Chapter 7.5, Problem 48E

To determine

To find: The height reached by the ball bearing.

Answer to Problem 48E

  $1\cdot 36m$

Explanation of Solution

Given information:

A $60g$ steel ball bearing is placed on a vertical spring whose force constant is $k=2\cdot 5\text{newtons/cm}$ . The spring is compressed $8\text{cm}$ and released. About how high does the ball bearing go $\text{?}$

The Force constant is $k=2\cdot 5\text{newtons/cm}$ and mass $\left(m\right)$ is $60g$ .

  $\begin{array}{l}k=2.5\text{N/cm}\\ \text{=2×5×100}\text{N/m}\\ \text{=250}\text{N/m}\\ \text{m=60}\text{g=}\frac{60}{1000}kg=0\cdot 06kg\end{array}$

The distance spring is compressed is $8\text{cm}$ .

  $s=8cm=\frac{8}{100}m=0\cdot 08m$

The Kinetic compression energy of the spring is given by:

  $\begin{array}{l}\frac{1}{2}k{s}^2=\frac{1}{2}\times 250\times {\left(0\cdot 08\right)}^2\\ =\frac{1}{2}\times 250\times 0\cdot 0064\\ =0\cdot 8\text{joule}\end{array}$

Let the height to which the steel ball rises be $h$ .

Then, gravitational potential energy is given by $mgh$ where $m$ is the mass of the ball and $g$ is acceleration due to gravity.

  $\begin{array}{l}mgh=0\cdot 06\times 9\cdot 8\times h\\ =0\cdot 588h\text{joule}\end{array}$

Since the Kinetic compression energy of the spring equals the gravitational potential energy, it follows that:

  $\begin{array}{l}0\cdot 588h=0\cdot 8\\ h=\frac{0\cdot 8}{0\cdot 588}\approx 1\cdot 36m\end{array}$

Hence, the ball bearing goes to the height of $1\cdot 36m$ approximately.