If a force of 22 N stretches a spring 0.2 m beyond its natural length, how much work does...


If a force of {eq}22\, N {/eq} stretches a spring {eq}0.2\, m {/eq} beyond its natural length, how much work does it take to stretch the spring {eq}5.5\, m {/eq} beyond its natural length?

Hooke's Law:

Recall that Hooke's law tells us that the force required to move stretch or compress a spring {eq}x {/eq} units beyond its natural length is proportional to the length of the stretch or compression:

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

Recall also that work is the amount of energy moved to displace an object, and is given by

{eq}\begin{align*} W &= \int F\ dx \end{align*} {/eq}

So for springs, this looks like

{eq}\begin{align*} W &= \int kx\ dx \\ &= \frac12 k x^2 \end{align*} {/eq}

when we stretch or compress from its natural length.

Answer and Explanation:

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We know {eq}F = 22 {/eq} N when {eq}x = 0.2 {/eq} m, so the spring constant is

{eq}\begin{align*} F &= kx \\ 22 &= k(0.2) \\ k &=...

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.

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