From Small Displacements to Total Work
When a force changes as an object moves, one constant-force calculation cannot describe the entire motion. The force may be stronger at some positions and weaker at others, and it may even point partly across the direction of travel. The key idea is to divide the motion into short pieces. On each sufficiently small piece, the force can be treated approximately as constant; adding those local contributions and taking a limit leads to an integral.
We will consider motion along a straight coordinate axis. Let \(x\) denote position, and let \(F_{\parallel}(x)\) be the signed component of the force along the positive \(x\)-direction. A positive component points along increasing \(x\); a negative component points along decreasing \(x\). If the object moves from position \(a\) to position \(b\), the displacement is signed: a step toward larger \(x\) has positive displacement, while a step toward smaller \(x\) has negative displacement.
The integral records both the size and the direction of each contribution. It is not generally correct to replace the force component by its magnitude or to discard the orientation of the displacement.
Deriving the Work Integral
For a constant force component \(F_{\parallel}\) acting through a signed displacement \(\Delta x\), the one-dimensional work formula is \(W=F_{\parallel}\Delta x\). To use this local rule when the force varies, partition the interval of motion into small pieces. On each piece, use the force at a selected point as an approximation to the force throughout that piece.
Choose positions \(a=x_0<x_1<\cdots<x_n=b\), with subinterval widths \(\Delta x_i=x_i-x_{i-1}\). For motion toward decreasing \(x\), reverse the limits or, equivalently, retain signed displacement increments.
Choose a sample position \(x_i^*\) in each subinterval. Treat the force there as constant over that short piece, giving the approximation \(F_{\parallel}(x_i^*)\Delta x_i\).
The total approximate work is the sum of the local contributions. Smaller subintervals generally make this approximation more faithful to the changing force.
As the largest subinterval width tends to zero, the sum becomes the Riemann integral of the force component over the displacement interval.
For increasing motion from \(a\) to \(b\), the sum in the derivation has the form
If \(F_{\parallel}\) is continuous, it is Riemann integrable, so these sums converge to the same value as the partition is refined and its largest width tends to zero. Thus the local constant-force rule gives
This explains why the integral is the appropriate total: it is not an unrelated formula added to the local rule. It is the limit of adding that rule over smaller and smaller pieces.
Worked Example: Building a Work Integral from a Riemann Sum
Worked Example: Building a Work Integral from a Riemann Sum
A force component along a straight track is \(F_{\parallel}(x)=2x+1\) newtons, where \(x\) is measured in metres. Find the work as the object moves from \(x=0\) to \(x=2\), first using equal-width right-endpoint sums and then the integral.
Divide \([0,2]\) into \(n\) equal pieces. Each width is \(\Delta x=2/n\), and the right endpoint of piece \(i\) is \(x_i=2i/n\). The force at that endpoint is
The right-endpoint work sum is therefore
Taking the limit gives \(W=\lim_{n\to\infty}(6+4/n)=6\) joules. The integral gives the same result:
Check: the derivative of the antiderivative \(x^2+x\) is \(2x+1\), the stated force component. The Riemann-sum limit and the evaluated integral both give \(6\) joules.
Signed Displacement and Integration Limits
The limits describe the direction in which the path is traversed. For an integral, reversing the limits changes the sign:
This is consistent with the local formula. Reversing a displacement changes \(\Delta x\) to \(-\Delta x\), so the work contribution from the same force component changes sign. The sign of the force component and the sign of the displacement are distinct: either can be positive or negative.
Worked Example: Work with a Force That Changes Sign
Suppose \(F_{\parallel}(x)=6-2x\) newtons, and the object moves from \(x=1\) metre to \(x=4\) metres. Find the signed work.
The force component is positive for \(x<3\), zero at \(x=3\), and negative for \(x>3\). The work integral is
The result is a signed total: the positive contribution before \(x=3\) is partly offset by the negative contribution after \(x=3\). To check the evaluation, differentiate \(6x-x^2\); its derivative is \(6-2x\), as required. If the same path is traversed in reverse, the limits reverse and the work is \(\int_4^1(6-2x)\,dx=-3\) joules. The negative result follows from the reversed displacement, not from changing the force law.
When the Force Is Not Parallel to the Path
A force can have components both along and perpendicular to the motion. The dot product expresses the local rule compactly: the component perpendicular to a displacement contributes zero, while the component parallel to it determines the work. For motion along the \(x\)-axis, if the force vector is \(\mathbf{F}(x)=(F_x(x),F_y(x))\), then a displacement is \(\Delta\mathbf{r}=(\Delta x,0)\), and
Worked Example: Using Only the Component Along the Motion
An object moves along the \(x\)-axis from \(x=0\) to \(x=2\) metres while experiencing the force \(\mathbf{F}(x)=(x+1,4)\) newtons. Find the work.
The displacement is along the \(x\)-axis, so the force component \(x+1\) contributes and the constant perpendicular component \(4\) contributes zero. Therefore
Check: the derivative of \(x^2/2+x\) is \(x+1\), the component along the motion. Equivalently, the dot product with a small displacement \((dx,0)\) is \((x+1)dx+4(0)=(x+1)dx\), which gives exactly the integral used.
A Common Mistake: Using Force Magnitude Instead of Its Component
Suppose a constant force vector is \(\mathbf{F}=(3,4)\) newtons and an object moves \(2\) metres along the positive \(x\)-axis. A tempting but incorrect calculation is to use the force magnitude \(\sqrt{3^2+4^2}=5\) newtons and write \(W=5(2)=10\) joules. That treats the entire force as though it acted along the displacement.
The displacement has no \(y\)-component, so the correct calculation uses the \(x\)-component of the force:
The perpendicular component does not contribute. More generally, using the force magnitude in place of its component along the displacement gives the wrong work unless the force is parallel to that displacement.
Why the Derivation Is Useful
The derivation provides a reliable way to build a work calculation. It identifies what must be specified: the force component along the motion, the position interval, and the direction of traversal. The integral then accumulates the small signed contributions across the path. This same accumulation principle appears whenever a changing quantity contributes incrementally to a total.
For a continuous force component on a straight path, the central formula is
The assumptions and signs matter. Continuity is a sufficient condition for the Riemann-sum limit to exist. The integration variable describes position along the path, the integrand is the force component in that direction, and the limits follow the motion. Work is measured in joules when force is in newtons and displacement is in metres.
Check Your Understanding
Use the small-displacement definition and signed integration limits to answer these questions.
- Why does work from a variable force begin with a sum of local constant-force contributions?
- What condition on a force component guarantees the Riemann integral in the stated derivation exists?
- For motion along the \(x\)-axis, which component of a force vector contributes to work?
- What happens to the sign of work if the same force law is integrated with the limits reversed?
- Why is the magnitude of a force generally not the correct integrand for work?
