Let's break down the problem step by step and solve it: --- ### **Step 1: Draw the Situation and Free Body Diagram (FBD)** - **Situation**: A car is moving around a banked curve (like a racetrack) with a banking angle \(\theta = 20^\circ\). - **Radius of curve**: \(r = 150.0\) m. - **No friction**: Only normal force and gravity act on the car. #### **Free Body Diagram (FBD)** - **Forces acting**: - Gravity (\(mg\)), vertically downward. - Normal force (\(N\)), perpendicular to the surface of the bank. The components of the normal force: - \(N \cos\theta\): Balances the weight (\(mg\)). - \(N \sin\theta\): Provides the centripetal force (\(m v^2 / r\)). --- ### **Step 2: Write the Equations** #### **Vertical Forces (No vertical movement)** \[ N \cos\theta = mg \] #### **Horizontal Forces (Centripetal force)** \[ N \sin\theta = \frac{m v^2}{r} \] --- ### **Step 3: Formula Development** Divide the horizontal by the vertical equation: \[ \frac{N \sin\theta}{N \cos\theta} = \frac{\frac{m v^2}{r}}{mg} \] \[ \tan\theta = \frac{v^2}{rg} \] --- ### **Step 4: Solve for Maximum Speed \(v\)** \[ v^2 = rg \tan\theta \] \[ v = \sqrt{rg \tan\theta} \] --- ### **Step 5: Substitute the Values** Given: - \(r = 150.0\) m - \(g = 9.8\) m/s² - \(\theta = 20^\circ\) \[ \tan 20^\circ \approx 0.3640 \] \[ v = \sqrt{150.0 \times 9.8 \times 0.3640} \] \[ v = \sqrt{150.0 \times 3.5672} \] \[ v = \sqrt{535.08} \] \[ v \approx 23.1 \text{ m/s} \] --- ### **Step 6: Final Answer** **The maximum speed at which a vehicle can negotiate the curve without friction is approximately \( \boxed{23.1 \text{ m/s}} \).** --- **Summary of steps:** 1. Drew the free body diagram (FBD). 2. Wrote force balance equations. 3. Developed the formula for speed. 4. Substituted values and solved for \(v\). Let me know if you'd like a sketch of the situation and FBD!