91Ó°ÊÓ

The spring constant, denoted as \( k \), is a key parameter in Hooke's Law. It represents the stiffness of a spring. A higher spring constant signifies a stiffer spring, which requires more force to stretch or compress it by a given amount. Hooke's Law can be expressed by the equation \( F = kx \), where:

• \( F \) is the force applied to the spring.
• \( x \) is the distance the spring is stretched or compressed from its normal length.
• \( k \) is the spring constant.

In the given exercise, a force of 6 pounds stretches the spring by 0.5 feet. By rearranging the formula to solve for \( k \), we use the equation \( k = \frac{F}{x} = \frac{6}{0.5} \). Calculating this gives us a spring constant \( k \) of 12 pounds per foot.
This means it takes 12 pounds of force to stretch the spring 1 foot beyond its natural length.

Work done on a spring refers to the energy required to stretch or compress the spring by a certain distance. In terms of physics, work is calculated as the integral of force over displacement. For a spring, the equation for work \( W \) is expressed as:

• \( W = \int_0^x kx \, dx \)

In this formula:

• \( W \) is the work done.
• \( k \) is the spring constant.
• \( x \) is the distance.

The work required to stretch the spring 0.5 feet using our spring constant \( k = 12 \) pounds per foot is calculated by integrating \(12x\) with respect to \(x\) from 0 to 0.5.
The computation involves evaluating the integral \( \int_0^{0.5} 12x \, dx \). Performing this definite integration results in 1.5 foot-pounds of work.

Understanding the definite integral is crucial when calculating work done, as it's used to evaluate the accumulated quantity over an interval. A definite integral, such as \( \int_a^b f(x) \, dx \), represents the area under the curve of function \( f(x) \) from \( x = a \) to \( x = b \).
For our spring problem, we use the definite integral to integrate the expression for work \( 12x \) from 0 to 0.5. The integration process involves finding the antiderivative of the function, which in this case is \( 6x^2 \). We then evaluate this antiderivative at the upper and lower limits of the interval:

• \( W = \left[ 6x^2 \right]_0^{0.5} = 6(0.5)^2 - 6(0)^2 \)

Solving this gives us the work done, which is 1.5 foot-pounds. Thus, the definite integral helps calculate the total work needed to stretch the spring by half a foot.