The length a spring is stretched from its natural length with work, W foot- pounds, is given by $\mathrm{L}=\sqrt{\frac{2 \mathrm{~W}}{\mathrm{k}}}$
Where $\mathrm{k}$ is a constant for the given spring. If a certain spring has a constant of 65.7, and the spring is to be stretched 2.6 feet from its natural length, how much work will be necessary? Round to the nearest tenth of a foot- pound.
A) 222.1 foot- pounds
B) 53 foot- pounds
C) 444.1 foot- pounds
D) 85.4 foot- pounds

Question

Quizplus · Accepted Answer

To find the work necessary, we can rearrange the given formula to solve for $W$: $W = \frac{kL^2}{2}$. Substituting $k = 65.7$ and $L = 2.6$ feet into the formula gives $W = \frac{65.7 \times (2.6)^2}{2}$, which calculates to approximately 222.1 foot-pounds.

The Length a Spring Is Stretched from Its Natural Length L=2 Wk\mathrm{L}=\sqrt{\frac{2 \mathrm{~W}}{\mathrm{k}}}L=k2 W​​

The Length a Spring Is Stretched from Its Natural Length $\mathrm{L}=\sqrt{\frac{2 \mathrm{~W}}{\mathrm{k}}}$