Chapter 8.5, Problem 49E

To determine

To find : How high the ball bearing go.

Answer to Problem 49E

The ball bearing goes $4.5$feet high.

Explanation of Solution

Given information :

Weight = $2$ oz

  $k=18\text{ lb/sec}$

The spring compressed $\left(s\right)=3$ inch $=\frac{3}{12}\text{ ft}$

Formula used :

  $mgh=\frac{1}{2}k{s}^2$

Calculation :

Here, $m=\frac{2}{16\times 32}$ , $g=32{\text{ ft/sec}}^2$

  $h$ is the required height.

  $mgh=\frac{1}{2}k{s}^2$

  $\begin{array}{l}\Rightarrow \frac{2}{32\times 16}\times 32\times h=\frac{1}{2}\times 18\times {\left(\frac{3}{12}\right)}^2\\ \Rightarrow h=4.5\end{array}$

So, the ball bearing goes $4.5$feet high.