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 navigate a horizontal turn without skidding, 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 km/h. To convert this to meters per second (m/s), we use the conversion factor:
\[ 1 \text{ km/h} = \frac{1}{3.6} \text{ m/s} \]
Therefore,
\[ v = 18 \text{ km/h} \times \frac{1}{3.6} = 5 \text{ m/s} \]

2. **Identify the Radius of the Turn**:
The radius of the turn is given as \( r = 10 \) meters.

3. **Use the Formula for Circular Motion**:
The formula relating the velocity, friction coefficient (\( \mu \)), radius (\( r \)), and gravitational acceleration (\( g \)) is:
\[ v = \sqrt{\mu \cdot r \cdot g} \]
Here, \( g \) (acceleration due to gravity) is approximately \( 10 \, \text{m/s}^2 \).

4. **Rearrange the Formula to Solve for \( \mu \)**:
Squaring both sides of the equation gives:
\[ v^2 = \mu \cdot r \cdot g \]
Rearranging for \( \mu \) gives:
\[ \mu = \frac{v^2}{r \cdot g} \]

5. **Substitute the Known Values**:
Now, substituting the values we have:
- \( v = 5 \, \text{m/s} \)
- \( r = 10 \, \text{m} \)
- \( g = 10 \, \text{m/s}^2 \)

We get:
\[ \mu = \frac{5^2}{10 \cdot 10} \]
\[ \mu = \frac{25}{100} = 0.25 \]

6. **Conclusion**:
The minimum friction coefficient required for the scooter to navigate the turn without skidding is \( \mu = 0.25 \).