A automobile moving at 90.0 km/hr travels around a level curve of radius 100.0 m. The minimum possible coefficient of friction required for the auto not to slip sideways is
A) 0.
B) 0.02.
C) 0.64.
D) 0.83.

Question

Quizplus · Accepted Answer

The minimum coefficient of friction required to prevent slipping is calculated using the formula for centripetal force, $F = \frac{mv^2}{r}$, where $F$ is the force required to keep the car moving in a circle, $m$ is the mass of the car, $v$ is the velocity, and $r$ is the radius of the curve. The force of friction, which prevents the car from slipping, is given by $F_{friction} = \mu mg$, where $\mu$ is the coefficient of friction, and $g$ is the acceleration due to gravity. Setting the centripetal force equal to the force of friction and solving for $\mu$ gives $\mu = \frac{v^2}{rg}$. Substituting $v = 90.0 \, \text{km/hr} = 25.0 \, \text{m/s}$, $r = 100.0 \, \text{m}$, and $g = 9.8 \, \text{m/s}^2$, we find $\mu = \frac{(25.0)^2}{100.0 \times 9.8} = 0.64$.

A Automobile Moving at 90