1. Problem: A 300-kg roller-coaster car starts from rest at the top of a 15.0-m-tall hill and rolls down a frictionless track. We need to find the speed at the bottom of the hill using conservation of energy. 2. Formula: The mechanical energy conservation principle states that the potential energy at the top converts to kinetic energy at the bottom: $$mgh = \frac{1}{2}mv^2$$ where $m$ is mass, $g$ is acceleration due to gravity (9.8 m/s²), $h$ is height, and $v$ is velocity. 3. Calculation: - Cancel $m$ from both sides: $$gh = \frac{1}{2}v^2$$ - Solve for $v$: $$v = \sqrt{2gh}$$ - Substitute values: $$v = \sqrt{2 \times 9.8 \times 15.0} = \sqrt{294} \approx 17.15\,\text{m/s}$$ 4. Explanation: The roller coaster converts all its potential energy at the top into kinetic energy at the bottom, so the speed is about 17.15 m/s. --- 1. Problem: The car slows down on a flat section with a constant deceleration of 1.50 m/s² until it stops. Find the distance traveled before stopping using kinematics. 2. Formula: Use the kinematic equation: $$v^2 = v_0^2 + 2a\Delta x$$ where $v$ is final velocity (0 m/s), $v_0$ is initial velocity (speed at bottom), $a$ is acceleration (negative for deceleration), and $\Delta x$ is distance. 3. Calculation: - Rearrange for $\Delta x$: $$\Delta x = \frac{v^2 - v_0^2}{2a}$$ - Substitute values: $$\Delta x = \frac{0 - (17.15)^2}{2 \times (-1.50)} = \frac{-294}{-3} = 98\,\text{m}$$ 4. Explanation: The car travels 98 meters on the flat section before coming to rest due to the brakes. Final answers: - Speed at bottom: $17.15$ m/s - Distance traveled before stopping: $98$ m