Chapter 14.1, Problem 2PP

To determine

Potential energy of the elastic stored in a spring

Answer to Problem 2PP

  $1.96J$

Explanation of Solution

Given information:

Spring at the position stretched by $16.5cm$ .

Spring constant $=144N/m$

Formula used:

The potential energy in a spring:

  $P{E}_{sp}=\frac{1}{2}k{x}^2$

Here, k is the spring constant and x is the stretch in the spring.

Calculation:

As per given information,

Here, $k$

  $=144N/m$

  $x$

  $=16.5cm=0.165m$

Now, substitute the values in

  $P{E}_{sp}=\frac{1}{2}k{x}^2$

  $\begin{array}{l}=\frac{1}{2}\left(144N/m\right)\left(0.165m\right)²\\ =1.96J\end{array}$

Conclusion:

Thus, the spring potential energy is $1.96J$ .