17. I A 1500 kg car drives around a flat 200-m-diameter circular track at 25 m/s. What are the magnitude and direction of the net force on the car? What causes this force?
17. I A 1500 kg car drives around a flat 200-m-diameter circular track at 25 m/s. What are the magnitude and direction of the net force on the car? What causes this force?
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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![**Physics Problems for Educational Exercise**
**Problem 17:**
A 1500 kg car drives around a flat 200-m-diameter circular track at 25 m/s. What are the magnitude and direction of the net force on the car? What causes this force?
**Problem 18:**
A fast pitch softball player does a “windmill” pitch, illustrated...
(Note: The image cuts off the rest of Problem 18's description, so further details are not available.)
---
**Detailed Explanation for Problem 17:**
To solve this problem, we need to determine the net force acting on the car as it moves in a circular path.
1. **Given Data:**
- Mass of the car, \( m = 1500 \) kg
- Diameter of the circular track, \( d = 200 \) m
- Speed of the car, \( v = 25 \) m/s
2. **Determine the Radius:**
- Radius, \( r = \frac{d}{2} = \frac{200}{2} = 100 \) m
3. **Centripetal Force Calculation:**
The net force on the car moving in a circular path is the centripetal force, which can be calculated using the formula:
\[
F_c = \frac{mv^2}{r}
\]
Substituting the given values:
\[
F_c = \frac{1500 \times (25)^2}{100} = \frac{1500 \times 625}{100} = 9375 \text{ N}
\]
4. **Direction of the Force:**
The direction of the net force is towards the center of the circular path, which is the direction of the centripetal force.
5. **Cause of the Force:**
This force is caused by the frictional force between the car's tires and the track, which provides the necessary centripetal force to keep the car moving in a circular path.
---
**Note:** This section contains three problems from a set, each designed to enhance the understanding of dynamics and forces in circular motion for students.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7b47dd07-9dca-411e-981f-312dfe7288cb%2F8768f6f5-3640-4ccf-adf9-28bcc87b108b%2F76e90mp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Physics Problems for Educational Exercise**
**Problem 17:**
A 1500 kg car drives around a flat 200-m-diameter circular track at 25 m/s. What are the magnitude and direction of the net force on the car? What causes this force?
**Problem 18:**
A fast pitch softball player does a “windmill” pitch, illustrated...
(Note: The image cuts off the rest of Problem 18's description, so further details are not available.)
---
**Detailed Explanation for Problem 17:**
To solve this problem, we need to determine the net force acting on the car as it moves in a circular path.
1. **Given Data:**
- Mass of the car, \( m = 1500 \) kg
- Diameter of the circular track, \( d = 200 \) m
- Speed of the car, \( v = 25 \) m/s
2. **Determine the Radius:**
- Radius, \( r = \frac{d}{2} = \frac{200}{2} = 100 \) m
3. **Centripetal Force Calculation:**
The net force on the car moving in a circular path is the centripetal force, which can be calculated using the formula:
\[
F_c = \frac{mv^2}{r}
\]
Substituting the given values:
\[
F_c = \frac{1500 \times (25)^2}{100} = \frac{1500 \times 625}{100} = 9375 \text{ N}
\]
4. **Direction of the Force:**
The direction of the net force is towards the center of the circular path, which is the direction of the centripetal force.
5. **Cause of the Force:**
This force is caused by the frictional force between the car's tires and the track, which provides the necessary centripetal force to keep the car moving in a circular path.
---
**Note:** This section contains three problems from a set, each designed to enhance the understanding of dynamics and forces in circular motion for students.
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