Work Done by a Force: Integral Formula and Examples

What Is Work Done By A Force?

Work done by a force is the energy a force transfers to an object as it moves the object along a path. When the force is constant and points along the motion, the familiar work formula gives $W = Fd$. When the force changes with position, the work done by a force moving an object along the $x$-axis from $x = a$ to $x = b$ is a definite integral:

$$W = \int_{a}^{b} F(x),dx$$

Every symbol has a job:

• $W$ is the work. In SI units it is measured in joules, where $1,\text{J} = 1,\text{N}\cdot\text{m}$; in US customary units it is measured in foot-pounds ($\text{ft}\cdot\text{lb}$).

• $F(x)$ is the force component along the direction of motion when the object sits at position $x$, in newtons or pounds.

• $x$ is position along a straight line, in metres or feet; $a$ is the start and $b$ is the finish.

The formula needs three conditions. The motion runs along a straight line, the force is measured along that line, and $F$ is integrable on $[a, b]$ (any continuous force qualifies). Where the force pushes against the motion, $F(x)$ is negative and that stretch of the path contributes negative work.

The constant-force rule is a special case, not a separate law. If $F(x) = F$ for every $x$, then

$$\int_{a}^{b} F,dx = F(b - a) = Fd,$$

which is the textbook $W = Fd$ with $d = b - a$. The joule and the foot-pound are the standard units for both cases, as the OpenStax University Physics section on work sets out.

Why Is Integration Used To Find The Work Done By A Variable Force?

Integration is used because $W = Fd$ only works when the force stays the same over the whole distance. When the force changes, you split the path into pieces short enough that the force is nearly constant on each one, use $W = Fd$ on each piece, and add the results. The integral is what that adding becomes when the pieces shrink to zero width.

Here is the derivation in four moves:

1. Split $[a, b]$ into $n$ equal pieces of width $\Delta x = \frac{b - a}{n}$.

2. On the $i$-th piece, pick a point $x_i^$. The force there is about$F(x_i^)$, so the work on that piece is about$F(x_i^*),\Delta x$.

3. Add the pieces, written in summation notation: $W \approx \sum_{i=1}^{n} F(x_i^*),\Delta x$.

4. Let the pieces shrink. The approximation becomes exact in the limit:

$$W = \lim_{n \to \infty} \sum_{i=1}^{n} F(x_i^*),\Delta x = \int_{a}^{b} F(x),dx$$

To evaluate the integral, the Fundamental Theorem of Calculus lets you skip the limit. Find a function $G$ with $G'(x) = F(x)$, then compute $W = G(b) - G(a)$.

What Does Work Look Like On A Graph?

Geometrically, work is the signed area between a force-displacement graph and the horizontal axis, the same area under a curve idea that integration measures everywhere else. Plot $F$ on the vertical axis and position $x$ on the horizontal axis. The region between the graph and the axis, from $x = a$ to $x = b$, has an area equal to the work, with any part below the axis counting as negative.

The sign convention took this writer a while to settle. "Work done by the spring" and "work done on the spring" are equal in size and opposite in sign, and textbooks switch between the two without announcing it. Every example on this page computes the work done by the outside agent (your hand, the winch, the pump), so the answers come out positive.

How Do You Find Work Done By A Force Using Integration?

You find it by turning the physical situation into "force on a piece times distance for that piece", then integrating. The same five-step setup works for every problem on this page:

1. Choose an axis. Decide where $x = 0$ (or $y = 0$) sits and which direction is positive, and write it on a sketch.

2. Slice. For one object pushed by a changing force, the slices are short stretches of the path. For a chain or a tank of water, the slices are thin pieces of the material itself, because different pieces move different distances.

3. Find the force on one slice. For a material, this is its weight: mass times $g$, where $g = 9.8,\text{m/s}^2$.

4. Find the distance that slice moves.

5. Multiply, integrate and check units. Newtons times metres gives joules; pounds times feet gives foot-pounds.

Problems split into two families. In the first, one object moves and the force changes with position, so $F(x)$ is given or comes from a law such as Hooke's law. In the second, the force on each slice stays fixed but each slice travels its own distance, so the integral adds up $(\text{weight of slice}) \times (\text{distance moved})$.

