To find the maximum speed of a car on a road turn of radius 30 m with a coefficient of friction of 0.4, we can follow these steps:
### Step-by-Step Solution:
1. **Identify the Given Values:**
- Radius of the turn, \( r = 30 \, \text{m} \)
- Coefficient of friction, \( \mu = 0.4 \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)
2. **Understand the Forces Involved:**
- When a car goes around a curve, the frictional force provides the necessary centripetal force to keep the car moving in a circular path.
- The maximum frictional force can be expressed as:
\[
F_{\text{friction}} = \mu \cdot N
\]
where \( N \) is the normal force. For a car on a flat surface, the normal force \( N \) equals the weight of the car, \( mg \).
3. **Set Up the Equation for Centripetal Force:**
- The centripetal force required to keep the car moving in a circle is given by:
\[
F_{\text{centripetal}} = \frac{mv^2}{r}
\]
- Setting the frictional force equal to the centripetal force gives:
\[
\mu mg = \frac{mv^2}{r}
\]
4. **Cancel Mass \( m \) from Both Sides:**
- Since mass \( m \) appears on both sides of the equation, we can cancel it out:
\[
\mu g = \frac{v^2}{r}
\]
5. **Rearrange to Solve for Maximum Speed \( v \):**
- Rearranging the equation gives:
\[
v^2 = \mu g r
\]
- Taking the square root of both sides, we find:
\[
v = \sqrt{\mu g r}
\]
6. **Substitute the Known Values:**
- Now, substitute \( \mu = 0.4 \), \( g = 9.8 \, \text{m/s}^2 \), and \( r = 30 \, \text{m} \):
\[
v = \sqrt{0.4 \times 9.8 \times 30}
\]
7. **Calculate the Value:**
- First, calculate the product:
\[
0.4 \times 9.8 = 3.92
\]
\[
3.92 \times 30 = 117.6
\]
- Now, take the square root:
\[
v = \sqrt{117.6} \approx 10.84 \, \text{m/s}
\]
8. **Conclusion:**
- The maximum speed of the car on the road turn is approximately \( 10.84 \, \text{m/s} \).
