Final answer:
The potential energy stored in a spring stretched by 15 cm with a force of 72 N is 5.4 Joules, calculated using the formula PE = 1/2 k x² and first determining the spring constant with k = F / x.
Step-by-step explanation:
To calculate the potential energy stored in a spring using Hooke's Law, we use the formula PE = 1/2 k x², where PE is the potential energy, k is the spring constant, and x is the displacement of the spring from its equilibrium position. However, in this question, instead of giving us the spring constant directly, we are provided with the force (72 N) required to stretch the spring by 15 cm (0.15 m). First, we'll determine the spring constant (k) using Hooke's Law: F = k x, which rearranges to k = F / x. Therefore, the spring constant k = 72 N / 0.15 m = 480 N/m.
Now, substituting k into the potential energy formula, we get: PE = 1/2 (480 N/m) (0.15 m)². Performing the calculation, the stored energy in the spring is: PE = 1/2 x 480 x 0.0225 = 5.4 Joules.