# A force of 8 lb is required to hold a spring stretched 8 inches beyond its natural length. How...

## Question:

A force of 8 lb is required to hold a spring stretched 8 inches beyond its natural length. How much work is done in stretching it from its natural length to 10 inches beyond its natural length?

## Work:

Work is the amount of energy it takes to move an object from one place to another. We compute it using a line integral:

{eq}W = \int_C \vec F \cdot d\vec r {/eq}

For linear displacement, this simplifies to

{eq}\begin{align*} W = \int_a^b F\ dx \end{align*} {/eq}

Recall Hooke's Law which says that for springs, the force is proportional to the displacement, i.e.

{eq}\begin{align*} F = kx \end{align*} {/eq}

## Answer and Explanation: 1

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We know that {eq}F = 8 {/eq} lbs when {eq}x=8 {/eq} in. Inches are not very standard, so we'll use the equivalent {eq}x = \frac23 {/eq} ft instead....

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#### Learn more about this topic: Hooke's Law & the Spring Constant: Definition & Equation

from

Chapter 4 / Lesson 19
202K

After watching this video, you will be able to explain what Hooke's Law is and use the equation for Hooke's Law to solve problems. A short quiz will follow.