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
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.
To determine how many inches the spring would stretch when the weight increases from 10 pounds to 15 pounds, you can use Hooke's Law, which states that the amount a spring stretches is directly proportional to the force applied.
Hooke's Law formula: F = k * x
Where:
F is the force applied (weight)
k is the spring constant
x is the displacement (stretch)
From the given information, we know that the spring stretches 0.5 inches when a 10-pound weight is applied. Now, we need to find the spring constant (k) to calculate the new displacement for a 15-pound weight.
To find the spring constant (k), we can rearrange the formula:
k = F / x
k = 10 lbs / 0.5 inches
k = 20 lbs/inch
Now, we can use the spring constant to calculate the new displacement (stretch) for a 15-pound weight:
F = 15 lbs (new weight)
k = 20 lbs/inch (spring constant)
Using the formula:
x = F / k
x = 15 lbs / 20 lbs/inch
x = 0.75 inches
Therefore, if you were to increase the weight to 15 pounds, the spring would stretch 0.75 inches.