A car body settles 1.50 cm when a 90.0 kg passenger climbs in. Taking g as 9.80 m/s squared, what is the force constant of the suspension?

1. While the car sits still, the restoring force from the springs equals the passenger's weight. 2. Weight w = m g = (90.0 kg)(9.80 m/s squared) = 882 N. 3. Convert the settling distance: 1.50 cm = 1.50 x 10^-2 m. 4. From k = F / x, k = 882 N / (1.50 x 10^-2 m) = 5.88 x 10^4 N/m. 5. Dividing 882 N by 1.50 without converting to meters gives 5.88 x 10^2 N/m, which is the trap in one option. 6. Multiplying weight by displacement instead of dividing gives 13.2 N m, an energy, not a force constant. _Source: OpenStax College Physics (CC BY 4.0), Ch 16 "Oscillatory Motion and Waves", section 16.1 Hooke's Law: Stress and Strain Revisited_