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A car takes a turn on a slippery ro...

A car takes a turn on a slippery road at a safe speed of 9.8 m/s .If the coefficient of friction is 0.2 the manimum radius of the are in which the car takes a trun is

A

20 m

B

49 m

C

24.5 m

D

80 m

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The correct Answer is:
To find the minimum radius of the arc in which the car can safely take a turn on a slippery road, we can follow these steps: ### Step 1: Understand the Forces Involved When a car takes a turn, it experiences a centripetal force that keeps it moving in a circular path. On a slippery road, this centripetal force is provided by the frictional force between the tires and the road. ### Step 2: Write the Equation for Centripetal Force The centripetal force \( F_c \) required to keep the car moving in a circle is given by the formula: \[ F_c = \frac{mv^2}{r} \] where: - \( m \) is the mass of the car, - \( v \) is the speed of the car, - \( r \) is the radius of the turn. ### Step 3: Determine the Frictional Force The maximum frictional force \( F_f \) that can act on the car is given by: \[ F_f = \mu N \] where: - \( \mu \) is the coefficient of friction, - \( N \) is the normal force. On a flat surface, the normal force \( N \) is equal to the weight of the car, \( mg \): \[ N = mg \] Thus, the frictional force becomes: \[ F_f = \mu mg \] ### Step 4: Set the Forces Equal For the car to safely make the turn, the frictional force must be equal to the centripetal force: \[ \mu mg = \frac{mv^2}{r} \] ### Step 5: Simplify the Equation We can cancel \( m \) from both sides (assuming \( m \neq 0 \)): \[ \mu g = \frac{v^2}{r} \] Rearranging this gives us: \[ r = \frac{v^2}{\mu g} \] ### Step 6: Substitute the Known Values Given: - \( v = 9.8 \, \text{m/s} \) - \( \mu = 0.2 \) - \( g = 9.8 \, \text{m/s}^2 \) Substituting these values into the equation: \[ r = \frac{(9.8)^2}{0.2 \times 9.8} \] ### Step 7: Calculate the Radius Calculating the right-hand side: \[ r = \frac{96.04}{1.96} = 49 \, \text{m} \] ### Conclusion The minimum radius of the arc in which the car can safely take a turn is: \[ \boxed{49 \, \text{m}} \]

To find the minimum radius of the arc in which the car can safely take a turn on a slippery road, we can follow these steps: ### Step 1: Understand the Forces Involved When a car takes a turn, it experiences a centripetal force that keeps it moving in a circular path. On a slippery road, this centripetal force is provided by the frictional force between the tires and the road. ### Step 2: Write the Equation for Centripetal Force The centripetal force \( F_c \) required to keep the car moving in a circle is given by the formula: \[ ...
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