91Ó°ÊÓ

$\mathrm{Driver}\text{'}\mathrm{s}\mathrm{mass}=70.0\mathrm{kg}$

Use the concept of uniform circular motion. Uniform circular motion is a motion of a particle moving at a constant speed on a circle.

Formulae are as follow:

${\mathrm{F}}_{\mathrm{N}}=\mathrm{m}(\mathrm{g}-{\mathrm{v}}^2/\mathrm{R})$

where, v is the velocity, g is an acceleration due to gravity, m is mass, ${}^{{\mathrm{F}}_{\mathrm{N}}}$is force and R is the radius.

At the top of the hill, normal force is directed upwards, gravitational force and the centripetal acceleration directed downward. Then by using the Newton’s 2nd law of motion,

${\mathrm{F}}_{\mathrm{N}}=\mathrm{m}(\mathrm{g}-{\mathrm{v}}^2/\mathrm{R})$

Since, ${\mathrm{F}}_{\mathrm{N}}=0$, as stated in the problem, then${\mathrm{v}}^2=\mathrm{gR}$. Later, at the bottom of the valley, reverse both the normal force direction and the acceleration direction.

Thus,

$$\begin{gathered} {\mathrm{F}}_{\mathrm{N}}=\mathrm{m}(\mathrm{g}+{\mathrm{v}}^2/\mathrm{R}) \\ =2\mathrm{mg} \\ =1372\mathrm{N} \\ ≈1.37×{10}^3\mathrm{N} \end{gathered}$$

Therefore, the magnitude of the normal force on the driver from the seat when the car passes through the bottom of the valley is$1.37×{10}^3\mathrm{N}$