Answer
$k = 1.4 \times 10^5 N/m$.
Work Step by Step
The car’s mechanical energy is conserved. The initial KE + initial PE equals the final KE + final PE. The car comes to a momentary stop at the end, so all of the initial kinetic energy of the car becomes potential energy of the compressed spring.
$$\frac{1}{2}mv_i^2 + 0 = 0 +\frac{1}{2}kx^2 $$
$$\frac{1}{2}(1200 kg)(23.6m/s)^2 + 0 = 0 +\frac{1}{2}k(2.2m)^2 $$
Solve for $k = 1.4 \times 10^5 N/m$.
