First, convert the design speed to SI units:

$$v = 60 \frac{\text{km}}{\text{h}} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = \frac{50}{3} \text{ m/s}$$

[Q1] For a correctly banked curve (no friction needed), the horizontal component of the normal force provides the centripetal force, while the vertical component balances gravity. Vertical forces: $N\cos\theta = mg$ Horizontal forces: $N\sin\theta = \frac{mv^2}{r}$ Dividing the horizontal equation by the vertical one gives:

$$\tan\theta = \frac{v^2}{gr}$$ $$\theta = \arctan\left(\frac{v^2}{gr}\right) = \arctan\left(\frac{(50/3 \text{ m/s})^2}{(9.8 \text{ m/s}^2)(150 \text{ m})}\right)$$

[Q2] For an unbanked curve, the static friction force $f_s$ must provide the entire centripetal force.

$$f_s = \frac{mv^2}{r}$$

The maximum available static friction is $f_{s, \text{max}} = \mu_s N$. On a flat road, the normal force $N = mg$. The minimum coefficient of friction required is when the needed centripetal force equals the maximum static friction.

$$\mu_s mg = \frac{mv^2}{r}$$ $$\mu_s = \frac{v^2}{gr} = \frac{(50/3 \text{ m/s})^2}{(9.8 \text{ m/s}^2)(150 \text{ m})}$$