Answer
$0.32$
Work Step by Step
Please see the attached image first.
The road is designed without considering the friction force acting on the car. Here, considers the normal force as the only force that affects the centripetal force on the car.
So let's apply equation $F=ma$ to the car - figure (1)/ no friction
$\rightarrow F=ma$ ; Let's plug known values into this equation.
$Rcos(90-\theta)=m\frac{V^{2}}{r}=\gt Rsin\theta=m\frac{V^{2}}{r}-(1)$
$ \angle \theta\nearrow F=ma$ ; Let's plug known values into this equation.
$Rsin(90-\theta)-mg=0$
$Rcos\theta=mg-(2)$
$(1)\div (2)=\gt tan\theta=\frac{V^{2}}{rg}$ ; Let's plug known values into this equation
$tan\theta=\frac{{(\frac{45\space km}{h}\times \frac{10^{3}\space m}{km}\times\frac{h}{3600\space s})^{2}}}{95m\times9.8\space m/s^{2}}= 0.17$
$\theta=tan^{-1}0.17=9.65^{\circ}$
So, let's apply equation $F=ma$ to the car - figure (2)/ with friction
$\rightarrow F=ma$ ; Let's plug known values into this equation
$Rsin\theta+Fcos\theta = m\frac{V^{2}}{r}$
$\frac{mg}{cos\theta}sin\theta+\mu \frac{mg}{cos\theta}=m\frac{V^{2}}{R}$
$\mu=\frac{V^{2}}{rg}-tan\theta$
$\mu=\frac{(\frac{77\times1000}{3600}m/s)^{2}}{(95m\times 9.8m/s^{2})}-0.17=0.32$
