Chapter 13, Problem 1P

Work Done by a Winch A motorized winch is being used to pull a felled tree to a logging truck. The motor exerts a force of $f(x)=1500+10x-\frac{1}{2}{x}^2\text{lb}$ on the tree at the instant when the tree has moved x ft. The tree must be moved a distance of 40 ft, from x = 0 to x = 40. How much work is done by the winch in moving the tree?

To determine

The work done by the winch to move a tree from $x=0$ to $x=40$, if a motor exerts a force $f\left(x\right)=1500+10x-\frac{1}{2}{x}^2\text{ lb}$ on the tree.

Answer to Problem 1P

The work done by the winch to move a tree is $57,333\frac{1}{3}\text{ ft-lb}$

Explanation of Solution

Given:

The motor exerts a force on the tree is,

$f\left(x\right)=1500+10x-\frac{1}{2}{x}^2\text{ lb}$

Calculation:

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

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

$x_k=a+k\Delta x$ (3)

Substitute 0 for a and 40 for b in equation (2) to find the value of $\Delta x$,

$\begin{array}{c}\Delta x=\frac{40-0}{n}\\ =\frac{40}{n}\end{array}$

Substitute 0 for a and $\frac{40}{n}$ for $\Delta x$ in equation (3) to find the value of $x_k$,

$\begin{array}{c}{x}_k=0+k\cdot \frac{40}{n}\\ =\frac{40k}{n}\end{array}$

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

$\begin{array}{c}f\left(x_k\right)=1500+10x_k-\frac{1}{2}{x}_k^2\\ =1500+10\left(\frac{40k}{n}\right)-\frac{1}{2}{\left(\frac{40k}{n}\right)}^2\end{array}$

Substitute $\frac{40}{n}$ for $\Delta x$, $\frac{40k}{n}$ for $x_k$ and $1500+10\left(\frac{40k}{n}\right)-\frac{1}{2}{\left(\frac{40k}{n}\right)}^2$ for $f\left(x_k\right)$

in equation (1) to find the value of work done,

$\begin{array}{c}W=\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\left[1500+10\left(\frac{40k}{n}\right)-\frac{1}{2}{\left(\frac{40k}{n}\right)}^2\right]}\left(\frac{40}{n}\right)\\ =\underset{n\to \infty }{\lim }\left(\frac{40}{n}{\displaystyle \sum _{k=1}^{n}1500+\frac{400}{n}\cdot \frac{40}{n}{\displaystyle \sum _{k=1}^{n}k}-\frac{800}{n^2}\cdot \frac{40}{n}{\displaystyle \sum _{k=1}^n{k}^2}}\right)\end{array}$

The value of ${\displaystyle \sum _{k=1}^{n}1=n}$, ${\displaystyle \sum _{k=1}^{n}k}=\frac{n\left(n+1\right)}{2}$ and ${\displaystyle \sum _{k=1}^n{k}^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}}$

Substitute n for ${\displaystyle \sum _{k=1}^{n}1}$, $\frac{n\left(n+1\right)}{2}$ for ${\displaystyle \sum _{k=1}^{n}k}$, $\frac{n\left(n+1\right)\left(2n+1\right)}{6}$ for ${\displaystyle \sum _{k=1}^n{k}^2}$ in the above value of W,

$\begin{array}{c}W=\underset{n\to \infty }{\lim }\left(\frac{40}{n}\cdot 1500n+\frac{16000}{n^2}\cdot \frac{n\left(n+1\right)}{2}-\frac{32000}{n^3}\cdot \frac{n\left(n+1\right)\left(2n+1\right)}{6}\right)\\ =\underset{n\to \infty }{\lim }\left(60000+8000\cdot \frac{\left(n+1\right)}{n}-\frac{16000}{3}\cdot \frac{\left(n+1\right)\left(2n+1\right)}{n^2}\right)\\ =\underset{n\to \infty }{\lim }\left(60000+8000\cdot \left(\frac{n}{n}+\frac{1}{n}\right)-\frac{16000}{3}\cdot \frac{\left(n+1\right)}{n}\cdot \frac{\left(2n+1\right)}{n}\right)\\ =\underset{n\to \infty }{\lim }\left(60000+8000\cdot \left(1+\frac{1}{n}\right)-\frac{16000}{3}\cdot \left(1+\frac{1}{n}\right)\cdot \left(2+\frac{1}{n}\right)\right)\end{array}$

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

$\begin{array}{c}W=60000+8000\left(1\right)-\frac{16000}{3}\cdot 1\cdot 2\\ =60000+8000-\frac{32000}{3}\\ =\frac{180000+24000-32000}{3}\\ =\frac{172,000}{3}\end{array}$

The above calculation gives the value of W,

$\begin{array}{c}W=\frac{172000}{3}\\ =57,333\frac{1}{3}\end{array}$

Thus, work done by the winch to move a tree is $57,333\frac{1}{3}\text{ ft-lb}$