Answer
(a) $\mu_s=0.912$
(b) $v_{max}=37.8m/s$
Work Step by Step
(a) Static friction $f_s$ acts as the centripetal force $F_c$ to keep the car from slipping. We have $$f_s^{max}=F_c=\frac{mv^2_{max}}{r} (1)$$
$f_s^{max}$ can be found by $f_s^{max}=\mu_sF_N$. To find $F_N$, we look at the vertical forces.
There are 2 downward forces: gravity $mg$ and downforce $D$. Normal force $F_N$ is the only upward force. As there is no vertical acceleration, $F_N=mg+D=830\times9.8+11000=19134N$
So, $f_s^{max}=19134\mu_s$
Plug this back to (1): $$\mu_s=\frac{mv^2_{max}}{19134r}$$
We know the car's mass $m=830kg$, $v_{max}=58m/s$ and radius $r=160m$. Therefore, $$\mu_s=0.912$$
(b) If there is no downforce, $F_N=mg$ only and $f_s^{max}=\mu_smg$
Plug this back to (1), we can eliminate $m$ on both sides: $$\mu_sg=\frac{v^2_{max}}{r}$$ $$v_{max}=\sqrt{\mu_sgr}$$
Again, plug back $\mu_s=0.912$, $g=9.8m/s^2$ and $r=160m$: $$v_{max}=37.8m/s$$
