Answer
$2.38%$
Work Step by Step
Special Theory of Relativity
$\gamma=\frac{1}{\sqrt{1 - \frac{(0.15c)^2}{c^2}}}=1.011$
$E_k=(\gamma-1)mc^2=(1.011-1)(17,000kg)c^2=1.68\times10^{19}J$
Classical Formulas:
$E_k=\frac{mv^2}{2}$
$E_k=\frac{(17,000kg)(0.15c)^2}{2}=1.72\times10^{19}J$
Percent Error
$E_k=\frac{E_{kc}-E_{ks}}{E_{ks}}=\frac{1.72\times10^{19}J-1.68\times10^{19}J}{1.68\times10^{19}J}\times100\%=2.38\%$
