A car travels at a steady 40.0 m/s around a horizontal curve of radius 200 m.What is the minimum coefficient of static friction between the road and the car's tires that will allow the car to travel at this speed without sliding?
A) 1.23
B) 0.816
C) 0.736
D) 0.952
E) 0.662

Question

Quizplus · Accepted Answer

The minimum coefficient of static friction ($μ_s$) required to prevent the car from sliding can be found using the formula for centripetal force ($F_c$) required to keep an object moving in a circle of radius $r$ at a constant speed $v$, which is $F_c = \frac{mv^2}{r}$, where $m$ is the mass of the object. The frictional force ($F_f$) that provides this centripetal force is given by $F_f = μ_sN$, where $N$ is the normal force. In this case, since the car is on a horizontal surface, $N = mg$, where $g$ is the acceleration due to gravity (approximately $9.8 m/s^2$). Setting $F_c = F_f$, we get $\frac{mv^2}{r} = μ_smg$. The mass ($m$) cancels out, and solving for $μ_s$ gives $μ_s = \frac{v^2}{rg}$. Substituting the given values ($v = 40.0 m/s$, $r = 200 m$, and $g = 9.8 m/s^2$) into the equation yields $μ_s = \frac{(40.0)^2}{200 \times 9.8} = \frac{1600}{1960} = 0.816$, which matches choice B.

A Car Travels at a Steady 40