When a force varies with position, the work done ($W$) is calculated by integrating the force over the displacement . For a force $F(y)$ acting in the $y$-direction, the formula is:

$W = \int_{y_1}^{y_2} F(y) dy$

Given the force $F = (20 + 10y)$ and the limits from $y = 0$ to $y = 1$ m:

1. Set up the integral: $W = \int_{0}^{1} (20 + 10y) dy$

2. Integrate the expression: $W = \left[ 20y + \frac{10y^2}{2} \right]_{0}^{1} = [20y + 5y^2]_{0}^{1}$

3. Apply the limits: $W = (20(1) + 5(1)^2) - (20(0) + 5(0)^2)$ $W = 20 + 5 = 25 \text{ J}$

Thus, the total work done is 25 J.