Chapter 5 - Dynamics of Uniform Circular Motion - Problems - Page 140: 55

$v=14m/s$

At the bottom of the roller coaster, $F_N$ and $mg$ oppose each other. Since $F_c$ is the net force of these two forces, we have $$F_c=F_N-mg$$ From the given information, $F_N=2mg$. So, $$F_c=2mg-mg=mg$$ Now we expand $F_c$: $$\frac{mv^2}{r}=mg$$ $$v^2=\frac{mgr}{m}=gr$$ $$v=\sqrt{gr}$$ We know $g=9.8m/s^2$ and $r=20m$ $$v=14m/s$$