1. **State the problem:** A 5 kg block slides on a horizontal surface with a coefficient of kinetic friction $\mu_k = 0.3$. The block's initial speed is 10 m/s. We need to find the distance it travels before stopping. 2. **Relevant formula:** The friction force causes deceleration. The friction force $f_k$ is given by: $$f_k = \mu_k mg$$ where $m$ is mass, $g$ is acceleration due to gravity (9.8 m/s²). 3. **Calculate friction force:** $$f_k = 0.3 \times 5 \times 9.8 = 14.7 \text{ N}$$ 4. **Calculate acceleration due to friction:** Using Newton's second law, $f = ma$, the acceleration $a$ (actually deceleration) is: $$a = \frac{f_k}{m} = \frac{14.7}{5} = 2.94 \text{ m/s}^2$$ Since friction opposes motion, acceleration is negative: $$a = -2.94 \text{ m/s}^2$$ 5. **Use kinematic equation to find distance $d$:** Initial velocity $v_0 = 10$ m/s, final velocity $v = 0$ m/s, acceleration $a = -2.94$ m/s². The equation: $$v^2 = v_0^2 + 2ad$$ Rearranged to solve for $d$: $$d = \frac{v^2 - v_0^2}{2a} = \frac{0 - 10^2}{2 \times (-2.94)} = \frac{-100}{-5.88} = 17.01 \text{ m}$$ 6. **Answer:** The block travels approximately 17.01 meters before stopping.