Final answer:
The spring constant of a spring stretched 6 inches by an 8lb mass is approximately 233.47 N/m. You find this by converting the mass to a force with F = mg and then using Hooke's Law (F = kx) to solve for the spring constant (k = F/x).
Step-by-step explanation:
To calculate the spring constant of a spring, you can use Hooke's Law, which states that the force (F) needed to stretch or compress a spring by a certain distance (x) is proportional to that distance. The law is usually expressed as F = kx, where k is the spring constant (measured in Newtons per meter, N/m) and x is the distance in meters. To find the spring constant, we rearrange the formula to k = F/x.
First, we need to convert the mass to a force, which is done using the formula F = mg, where m is the mass (in kilograms), and g is the acceleration due to gravity (9.8 m/s^2 on Earth). A mass that weighs 8lb is approximately 3.63 kg (since 1 pound is roughly 0.453592 kg).
Now calculate the force as F = 3.63 kg × 9.8 m/s^2 = 35.574 N. The distance the spring is stretched is given in inches, so we need to convert 6 inches to meters, which is about 0.1524 m (since 1 inch is 0.0254 meters).
Calculating the spring constant, we get k = 35.574 N / 0.1524 m ≈ 233.47 N/m. Therefore, the spring constant of the spring is approximately 233.47 N/m.