10. A curve in a road forms part of a horizontal circle. As a car of mass \( 1,500 \mathrm{~kg} \) goes around it at constant speed \( 14.0 \mathrm{~m} / \mathrm{s} \), the total force on the driver has magnitude 130 N . What is the total vector force on the driver if the speed is \( 18.0 \mathrm{~m} / \mathrm{s} \) instead?

To determine the total vector force on the driver when the speed increases from \( 14.0 \, \text{m/s} \) to \( 18.0 \, \text{m/s} \), we'll analyze the centripetal force involved. ### Given: - **Mass of the car (and driver):** \( m = 1,500 \, \text{kg} \) - **Initial speed (\( v_1 \)):** \( 14.0 \, \text{m/s} \) - **Initial total force (\( F_1 \)):** \( 130 \, \text{N} \) - **New speed (\( v_2 \)):** \( 18.0 \, \text{m/s} \) ### Understanding the Forces: The force you're referring to is the **centripetal force** required to keep the car moving in a circular path. The formula for centripetal force (\( F_c \)) is: \[ F_c = \frac{m v^2}{r} \] where: - \( m \) is the mass, - \( v \) is the speed, and - \( r \) is the radius of the circular path. ### Step 1: Determine the Radius of the Curve Using the initial conditions: \[ 130 \, \text{N} = \frac{1,500 \times (14.0)^2}{r} \] Solving for \( r \): \[ r = \frac{1,500 \times 196}{130} \approx 2,261.54 \, \text{m} \] ### Step 2: Calculate the New Centripetal Force at \( 18.0 \, \text{m/s} \) Using the same radius: \[ F_2 = \frac{1,500 \times (18.0)^2}{2,261.54} \] \[ F_2 = \frac{1,500 \times 324}{2,261.54} \approx 215 \, \text{N} \] ### Conclusion: When the speed increases to \( 18.0 \, \text{m/s} \), the total vector (centripetal) force on the driver increases to approximately **215 N** directed toward the center of the circular path. **Answer:** At 18 m/s, the total force on the driver is approximately 215 newtons directed inward.