Ten and one-half foot-pounds of work is required to compress a spring 2 inches from its natural...

Question:

Ten and one-half foot-pounds of work is required to compress a spring {eq}2 {/eq} Inches from its natural length. Find the work required to compress the spring an additional one-half Inch. (Round the answer to two decimal places.)

{eq}\rule{20mm}{.5pt} {/eq}ft-lb

Hooke's Law And Work Done To Compress The Spring:


Hooke's law is defined as within the elastic limit the restoring force is directly proportional to the displacement(x) by stretch spring.

{eq}F = kx {/eq}

Formula used for finding work done by spring is

{eq}\displaystyle W = \int kxdx {/eq}

Work done measured in Joules in SI system.

Answer and Explanation:

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To find work done to compress the spring

Change work done from foot-pounds into inch-lbs

{eq}W = 10.5 \ \ ft-lbs \ \implies 10.5(12) = 126...

See full answer below.


Learn more about this topic:

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Hooke's Law & the Spring Constant: Definition & Equation

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Chapter 4 / Lesson 19
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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.


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