Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.8 - Modeling Using Variation - Exercise Set - Page 424: 24


The force required is $20$ pounds.

Work Step by Step

Consider that the distance by which the spring will stretch is D and the force required to stretch the spring is F. It is provided that the variable D varies directly as F, that is $D\ =\ kF$. Thus, $\begin{align} & 9=12k \\ & k=\frac{9}{12} \\ & k=\frac{3}{4} \end{align}$ Substitute D = 15 and $k=\frac{3}{4}$ in $D\ =\ kF$. $\begin{align} & D=kF \\ & 15=\frac{3}{4}F \\ & F=\frac{15\times 4}{3} \\ & F=20 \end{align}$ Thus, a force of 20 pounds is required to stretch the spring by 15 inches.  
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.