Answer
$0.2\%$
Work Step by Step
We know, Energy in Simple Harmonic Motion $=\frac{1}{2}KA^{2}$ Where, K - spring constant of the shock absorber, A - Amplitude of the shock absorber.
$E_{SHM}=\frac{1}{2}KA^{2}-(1)$
We know, frequency $(f)=\frac{1}{2\pi}\sqrt {\frac{K}{m}}$ in S.H.M. where, K - Spring constant, m - mass on the shock absorber.
Let's plug known values into this equation.
$1\space s^{-1}=\frac{1}{2\pi}\sqrt {\frac{K}{500\space kg}}=\gt1\space s^{-2}=\frac{1}{4}\times\frac{K}{500\space kg}$
$2000\space \pi^{2}\space N/m=K$
$(1)=\gt$
$E_{SHM}=\frac{1}{2}\times2000\space \pi^{2}\space N/m\times(0.2\space m)^{2}=394.78\space J$
Total energy in all shock absorbers $(E_{T})=E_{SHM}\times4$
$E_{T}=394.78\space J\times 4=1580\space J$
Let's find the Kinetic energy in the car,
$E_{K}=\frac{1}{2}mV^{2}=\frac{1}{2}\times2000\space kg\times[(\frac{100\space km}{h})(\frac{1000\space m}{km})(\frac{h}{3600\space s})]^{2}$
$E_{K}=771.6\space kJ$
Energy percentage $=\frac{E_{T}}{E_{K}}\times100\%=\frac{1580\space J}{771.6\times10^{3}\space J}\times100\%=0.2\%$