Final answer:
To store 0.143 J of potential energy, a spring with a constant of 400 N/m, calculated from its response to a 12 N force, must be stretched approximately 2.7 cm.
Step-by-step explanation:
The question pertains to the potential energy stored in a spring, which can be described by the formula U = ½ kx², where U represents the potential energy, k is the spring constant, and x is the displacement of the spring from its equilibrium position. To find the displacement required to produce a certain amount of potential energy, we can rearrange the formula to solve for x. Given that a force of 12 N stretches the spring by 3 cm (0.03 m), we can calculate the spring constant (k) using Hooke's law (F = kx). From this, k = 12 N / 0.03 m = 400 N/m. Thus, to store 0.143 J of potential energy, we use the formula U = ½ kx² to solve for x: x = √(2U/k) = √(2 × 0.143 J / 400 N/m) = √(0.000715 m²) ≈ 0.027 m or 2.7 cm. This means to store 0.143 J of potential energy, the spring would have to be stretched approximately 2.7 cm.