One Method, Several Physical Models
Work problems involving variable forces share a common mathematical structure, but the physical force model changes from one situation to another. A spring force depends on its stretch. A hanging cable has different portions lifted by different distances. In a liquid-pumping problem, each slice has its own weight and lift distance. A piecewise force law may require separate integrals on separate intervals.
The central modeling task is to describe the motion with a coordinate and determine the force component acting along that motion at each position. Once those choices are clear, work is accumulated by integration. The integral is only as reliable as the coordinate, limits, and force model used to construct it.
In physical applications, the integrand has units of force and the differential has units of length. Thus work has units of force multiplied by distance: joules when force is measured in newtons and distance in metres, or foot-pounds when force is measured in pounds and distance in feet.
Why the Work Integral Applies
For a small displacement \(\Delta x\), a force that is nearly constant over that displacement does approximately \(F(x)\Delta x\) units of work. Divide the full interval into small pieces, evaluate the force at a sample point \(x_i^*\) in each piece, and add the approximate contributions:
As the pieces become shorter, this sum approaches the definite integral of the force over the displacement interval. This explains why the integral uses the force component along the motion and the coordinate’s displacement. It also clarifies why a force that varies with position cannot generally be replaced by one value multiplied by the entire distance.
Worked Example: Stretching a Spring Through a Range
Worked Example: Stretching a Spring Through a Range
A spring has spring constant \(k=120\text{ N/m}\). It is stretched from \(x=0.10\text{ m}\) to \(x=0.25\text{ m}\), where \(x\) is measured from the spring’s unstretched length. Find the work done by the external force, assuming the spring is stretched slowly enough that the applied force balances the spring force.
Coordinate and limits: Let \(x\) be the spring’s stretch in metres. The motion goes from \(0.10\text{ m}\) to \(0.25\text{ m}\).
Force model: By Hooke’s law, the spring force has magnitude \(kx\) and points opposite the stretch. The external force needed to stretch it slowly points in the positive direction and has magnitude \(F(x)=kx=120x\text{ N}\).
Set up and evaluate:
The units are \((\text{N/m})(\text{m})(\text{m})=\text{N}\cdot\text{m}=\text{J}\). To check, differentiate the antiderivative: \(\frac{d}{dx}(60x^2)=120x\), the stated external force. Substituting the two limits gives \(3.75-0.60=3.15\text{ J}\). The positive result is consistent with an external force doing work while stretching the spring.
Worked Example: Lifting a Hanging Cable
Worked Example: Lifting a Hanging Cable
A uniform cable is \(30\text{ m}\) long and has total mass \(80\text{ kg}\). It hangs vertically, and its top end is pulled up until the entire cable is at the top. Find the work required against gravity. Use \(g=9.8\text{ m/s}^2\).
Coordinate and limits: Let \(y\) measure distance down the hanging cable from its top. A small piece at coordinate \(y\) must be lifted \(y\) metres. The cable pieces range from \(y=0\) to \(y=30\text{ m}\).
Mass and force model: The cable’s uniform mass per unit length is \(\lambda=\frac{80\text{ kg}}{30\text{ m}}=\frac{8}{3}\text{ kg/m}\). A short piece of length \(dy\) has mass \(\lambda\,dy\), so its weight is \(\lambda g\,dy\). Its contribution to the lifting work is its weight multiplied by its lift distance \(y\).
Set up and evaluate:
Here \(\lambda g\) has units \(\text{N/m}\), and multiplying by \(y\,dy\) gives \(\text{N}\cdot\text{m}\). For a check, the antiderivative of \(\lambda g y\) is \(\frac{\lambda g}{2}y^2\); differentiating it recovers \(\lambda g y\). Evaluating at \(30\) and \(0\) gives the displayed \(11760\text{ J}\). As a second check, the cable’s centre is initially \(15\text{ m}\) below the top, so lifting the whole \(80\text{ kg}\) cable through \(15\text{ m}\) requires \(80\cdot9.8\cdot15=11760\text{ J}\), in agreement.
Worked Example: Pumping Liquid from a Cylindrical Tank
Worked Example: Pumping Liquid from a Cylindrical Tank
A cylindrical tank has radius \(1.2\text{ m}\) and height \(3.0\text{ m}\). It contains water to a depth of \(2.5\text{ m}\), and the water is pumped over the top edge. Find the work required. Use water density \(1000\text{ kg/m}^3\) and \(g=9.8\text{ m/s}^2\).
Coordinate and limits: Let \(y\) be the height in metres above the tank bottom. A thin horizontal slice of water at height \(y\) must be lifted \(3-y\) metres to the tank top. Water occupies \(0\leq y\leq2.5\).
