# A force of 25 lb. is required to compress a spring of natural length 0.80 ft. to a length of 0.75...

## Question:

A force of {eq}25\ lb {/eq}. is required to compress a spring of natural length {eq}0.80\ ft {/eq} to a length of {eq}0.75\ ft {/eq}. Find the work done in compressing the spring from its natural length to a length of {eq}0.70\ ft {/eq}.

## Work Done On A Spring

The work done to compress the spring, which depends on the spring constant {eq}k {/eq} as well as the distance of deformation {eq}x {/eq}, is described by the equation below:

$$\displaystyle W= \int(kx)\,dx\\$$

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Starting with the equation of work, we can see that the value of {eq}k {/eq} is unknown, so we will solve for that first. We will use the values...

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

from

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.