Answer
The normal force at the bottom of the valley is $~~1370~N$
Work Step by Step
At the top of the hill, since the normal force is zero, the gravitational force is equal to the centripetal force.
We can find an expression for the speed:
$F = \frac{mv^2}{r}$
$mg = \frac{mv^2}{r}$
$g = \frac{v^2}{r}$
$v^2 = g~r$
$v = \sqrt{g~r}$
We can find the normal force at the bottom of the valley:
$F_N-mg = \frac{mv^2}{r}$
$F_N = mg+\frac{m(\sqrt{g~r})^2}{r}$
$F_N = mg+mg$
$F_N = 2mg$
$F_N = (2)(70.0~kg)(9.8~m/s^2)$
$F_N = 1370~N$
The normal force at the bottom of the valley is $~~1370~N$
