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A spring has a natural length of 15 inches. If a force of 5 pounds is required to keep the spring compressed to a length of 12 inches, then how much work is done in compressing the spring from its natural length to a length of 9 inches? Express your answer in foot-pounds

User Brenden
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Answer:

Explanation:

To calculate the work done in compressing the spring from its natural length to a length of 9 inches, you can use Hooke's Law, which states that the force required to compress or extend a spring is directly proportional to the displacement from its natural length.

Hooke's Law can be expressed as:

F = k * x

Where:

F is the force (in pounds)

k is the spring constant (a measure of stiffness)

x is the displacement from the natural length (in inches)

First, let's find the spring constant (k) using the information provided:

When the spring is compressed to 12 inches, a force of 5 pounds is required.

F = 5 pounds

x = 12 inches

Now, solve for k:

5 = k * 12

k = 5 / 12

Now that you have the spring constant (k), you can calculate the work done to compress the spring to 9 inches from its natural length (15 inches). The displacement (x) is 15 - 9 = 6 inches.

W = (1/2) * k * x^2

W = (1/2) * (5/12) * (6^2)

W = (1/2) * (5/12) * 36

W = (5/24) * 36

W = (5/24) * 36

W = 5 * 3

W = 15 foot-pounds

So, the work done in compressing the spring from its natural length to a length of 9 inches is 15 foot-pounds.

User Denisha
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