Skip to main content
\(\require{cancel}\newcommand{\highlight}[1]{{\color{blue}{#1}}} \newcommand{\apex}{A\kern -1pt \lower -2pt\mbox{P}\kern -4pt \lower .7ex\mbox{E}\kern -1pt X} \newcommand{\colorlinecolor}{blue!95!black!30} \newcommand{\bwlinecolor}{black!30} \newcommand{\thelinecolor}{\colorlinecolor} \newcommand{\colornamesuffix}{} \newcommand{\linestyle}{[thick, \thelinecolor]} \newcommand{\bmx}[1]{\left[\hskip -3pt\begin{array}{#1} } \newcommand{\emx}{\end{array}\hskip -3pt\right]} \newcommand{\ds}{\displaystyle} \newcommand{\fp}{f'} \newcommand{\fpp}{f''} \newcommand{\lz}[2]{\frac{d#1}{d#2}} \newcommand{\lzn}[3]{\frac{d^{#1}#2}{d#3^{#1}}} \newcommand{\lzo}[1]{\frac{d}{d#1}} \newcommand{\lzoo}[2]{{\frac{d}{d#1}}{\left(#2\right)}} \newcommand{\lzon}[2]{\frac{d^{#1}}{d#2^{#1}}} \newcommand{\lzoa}[3]{\left.{\frac{d#1}{d#2}}\right|_{#3}} \newcommand{\plz}[2]{\frac{\partial#1}{\partial#2}} \newcommand{\plzoa}[3]{\left.{\frac{\partial#1}{\partial#2}}\right|_{#3}} \newcommand{\inflim}[1][n]{\lim\limits_{#1 \to \infty}} \newcommand{\infser}[1][1]{\sum_{n=#1}^\infty} \newcommand{\Fp}{F\primeskip'} \newcommand{\Fpp}{F\primeskip''} \newcommand{\yp}{y\primeskip'} \newcommand{\gp}{g\primeskip'} \newcommand{\dx}{\Delta x} \newcommand{\dy}{\Delta y} \newcommand{\ddz}{\Delta z} \newcommand{\thet}{\theta} \newcommand{\norm}[1]{\left\lVert#1\right\rVert} \newcommand{\vnorm}[1]{\left\lVert\vec #1\right\rVert} \newcommand{\snorm}[1]{\left|\left|\ #1\ \right|\right|} \newcommand{\la}{\left\langle} \newcommand{\ra}{\right\rangle} \newcommand{\dotp}[2]{\vec #1 \cdot \vec #2} \newcommand{\proj}[2]{\text{proj}_{\,\vec #2}{\,\vec #1}} \newcommand{\crossp}[2]{\vec #1 \times \vec #2} \newcommand{\veci}{\vec i} \newcommand{\vecj}{\vec j} \newcommand{\veck}{\vec k} \newcommand{\vecu}{\vec u} \newcommand{\vecv}{\vec v} \newcommand{\vecw}{\vec w} \newcommand{\vecx}{\vec x} \newcommand{\vecy}{\vec y} \newcommand{\vrp}{\vec r\, '} \newcommand{\vsp}{\vec s\, '} \newcommand{\vrt}{\vec r(t)} \newcommand{\vst}{\vec s(t)} \newcommand{\vvt}{\vec v(t)} \newcommand{\vat}{\vec a(t)} \newcommand{\px}{\partial x} \newcommand{\py}{\partial y} \newcommand{\pz}{\partial z} \newcommand{\pf}{\partial f} \newcommand{\mathN}{\mathbb{N}} \newcommand{\zerooverzero}{\ds \raisebox{8pt}{\text{``\ }}\frac{0}{0}\raisebox{8pt}{\textit{ ''}}} \newcommand{\deriv}[2]{\myds\frac{d}{dx}\left(#1\right)=#2} \newcommand{\myint}[2]{\myds\int #1\ dx= {\ds #2}} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \newcommand{\primeskip}{\hskip.75pt} \newcommand{\plotlinecolor}{blue} \newcommand{\colorone}{blue} \newcommand{\colortwo}{red} \newcommand{\coloronefill}{blue!15!white} \newcommand{\colortwofill}{red!15!white} \newcommand{\abs}[1]{\left\lvert #1\right\rvert} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} \)


Work is the scientific term used to describe the action of a force which moves an object. When a constant force \(F\) is applied to move an object a distance \(d\text{,}\) the amount of work performed is \(W=F\cdot d\text{.}\)

The SI unit of force is the Newton, (kg\(\cdot\)m/s\(^2\)), and the SI unit of distance is a meter (m). The fundamental unit of work is one Newton–meter, or a joule (J). That is, applying a force of one Newton for one meter performs one joule of work. In Imperial units (as used in the United States), force is measured in pounds (lb) and distance is measured in feet (ft), hence work is measured in ft–lb.

Mass and weight are closely related, yet different, concepts. The mass \(m\) of an object is a quantitative measure of that object's resistance to acceleration. The weight \(w\) of an object is a measurement of the force applied to the object by the acceleration of gravity \(g\text{.}\)

Since the two measurements are proportional, \(w=m\cdot g\text{,}\) they are often used interchangeably in everyday conversation. When computing work, one must be careful to note which is being referred to. When mass is given, it must be multiplied by the acceleration of gravity to reference the related force.

When force is constant, the measurement of work is straightforward. For instance, lifting a 200 lb object 5 ft performs \(200\cdot 5 = 1000\) ft–lb of work.

What if the force applied is variable? For instance, imagine a climber pulling a 200 ft rope up a vertical face. The rope becomes lighter as more is pulled in, requiring less force and hence the climber performs less work.

