing with here is we are just going to only look at the brackets with the 'h's' in them that's what we are going to get the sum of. And then we will take our answer and multiply it by 'mg' and we are actually going to put a little zero in front of it. Because if you had 1 books in your stack then you would have zero in as far as these brackets go, there would be no height lifted at all. If you have 2 books use 'h', if you have 3 books there is 'h' plus '2h'. Basically what I'm repeating is here. So if you have 3 books its 'h' plus '2h'. 2 books is just 'h', that's getting multiplied by 'mg' and 1 book while there is no 'h's' at all. And 4 books would be 'h' plus '2h' plus '3h'. Okay. All that to say that the first term equals zero and the last term equals 'n' minus 1 times 'h'. So in this case the last term was '4h' and that is 5, which is the number stacked, minus 1. So we have the number stacked minus 1 times 'h' is the last term.

Okay. So in this case we have the sum of the brackets here. all the 'h's' added together is going to be the number of books that are in the stack divided by two times zero, because that is the first term, plus 'n' minus 1 times 'h', that the last term. And making that look a little cleaner 'nh' times 'n' minus one all over 2. And the work done is going to be 'mg' times that. So that's 'mg' times 'hn' 'n' minus 1 all over 2. So that's the general formula then we just did that for fun, it's kind of interesting. This is the formula for the work done to stack books, any number of books. And if we have to stack 8 books, which is what the actual question is, we would substitute in the mass of each book. 1.7 kilograms for the sixth edition time's 9.8 newtons per kilogram times 0.043 meters the thickness of one book times 8 books times 8 minus 1 because that's what our formula says and divide all that by 2, and we get an answer of SOLUTION