Final answer:
Using Hooke's Law and understanding that the weight of the person acts as the force on the bungee cord, it's evident that if the cord has a stiffness of exactly 28 pounds per foot, a 220-pound person would cause it to stretch by significantly less than 37 feet.
Step-by-step explanation:
The question at hand involves a 220-pound person planning to bungee jump off a ledge, using a cord with a stiffness or spring constant no less than 28 pounds per foot. To find out how much the cord will stretch, one must understand the relationship between force, spring constant (k), and displacement (x) as described by Hooke's Law, which states F = kx. In this scenario, the weight of the person (W) is the force exerted on the bungee cord by gravity, and it equals to their mass multiplied by the acceleration due to gravity (W = mg).
Converting the person's weight to newtons (1 pound = 4.44822 newtons) and using the minimum stiffness given (k=28 pounds per foot), we can calculate the stretch (x) with the formula x = W/k. However, the actual numbers are not provided in the details. Nonetheless, one can infer that if the stiffness is exactly 28 pounds per foot, and the person weighs 220 pounds, the cord would theoretically stretch exactly 220/28 = 7.8571 feet, which is clearly less than 37 feet. Hence, the answer to the question is (b) less than 37 feet.