#### Answer

(a) The speed at the top of the loop is $19.8~m/s$
(b) At the top of the loop, the force exerted on the car by the track is $29,050~N$
(c) The minimum starting height is 25.0 meters above the bottom of the loop.

#### Work Step by Step

(a) We can use conservation of energy to find the speed at the top of the loop:
$\frac{1}{2}mv_2^2+mgh_2 = mgh_1$
$v_2^2 = 2g~(h_1-h_2)$
$v_2 = \sqrt{2g~(h_1-h_2)}$
$v_2 = \sqrt{(2)(9.80~m/s^2)~(40.0~m-20.0~m)}$
$v_2 = 19.8~m/s$
The speed at the top of the loop is $19.8~m/s$
(b) The sum of the normal force from the track and the gravitational force provide the centripetal force to keep the roller coaster moving in a circle:
$F_N+mg = \frac{mv^2}{r}$
$F_N = \frac{mv^2}{r}-mg$
$F_N = \frac{(988~kg)(19.8~m/s)^2}{10.0~m}-(988~kg)(9.80~m/s^2)$
$F_N = 29,050~N$
At the top of the loop, the force exerted on the car by the track is $29,050~N$
(c) To find the minimum height where the car does not lose contact with the track, we can assume that the normal force is zero and the gravitational force provides the centripetal force to keep the car moving in a circle. We can find the speed at the top of the loop:
$\frac{mv^2}{r} = mg$
$v = \sqrt{gr}$
We can use conservation of energy to find the minimum starting height $h_1$:
$mgh_1 = mgh_2+\frac{1}{2}mv^2$
$h_1 = h_2+\frac{v^2}{2g}$
$h_1 = 2r+\frac{(\sqrt{gr})^2}{2g}$
$h_1 = 2r+\frac{gr}{2g}$
$h_1 = 2r+\frac{r}{2}$
$h_1 = \frac{5r}{2}$
$h_1 = \frac{(5)(10.0~m)}{2}$
$h_1 = 25.0~m$
The minimum starting height is 25.0 meters above the bottom of the loop.