Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 7

Applications of the Definite Integral

Work

To find work, we must find the corresponding force function.

Study the section titled “Work” on pages 326-329 of the textbook.

The following table gives the units we use when dealing with work.

The conversion factors are given below.

Example 7.13. In 1976 Vasili Alexeev lifted a record-breaking $562\text{lb}$ from the floor to above his head (about $2\text{m}$). In 1957, Paul Anderson braced himself against the floor and used his back to lift $6720\text{lb}$ of lead and automobile parts a distance of $1\text{cm}$. Who did more work?

The force that the Earth exert on an object is the object’s weight; thus, Alexeev applied a force of $562\text{lb}$ over a distance of $2\text{m}$ and Anderson applied a force of $6720\text{lb}$ over a distance of $1\text{cm}$.

We choose SI units. Converting, we have

${\displaystyle 562}$ lb ${\displaystyle \approx 562}$ lb ${\displaystyle \times 4.45}$ N/lb ${\displaystyle \approx 2500}$ N,

and

${\displaystyle 6270}$ lb ${\displaystyle \approx 6270\times 4.45}$ N/lb ${\displaystyle \approx 27900}$ N.

Since $1\text{cm}=0.01\text{m}$, we have

Alexeev’s work $=(2500\text{N/lb})\times (2\text{m})=5000\text{J}$,

and

Anderson’s work $=(27900\text{N})\times (0.01\text{m})=279\text{J}$.

Therefore, Alexeev did more work. For Anderson to have done as much work as Alexeev, he would have needed to move the ${\displaystyle 6270}$ lb a distance of

${\displaystyle d=\frac{5000}{27900}\approx 0.179\text{m}}$;

that is, about ${\displaystyle 18\text{cm}}$.

Example 7.14. [See Exercise 16 on page 329 of the textbook.]

A bucket that weighs ${\displaystyle 4}$ lb and a rope of negligible weight are used to draw water from a well that is ${\displaystyle 80}$ ft deep. The bucket is filled with ${\displaystyle 40}$ lb of water and is pulled up at a rate of ${\displaystyle 2}$ ft/s, but water leaks out of a hole in the bucket at a rate of ${\displaystyle 0.2}$ lb/s. Find the work done by pulling the bucket to the top of the well.

Observe that the work needed to lift the bucket is ${\displaystyle 4\text{lb}\left(80\text{ft}\right)=320}$ ft-lb. Thinking through the process we, see that at time ${\displaystyle t}$ seconds the bucket is ${\displaystyle 2t}$ ft above the original ${\displaystyle 80}$ ft depth of the well. Hence, ${\displaystyle x_i*=2t\text{,}}$ and at this instant the bucket holds ${\displaystyle \left(40-0.2t\right)}$ lb of water. When the bucket is ${\displaystyle x_i*}$ ft above its original ${\displaystyle 80}$ ft, the bucket holds

${\displaystyle \left[40-0.2\left(\frac{1}{2}\right)x_i\text{*}\right]\text{ lb of water}.}$

To move this water by a distance of ${\displaystyle \text{Δ}x\text{,}}$ requires

${\displaystyle \left(40-\frac{1}{10}{x}_i\text{*}\right)\text{Δ}x\text{ ft-lb of work}.}$

Thus, the work needed to lift the water is

$\begin{array}{rl}W& =\underset{x\to \infty }{\lim }{\displaystyle {\sum }_{i=1}^n\left(40-\left({\displaystyle \frac{1}{10}}\right)x_i\text{*}\right)\Delta x}\\ & ={\displaystyle {\int }_0^{80}40-{\displaystyle \frac{x}{10}}dx}\\ & ={40x-{\displaystyle \frac{x^2}{20}}|}_0^{80}\\ & =3200-320.\end{array}$

Adding the work of lifting the bucket gives a total of ${\displaystyle 3200}$ ft-lb.