Question

the length that a hanging spring stretches varies directly with the weight placed at the end of the spring. if a weight of 8 lb stretches a certain spring 2 in., how far will the spring stretch if the weight is increased.to 42 lb? (leave the variation constant in fraction form. round off your final answer to the nearest in.)
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Explanation:

Step1: Find the variation constant \(k\)

Since the length \(y\) varies directly with the weight \(x\), the equation is \(y = kx\). Given \(x = 8\) lb and \(y=2\) in, substitute into the equation: \(2=k\times8\). Solve for \(k\): \(k=\frac{2}{8}=\frac{1}{4}\).

Step2: Find the length when \(x = 42\) lb

Use the equation \(y = kx\) with \(k=\frac{1}{4}\) and \(x = 42\). Then \(y=\frac{1}{4}\times42=\frac{42}{4}=10.5\)

Answer:

\(10.5\)