Final answer:
The work required to compress a non-linear spring by an additional 10 cm after a 20 cm compression is relative to the square of the displacement from the equilibrium. It implies calculating the difference between the potential energy stored in the spring at 20 cm and at 30 cm.
Step-by-step explanation:
The question is asking how much work is required to compress a spring by an additional 10 cm after it has already been compressed by 20 cm by a 6 kg block. The spring is evidently non-linear since it is not following Hooke's Law; otherwise, the information would be insufficient to pinpoint the additional work required without the spring constant value. However, the relationships provided suggest that the work required to compress a spring relates to the square of the displacement from the equilibrium position. Given that the work done for an additional displacement follows the relationship W(6 cm to 12 cm) = 3W(0 cm to 6 cm), where W represents the work done, we can use this pattern to deduce the relative work done for additional compression of the spring.
Similarly, if we were to compress the spring by 10 cm more from 20 cm to 30 cm, the work required follows a similar pattern, where this additional work is based on the difference in the squares of the final and initial compression amounts. Thus, it involves calculating the quantities for 1/2 kx2 at the initial compression of 20 cm and then at the new compression of 30 cm, and taking the difference between these quantities.