Now, suppose that the curve is level (θ=0θ=0) and that the ice has melted, so that there is a coefficient of static friction μμ between the road and the car's tires as shown in (Figure 2). What is μminμmin, the minimum value of the coefficient of static friction between the tires and the road required to prevent the car from slipping? Assume that the car's speed is still 50.0 km/hourkm/hour and that the radius of the curve is 54.1 mm .

To solve for the minimum coefficient of static friction (\(\mu_{\text{min}}\)) that will prevent the car from slipping on a flat curve, we can use the following principles of circular motion and friction:

### Step 1: Analyze the forces involved.
- **Centripetal Force**: The car must experience a centripetal force to move in a circular path. The frictional force between the tires and the road provides this centripetal force.
- **Frictional Force**: This is given by \( F_{\text{friction}} = \mu \cdot N \), where:
  - \(\mu\) is the coefficient of static friction.
  - \(N\) is the normal force, which equals the weight of the car (\(N = mg\)) since the road is level.
  
  The centripetal force \( F_{\text{centripetal}} \) required for circular motion is given by:
  \[
  F_{\text{centripetal}} = \frac{mv^2}{r}
  \]
  where:
  - \(m\) is the mass of the car,
  - \(v\) is the speed of the car, and
  - \(r\) is the radius of the curve.

Since the frictional force provides the necessary centripetal force, we have:
\[
\mu \cdot N = \frac{mv^2}{r}
\]
Since \(N = mg\), this simplifies to:
\[
\mu mg = \frac{mv^2}{r}
\]
The mass \(m\) cancels out from both sides, leaving:
\[
\mu = \frac{v^2}{rg}
\]

### Step 2: Convert speed to meters per second.
The speed of the car is given as 50.0 km/h. To convert this to meters per second (m/s):
\[
v = 50.0 \, \text{km/h} \times \frac{1000 \, \text{m}}{1 \, \text{km}} \times \frac{1 \, \text{h}}{3600 \, \text{s}} = 13.89 \, \text{m/s}
\]

### Step 3: Plug in the values.
Now substitute the given values:
- \(v = 13.89 \, \text{m/s}\),
- \(r = 54.1 \, \text{m}\),
- \(g = 9.81 \, \text{m/s}^2\).

Using the formula for \(\mu_{\text{min}}\):
\[
\mu_{\text{min}} = \frac{(13.89)^2}{54.1 \times 9.81}
\]

Let's calculate this.The minimum coefficient of static friction (\(\mu_{\text{min}}\)) required to prevent the car from slipping is approximately \(0.364\).

Would you like more details or have any questions about the process? Here are five related questions for further exploration:

1. How does increasing the car's speed affect the required coefficient of friction?
2. What happens if the road is banked instead of level?
3. How would the problem change if the car was driving on an icy road with very low friction?
4. Can this method be used for cars of different masses?
5. What other forces might affect the car's ability to stay on the curve?

**Tip**: Always ensure to convert speeds to SI units (meters per second) when dealing with equations that use standard physical constants like \(g = 9.81 \, \text{m/s}^2\).

Now, suppose that the curve is level (θ=0) and that the ice has melted, so that there is a coefficient of static friction μ between the road and the car's tires as shown in (Figure 2). What is μmin, the minimum value of the coefficient of static friction between the tires and the road required to prevent the car from slipping? Assume that the car's speed is still 50.0 km/hour and that the radius of the curve is 54.1 m.

Calculate the minimum coefficient of static friction (μmin) required to prevent a car from slipping on a flat curve at 50 km/h with a curve radius of 54.1 meters. Explore concepts of circular motion, friction, and Newton's laws with detailed explanations.

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