Chapter 10, Problem 60A

To determine

The work required to stretch the spring from $0.12\text{m}$ to $0.28\text{m}$ .

Answer to Problem 60A

  $0.8\text{J}$

Explanation of Solution

Given:

The Force v. Elongation graph of the spring in figure 22.

Formula used:

Work done is given by,

  $W=Fd\cos \theta$

Where $F$ is the force done, $d$ is the distance, and $\theta$ is the angle between the distance and force.

Calculation:

Consider the given graph,

The force at $d_1=0.12\text{m}$ is, $F_1=3.0\text{N}$

The force at $d_2=0.28\text{m}$ is, $F_2=7.0\text{N}$

Now, the work required to stretch the spring from $0.12\text{m}$ to $0.28\text{m}$ will be equal to the area of the curve between $x=0.12\text{m}$ and $x=0.28\text{m}$ as,

  $\begin{array}{c}W=\text{Area of triangle + Area of rectangle}\\ =\left(\frac{1}{2}\left(d_2-d_1\right)\left(F_2-F_1\right)\right)+\left(\left(d_2-d_1\right)\left(F_1-0\right)\right)\end{array}$

Using given values in above,

  $\begin{array}{c}W=\left(\frac{1}{2}\left(d_2-d_1\right)\left(F_2-F_1\right)\right)+\left(\left(d_2-d_1\right)\left(F_1-0\right)\right)\\ =\left(\frac{1}{2}\left(0.28-0.12\right)\left(7.0-3.0\right)\right)+\left(\left(0.28-0.12\right)\left(3.0-0\right)\right)\\ =\left(0.32+0.48\right)\text{J}\\ =0.8\text{J}\end{array}$

Conclusion:

Thus, the work required to stretch the spring from $0.12\text{m}$ to $0.28\text{m}$

  $0.8\text{J}$ .