Chapter 10, Problem 59A

(a)

To determine

The magnitude of slope of the graph.

To show: The relation $F=kd$ for the value of k

Answer to Problem 59A

The slope of the graph $k=25\text{N}/\text{m}$ .

The force $F=kd$ is proved for $k=25\text{N}/\text{m}$ .

Explanation of Solution

Given:

The given force-elongation graph is shown below.

  

Formula used:

For the graph, the expression for the slope is,

  $k=\frac{\Delta y}{\Delta x}$

Here, $\Delta y$ is the loading change and $\Delta x$ is the changing distance.

Calculation:

Consider the given graph. The slope of the graph is,

  $\begin{array}{l}k=\frac{\Delta y}{\Delta x}\\ k=\frac{\left(5.00\text{ N}\right)-\left(0.00\text{ N}\right)}{\left(0.20\text{ m}\right)-\left(0.00\text{ m}\right)}\\ k=25\text{N}/\text{m}\end{array}$

From the graph, the distance $d=0.20\text{ m}$ and the force at a distance of $0.20\text{ m}$ is $F=5.00\text{ N}$

Substitute the value of k and d in $F=kd$

  $\begin{array}{l}F=kd\\ F=\left(25\text{N}/\text{m}\right)\times \left(0.20\text{ m}\right)\\ F=5.00\text{ N}\end{array}$

Hence, the force $F=kd$ is proved.

Conclusion:

Thus, the slope of the graph is $k=25\text{N}/\text{m}$ . Also, the force $F=kd$ is proved.

(b)

To determine

The work done in stretching the spring using the graph.

Answer to Problem 59A

The work done in stretching the spring is $0.50\text{ J}$ .

Explanation of Solution

Given:

The given force-elongation graph is shown below.

  

Formula used:

The expression for the graph is,

  $\text{Work done}=\text{Area under the graph}$

Calculation:

Consider the given graph, the work done in stretching the spring is

  $\begin{array}{l}\text{Work done}=\text{Area}\\ \text{Work done}=\frac{1}{2}\times Fd\\ \text{Work done}=\frac{1}{2}\times \left(5.00\text{ N}\right)\times \left(0.20\text{ m}\right)\\ \text{Work done}=0.50\text{ J}\end{array}$

Conclusion:

Thus, the work done in stretching the spring is $0.50\text{ J}$ .

(c)

To determine

To show:The work done obtained in part (b) is valid for the given values.

Answer to Problem 59A

The work done in stretching the spring is $0.50\text{ J}$ is same as given value.

Explanation of Solution

Given:

The given force-elongation graph is shown below.

  

The distance is 0.20 m.

Spring constant is $k=25\text{N}/\text{m}$

Formula used:

The expression for the graph using the graph as follows:

  $W=\frac{1}{2}k{d}^2$

Here, $k$ is the slope of the graph and $d$ is the distance the spring is stretched.

Calculation:

Work done obtained in part (b) is 0.50 J.

For the given values, work done in stretching the spring is,

  $\begin{array}{l}W=\frac{1}{2}k{d}^2\\ W=\frac{1}{2}\times \left(25\text{N}/\text{m}\right)\times {\left(0.20\text{ m}\right)}^2\\ W=0.50\text{ J}\end{array}$

So, the work done in stretching the spring is $W=0.50\text{ J}$ .

Conclusion:

Thus, the work done in stretching the spring is same as the given value.