Chapter 13, Problem 2P

To determine

The work done when the spring is stretched from $x=0$ to $x=2\text{ ft}\text{.}$ by exerting a force $f\left(x\right)=kx$.

Answer to Problem 2P

The work done in pulling the spring is $\text{40 ft-lb}$

Explanation of Solution

Given:

The spring with spring constant $k=20\text{ lb/ft}\text{.}$ exerts a force is,

$f\left(x\right)=kx$

Calculation:

The work done in moving an object from a and b is defined as the limit of this quantity as $n\to \infty$,

$W=\underset{n\to \infty }{\lim }{\displaystyle \sum _{m=1}^{n}f\left(x_m\right)}\Delta x$ (1)

$\Delta x$ can be calculated by below formula,

$\Delta x=\frac{b-a}{n}$ (2)

Formula to calculate the value of $x_k$ is,

$x_m=a+m\Delta x$ (3)

Substitute 0 for a and 2 for b in equation (2) to find the value of $\Delta x$,

$\begin{array}{c}\Delta x=\frac{2-0}{n}\\ =\frac{2}{n}\end{array}$

Substitute 0 for a and $\frac{2}{n}$ for $\Delta x$ in equation (3) to find the value of $x_m$,

$\begin{array}{c}{x}_m=0+m\cdot \frac{2}{n}\\ =\frac{2m}{n}\end{array}$

Substitute the value of $x_m$ for x in function $f\left(x\right)$

$\begin{array}{c}f\left(x_m\right)=k{x}_m\\ =k\cdot \frac{2m}{n}\\ =\frac{2k}{n}\cdot m\end{array}$

Substitute $\frac{2}{n}$ for $\Delta x$, $\frac{2m}{n}$ for $x_m$ and $\frac{2k}{n}\cdot m$ for $f\left(x_m\right)$ in equation (1) to find the value of work done,

$\begin{array}{c}W=\underset{n\to \infty }{\lim }{\displaystyle \sum _{m=1}^n\frac{2k}{n}\cdot m\cdot }\left(\frac{2}{n}\right)\\ =\underset{n\to \infty }{\lim }\left({\displaystyle \sum _{m=1}^n\frac{4k}{n^2}\cdot m}\right)\\ =\underset{n\to \infty }{\lim }\frac{4k}{n^2}{\displaystyle \sum _{m=1}^{n}m}\end{array}$

The value of ${\displaystyle \sum _{m=1}^{n}m}=\frac{n\left(n+1\right)}{2}$,

Substitute $\frac{n\left(n+1\right)}{2}$  for ${\displaystyle \sum _{m=1}^{n}m}$ in the above value of W,

$\begin{array}{c}W=\underset{n\to \infty }{\lim }\frac{4k}{n^2}\cdot \frac{n\left(n+1\right)}{2}\\ =\underset{n\to \infty }{\lim }2k\cdot \left(1+\frac{1}{n}\right)\end{array}$

Substitute $\infty$ for n in above values of W because the $n=\infty$ is limit,

$\begin{array}{c}W=2k\cdot 1\cdot 1\\ =2k\end{array}$

Spring constant $k=20\text{ lb/ft}\text{.}$

Substitute 20 for k in the value of W,

$\begin{array}{c}W=2\cdot 20\\ =40\end{array}$

Thus, work done by the winch to move a tree is $\text{40 ft-lb}$