Chapter 14, Problem 47A

To determine

To Calculate: The compression of the string.

Answer to Problem 47A

  $0.29m$

Explanation of Solution

Given:

Spring constant, $k=35\text{ N/m}$

Energy stored in spring is $P{E}_{sp}=1.5\text{J}$ .

Formula Used:

Potential energy stored in the spring:

  $P{E}_{sp}=\frac{1}{2}k{x}^2$

  $x=\sqrt{\frac{2P{E}_{sp}}{k}}$

Calculation:

Substituting the values in the equation and solve:

  $\begin{array}{l}x=\sqrt{\frac{\left(2\right)\left(1.5\text{J}\right)}{35\text{N/m}}}\\ x=0.29\text{m}\end{array}$

Conclusion: Hence,the spring must be compressed $0.29m$ long to store $1.5J$ of energy.