Chapter 14, Problem 46A

To determine

The Potential energy of the spring.

Answer to Problem 46A

  $0.35\text{J}$

Explanation of Solution

Given:

Spring constant of spring is $k=27\text{N/m}$ .

The string is stretched at a distance of $x=16\text{cm}$ .

Formula Used:

Potential Energy of spring is:

  $P{E}_{sp}=\frac{1}{2}k{x}^2$

Where,

  $k$ : spring constant

x: stretch in spring

Calculation:

Put the given values:

  $\begin{array}{l}P{E}_{sp}=\frac{1}{2}k{x}^2\\ P{E}_{sp}=\frac{1}{2}\left(27\text{Nm}\right){\left(0.16\text{m}\right)}^2\\ P{E}_{sp}=0.35\text{J}\end{array}$

Conclusion: Hence, the potential energy of the spring is $0.35\text{J}$ .