Units cause more lost marks than the calculus does, so keep this table beside you:

One foot-pound is about $1.3558,\text{J}$.

How Do You Find The Work Done On A Spring?

You find it with Hooke's law, which says the force needed to hold a spring stretched (or compressed) a distance $x$ beyond its natural length is

$$F(x) = kx,$$

where $k > 0$ is the spring constant in newtons per metre. The law holds only within the spring's elastic limit, the range in which the spring returns to its natural length once released. Integrating gives the work to stretch the spring from $x = a$ to $x = b$:

$$W = \int_{a}^{b} kx,dx = \frac{k}{2}\left(b^2 - a^2\right)$$

From the natural length ($a = 0$) this becomes $W = \frac{1}{2}kb^2$, the energy stored in the stretched spring. The $x$ in Hooke's law is the stretch beyond the natural length, never the spring's total length.

How Do You Find The Work Done Lifting A Chain Or Cable?

You slice the chain into short pieces, find the weight of one piece and how far it rises, then integrate along the chain. Take a chain of length $L$ hanging from the top of a building, with mass $\lambda$ kilograms per metre. A piece at depth $y$ below the top, of length $dy$, weighs $\lambda g,dy$ newtons and rises $y$ metres:

$$W = \int_{0}^{L} \lambda g,y,dy = \frac{1}{2}\lambda g L^2$$

A second setup gives the same answer. Let $x$ be the length already pulled up; the hanging part then weighs $\lambda g (L - x)$, and $W = \int_{0}^{L} \lambda g (L - x),dx$, which also equals $\frac{1}{2}\lambda g L^2$. Pick whichever setup you can sketch more clearly.

There is a shortcut worth knowing. The chain's total weight is $\lambda g L$, and its centre of mass rises $\frac{L}{2}$, so the work is the total weight times that rise. The idea comes from moments and the centre of mass, and it makes a quick check on any lifting answer. A bucket hanging on the end adds its own weight times the full lift distance.

How Do You Find The Work Required To Pump Water Out Of A Tank?

You slice the water into thin horizontal layers, find each layer's weight and the distance it travels to the outlet, then integrate over the depth of the water. A layer at height $y$ with cross-sectional area $A(y)$ and thickness $dy$ has volume $A(y),dy$, mass $\rho A(y),dy$ and weight $\rho g A(y),dy$. It must rise to the outlet height $h_{\text{out}}$, so

$$W = \int_{y_1}^{y_2} \rho,g,A(y),\left(h_{\text{out}} - y\right)dy$$

Here $\rho = 1000,\text{kg/m}^3$ for water, $g = 9.8,\text{m/s}^2$, and $y_1$ to $y_2$ covers only the heights where water actually sits. The shape of the tank changes only $A(y)$:

• Upright cylinder: every layer is the same circle, so $A(y) = \pi r^2$. The right circular cylinder is the easiest tank.

• Cone with its point down: the radius grows in proportion to the height, so $A(y) = \pi\left(\frac{R}{H}y\right)^2$ for a cone of top radius $R$ and height $H$.

• Sphere of radius $R$ centred at the origin: the layer radius satisfies $r^2 = R^2 - y^2$, so $A(y) = \pi\left(R^2 - y^2\right)$. A hemisphere uses the same formula over half the range of $y$.

• Horizontal cylinder of radius $R$ and length $\ell$, axis at $y = 0$: each layer is a rectangle, $\ell$ long and $2\sqrt{R^2 - y^2}$ wide, so $A(y) = 2\ell\sqrt{R^2 - y^2}$.

Is Work The Same As Energy?

Work is energy in transit: the amount of energy a force moves into or out of an object. The work-energy theorem makes this exact. The net work done by all forces on an object equals its change in kinetic energy, $W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$.

Work done on a spring is stored as elastic potential energy, which is why the $\frac{1}{2}kx^2$ from the spring section reappears in physics as the spring's energy formula. When the path curves through the plane or through space, the same idea becomes a line integral of the force along the path, a topic for multivariable calculus.

