Chapter 6.6, Problem 9E

To determine

To Compare: The work done to stretch a spring by a given length at different elongation levels. regimes.

Answer to Problem 9E

  $W_{30\to 40}$ , $W_{20\to 30}$ are in the ratio $3:1$

Explanation of Solution

Given Information: The natural length of the spring is $20cm$ . First time, the spring is stretched from $20cm$ length to $30cm$ and the second time, the spring is stretched from $30cm$ length to $40cm$ .

Method Used:

• The work done by a force $F$ in moving a particle from $x_1$ to $x_2$ is given by

  $W={{\displaystyle \int }}_{x_1}^{x_2}F\left(x\right)dx$

• Hooke’s Law: $f\left(x\right)=kx$ , where $x$ is the amount by which the spring is elongated, $k$ is the spring constant and $f\left(x\right)$ is the force required to maintain the spring at that elongation length.

Therefore, the work done in stretching a spring of spring constant $k$ from elongation length $x_1$ to $x_2$ is

  $W={{\displaystyle \int }}_{x_1}^{x_2}kx\cdot dx={\frac{{\text{kx}}^2}{2}]}_{{\text{x}}_1}^{{\text{x}}_2}$

  $\Rightarrow W=\frac{1}{2}\text{k}\left({\text{x}}_2{}^2-{\text{x}}_1{}^2\right)$

Calculation: Let the spring constant be $k$ . Calculate the work required in both the cases as follows

  $\underset{\_}{20cm\to 30cm}:$

  $x_1=20-20=0$

  $x_2=30-20=10cm=0.1m$

Therefore

  $W_{20\to 30}=\frac{1}{2}k\left({0.1}^2-0^2\right)$

  $\Rightarrow W_{20\to 30}=\frac{1}{2}k\times 0.01$

  $\underset{\_}{30cm\to 40cm}:$

  $x_1=30-20=10cm=0.1m$

  $x_2=40-20=20cm=0.2m$

Therefore

  $W_{30\to 40}=\frac{1}{2}k\left({0.2}^2-{0.1}^2\right)$

  $\Rightarrow W_{30\to 40}=\frac{1}{2}k\times 0.03$

Now, divide the work done in both the cases to write

  $\frac{W_{30\to 40}}{W_{20\to 30}}=\frac{\frac{1}{2}k\times 0.03}{\frac{1}{2}k\times 0.01}=\frac{0.03}{0.01}$

  $\Rightarrow \frac{W_{30\to 40}}{W_{20\to 30}}=3$

It means it requires $3$times more work to stretch the spring from $30cm$ to $40cm$ than to stretch it from $20cm$ to $30cm$ .