From Force to Work
A force can transfer energy when it moves an object. If the force is constant and acts along the direction of motion, the work is the force multiplied by the distance moved. But when the force changes with position, no single force value describes the whole trip. The work must account for the force at each part of the path.
The definite integral provides a precise way to do that. We divide the path into small intervals, estimate the work over each interval using the force there, and add the contributions. Taking the limit as the intervals become arbitrarily small gives the work integral.
These assumptions keep the force aligned with the motion throughout the path. The integral developed here is for this one-dimensional situation; it does not require us to resolve changes in direction.
Small Force-Displacement Contributions
For a constant force \(F\), moving an object a distance \(\Delta x\) in the force’s direction contributes \(F\Delta x\) units of work. If the force varies, we can still use this rule approximately on a sufficiently short interval: the force changes little over that interval, so we can represent it by a value at a sample position.
To make this precise, partition the path \([a,b]\) into \(n\) subintervals with endpoints \(a=x_0<x_1<\cdots<x_n=b\). The width of the \(i\)th subinterval is \(\Delta x_i=x_i-x_{i-1}\). Choose a sample point \(\xi_i\) in that subinterval. The work on the subinterval is approximated by \(F(\xi_i)\Delta x_i\), and the total approximate work is the sum of these contributions.
As the partition is refined, the subintervals become shorter. Since \(F\) is continuous on the closed interval \([a,b]\), it is uniformly continuous there: on sufficiently short subintervals, the force values are uniformly close. Thus the approximation on each subinterval improves, and the sums converge to the definite integral of \(F\) from \(a\) to \(b\).
Why the Sum Becomes an Integral
The connection to the definition of a definite integral is direct. For any chosen partition and sample points, \(F(\xi_i)\Delta x_i\) is a rectangle’s height times width. The sum adds the areas of all these rectangles. The definite integral is the limit of these Riemann sums as the largest subinterval width tends to zero.
Continuity ensures that this limit is well-defined. More explicitly, on each subinterval, the difference between any two force values is no greater than the largest variation of \(F\) over intervals of that width. Uniform continuity says this largest variation tends to zero as the width tends to zero. Multiplying that variation by the total path length \(b-a\) bounds the total error between choices of force values on the subintervals. Because \(b-a\) is fixed, that error tends to zero as well. Therefore the limit of the small force-displacement sums is precisely \(\int_a^b F(x)\,dx\).
This argument explains why the integral is not an arbitrary formula attached to work. It is the limiting total of the local contributions \(F(\xi_i)\Delta x_i\).
Constant Force
If the force has the same value \(F_0\) at every position on the path, it can be taken outside the integral. The result is the familiar force-times-distance formula.
Here, \(b-a\) is the distance traveled because \(b>a\). The formula applies when the force is constant and acts along the direction of motion over the stated interval. For a varying force, the force value must remain inside the integral.
Worked Example: Constant Force Over a Measured Distance
A constant force of \(12\,\mathrm{N}\) acts along the direction of motion as an object moves from \(x=2\,\mathrm{m}\) to \(x=7\,\mathrm{m}\). Find the work.
The force is constant, so \(F(x)=12\,\mathrm{N}\). The distance traveled is \(7\,\mathrm{m}-2\,\mathrm{m}=5\,\mathrm{m}\). Set up the work integral:
Evaluate it:
Check: The force multiplied by the distance is \(12\,\mathrm{N}\cdot5\,\mathrm{m}=60\,\mathrm{J}\), agreeing with the integral. The units are joules because \(1\,\mathrm{J}=1\,\mathrm{N\,m}\).
Position-Dependent Force
If the force depends on position, use its value \(F(x)\) at each position as the integrand. The bounds are the starting and ending positions, and \(dx\) indicates that the contributions are being accumulated over distance.
Before evaluating, check that the force model has units of newtons at every position. Since \(dx\) has units of metres, the integral then has units of newton-metres, or joules. A correct numerical answer with incorrect units is a warning that the model or calculation may be wrong.
Worked Example: A Force That Increases With Position
On the path from \(x=0\,\mathrm{m}\) to \(x=4\,\mathrm{m}\), suppose the force is \(F(x)=2\,\mathrm{N}+(3\,\mathrm{N/m})x\). Find the work.