In general, let \(F(x)\) be a force function on an interval \([a,b]\text{.}\) We want to measure the amount of work done applying the force \(F\) from \(x=a\) to \(x=b\text{.}\) We can approximate the amount of work being done by partitioning \([a,b]\) into subintervals \(a=x_1\lt x_2 \lt \cdots \lt x_{n+1}=b\) and assuming that \(F\) is constant on each subinterval. Let \(c_i\) be a value in the \(i^{\text{ th } }\) subinterval \([x_i,x_{i+1}]\text{.}\) Then the work done on this interval is approximately \(W_i\approx F(c_i)\cdot(x_{i+1}-x_i) = F(c_i)\dx_i\text{,}\) a constant force × the distance over which it is applied. The total work is \begin{equation*} W = \sum_{i=1}^n W_i \approx \sum_{i=1}^n F(c_i)\dx_i. \end{equation*}

This, of course, is a Riemann sum. Taking a limit as the subinterval lengths go to zero give an exact value of work which can be evaluated through a definite integral.

Key Idea7.5.1Work

Let \(F(x)\) be a continuous function on \([a,b]\) describing the amount of force being applied to an object in the direction of travel from distance \(x=a\) to distance \(x=b\text{.}\) The total work \(W\) done on \([a,b]\) is \begin{equation*} W = \int_a^b F(x)\ dx. \end{equation*}

Example7.5.2Computing work performed: applying variable force

A 60m climbing rope is hanging over the side of a tall cliff. How much work is performed in pulling the rope up to the top, where the rope has a mass of 66g/m?

Example7.5.3Computing work performed: applying variable force

Consider again pulling a 60 m rope up a cliff face, where the rope has a mass of 66 g/m. At what point is exactly half the work performed?

Example7.5.4Computing work performed: applying variable force

A box of 100 lb of sand is being pulled up at a uniform rate a distance of 50 ft over 1 minute. The sand is leaking from the box at a rate of 1 lb/s. The box itself weighs 5 lb and is pulled by a rope weighing .2 lb/ft.

  1. How much work is done lifting just the rope?

  2. How much work is done lifting just the box and sand?

  3. What is the total amount of work performed?


Hooke's Law and Springs Hooke's Law states that the force required to compress or stretch a spring \(x\) units from its natural length is proportional to \(x\text{;}\) that is, this force is \(F(x) = kx\) for some constant \(k\text{.}\) For example, if a force of 1 N stretches a given spring 2 cm, then a force of 5 N will stretch the spring 10 cm. Converting the distances to meters, we have that stretching this spring 0.02 m requires a force of \(F(0.02) = k(0.02) = 1\) N, hence \(k = 1/0.02 = 50\) N/m.

Example7.5.5Computing work performed: stretching a spring

A force of 20 lb stretches a spring from a natural length of 7 inches to a length of 12 inches. How much work was performed in stretching the spring to this length?


Subsection7.5.1Pumping Fluids

Another useful example of the application of integration to compute work comes in the pumping of fluids, often illustrated in the context of emptying a storage tank by pumping the fluid out the top. This situation is different than our previous examples for the forces involved are constant. After all, the force required to move one cubic foot of water (about 62.4 lb) is the same regardless of its location in the tank. What is variable is the distance that cubic foot of water has to travel; water closer to the top travels less distance than water at the bottom, producing less work.

Fluid lb/ft\(^3\) kg/m\(^3\)
Concrete \(150\) \(2400\)
Fuel Oil \(55.46\) \(890.13\)
Gasoline \(45.93\) \(737.22\)
Iodine \(307\) \(4927\)
Methanol \(49.3\) \(791.3\)
Mercury \(844\) \(13546\)
Milk 63.6–65.4 1020 — 1050
Water \(62.4\) \(1000\)
Table7.5.7Weight and Mass densities

We demonstrate how to compute the total work done in pumping a fluid out of the top of a tank in the next two examples.

Example7.5.8Computing work performed: pumping fluids

A cylindrical storage tank with a radius of 10 ft and a height of 30 ft is filled with water, which weighs approximately 62.4 lb/ft\(^3\text{.}\) Compute the amount of work performed by pumping the water up to a point 5 feet above the top of the tank.


We can “streamline” the above process a bit as we may now recognize what the important features of the problem are. Figure 7.5.10 shows the tank from Example 7.5.8 without the \(i^\text{ th }\) subinterval identified.

<<SVG image is unavailable, or your browser cannot render it>>

Figure7.5.10A simplified illustration for computing work.

Instead, we just draw one differential element. This helps establish the height a small amount of water must travel along with the force required to move it (where the force is volume × density).

We demonstrate the concepts again in the next examples.

Example7.5.11Computing work performed: pumping fluids

A conical water tank has its top at ground level and its base 10 feet below ground. The radius of the cone at ground level is 2 ft. It is filled with water weighing 62.4 lb/ft\(^3\) and is to be emptied by pumping the water to a spigot 3 feet above ground level. Find the total amount of work performed in emptying the tank.

Example7.5.13Computing work performed: pumping fluids

A rectangular swimming pool is 20 ft wide and has a 3 ft “shallow end” and a 6 ft “deep end.” It is to have its water pumped out to a point 2 ft above the current top of the water. The cross–sectional dimensions of the water in the pool are given in Figure 7.5.14; note that the dimensions are for the water, not the pool itself. Compute the amount of work performed in draining the pool.


The next section introduces one final application of the definite integral, the calculation of fluid force on a plate.