Q.) A student has six textbooks, each 4.0 cm thick and 30 N inweight. What is the minimum work the student would have to do toplace the books in a vertical stack, starting with all the books onthe surface of the table?

Q.) A student has six textbooks, each 4.0 cm thick and 30 N inweight. What is the minimum work the student would have to do toplace the books in a vertical stack, starting with all the books onthe surface of the table?

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asked 2021-05-17
Q.) A student has six textbooks, each 4.0 cm thick and 30 N inweight. What is the minimum work the student would have to do toplace the books in a vertical stack, starting with all the books onthe surface of the table?

Answers (1)

2021-05-19
Let mg be the weight of each book
let t be the thichness of each book.
The 1st book is aready on the table, so no work needs to be done.
The 2nd book has to be lifted through a height of t to beplaced on top of the 1st book.
The 3rd book has to be lifted through a height of 2t to beplaced on top of the 2nd book. etc.
The general case can be given as,
\(\displaystyle{W}_{{1}}={m}{g}\times{t}\)
\(\displaystyle{W}_{{2}}={m}{g}\times{2}{t}\)
\(\displaystyle{W}_{{3}}={m}{g}\times{3}{t}\)
\(\displaystyle{W}_{{{n}-{1}}}={m}{g}\times{\left({n}-{1}\right)}{t}\)
So, if you have n books, of thickness t, then the work done inlifting (n-1) of them on top of each other is given by,
\(\displaystyle{W}={m}{>}{\left({1}+{2}+{3}+\ldots+{\left({n}-{1}\right)}\right)}\)
You have a simple summation of the 1st (n-1) integers which is \(\displaystyle{\frac{{{1}}}{{{2}}}}{\left({n}-{1}\right)}{n}\) so,
\(\displaystyle{W}={\frac{{{1}}}{{{2}}}}{m}{>}{\left({n}-{1}\right)}{n}\)
\(\displaystyle{m}{g}={30}{N}\)
\(\displaystyle{t}={0.04}{m}\)
\(\displaystyle{n}={6}\)
\(\displaystyle{W}={\frac{{{1}}}{{{2}}}}\times{30}\times{0.04}\times{5}\times{6}\)
\(\displaystyle{W}={15}\times{0.2}\times{6}\)
\(\displaystyle{W}={18}{J}\)
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