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


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:

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

Answer and Explanation:

Become a Study.com member to unlock this answer! Create your account

View this answer

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...

See full answer below.

Learn more about this topic:

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


Chapter 4 / Lesson 19

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.

Related to this Question

Explore our homework questions and answers library