$A$ block of mass $m=1 \; kg$ moving on a horizontal surface with speed $v_{i}=2 \; m \; s^{-1}$ enters a rough patch ranging from $x=0.10 \; m$ to $x=2.01 \; m$. The retarding force $F$ on the block in this range is inversely proportional to $x$ over this range,$F_{r} = -k/x$ for $0.1 < x < 2.01 \; m$,and $F_{r} = 0$ for $x < 0.1 \; m$ and $x > 2.01 \; m$,where $k=0.5 \; J$. What is the final kinetic energy $K_{f}$ and speed $v_{f}$ of the block as it crosses this patch?

(A) According to the Work-Energy Theorem,the change in kinetic energy is equal to the work done by the net force: $K_{f} - K_{i} = W$.
$W = \int_{0.1}^{2.01} F_{r} \, dx = \int_{0.1}^{2.01} \left( -\frac{k}{x} \right) dx = -k [\ln(x)]_{0.1}^{2.01} = -k \ln\left( \frac{2.01}{0.1} \right) = -k \ln(20.1)$.
Given $m = 1 \; kg$,$v_{i} = 2 \; m \; s^{-1}$,and $k = 0.5 \; J$:
$K_{i} = \frac{1}{2} m v_{i}^{2} = \frac{1}{2} \times 1 \times (2)^{2} = 2 \; J$.
$K_{f} = K_{i} - k \ln(20.1) = 2 - 0.5 \times \ln(20.1)$.
Using $\ln(20.1) \approx 3.00$,$K_{f} = 2 - 0.5 \times 3.00 = 2 - 1.5 = 0.5 \; J$.
Now,$v_{f} = \sqrt{\frac{2 K_{f}}{m}} = \sqrt{\frac{2 \times 0.5}{1}} = \sqrt{1} = 1 \; m \; s^{-1}$.