#### Answer

The speed at the top is 19.8 m/s

#### Work Step by Step

Note that at the top of the loop, the normal force from the track is directed straight down. We can use an equation for the centripetal force to find the speed.
$\sum F = \frac{mv^2}{r}$
$F_N+mg = \frac{mv^2}{r}$
$mg+mg = \frac{mv^2}{r}$
$v^2 = 2~g~r$
$v = \sqrt{2~g~r}$
$v = \sqrt{(2)(9.80~m/s^2)(20~m)}$
$v = 19.8~m/s$
The speed at the top is 19.8 m/s.