Solve the problem.
-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 66.5, and the spring is to be stretched 4.5 feet from its natural length, how much work will be necessary?
A) 673.3 foot-pounds
B) 70.5 foot-pounds
C) 1346.6 foot-pounds
D) 149.6 foot-pounds

Question

Quizplus · Accepted Answer

To find the work $W$, use the formula $L = \sqrt{\frac{2W}{k}}$. Given $L = 4.5$ and $k = 66.5$, solve for $W$:  $4.5 = \sqrt{\frac{2W}{66.5}}$  Square both sides:  $20.25 = \frac{2W}{66.5}$  Multiply both sides by 66.5:  $1346.625 = 2W$  Divide by 2:  $W = 673.3$ foot-pounds.

Solve the Problem WWW Foot-Pounds, Is Given By L=2 Wk\mathrm{L}=\sqrt{\frac{2 \mathrm{~W}}{\mathrm{k}}}L=k2 W​​ Where k\mathrm{k}k Is a Constant for the Given Spring

Solve the Problem $W$ Foot-Pounds, Is Given By $\mathrm{L}=\sqrt{\frac{2 \mathrm{~W}}{\mathrm{k}}}$ Where $\mathrm{k}$ Is a Constant for the Given Spring