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m 70.0 m A car goes over a hill at a speed of 21.0 m/s. The shape of the hill is approximately circular, with a radius of 70.0 m, as in the figure. When the car is at the highest point of the hill, the normal force of the road on the car is 7000N. What is the mass of the car? O 2000 kg 736.8 kg O 1111.1 kg O 714.3 kg 1500 kg

m 70.0 m A car goes over a hill at a speed of 21.0 m/s. The shape of the hill is approximately circular, with a radius of 70.0 m, as in the figure. When the car is at the highest point of the hill, the normal force of the road on the car is 7000N. What is the mass of the car? O 2000 kg 736.8 kg O 1111.1 kg O 714.3 kg 1500 kg

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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The last answer choice given is 434.8 kg

### Physics Problem: Finding the Mass of a Car on a Circular Hill #### Problem Statement: A car goes over a hill at a speed of 21.0 m/s. The shape of the hill is approximately circular, with a radius of 70.0 m, as illustrated in the figure below. When the car is at the highest point of the hill, the normal force of the road on the car is 7000 N. What is the mass of the car? #### Diagram: - The diagram shows a car at the top of a hill. - The hill has a circular shape with a radius of 70.0 m, denoted as \( R \). - The car's speed is depicted as \( \vec{v} \), which is 21.0 m/s. - There is a force vector pointing downward, which represents the normal force (7000 N) at the highest point. #### Given Data: - Speed of the car, \( v = 21.0 \, \text{m/s} \) - Radius of the hill, \( R = 70.0 \, \text{m} \) - Normal force at the highest point, \( F_N = 7000 \, \text{N} \) #### Question: What is the mass of the car? #### Answer Options: - \( \) 2000 kg - \( \) 736.8 kg - \( \) 1111.1 kg - \( \) 714.3 kg - \( \) 1500 kg ### Explanation: To solve for the mass of the car, we need to apply the concepts of circular motion and forces. The centripetal force acting on the car at the highest point of the circular hill is provided by the difference between the gravitational force and the normal force. The gravitational force can be expressed as \( F_g = mg \). When at the highest point, the net centripetal force \( F_c \) acting towards the center of the circular path is given by: \[ F_c = F_g - F_N \] \[ F_c = mg - F_N \] The centripetal force is also given by: \[ F_c = \frac{mv^2}{R} \] By equating the two expressions for the centripetal force, we get: \[ mg - F_N = \frac{mv^2}{R} \]
Transcribed Image Text:### Physics Problem: Finding the Mass of a Car on a Circular Hill #### Problem Statement: A car goes over a hill at a speed of 21.0 m/s. The shape of the hill is approximately circular, with a radius of 70.0 m, as illustrated in the figure below. When the car is at the highest point of the hill, the normal force of the road on the car is 7000 N. What is the mass of the car? #### Diagram: - The diagram shows a car at the top of a hill. - The hill has a circular shape with a radius of 70.0 m, denoted as \( R \). - The car's speed is depicted as \( \vec{v} \), which is 21.0 m/s. - There is a force vector pointing downward, which represents the normal force (7000 N) at the highest point. #### Given Data: - Speed of the car, \( v = 21.0 \, \text{m/s} \) - Radius of the hill, \( R = 70.0 \, \text{m} \) - Normal force at the highest point, \( F_N = 7000 \, \text{N} \) #### Question: What is the mass of the car? #### Answer Options: - \( \) 2000 kg - \( \) 736.8 kg - \( \) 1111.1 kg - \( \) 714.3 kg - \( \) 1500 kg ### Explanation: To solve for the mass of the car, we need to apply the concepts of circular motion and forces. The centripetal force acting on the car at the highest point of the circular hill is provided by the difference between the gravitational force and the normal force. The gravitational force can be expressed as \( F_g = mg \). When at the highest point, the net centripetal force \( F_c \) acting towards the center of the circular path is given by: \[ F_c = F_g - F_N \] \[ F_c = mg - F_N \] The centripetal force is also given by: \[ F_c = \frac{mv^2}{R} \] By equating the two expressions for the centripetal force, we get: \[ mg - F_N = \frac{mv^2}{R} \]
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