Examples Of Work Done By A Force

Example 1

A constant force of 50 N pushes a crate along the floor from x = 0 m to x = 4 m. Find the work using the integral.

$$W = \int_{0}^{4} 50,dx$$

$$= \left[50x\right]_{0}^{4}$$

$$= 200 - 0$$

$$= 200,\text{J}$$

Final answer: $W = 200,\text{J}$, which matches $Fd = 50 \times 4$.

Example 2

A force F(x) = 3x² + 2 newtons acts on an object as it moves from x = 1 m to x = 3 m. Find the work done.

$$W = \int_{1}^{3} \left(3x^2 + 2\right)dx$$

$$= \left[x^3 + 2x\right]_{1}^{3}$$

$$= (27 + 6) - (1 + 2)$$

$$= 30,\text{J}$$

The antiderivative checks out: with $G(x) = x^3 + 2x$, $G'(x) = 3x^2 + 2$.

Final answer: $W = 30,\text{J}$, the shaded area in the graph above.

Example 3

A spring has a natural length of 40 cm. A force of 30 N holds it stretched to 55 cm. How much work does it take to stretch the spring from 50 cm to 70 cm?

The tempting route is to find the force at the end and multiply by the distance moved. The stretch at 55 cm is $0.15,\text{m}$, so $30 = k(0.15)$ and $k = 200,\text{N/m}$. At 70 cm the stretch is $0.30,\text{m}$, so the force there is $200 \times 0.30 = 60,\text{N}$, and $60 \times 0.20 = 12,\text{J}$.

That answer cannot be right. The spring pulls back with 60 N only at the very end; at 50 cm it pulls back with just $200 \times 0.10 = 20,\text{N}$. Using the largest force for the whole 0.20 m counts too much work. The force changes along the way, so the integral has to do the adding.

Rescue: convert both positions to stretches beyond the natural length, then integrate.

$$k(0.15) = 30, \quad k = 200,\text{N/m}$$

$$50,\text{cm} \rightarrow x = 0.10,\text{m}, \quad 70,\text{cm} \rightarrow x = 0.30,\text{m}$$

$$W = \int_{0.10}^{0.30} 200x,dx$$

$$= \left[100x^2\right]_{0.10}^{0.30}$$

$$= 100,(0.09 - 0.01)$$

$$= 8,\text{J}$$

Check: the force rises evenly from 20 N to 60 N, so its average is 40 N, and $40 \times 0.20 = 8,\text{J}$.

Final answer: $W = 8,\text{J}$.

Example 4

A 10 m chain with a mass of 2 kg per metre hangs from the top of a building. How much work is needed to pull the whole chain up to the top? Use g = 9.8 m/s².

Let $y$ be the depth of a piece below the top. A piece of length $dy$ weighs $2(9.8),dy = 19.6,dy$ newtons and rises $y$ metres.

$$W = \int_{0}^{10} 19.6,y,dy$$

$$= 19.6\left[\frac{y^2}{2}\right]_{0}^{10}$$

$$= 19.6 \times 50$$

$$= 980,\text{J}$$

Check with the shortcut: the chain weighs $2 \times 10 \times 9.8 = 196,\text{N}$ and its centre of mass rises 5 m, so $196 \times 5 = 980,\text{J}$.

Final answer: $W = 980,\text{J}$.

Example 5

An 8 kg bucket hangs at the bottom of a 20 m well on a rope with a mass of 0.5 kg per metre. Find the work needed to wind the bucket and rope to the top.

Let $x$ be the length of rope already wound up. The load still hanging is the bucket plus $20 - x$ metres of rope, so the force is $9.8\left(8 + 0.5(20 - x)\right)$ newtons.

$$W = \int_{0}^{20} 9.8\left(8 + 0.5(20 - x)\right)dx$$

$$= 9.8\int_{0}^{20} \left(18 - 0.5x\right)dx$$

$$= 9.8\left[18x - 0.25x^2\right]_{0}^{20}$$

$$= 9.8,(360 - 100)$$

$$= 2548,\text{J}$$

Check by parts: the bucket alone needs $8 \times 9.8 \times 20 = 1568,\text{J}$, the rope alone needs $\frac{1}{2}(0.5)(9.8)(20^2) = 980,\text{J}$, and $1568 + 980 = 2548$.

