Given an initial force F1 = 2 N and an initial extension x1 = 30 mm.

Given an initial force $F_1 = 2 \, \text{N}$ and an initial extension $x_1 = 30 \, \text{mm}$.

$$k = \frac{F_1}{x_1} = \frac{2 \, \text{N}}{30 \, \text{mm}} = \frac{1}{15} \, \text{N/mm}$$

Using the calculated spring constant $k$ and the new total extension $x_2 = 35 \, \text{mm}$:

$$F_2 = kx_2 = \left(\frac{1}{15} \, \text{N/mm}\right) \times (35 \, \text{mm})$$ $$F_2 = \frac{35}{15} \, \text{N} = \frac{7}{3} \, \text{N}$$

The additional force is the difference between the total force required for $35 \, \text{mm}$ extension and the initial force that caused $30 \, \text{mm}$ extension.

$$\text{Additional Force} = F_2 - F_1 = \frac{7}{3} \, \text{N} - 2 \, \text{N}$$ $$\text{Additional Force} = \frac{7}{3} \, \text{N} - \frac{6}{3} \, \text{N} = \frac{1}{3} \, \text{N}$$

The additional force required is $\frac{1}{3} \, \text{N}$.