Final answer:
The work required to stretch a spring with a 31 N/m spring constant an additional 0.12 m from a starting stretch of 0.18 m is approximately 0.892 Joules.
Step-by-step explanation:
To calculate the work required to stretch a spring an additional 0.12 m after it has already been stretched by 0.18 m, we can use Hooke's Law and the work-energy principle. The work done on a spring is given by the equation W = ½ k (x_f^2 - x_i^2), where k is the spring constant and x_f and x_i are the final and initial stretches from the equilibrium position, respectively.
In this scenario, the spring constant k is given as 31 N/m. The initial stretch x_i is 0.18 m, and the spring will be stretched an additional 0.12 m to a final stretch x_f of 0.18 m + 0.12 m = 0.30 m.
Using these values, we can calculate the work:
W = ½ k (x_f^2 - x_i^2) = ½ (31 N/m) ((0.30 m)^2 - (0.18 m)^2).
W = ½ (31 N/m) (0.09 m^2 - 0.0324 m^2)
W = ½ (31 N/m) (0.0576 m^2)
W = 0.8916 J
Therefore, the amount of work required to stretch the spring an additional 0.12 m is approximately 0.892 Joules.