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
