A spring scale stretches 5.00 cm when a 2.00 kg fish hangs at rest from it. With g taken as 9.80 m/s squared, what is the force constant of its spring?

1. The fish hangs at rest, so the restoring force of the spring equals the weight it supports. 2. Weight w = m g = (2.00 kg)(9.80 m/s squared) = 19.6 N. 3. Convert the stretch: 5.00 cm = 5.00 x 10^-2 m. 4. k = F / x = 19.6 N / (5.00 x 10^-2 m) = 392 N/m. 5. Dividing by 5.00 cm without converting to meters gives 3.92 N/cm, which is the same thing written badly, while forgetting g altogether gives 40.0 N/m. _Source: OpenStax College Physics (CC BY 4.0), Ch 16 "Oscillatory Motion and Waves", section 16.1 Hooke's Law: Stress and Strain Revisited_