A 60.0-kg person drops from rest a distance of 1.20 m to a platform of negligible mass supported by an ideal stiff spring of negligible mass. the platform drops 6.00 cm before the person comes to rest. what is the spring constant of the spring?

Draw a diagram to illustrate the problem s shown in the figure below.

At state A, the person has potential energy relative to the spring of

PE = (60 kg) * (9.8 m/s²) * (1.2 m)

= 705.6 J

At state B, the PE is converted to kinetic energy which is equal to the potential energy.

Assume that aerodynamic losses are ignored during the free fall.

At state C, the potential energy (or kinetic energy) is converted into strain energy in the spring. Assume that energy losses are negligible.

The spring deflects by 6 cm = 6x10⁻² m, and its spring constant = k N/m.

Therefore

(1/2) (k N/m) (6x10⁻² m) ² = 705.6 J

k = 705.6/0.03 = 23520 N/m = 23.52 kN/m

Answer: 23.52 kN/m