The distance a spring stretches varies directly with the force applied to it. If a 7-pound weight stretches a spring a distance of 24.5 inches, how far will the spring stretch if a 12-pound weight is applied?
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To solve this problem, we can use the concept of direct variation. Direct variation means that the two variables (distance stretched and force applied) are directly proportional to each other.
Let's denote the distance stretched by "d" and the force applied by "f". We can set up a proportion using the information given:
force applied (f) / distance stretched (d) = force applied (f) / distance stretched (d)
We are given that a 7-pound weight stretches the spring a distance of 24.5 inches. Let's plug in these values:
7 pounds / 24.5 inches = 12 pounds / x
Now, we can solve for "x", which represents the distance the spring will stretch when a 12-pound weight is applied.
To isolate "x", we can cross-multiply:
7 pounds * x = 12 pounds * 24.5 inches
7x = 294
Divide both sides of the equation by 7:
x = 294 / 7
x ≈ 42
Therefore, when a 12-pound weight is applied, the spring will stretch approximately 42 inches.
To solve this problem, we need to use the concept of direct variation, which states that when two variables are directly proportional, their ratio remains constant.
In this case, the distance a spring stretches (let's call it "d") is directly proportional to the force applied to it (let's call it "f"). Mathematically, we can express this relationship as:
d ∝ f
We can further rewrite this as:
d = kf
where k is the constant of variation.
To find the value of k, we can use the information given in the problem. It states that when a 7-pound weight is applied, the spring stretches a distance of 24.5 inches. So we can set up the equation:
24.5 = k * 7
To solve for k, divide both sides of the equation by 7:
k = 24.5 / 7
Now that we have the value of k, we can use it to find the distance the spring will stretch when a 12-pound weight is applied. Let's call this distance "d2". We can use the equation:
d2 = k * 12
Substituting the value of k:
d2 = (24.5 / 7) * 12
Calculating this expression, we find:
d2 ≈ 41.71 inches
Therefore, the spring will stretch approximately 41.71 inches when a 12-pound weight is applied.
since d = kf, d/f = k and is constant. So, you want d such that
d/12 = 24.5/7