1. A spring bearing a 10-pound weight has stretched 0.5 inches. If you were to increase the weight to 15 pounds, how many inches would the spring stretch

Answer 1

To find the answer, we can first determine the spring constant (k) using Hooke's Law. Hooke's Law states that the force applied to a spring is directly proportional to the extension or compression of the spring. Mathematically, it is represented as F = kx, where F is the force applied, k is the spring constant, and x is the displacement or stretch of the spring.

Given that a 10-pound weight has stretched the spring by 0.5 inches, we can plug this into Hooke's Law equation:

10 lbs = k * 0.5 inches

To find the spring constant (k), we divide both sides of the equation by 0.5:
k = (10 lbs) / (0.5 inches)
k = 20 lbs/inch

Now, we can use the spring constant to determine how many inches the spring would stretch when a 15-pound weight is applied:

15 lbs = k * x

Rearranging the equation to solve for x:
x = (15 lbs) / (k)
x = (15 lbs) / (20 lbs/inch)
x = 0.75 inches

Therefore, if the weight is increased to 15 pounds, the spring would stretch by 0.75 inches.