To find the final speed of a block moving through a rough region, consider the following steps:

Given Parameters:

Initial speed of the block, $ v_i = 10 \, \text{m/s} $.

Mass of the block, $ m = 1 \, \text{kg} $.

Force of friction acting on the block, $ F_r = -k \cdot x $.

$ k = 10 \, \text{N/m} $.

Forces and Acceleration:

The acceleration $ a $ of the block is given by $ a = \frac{F}{m} = -kx $.

So, we have: $ a = -10x $.

Using the Relationship:

We use the relationship between velocity, acceleration, and position: $ v \frac{dv}{dx} = a $.

Substituting $ a $ gives: $ v \frac{dv}{dx} = -10x $.

Integrating the Equation:

Integrate both sides with respect to their variables:

$ \int_{10}^{v} v \, dv = -10 \int_{0.1}^{1.9} x \, dx $

Solving the Integrals:

The left side becomes $ \frac{v^2}{2} - \frac{100}{2} $.

The right side evaluates to $-10 \cdot \frac{(1.9^2 - 0.1^2)}{2}$.

Therefore:

$ \frac{v^2 - 100}{2} = -10 \left(\frac{1.9^2 - 0.1^2}{2}\right) $

Final Calculation:

Solve the above equation for $ v $:

$ \frac{v^2 - 100}{2} = -10 \cdot \frac{(3.61 - 0.01)}{2} $

Simplifying and solving gives:

$ v^2 - 100 = -10 \cdot 1.8 $

Solving further:

$ v^2 = 100 - 18 $

$ v^2 = 82 $

$ v = \sqrt{82} \approx 8 \, \text{m/s} $

Therefore, the final speed of the block as it exits the rough region is approximately $ 8 \, \text{m/s} $.