Final answer:
The spring constant for a spring stretched 2 feet by a 14-pound mass is approximately 102.13 N/m, calculated using Hooke's Law after converting the given units to Newtons and meters.
Step-by-step explanation:
To find the spring constant of the spring, we can use Hooke's Law, which states that the force (F) exerted by a spring is directly proportional to the displacement (x) from its equilibrium position, according to the equation F = kx, where k is the spring constant. In this case, we have a mass of 14 pounds stretching the spring 2 feet. First, we need to convert the weight into a force in Newtons (N), knowing that 1 pound is equivalent to 4.44822 N, and the distance into meters (m), knowing that 1 foot is 0.3048 m.
The force in Newtons is 14 pounds × 4.44822 N/pound = 62.27508 N. The displacement is 2 feet × 0.3048 m/foot = 0.6096 m. Using Hooke's Law, we divide the force by the displacement to find the spring constant:
k = F / x = 62.27508 N / 0.6096 m = 102.13 N/m.
So, the spring constant for this spring is approximately 102.13 N/m.