Force model: Each slice has thickness \(dy\) and cross-sectional area \(A=\pi(1.2)^2=1.44\pi\text{ m}^2\). Its volume is \(1.44\pi\,dy\), so its weight is \(\rho gA\,dy\). The work for that slice is its weight times its lift distance \(3-y\).
Set up and evaluate:
The approximation is rounded to the nearest joule. The coefficient \(14112\pi\) has units \(\text{N/m}\), while the integral of the lift distance with respect to height has units \(\text{m}^2\); their product is joules. Differentiating \(3y-\frac{y^2}{2}\) gives \(3-y\), the lift-distance factor. The antiderivative evaluated at \(2.5\) is \(4.375\), and at \(0\) it is \(0\), confirming the limits and exact result. The work is positive because the pump raises the water against gravity.
Worked Example: A Piecewise Force Law
Worked Example: A Piecewise Force Law
An object moves in the positive \(x\)-direction from \(x=0\text{ m}\) to \(x=5\text{ m}\). The applied force is \(F(x)=4x\text{ N}\) for \(0\leq x\leq2\text{ m}\), and \(F(x)=8\text{ N}\) for \(2<x\leq5\text{ m}\). Find the work done by this force.
Coordinate and limits: The coordinate \(x\) is the object’s position in metres; the complete interval is \(0\leq x\leq5\).
Force model and split: The formula for the force changes at \(x=2\text{ m}\), so the integral must be evaluated separately on the two intervals. At \(x=2\), the first formula gives \(4(2)=8\text{ N}\), matching the second formula.
On the first interval, differentiating \(2x^2\) gives \(4x\); on the second, differentiating \(8x\) gives \(8\). Evaluating at the stated endpoints gives \(8\text{ J}\) and \(24\text{ J}\), respectively, for a total of \(32\text{ J}\). The units are newton-metres, or joules.
How to Choose and Check a Model
These problems are easier to organize when each has the same modeling record. The physical situation supplies the details, but the questions remain consistent.
| Modeling choice | Question to answer |
|---|---|
| Coordinate | What position or slice location does the variable measure? |
| Limits | Which positions or slices undergo the motion being counted? |
| Force | What force acts on each piece, and what is its component along the motion? |
| Distance | How far does each piece move? Is that distance constant or position-dependent? |
| Units | Does force times displacement produce the appropriate unit of work? |
| Interpretation | Should the work be positive or negative, and does its size fit the situation? |
The coordinate need not measure the same physical quantity in every problem. For a spring it can measure stretch; for a hanging cable it can measure distance down from the top; for a liquid it can measure height above the tank bottom. The coordinate is useful when it makes both the slice’s force and its displacement straightforward to express.
A Common Modeling Pitfall
A frequent error in pumping problems is to assign every slice the same lift distance. In the cylindrical-tank example, a slice at height \(y\) travels \(3-y\) metres, not \(3\) metres. A slice at the bottom travels \(3\text{ m}\), while one near the water surface travels only about \(0.5\text{ m}\).
For instance, incorrectly using a \(3\text{ m}\) lift for all the water would give
That treats each slice as if it began at the tank bottom. The correct lift distance varies with \(y\), so the correct work is \(14112\pi\int_0^{2.5}(3-y)\,dy=61740\pi\text{ J}\). The wrong answer is larger because it overestimates the distance travelled by every slice except the one at the bottom.
Interpreting the Result
A work integral is not merely an instruction to integrate a formula. It encodes a physical account of how force acts across displacement. In a spring problem, the force changes as the stretch changes. In a cable problem, pieces have different distances to travel. In pumping, both the slice’s location and its lift distance matter. In a piecewise problem, each force formula applies only on its specified interval.
Before accepting a result, check that the integration limits cover exactly the material or motion being counted, that every changing quantity is expressed in terms of the chosen coordinate, and that the units reduce to force times distance. Finally, consider the sign and interpretation: positive work may be required to lift or stretch an object, while a force opposing the displacement does negative work.
Check Your Understanding
For each question, identify the modeling choice that determines the work integral before attempting to evaluate it.
- A spring is stretched from a nonzero initial stretch to a larger final stretch. What coordinate and limits should be used, and why is the external force not constant?
- For a uniform hanging cable pulled to its top, why does a piece at distance \(y\) below the top travel \(y\) metres?
- In a liquid-pumping problem, what quantities determine the weight of a thin horizontal slice?
- A force law changes formula at a specified position. How should the work integral be organized?
- What units must result when force is measured in newtons and displacement in metres?