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![### Physics Problem: Circular Motion and Static Friction
**Problem Statement:**
A 30 kg cart is moving in a circle of radius 10 m with a tangential velocity of 5 m/s. Find the coefficient of static friction between the tires and the road.
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**Solution Approach:**
To solve for the coefficient of static friction (μ), we need to consider the forces acting on the cart as it moves in circular motion.
1. **Centripetal Force**: This is the force that keeps the cart moving in a circular path and is provided by the static friction between the tires and the road.
2. **Static Friction Force**: This force must be equal to the centripetal force for the cart to remain in circular motion without slipping.
The centripetal force (\( F_c \)) needed to keep the cart moving in a circle is given by:
\[ F_c = \frac{mv^2}{r} \]
where
- \( m \) is the mass of the cart (30 kg),
- \( v \) is the tangential velocity (5 m/s),
- \( r \) is the radius of the circle (10 m).
Substituting the given values:
\[ F_c = \frac{30 \ \text{kg} \times (5 \ \text{m/s})^2}{10 \ \text{m}} \]
\[ F_c = \frac{30 \times 25}{10} \]
\[ F_c = 75 \ \text{N} \]
The static frictional force (\( F_f \)) is also given by:
\[ F_f = \mu N \]
where
- \( \mu \) is the coefficient of static friction,
- \( N \) is the normal force, which equals \( mg \) for a horizontal surface.
Given \( g \approx 9.81 \ \text{m/s}^2 \):
\[ N = mg \]
\[ N = 30 \ \text{kg} \times 9.81 \ \text{m/s}^2 \]
\[ N = 294.3 \ \text{N} \]
Since \( F_f = F_c \):
\[ \mu \times 294.3 \ \text{N} = 75 \ \text{N} \]
Solving for \( \mu \):
\[ \mu = \frac{75 \](https://content.bartleby.com/qna-images/question/3badc987-71ef-407e-94db-ec928f27bcbd/5ce2339f-fa2b-4f57-941c-c2eb17e5d47e/050zc2_processed.jpeg)
Transcribed Image Text:### Physics Problem: Circular Motion and Static Friction
**Problem Statement:**
A 30 kg cart is moving in a circle of radius 10 m with a tangential velocity of 5 m/s. Find the coefficient of static friction between the tires and the road.
---
**Solution Approach:**
To solve for the coefficient of static friction (μ), we need to consider the forces acting on the cart as it moves in circular motion.
1. **Centripetal Force**: This is the force that keeps the cart moving in a circular path and is provided by the static friction between the tires and the road.
2. **Static Friction Force**: This force must be equal to the centripetal force for the cart to remain in circular motion without slipping.
The centripetal force (\( F_c \)) needed to keep the cart moving in a circle is given by:
\[ F_c = \frac{mv^2}{r} \]
where
- \( m \) is the mass of the cart (30 kg),
- \( v \) is the tangential velocity (5 m/s),
- \( r \) is the radius of the circle (10 m).
Substituting the given values:
\[ F_c = \frac{30 \ \text{kg} \times (5 \ \text{m/s})^2}{10 \ \text{m}} \]
\[ F_c = \frac{30 \times 25}{10} \]
\[ F_c = 75 \ \text{N} \]
The static frictional force (\( F_f \)) is also given by:
\[ F_f = \mu N \]
where
- \( \mu \) is the coefficient of static friction,
- \( N \) is the normal force, which equals \( mg \) for a horizontal surface.
Given \( g \approx 9.81 \ \text{m/s}^2 \):
\[ N = mg \]
\[ N = 30 \ \text{kg} \times 9.81 \ \text{m/s}^2 \]
\[ N = 294.3 \ \text{N} \]
Since \( F_f = F_c \):
\[ \mu \times 294.3 \ \text{N} = 75 \ \text{N} \]
Solving for \( \mu \):
\[ \mu = \frac{75 \
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