Answer
$3920 \ J$
Work Step by Step
The force of one layer is equal to:
$\ Force= \ Mass \times \ gravity = 78.4 \Delta y \ N$
The work done on the $i$th segment is the sum of the $W_i$; that is,
$W=\Sigma_{j=1}^N W_i \approx \Sigma_{j=1}^N 78.4 y_j \Delta y$
Therefore, the work done when $N \to \infty $ can be computed as:
$ W=\int_{0}^{10} 78.4 \ y \ dy \\=39.2 [y^2]_{0}^{10} \\ =3920 \ J$
