0. Car A is moving in the direction \( 350^{\circ} \) and the engine of the car exerts a force of 300 kN . Car B , moving in the direction \( 80^{\circ} \), collides with car A . Car B exerts a force of \( 1,77 \times 10^{5} \mathrm{~N} \) on car A . Calculate the magnitude of the resultant force on car A

To calculate the magnitude of the resultant force on Car A, we need to consider both the force exerted by the engine and the force exerted by Car B during the collision. Since these forces act in different directions, we'll break them down into their horizontal (x) and vertical (y) components, sum these components, and then determine the resultant vector. ### Given: 1. **Engine Force (\( \mathbf{F}_1 \))**: - Magnitude: 300 kN = \( 3.00 \times 10^5 \) N - Direction: \( 350^{\circ} \) 2. **Collision Force (\( \mathbf{F}_2 \))**: - Magnitude: \( 1.77 \times 10^5 \) N - Direction: \( 80^{\circ} \) ### Step 1: Break Down Forces into Components **For \( \mathbf{F}_1 \) (350°):** - \( F_{1x} = F_1 \cos(350^\circ) \) - \( F_{1y} = F_1 \sin(350^\circ) \) Using: - \( \cos(350^\circ) \approx 0.98480775 \) - \( \sin(350^\circ) \approx -0.17364818 \) Calculations: - \( F_{1x} = 3.00 \times 10^5 \times 0.98480775 \approx 295,442 \) N - \( F_{1y} = 3.00 \times 10^5 \times (-0.17364818) \approx -52,094 \) N **For \( \mathbf{F}_2 \) (80°):** - \( F_{2x} = F_2 \cos(80^\circ) \) - \( F_{2y} = F_2 \sin(80^\circ) \) Using: - \( \cos(80^\circ) \approx 0.17364818 \) - \( \sin(80^\circ) \approx 0.98480775 \) Calculations: - \( F_{2x} = 1.77 \times 10^5 \times 0.17364818 \approx 30,727 \) N - \( F_{2y} = 1.77 \times 10^5 \times 0.98480775 \approx 174,757 \) N ### Step 2: Sum the Components - **Total Horizontal Component (\( F_{x} \))**: \[ F_x = F_{1x} + F_{2x} = 295,442 \, \text{N} + 30,727 \, \text{N} = 326,169 \, \text{N} \] - **Total Vertical Component (\( F_{y} \))**: \[ F_y = F_{1y} + F_{2y} = -52,094 \, \text{N} + 174,757 \, \text{N} = 122,663 \, \text{N} \] ### Step 3: Calculate the Magnitude of the Resultant Force Using the Pythagorean theorem: \[ |\mathbf{F}_{\text{resultant}}| = \sqrt{F_x^2 + F_y^2} \] \[ |\mathbf{F}_{\text{resultant}}| = \sqrt{(326,169)^2 + (122,663)^2} \] \[ |\mathbf{F}_{\text{resultant}}| \approx \sqrt{1.0637 \times 10^{11} + 1.5032 \times 10^{10}} = \sqrt{1.2140 \times 10^{11}} \approx 348,365 \, \text{N} \] ### Conclusion The magnitude of the resultant force on Car A is approximately **348 kN**. **Final Answer:** The resultant force on Car A has a magnitude of approximately 348 kN.