#### Answer

(a) $k = 21,560~N/m$
(b) $v = 18.6~m/s$

#### Work Step by Step

(a) We can find the minimum spring constant required to send the car to the top of the 10 meter hill:
$U_e = \Delta U_g$
$\frac{1}{2}kx^2 = mgh$
$k = \frac{2mgh}{x^2}$
$k = \frac{(2)(400~kg)(9.8~m/s^2)(10~m)}{(2.0~m)^2}$
$k = 19,600~N/m$
Since the actual spring constant needs to be 10% larger, the spring constant should be $~~k = 21,560~N/m$
(b) The maximum speed will occur at the track's lowest point which is 5 meters below the starting point.
We can use conservation of energy to find the maximum speed of a 350-kg car:
$U_{g2}+K_2 = U_{g1}+U_e$
$K_2 = U_{g1}-U_{g2}+U_e$
$\frac{1}{2}mv^2 = mgh_1-mgh_2+\frac{1}{2}kx^2$
$v^2 = 2g~(h_1-h_2)+\frac{kx^2}{m}$
$v = \sqrt{2g~(h_1-h_2)+\frac{kx^2}{m}}$
$v = \sqrt{(2)(9.8~m/s^2)~(5~m-0)+\frac{(21,560~N/m)(2.0~m)^2}{350~kg}}$
$v = 18.6~m/s$