Chapter 14, Problem 48A

(a)

To determine

The spring constant of the spring.

Answer to Problem 48A

The spring constant of the spring is $20$ .

Explanation of Solution

Given: $x$ and $y$ co-ordinates are given below:

  $\left(0,0\right)$ and $\left(0.8,16\right)$ .

Formula used:

The expression used to find the spring constant is:

  $k=\frac{y_2-y_1}{x_2-x_1}$

Where, $k$ : spring constant.

Calculation:

The following expression is used to find the spring constant:

  $k=\frac{y_2-y_1}{x_2-x_1}$

Put the given values:

  $\begin{array}{l}k=\frac{y_2-y_1}{x_2-x_1}\\ k=\frac{\left(16-0\right)}{\left(0.8-0\right)}\end{array}$

  $k=20$

Conclusion:

The spring constant of the spring is $k=20$ .

(b)

To determine

The spring potential energy when it is stretched.

Answer to Problem 48A

The spring potential energy is $2.5$ J.

Explanation of Solution

Given:

Length: $0.50$ m

Spring constant: $20$

Formula used:

The following expression is used to find the spring potential energy:

  $E=\frac{1}{2}k{x}^2$

Where,

  $E$ : Spring potential energy.

  $k$ : spring constant.

  $x$ : Length.

Calculation:

The expression used to calculate thespring potential energy:

  $E=\frac{1}{2}k{x}^2$

Put the given values.

  $\begin{array}{l}E=\frac{1}{2}k{x}^2\\ E=\frac{1}{2}\times 20\times {\left(0.50\right)}^2\\ E=2.5\text{J}\end{array}$

Conclusion:

The potential energy of the spring is $E=2.5$ J.