4. 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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### 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 \
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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