If the road of the previous problem is horizontal (no banking), what should be the minimum friction coefficient so that a scooter going at `18km//h` does not skid?

Explanation

To solve the problem of finding the minimum friction coefficient required for a scooter to make a turn without skidding on a horizontal road, we can follow these steps:

### Step-by-Step Solution:

1. **Convert the Speed from km/h to m/s**:
The speed of the scooter is given as \( 18 \, \text{km/h} \). To convert this to meters per second (m/s), we use the conversion factor:
\[ 1 \, \text{km/h} = \frac{5}{18} \, \text{m/s} \]
Therefore,
\[ v = 18 \, \text{km/h} \times \frac{5}{18} = 5 \, \text{m/s} \]

2. **Identify the Forces Acting on the Scooter**:
When the scooter is turning, the centripetal force required to keep it moving in a circular path is provided by the frictional force. The frictional force can be expressed as:
\[ F_{\text{friction}} = \mu \cdot N \]
where \( \mu \) is the coefficient of friction and \( N \) is the normal force. On a horizontal road, the normal force \( N \) equals the weight of the scooter:
\[ N = mg \]
where \( m \) is the mass of the scooter and \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

3. **Set Up the Equation for Centripetal Force**:
The centripetal force required to keep the scooter in circular motion is given by:
\[ F_{\text{centripetal}} = \frac{mv^2}{r} \]
where \( r \) is the radius of the turn.

4. **Equate the Forces**:
For the scooter to not skid, the frictional force must be equal to the centripetal force:
\[ \mu mg = \frac{mv^2}{r} \]

5. **Cancel the Mass \( m \)**:
Since \( m \) appears on both sides of the equation, we can cancel it out:
\[ \mu g = \frac{v^2}{r} \]

6. **Solve for the Coefficient of Friction \( \mu \)**:
Rearranging the equation gives:
\[ \mu = \frac{v^2}{gr} \]

7. **Substitute the Values**:
We know:
- \( v = 5 \, \text{m/s} \)
- \( g = 10 \, \text{m/s}^2 \)
- Assume a radius \( r \) (for example, \( r = 10 \, \text{m} \) for calculation purposes):
\[ \mu = \frac{(5)^2}{10 \cdot 10} = \frac{25}{100} = 0.25 \]

### Final Answer:
The minimum coefficient of friction required for the scooter to not skid while turning is \( \mu = 0.25 \).