The force varies with position, so keep the full expression inside the integral:
Integrate each term with respect to \(x\):
At the upper bound, the first term is \((2\,\mathrm{N})(4\,\mathrm{m})=8\,\mathrm{N\,m}\). The second is \(\frac{3\,\mathrm{N/m}}{2}(4\,\mathrm{m})^2 =\frac{3}{2}\cdot16\,\mathrm{N\,m} =24\,\mathrm{N\,m}\). At the lower bound, both terms are zero. Therefore:
Check: An antiderivative is \(A(x)=(2\,\mathrm{N})x+\frac{3\,\mathrm{N/m}}{2}x^2\). Differentiating gives \(A'(x)=2\,\mathrm{N}+(3\,\mathrm{N/m})x=F(x)\). Also, the force increases from \(2\,\mathrm{N}\) to \(14\,\mathrm{N}\), so a positive work value is consistent with the model.
Worked Example: A Force That Decreases With Position
Suppose the force along a path is \(F(x)=\dfrac{10\,\mathrm{N\,m}}{x+1\,\mathrm{m}}\) as the object moves from \(x=0\,\mathrm{m}\) to \(x=3\,\mathrm{m}\). Find the work. The factor \(x+1\,\mathrm{m}\) has units of metres, so the force law has units of newtons.
Set up the integral and use the logarithmic antiderivative:
The approximation is rounded to four decimal places. The logarithm’s argument is dimensionless, as required, because the numerator and denominator both have units of metres.
Check: Differentiating \(10\,\mathrm{N\,m}\ln\left(\frac{x+1\,\mathrm{m}}{1\,\mathrm{m}}\right)\) with respect to \(x\) gives \(\frac{10\,\mathrm{N\,m}}{x+1\,\mathrm{m}}=F(x)\). The force decreases from \(10\,\mathrm{N}\) to \(2.5\,\mathrm{N}\), while remaining positive, consistent with the positive work obtained.
A Reliable Setup
Before doing algebra, identify what the force model says and which part of the path is being considered. A useful setup process is:
Record the starting position \(a\) and ending position \(b\), with \(a<b\) for the forward motion considered here.
Use \(F_0\) for a constant force or \(F(x)\) for a force that changes along the path.
Write \(W=\int_a^b F(x)\,dx\), including units in the force law and position bounds.
Substitute the bounds into an antiderivative, confirm its derivative is \(F(x)\), and verify that the result has units of joules.
A Common Pitfall: Treating a Variable Force as Constant
Suppose the force in the decreasing-force example is mistakenly treated as though its initial value, \(10\,\mathrm{N}\), acted over the full \(3\,\mathrm{m}\) path. This gives \(10\,\mathrm{N}\cdot3\,\mathrm{m}=30\,\mathrm{J}\). That multiplication is not the work for this model: the force decreases as position increases, so \(10\,\mathrm{N}\) does not describe the entire path.
The correct calculation retains the changing force in the integral and gives \(10\ln4\,\mathrm{J}\approx13.8629\,\mathrm{J}\), not \(30\,\mathrm{J}\). Multiplication by distance is valid for a constant force. For a position-dependent force, use the integral unless a separate argument justifies another method.
Why the Work Integral Is Useful
The integral model lets us calculate accumulated work even when the force changes continuously during motion. Its structure is simple and general: each small distance contributes approximately force times distance, and the definite integral adds those contributions across the path.
This also explains why the bounds matter. They specify which portion of the force law is being accumulated. Changing the start or end position changes the set of local force-displacement contributions included in the total. Before evaluating an integral, make sure its bounds match the actual path described in the problem.
Finally, the units provide an independent check. Force has units of newtons, and position has units of metres, so force integrated with respect to position has units \(\mathrm{N\,m}=\mathrm{J}\). The result therefore measures work, rather than force alone.
Check Your Understanding
Use the definitions and methods in this tutorial to answer the following questions.
- In the sum \(\sum_{i=1}^{n}F(\xi_i)\Delta x_i\), what does each factor represent in the work approximation?
- A constant \(5\,\mathrm{N}\) force acts along the path from \(x=1\,\mathrm{m}\) to \(x=4\,\mathrm{m}\). Write the work integral and find its value with units.
- For \(F(x)=x\,\mathrm{N/m}\) on \(0\leq x\leq2\,\mathrm{m}\), write the work integral. What units should the result have?
- Why does continuity of \(F\) help justify replacing small contributions by a definite integral?
- What is wrong with using \(F(a)(b-a)\) as the work when \(F\) changes along the path?
