A 2000 kg car has to go over a turn whose radius is `750 m` and the abgle of slopw is `5^(@)`. The coefficient of friction between the wheels and the road is 0.5. What should be the maximum speed of the car so that it may go over the turn without slipping ?

Answer

To find the maximum speed of the car so that it may go over the turn without slipping, we can use the following steps:

### Step 1: Identify the given values
- Mass of the car (m) = 2000 kg (not directly needed for this calculation)
- Radius of the turn (R) = 750 m
- Angle of slope (θ) = 5 degrees
- Coefficient of friction (μ) = 0.5
- Acceleration due to gravity (g) = 9.8 m/s²

### Step 2: Calculate the forces acting on the car
When the car is going around the turn, the forces acting on it include:
- The gravitational force acting downwards (mg).
- The normal force (N) acting perpendicular to the road surface.
- The frictional force (f) that provides the centripetal force required for circular motion.

### Step 3: Resolve the forces
The normal force can be calculated using the angle of slope:
- The component of gravitational force acting perpendicular to the slope: \( N = mg \cos(θ) \)
- The component of gravitational force acting parallel to the slope: \( mg \sin(θ) \)

### Step 4: Calculate the maximum frictional force
The maximum frictional force can be calculated as:
- \( f_{max} = μN = μ(mg \cos(θ)) \)

### Step 5: Set up the equation for centripetal force
The centripetal force required to keep the car moving in a circle is given by:
- \( f_{centripetal} = \frac{mv^2}{R} \)

For the car to go around the turn without slipping, the maximum frictional force must equal the centripetal force:
- \( f_{max} = \frac{mv^2}{R} \)

### Step 6: Substitute the values into the equation
Substituting the expression for frictional force into the centripetal force equation:
- \( μ(mg \cos(θ)) = \frac{mv^2}{R} \)

### Step 7: Simplify the equation
Canceling the mass (m) from both sides:
- \( μg \cos(θ) = \frac{v^2}{R} \)

Rearranging gives:
- \( v^2 = μg \cos(θ) R \)

### Step 8: Calculate the maximum speed
Now substituting the known values into the equation:
- \( v^2 = 0.5 \times 9.8 \times \cos(5°) \times 750 \)

Calculating \( \cos(5°) \):
- \( \cos(5°) \approx 0.9962 \)

Now substituting this value:
- \( v^2 = 0.5 \times 9.8 \times 0.9962 \times 750 \)
- \( v^2 = 0.5 \times 9.8 \times 0.9962 \times 750 \approx 3656.85 \)

Taking the square root to find v:
- \( v \approx \sqrt{3656.85} \approx 60.5 \, \text{m/s} \)

### Final Answer
The maximum speed of the car so that it may go over the turn without slipping is approximately **60.5 m/s**.

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