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A force of 750 pounds compresses a 15 -inch spring 3 inches from its relaxed position. Find the work done in compressing the spring an additional 3 inches.
Hooke's Law applies to stretched and compressed springs.

A force of $750$ pounds compresses a $15$ -inch spring $3$ inches from its relaxed position. Find the work done in compressing the spring an additional $3$ inches. $\newline$ Hooke's Law applies to stretched and compressed springs.

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One-step inequalities: word problems

Full solution

Q. A force of

750

pounds compresses a

15

-inch spring

3

inches from its relaxed position. Find the work done in compressing the spring an additional

3

inches.

\newline

Hooke's Law applies to stretched and compressed springs.

Hooke's Law Explanation: Hooke's Law states that the force needed to compress or extend a spring by some distance $x$ from its natural length is directly proportional to $x$ . The work done on the spring is the integral of this force over the distance compressed.
Find Spring Constant: First, we need to find the spring constant $k$ . We know that a force of $750$ pounds compresses the spring $3$ inches from its relaxed position. So, $F = kx$ , where $F$ is the force and $x$ is the compression distance. $\newline$ $750 = k \times 3$ $\newline$ $k = \frac{750}{3}$ $\newline$ $k = 250$ pounds per inch.
Calculate Work Done: Now, we want to find the work done in compressing the spring an additional $3$ inches. The work done on a spring is given by the formula $W = \frac{1}{2} \cdot k \cdot x^2$ , where $x$ is the total distance compressed. $\newline$ We already compressed the spring $3$ inches, and we're compressing it an additional $3$ inches, so the total compression is $3 + 3 = 6$ inches.
Plug Values and Calculate: Plug the values into the work formula: $\newline$ $W = \frac{1}{2} \cdot k \cdot x^2$ $\newline$ $W = \frac{1}{2} \cdot 250 \cdot 6^2$ $\newline$ $W = \frac{1}{2} \cdot 250 \cdot 36$ $\newline$ $W = 125 \cdot 36$ $\newline$ $W = 4500$ pound-inches.

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