physics
posted by flame on .
I tried doing this problem but couldn't get the right answer. I don't get how the answer is 2.19cm. Is the distance referring to the vertical or horizontal distance
Four people, each with a mass of 73.5 kg, are in a car with a mass of 1190 kg. An earthquake strikes. The driver manages to pull of the road and stop, as the vertical oscillations of the ground surface make the car bounce up and down on its suspension springs. When the frequency of the shaking is 1.50 Hz, the car exhibits a maximum amplitude of vibration. The earthquake ends and the four people leave the car as fast as they can. By what distance does the car's undamaged suspension lift the car's body as the people get out?

Wo = sqrt(k/m)
2 pi * 1.5 = Wo = 9.42 radians/s
so
9.42 = sqrt (k/(1190+4*73.5) )
88.8 = k/1484
k = 131,780 Newtons/meter
newtons lost = 4 * 73.5 * 9.81 = weight (not mass) of people
newtons lost = 2884 Newtons
elongation = F/k = 2884/131,780 = .022 meters = 2.2 centimeters, about an inch 
why are we using the newton lost but not the newton remaining