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(A) The initial kinetic energy of the system is $K_i = \frac{1}{2}mv^2$,where $m = 40,000\, 	ext{kg}$ and $v = 72\, 	ext{km/h} = 20\, 	ext{m/s}$.$K

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$16$

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(A) The initial kinetic energy of the system is $K_i = \frac{1}{2}mv^2$,where $m = 40,000\, 	ext{kg}$ and $v = 72\, 	ext{km/h} = 20\, 	ext{m/s}$.$K_i = \frac{1}{2} 	imes 40,000 	imes (20)^2 = 80,00,000\, 	ext{J} = 80 	imes 10^5\, 	ext{J}$.According to the work-energy theorem,the total work done is equal to the change in kinetic energy: $W_{	ext{friction}} + W_{	ext{spring}} = K_f - K_i$.Given that $90\%$ of the energy is lost due to friction,$W_{	ext{friction}} = -0.9 K_i$.Since the system comes to rest,$K_f = 0$. Thus,$-0.9 K_i + W_{	ext{spring}} = -K_i$.$W_{	ext{spring}} = -0.1 K_i$.The work done by the spring is $W_{	ext{spring}} = -\frac{1}{2}kx^2$,where $x = 1.0\, 	ext{m}$.$-\frac{1}{2}k(1)^2 = -0.1 	imes (80 	imes 10^5)$.$\frac{1}{2}k = 8 	imes 10^5$.$k = 16 	imes 10^5\, 	ext{N/m}$.Therefore,the spring constant is $16 	imes 10^5\, 	ext{N/m}$.

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