Step 1: Understanding the Concept:
For a block sliding down an incline of length $s$, the time taken is $t = \sqrt{2s/a}$. We compare the acceleration on a smooth plane ($a_1$) versus a rough plane ($a_2$).
Step 2: Key Formula or Approach:
1. Smooth plane: $a_1 = g \sin \theta$.
2. Rough plane: $a_2 = g (\sin \theta - \mu \cos \theta)$.
3. Time relation: $t_r = t_s + 0.5t_s = 1.5 t_s = \frac{3}{2} t_s$.
4. Since $s$ is the same, $t \propto 1/\sqrt{a}$, so $\frac{t_r}{t_s} = \sqrt{\frac{a_1}{a_2}}$.
Step 3: Detailed Explanation:
1. Square the time ratio:
\[ \left( \frac{3}{2} \right)^2 = \frac{a_1}{a_2} \implies \frac{9}{4} = \frac{g \sin 45^{\circ}}{g (\sin 45^{\circ} - \mu \cos 45^{\circ})} \]
2. Since $\sin 45^{\circ} = \cos 45^{\circ} = 1/\sqrt{2}$, they cancel out:
\[ \frac{9}{4} = \frac{1}{1 - \mu} \]
3. Solve for $\mu$:
\[ 9(1 - \mu) = 4 \implies 9 - 9\mu = 4 \]
\[ 9\mu = 5 \implies \mu = 5/9 \]
Step 4: Final Answer:
The coefficient of kinetic friction is 5/9.