Final answer: $W = 2548,\text{J}$.

Example 6

An upright cylindrical tank has radius 2 m and height 5 m. It holds water 4 m deep. Find the work to pump all the water out through an outlet 1 m above the top of the tank. Use 1000 kg/m³ for water and g = 9.8 m/s².

Put $y = 0$ at the bottom of the tank, so the outlet sits at $y = 6$. A layer at height $y$ has volume $\pi(2)^2,dy = 4\pi,dy$, weight $1000 \times 9.8 \times 4\pi,dy = 39200\pi,dy$ newtons, and travels $6 - y$ metres. The water only fills $0 \leq y \leq 4$.

$$W = \int_{0}^{4} 39200\pi,(6 - y),dy$$

$$= 39200\pi\left[6y - \frac{y^2}{2}\right]_{0}^{4}$$

$$= 39200\pi,(24 - 8)$$

$$= 627200\pi,\text{J}$$

$$\approx 1970406.9123,\text{J}$$

Check with the centre of mass: the water's mass is $1000 \times 16\pi = 16000\pi,\text{kg}$, its centre sits at $y = 2$, and it rises 4 m to the outlet, so $16000\pi \times 9.8 \times 4 = 627200\pi,\text{J}$.

Final answer: $W = 627200\pi,\text{J} \approx 1.97 \times 10^{6},\text{J}$.

What Problem Does Work Done By A Force Solve?

Work done by a force solves one specific problem: force times distance fails when the force keeps changing. Engineers needed a single, honest number for the effort a machine delivers, even when the load shrinks, the spring stiffens or the water sits at many heights. The integral gives that number.

The three classic cases each break $W = Fd$ in a different way:

• Springs: the force grows with every millimetre of stretch, so no single force value describes the motion.

• Cables and chains: the load gets lighter as the cable winds up, so the force falls along the path.

• Pumping: every layer of water travels its own distance, so no single distance describes the job.

These are the headline applications in standard calculus courses, as in the OpenStax Calculus Volume 1 section on physical applications. Once you see that all three are "slice, weigh, measure the distance, add", the topic stops being three separate formulas.

Where Is Work Done By A Force Used In Real Life?

The work integral sits underneath any job where a load, a stretch or a lift changes along the way:

• Water supply: sizing the pumps that fill water towers and empty reservoirs is a pumping integral over the shape of the tank.

• Suspension and spring design: the energy a car spring or a shock absorber stores comes from integrating its force over its compression.

• Cranes, lifts and mine hoists: a long steel cable is heavy, so the motor's work depends on how much cable is still hanging.

• Archery: the energy stored in a drawn bow is the area under its draw-force curve, which is why archers plot force against draw length.

• Engines: gas pressure pushing a piston does work that is computed the same way, with pressure and volume in place of force and distance.

Where Do Students Trip Up On Work Done By A Force?

Mistake 1: Measuring The Spring From The Wrong Zero

Where it slips in: Spring problems that give the spring's total length (55 cm, 70 cm) instead of its stretch.

Don't do this: Plug the lengths straight into $F = kx$. In Example 3, using $x = 0.50$ to $x = 0.70$ as limits gives $100(0.49 - 0.25) = 24,\text{J}$, three times the true answer.

The correct way: Subtract the natural length first and convert to metres, so the limits become $x = 0.10$ and $x = 0.30$.

The rusher reads "70 cm" and writes 0.70 in the integral before checking what $x$ means. The first instinct on any spring question is to treat the lengths in the question as values of $x$. Rewriting every length as "stretch beyond natural length" before you write the integral removes the error at its source.

Mistake 2: Treating Mass As If It Were Force

Where it slips in: Chain problems given in kilograms per metre and tank problems given with water at $1000,\text{kg/m}^3$.

