Assume a spring obeys Hooke’s law and has length 1 meter. If a force of 100 newtons compresses it by 0.1 meters, how many Joules are required to compress it 0.5 meters? What is the length of the spring when 20 Joules of work have been expended?
Since the spring obeys Hooke’s law and we are given that 100 newtons compresses it by 0.1 meters, we have,
![\[ f(100) = 100c = 0.1 \quad \implies \quad c = 1000. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-76b95d8fab5fe229cafd8f1fa5902633_l3.png "Rendered by QuickLaTeX.com")
Then we compute how much work is required to compress the spring 0.5 meters,
![\[ W = \int_0^{.5} 1000 x \, dx = 1000 \left( \left. \frac{x^2}{2} \right|_0^{.5} \right) = 125 \text{ Joules}.\]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-7ef6e1134890a9d96ff7596b721c2ebd_l3.png "Rendered by QuickLaTeX.com")
Next, we are given an amount of work (20 Joules) and want to solve for a distance 
. From the formula for work we have,
![\[ \int_0^a 1000 x \, dx = 20 \quad \implies \quad \frac{1000a^2}{2} = 20 \quad \implies \quad a = 0.2m. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-7c19e45962852bb0b82992d1fd916a0b_l3.png "Rendered by QuickLaTeX.com")
This is the amount the spring is compressed by 20 Joules. Since it’s initial length is 1m we then have the spring’s length when compressed equal to 1m – 0.2m = 0.8m.