Answer
The total work required to put 28 books on each shelf is 1960 J.
Work Step by Step
Let's pick up one book and stand it on the first (bottom) shelf. We need to raise the center of mass of the book 0.260 m, because this is 0.150 m (the first shelf) plus 0.110 m (the center of mass of the book).
$Work = \Delta PE = mgy$
$Work = (1.40 ~kg)(9.80 ~m/s^2)(0.260 ~m)$
To put one book on each of the remaining four shelves, we need to increase the height $y$ by 0.380 m each time.
Therefore the work required to stand one book on each shelf is (in order from bottom to top):
$Work = (1.40 ~kg)(9.80 ~m/s^2)(0.260 ~m)$
$Work = (1.40 ~kg)(9.80 ~m/s^2)(0.260 ~m + 0.380 ~m)$
$Work = (1.40 ~kg)(9.80 ~m/s^2)(0.260 ~m + 2 \times 0.380 ~m)$
$Work = (1.40 ~kg)(9.80 ~m/s^2)(0.260 ~m + 3 \times 0.380 ~m)$$Work = (1.40 ~kg)(9.80 ~m/s^2)(0.260 ~m + 4 \times 0.380 ~m)$
Let's find the sum of all this work.
$Work = (1.40 ~kg)(9.80 ~m/s^2)(5\times 0.260 ~m + 10 \times 0.380 ~m)$
$Work = (1.40 ~kg)(9.80 ~m/s^2)(5.10 ~m)$
Since we need to put 28 books on each shelf, we need to multiply this work by 28 to get the total work required:
$Work = 28\times (1.40 ~kg)(9.80 ~m/s^2)(5.10 ~m)$
$Work = 1960 ~J$
The total work required to put 28 books on each shelf is 1960 J.
