Chapter 6, Problem 27RE

To determine

To calculate: the work done in stretching the spring from $12\text{cm}$ to $20\text{cm}$.

Answer to Problem 27RE

The required work done to stretch the spring from $12\text{cm}$ to $20\text{cm}$ is $\underset{\_}{3.2\text{J}}$.

Explanation of Solution

Given information:

The required force to maintain a spring stretched from its natural length of 12 cm to 20 cm is 30 N.

Apply Hooke’s Law to find the spring constant:

$F\left(x\right)=kx$ (1)

Here, $F\left(x\right)$ is force, k is spring constant, and x is amount of extension.

Calculation:

Calculate the spring constant using Equation (1).

Substitute $30\text{N}$ for F and $\left(15\text{cm}-\text{12}\text{cm}\right)$ for x in Equation (1).

$\begin{array}{c}30\text{N}=\left(15\text{cm}-12\text{cm}\right)\times k\\ =\left(3\text{cm}\times \frac{\text{1}\text{m}}{100\text{cm}}\right)\times k\\ =0.03k\end{array}$

$\begin{array}{c}k=\frac{30}{0.03}\\ =1,000\text{N/m}\end{array}$

Show the expression to find the work done to stretch the spring from 12 cm to 20 cm.

$W={\displaystyle \underset{a}{\overset{b}{\int }}kx}dx$ (2)

Here, a is the lower limit and b is the upper limit.

Consider that the lower limit as 0 cm. Hence, $a=0\text{cm}$

Calculate the upper limit as follows:

$\begin{array}{c}b=\text{Stretching}\text{length}-\text{Natural}\text{length}\\ =\text{20}\text{cm}-\text{12}\text{cm}\\ =\text{8}\text{cm}\times \frac{\text{1}\text{m}}{100\text{cm}}\\ =0.08\text{m}\end{array}$

Substitute 0 for a, 0.08 for b, and $1,000\text{N/m}$ for k in Equation (2).

$\begin{array}{c}W=1000{\displaystyle \underset{0}{\overset{0.08}{\int }}x}dx\\ =1000{\left[\frac{x^2}{2}\right]}_0^{0.08}\\ =500{\left(x^2\right)}_0^{0.08}\end{array}$

$\begin{array}{c}=500\left({0.08}^2-0\right)\\ =3.2\text{N}\cdot \text{m}\end{array}$

Show the unit conversion as follows:

$\begin{array}{c}W=3.2\text{N}\cdot \text{m}\times \left(\frac{1\text{J}}{1\text{N}\cdot \text{m}}\right)\\ =\text{3}\text{.2}\text{J}\end{array}$

Thus, The required work done to stretch the spring from $12\text{cm}$ to $20\text{cm}$ is $\underset{\_}{3.2\text{J}}$.