# The force, F , required to compress a spring by distance x meters is given by F=8x newtons....

## Question:

The force, {eq}F {/eq}, required to compress a spring by distance {eq}x {/eq} meters is given by {eq}F=8x {/eq} newtons.

Which will be more work?

A. The work to compress form {eq}0 {/eq} to {eq}1 {/eq}

B.The work to compress from {eq}4 {/eq} to 5

(Explain why this is the case!)

Find the work done in compressing the spring from {eq}x=0 {/eq} to {eq}x=1 {/eq} and in compressing the spring from {eq}x=4 {/eq} to {eq}x=5 {/eq}.

Work compressing from {eq}x=0 {/eq} to {eq}x=1: \rule{20mm}{.5pt} {/eq}

(include units)

Work compressing from {eq}x=4 {/eq} to {eq}x=5: \rule{20mm}{.5pt} {/eq}

(include units)

## Work:

Work is the amout of energy used to displace something. We compute it via a line integral

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

which reduces to {eq}\begin{align*} W = \int F\ dx \end{align*} {/eq}

when motion is restricted to one dimension, as in the case of springs. Recall also that for springs, the amount of force required to compress a spring obeys Hooke's Law:

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

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Just think about this. We've all dealt with springs. And we all know that the more its already compressed the harder is to compress more. So we the...

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.