Your little sister, Nicole, is excited about participating in Field Day at Clubview Elementary, and since you are a student at Columbus State, you take a break between classes to go watch her class participate for a bit. Nicole swings a ball on a string at constant speed in a circle that has a diameter of 1.9 m. (The kid with the biggest diameter circle wins, so you're cheering her on!) Suddenly, you remember the physics you were studying, and wonder: how much work is being done on the ball by the 10 N tension force in the string during one complete revolution of the ball?

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**Physics in Everyday Life: Analyzing Circular Motion at Field Day**

Your little sister, Nicole, is excited about participating in Field Day at Clubview Elementary, and since you are a student at Columbus State, you take a break between classes to go watch her class participate for a bit.

Nicole swings a ball on a string at constant speed in a circle that has a diameter of 1.9 m. (The kid with the biggest diameter circle wins, so you're cheering her on!) Suddenly, you remember the physics you were studying, and wonder: how much work is being done on the ball by the 10 N tension force in the string during one complete revolution of the ball?

---

**Explaining the Physics:**

When analyzing problems involving circular motion, it's important to remember that work is defined as the force applied in the direction of the displacement times the distance moved. However, in the case of circular motion, the force (tension in the string) is always perpendicular to the displacement of the ball at any point along its path.

Since the force and displacement are perpendicular, no work is done by the tension force over one complete revolution. 

Let's break this down:

1. **Circular Path**: Nicole swings the ball around in a circle, which means that despite the ball traveling along the path, it returns to its starting point after one complete revolution.
2. **Force and Displacement Relationship**: The tension force in the string is directed towards the center of the circle (centripetal force), while the ball’s displacement at any instant is tangential to the circle. As a result, these two vectors are perpendicular.
3. **Work Calculation**: Work done (W) by the force is calculated as:
   \[
   W = \mathbf{F} \cdot \mathbf{d} \cdot \cos(\theta)
   \]
   where \( \mathbf{F} \) is the force, \( \mathbf{d} \) is the displacement, and \( \theta \) is the angle between the force and displacement vectors. In this case, \( \theta = 90^{\circ} \), and \( \cos(90^{\circ}) = 0 \).

Therefore, \( W = 0 \).
Transcribed Image Text:**Physics in Everyday Life: Analyzing Circular Motion at Field Day** Your little sister, Nicole, is excited about participating in Field Day at Clubview Elementary, and since you are a student at Columbus State, you take a break between classes to go watch her class participate for a bit. Nicole swings a ball on a string at constant speed in a circle that has a diameter of 1.9 m. (The kid with the biggest diameter circle wins, so you're cheering her on!) Suddenly, you remember the physics you were studying, and wonder: how much work is being done on the ball by the 10 N tension force in the string during one complete revolution of the ball? --- **Explaining the Physics:** When analyzing problems involving circular motion, it's important to remember that work is defined as the force applied in the direction of the displacement times the distance moved. However, in the case of circular motion, the force (tension in the string) is always perpendicular to the displacement of the ball at any point along its path. Since the force and displacement are perpendicular, no work is done by the tension force over one complete revolution. Let's break this down: 1. **Circular Path**: Nicole swings the ball around in a circle, which means that despite the ball traveling along the path, it returns to its starting point after one complete revolution. 2. **Force and Displacement Relationship**: The tension force in the string is directed towards the center of the circle (centripetal force), while the ball’s displacement at any instant is tangential to the circle. As a result, these two vectors are perpendicular. 3. **Work Calculation**: Work done (W) by the force is calculated as: \[ W = \mathbf{F} \cdot \mathbf{d} \cdot \cos(\theta) \] where \( \mathbf{F} \) is the force, \( \mathbf{d} \) is the displacement, and \( \theta \) is the angle between the force and displacement vectors. In this case, \( \theta = 90^{\circ} \), and \( \cos(90^{\circ}) = 0 \). Therefore, \( W = 0 \).
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