Answer
The effective spring constant of the bumper must be $k = 2.3\times 10^7~N/m$
Work Step by Step
$v = (8~km/h)(\frac{1000~m}{1~km})(\frac{1~h}{3600~s}) = 2.22~m/s$
When the spring compresses 1.5 cm, then the energy stored as elastic potential energy in the spring must equal all of the car's original kinetic energy. Therefore,
$EPE = KE$
$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$
$k = \frac{mv^2}{x^2} = \frac{(1050~kg)(2.22~m/s)^2}{(0.015~m)^2}$
$k = 2.3\times 10^7~N/m$
The effective spring constant of the bumper must be $k = 2.3\times 10^7~N/m$.
