Answer:
Approximately [tex]1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}[/tex], assuming friction between the vehicle and the ground is negligible.
Explanation:
Let [tex]m[/tex] denote the mass of the vehicle. Let [tex]v[/tex] denote the initial velocity of the vehicle. Let [tex]k[/tex] denote the spring constant (needs to be found.) Let [tex]x[/tex] denote the maximum displacement of the spring.
Convert velocity of the vehicle to standard units (meters per second):
[tex]\begin{aligned}v &= 20.8\; {\rm km \cdot h^{-1}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \\ &\approx 1.908\; {\rm m \cdot s^{-1}}\end{aligned}[/tex].
Initial kinetic energy ([tex]{\rm KE}[/tex]) of the vehicle:
[tex]\begin{aligned}\frac{1}{2}\, m \, v^{2}\end{aligned}[/tex].
When the vehicle is brought to a rest, the elastic potential energy ([tex]\text{EPE}[/tex]) stored in the spring would be:
[tex]\displaystyle \frac{1}{2}\, k\, x^{2}[/tex].
By the conservation of energy, if the friction between the vehicle and the ground is negligible, the initial [tex]\text{KE}[/tex] of the vehicle should be equal to the [tex]{\rm EPE}[/tex] of the vehicle. In other words:
[tex]\begin{aligned}\frac{1}{2}\, m \, v^{2} &= \frac{1}{2}\, k\, x^{2}\end{aligned}[/tex].
Rearrange this equation to find an expression for [tex]k[/tex], the spring constant:
[tex]\begin{aligned}k &= \frac{m\, v }{x^{2}}\end{aligned}[/tex].
Substitute in the given values [tex]m = 1508\; {\rm kg}[/tex], [tex]v \approx 1.908\; {\rm m\cdot s^{-1}}[/tex], and [tex]x = 6.87\; {\rm m}[/tex]:
[tex]\begin{aligned}k &= \frac{m\, v }{x^{2}} \\ &\approx \frac{1508\; {\rm kg} \times 1.908\; {\rm m\cdot s^{-1}}}{(6.87\; {\rm m})^{2}} \\ &\approx 1.79 \times 10^{5}\; {\rm kg \cdot m \cdot s^{-2} \cdot m^{-3}}\\ &\approx 1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}\end{aligned}[/tex]