Don't do this: Integrate the mass directly. For the chain in Example 4, $\int_{0}^{10} 2y,dy = 100$, which is not a work in joules at all.

The correct way: Multiply mass by $g = 9.8,\text{m/s}^2$ to get weight in newtons, which gives $980,\text{J}$. US problems in $\text{lb/ft}$ or $62.4,\text{lb/ft}^3$ already give weight, so they need no extra $g$.

The memorizer learns the US version first, with $62.4$ already a weight, and carries the same template into SI problems without the $g$.

Mistake 3: Keeping The Layer Radius Fixed In A Curved Tank

Where it slips in: Spherical and conical tanks, where each layer of water has a different width.

Don't do this: Use $\pi R^2,dy$ for every layer of a sphere, as if the tank were a cylinder. A Physics Forums homework thread on a spherical tank shows exactly this setup going wrong.

The correct way: Write the layer radius as a function of height first: $r^2 = R^2 - y^2$ for a sphere, $r = \frac{R}{H}y$ for a cone with its point down.

Students who sketch one layer and label its radius in terms of $y$ before writing any integral stop making this mistake, because the sketch forces the question "how wide is this layer?"

Mistake 4: Integrating Over The Tank Instead Of The Water

Where it slips in: Tanks that are only partly full, or outlets that sit above the top of the tank.

Don't do this: Run the limits from the bottom to the top of the tank, or measure the distance to the top rim when the outlet sits higher.

The correct way: Set the limits where the water is (0 to 4 m in Example 6) and measure each layer's distance to the outlet ($6 - y$ there).

Practice Problems On Work Done By A Force

1. A constant force of 12 N moves a box from $x = 2,\text{m}$ to $x = 7,\text{m}$. Find the work.

2. A force $F(x) = \frac{6}{x^2}$ newtons acts from $x = 1,\text{m}$ to $x = 3,\text{m}$. Find the work.

3. A force of 24 N holds a spring stretched 8 cm beyond its natural length. Find the work to stretch it from its natural length to 20 cm beyond it.

4. A 15 m cable with a mass of 3 kg per metre hangs from a winch. Find the work to wind up the first 5 m of cable. Use $g = 9.8,\text{m/s}^2$.

5. A rectangular tank with a 3 m by 2 m base is 2 m deep and full of water. Find the work to pump all the water over the top edge.

Answer to Problem 1: $\int_{2}^{7} 12,dx = 60,\text{J}$.

Answer to Problem 2: $\int_{1}^{3} \frac{6}{x^2},dx = \left[-\frac{6}{x}\right]_{1}^{3} = -2 + 6 = 4,\text{J}$.

Answer to Problem 3: $k = \frac{24}{0.08} = 300,\text{N/m}$, so $W = \int_{0}^{0.2} 300x,dx = 150(0.04) = 6,\text{J}$.

Answer to Problem 4: $W = \int_{0}^{5} 3(9.8)(15 - x),dx = 29.4,(75 - 12.5) = 1837.5,\text{J}$.

Answer to Problem 5: $W = \int_{0}^{2} 1000(9.8)(6)(2 - y),dy = 58800 \times 2 = 117600,\text{J}$.

What To Remember About Work Done By A Force

• Work done by a force that varies along a straight path is $W = \int_{a}^{b} F(x),dx$, and $W = Fd$ is its constant-force special case.

• The integral comes from adding the work on short pieces and letting the pieces shrink, which is why work equals the area under the force-displacement graph.

• Springs use Hooke's law $F(x) = kx$, with $x$ measured from the natural length.

• Chains and tanks use "weight of a slice times the distance it moves", with mass multiplied by $g = 9.8,\text{m/s}^2$.

• Check every pumping answer for the right layer area, limits set where the water is, and distance measured to the outlet.

Your Next Move

Work through Practice Problems 1 to 5 in order; if the pumping setup in Problem 5 trips you up, return to the section on pumping water out of a tank and draw one layer before you write the integral. Want a live Bhanzu trainer to walk through more work problems with you? Book a free demo class.

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