Taking the gravitational constant to be 32 ft/sec${}^2$, the mass of the car is

m = $\frac{w}{g}$ = $\frac{3200}{32}$ = 100 slugs$$

The curvature of the track is the reciprocal of its radius, so

$\mathrm{\kappa }\=\frac{1}{r}$ = $\frac{1}{200}$ 

The speed $v$ is constant, so $\stackrel{\.}{v}\=0$.  Hence, the acceleration becomes

 a = 0 T + $\frac{1}{200}$ $v^2$ N$$

The acceleration is directed along N, which itself points towards the center of the circular track.  The force of friction also points towards the center of the track, and so must be along N.  Hence, F = 450 N.  But F = $m$ a, and a = $\frac{v^2}{200}$ N, so

450 N = 100 $\left[\frac{v^2}{200}\right]$ N

is to be solved for $v$. Thus, write $450\=v^2\/2$, so $v\=\pm \sqrt{900}\=\pm 30$. The positive answer, 30 feet per second, is the appropriate answer.  Note that the units in the example were fixed as soon as the gravitational constant $g$ was taken as 32 ft/sec${}^2$.