Answer
(a) $k = 37,000~N/m$
(b) The car is in contact with the spring for 0.50 seconds.
Work Step by Step
(a) We find the spring constant as:
$\frac{1}{2}kA^2 = \frac{1}{2}mv^2$
$k = \frac{mv^2}{A^2}$
$k = \frac{(950~kg)(25~m/s)^2}{(4.0~m)^2}$
$k = 37,000~N/m$
(b) We find the period of oscillation as:
$T = 2\pi~\sqrt{\frac{m}{k}}$
$T = 2\pi~\sqrt{\frac{950~kg}{37,000~N/m}}$
$T = 1.0~s$
Since the car is only in contact with the spring for half an oscillation, the car is in contact with the spring for 0.50 seconds.