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Tom enlists the help of his friend John to move his car. They apply forces to the car as shown in the diagram. Here F, = 444 N and F, = 343 N and friction is negligible. In the diagram below, the mass of the car = 3500 kg, 8, = -25° and e, = 12°. (Assume the car faces the positive x-axis before the forces are applied.) F, F, (b) What is the acceleration (in m/s?) of the car? magnitude m/s? direction (counterclockwise from the +x-axis)

Tom enlists the help of his friend John to move his car. They apply forces to the car as shown in the diagram. Here F, = 444 N and F, = 343 N and friction is negligible. In the diagram below, the mass of the car = 3500 kg, 8, = -25° and e, = 12°. (Assume the car faces the positive x-axis before the forces are applied.) F, F, (b) What is the acceleration (in m/s?) of the car? magnitude m/s? direction (counterclockwise from the +x-axis)

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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I just need part B explained. Thank you!

Tom enlists the help of his friend John to move his car. They apply forces to the car as shown in the diagram. Here \( F_1 = 444 \, \text{N} \) and \( F_2 = 343 \, \text{N} \) and friction is negligible. In the diagram below, the mass of the car is 3500 kg, \( \theta_1 = -25^\circ \) and \( \theta_2 = 12^\circ \). (Assume the car faces the positive x-axis before the forces are applied.) ![Diagram of Forces Acting on the Car](https://example.com/diagram.png) In the diagram: - \( \vec{F_1} \) is a force of 444 N applied at an angle \( \theta_1 = -25^\circ \) from the positive x-axis. - \( \vec{F_2} \) is a force of 343 N applied at an angle \( \theta_2 = 12^\circ \) from the positive x-axis. - The car is represented by a top-down view, indicating the direction of the applied forces. **(b) What is the acceleration (in m/s\(^2\)) of the car?** - Magnitude: \[ \boxed{ \dots } \, \text{m/s}^2 \ ] - Direction (counterclockwise from the +x-axis): \[ \boxed{ \dots }^\circ \] To calculate the acceleration: 1. Resolve each force into its x and y components. 2. Sum the components to find the net force. 3. Use Newton's second law, \( \vec{F} = m \vec{a} \), to find the acceleration. Calculation steps: 1. For \( F_1 \): - \( F_{1x} = F_1 \cos(\theta_1) \) - \( F_{1y} = F_1 \sin(\theta_1) \) 2. For \( F_2 \): - \( F_{2x} = F_2 \cos(\theta_2) \) - \( F_{2y} = F_2 \sin(\theta_2) \) 3. Sum the components: - Net force in the x-direction: \( F_{net,x} = F_{1x
Transcribed Image Text:Tom enlists the help of his friend John to move his car. They apply forces to the car as shown in the diagram. Here \( F_1 = 444 \, \text{N} \) and \( F_2 = 343 \, \text{N} \) and friction is negligible. In the diagram below, the mass of the car is 3500 kg, \( \theta_1 = -25^\circ \) and \( \theta_2 = 12^\circ \). (Assume the car faces the positive x-axis before the forces are applied.) ![Diagram of Forces Acting on the Car](https://example.com/diagram.png) In the diagram: - \( \vec{F_1} \) is a force of 444 N applied at an angle \( \theta_1 = -25^\circ \) from the positive x-axis. - \( \vec{F_2} \) is a force of 343 N applied at an angle \( \theta_2 = 12^\circ \) from the positive x-axis. - The car is represented by a top-down view, indicating the direction of the applied forces. **(b) What is the acceleration (in m/s\(^2\)) of the car?** - Magnitude: \[ \boxed{ \dots } \, \text{m/s}^2 \ ] - Direction (counterclockwise from the +x-axis): \[ \boxed{ \dots }^\circ \] To calculate the acceleration: 1. Resolve each force into its x and y components. 2. Sum the components to find the net force. 3. Use Newton's second law, \( \vec{F} = m \vec{a} \), to find the acceleration. Calculation steps: 1. For \( F_1 \): - \( F_{1x} = F_1 \cos(\theta_1) \) - \( F_{1y} = F_1 \sin(\theta_1) \) 2. For \( F_2 \): - \( F_{2x} = F_2 \cos(\theta_2) \) - \( F_{2y} = F_2 \sin(\theta_2) \) 3. Sum the components: - Net force in the x-direction: \( F_{net,x} = F_{1x
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