Why Variable-Force Work Models Go Wrong
A variable-force work problem can have a correct integral form and still be modeled incorrectly. The usual causes are mismatched choices: a force component is given a sign without reference to the coordinate axis, a slice is measured in one coordinate but its lifting distance is written in another, or a piecewise force is integrated over only part of the motion.
The reliable approach is to identify the direction of motion, define the coordinate, state the force component in that coordinate, and calculate each element's displacement. For lifting problems, the force needed is often the weight of a slice, while the work done on it depends on how far that slice travels. Those are different quantities.
Coordinate Direction Determines the Force Sign
In one-dimensional motion, choose a positive direction and measure position \(x\) along that axis. The force component \(F_x\) is positive when the force points in the positive \(x\)-direction and negative when it points in the negative \(x\)-direction. The displacement element \(dx\) is also signed: it is positive for motion in the positive direction and negative for motion in the negative direction.
Thus, a force that opposes the motion does negative work, but its component is not necessarily negative. The component's sign is determined by the coordinate axis, not by whether the force helps or resists the motion. This distinction prevents a common sign error.
Worked Example: Force Opposing Motion When Position Decreases
An object moves from \(x=1\text{ m}\) to \(x=0\text{ m}\) while a constant \(+1\text{ N}\) force acts in the positive \(x\)-direction. The force opposes the object's displacement. Find the work done by the force.
The force points in the positive coordinate direction, so its component is \(F_x=+1\text{ N}\). The initial and final positions are \(a=1\text{ m}\) and \(b=0\text{ m}\). Therefore,
The negative result is consistent with the force opposing the motion. The force component is positive, while the displacement is negative. A check using force times displacement gives \(W=(+1\text{ N})(-1\text{ m})=-1\text{ J}\).
Deriving the Work Integral and Keeping Its Sign
Over a small displacement \(\Delta x\), a force that is approximately constant does work approximately equal to its component times the signed displacement. Subdividing the motion into small intervals gives a sum of contributions. If the force component is continuous, the limit of these sums is the definite integral of the component over position.
Let \(x=a\) be the starting position and \(x=b\) the ending position. The order of these bounds records the direction of travel.
At each position, \(F_x(x)\) is positive or negative according to the chosen axis, not according to whether it helps the motion.
For a small interval, the contribution is approximately \(F_x(x_i^*)\Delta x_i\), where \(x_i^*\) is a point in that interval.
The sum approaches the definite integral, so \(W=\int_a^b F_x(x)\,dx\). For a piecewise-continuous force, integrate over its pieces and add the results.
This argument explains why reversing the bounds reverses the sign of the work for the same force function: \(\int_b^a F_x(x)\,dx=-\int_a^b F_x(x)\,dx\). A negative answer is not automatically an error. It may correctly indicate that the force does negative work during the stated motion.
Spring Work: Use Extension, Not an Unspecified Position
For an ideal spring with spring constant \(k\), let \(x\) denote its signed extension from its natural length, with positive \(x\) chosen in the direction of stretch. Hooke's law gives the spring's force component as \(F_{\text{spring}}(x)=-kx\). If an external agent slowly stretches the spring, its force is \(F_{\text{ext}}(x)=kx\) in the positive direction. Work by the spring and work by the external agent have opposite signs in this situation.
Worked Example: Work to Stretch a Spring from a Nonzero Extension
A spring has \(k=80\text{ N/m}\). It is stretched slowly from \(x=0.05\text{ m}\) to \(x=0.20\text{ m}\), where \(x\) is measured from the spring's natural length. Find the external work required.
The external force balances the spring's restoring force, so \(F_{\text{ext}}(x)=kx=80x\text{ N}\). The extension values, rather than distances measured from an unrelated origin, are the integration bounds:
At the upper bound, \(40(0.20)^2=40(0.04)=1.6\text{ J}\). At the lower bound, \(40(0.05)^2=40(0.0025)=0.1\text{ J}\). Thus,
Check the units: \((\text{N/m})(\text{m})(\text{m})=\text{N}\cdot\text{m}=\text{J}\). The spring's work is \(-1.5\text{ J}\), since its force opposes the stretch. The external work is positive. This also checks the sign convention.
Cable Lifting: Each Slice Has Its Own Travel Distance
For a uniform cable, its weight per unit length, denoted \(\lambda\), has units of \(\text{N/m}\). A slice of length \(dy\) weighs \(\lambda\,dy\). To calculate work, multiply that weight by the distance that particular slice moves. A frequent mistake is to use the total cable length as the lifting distance for every slice.
Worked Example: Pulling a Uniform Hanging Cable onto a Platform
A uniform \(10\text{ m}\) cable hangs vertically from a platform. Its total weight is \(100\text{ N}\). Find the work required to pull the entire cable onto the platform, neglecting friction.
The weight per unit length is
Let \(y\) be the initial distance of a slice below the platform. Then \(y\) ranges from \(0\text{ m}\) at the top to \(10\text{ m}\) at the bottom. A slice at distance \(y\) must be lifted \(y\) metres. Its weight is \((10\text{ N/m})\,dy\), so its work is its weight multiplied by \(y\):
Evaluating gives
Check: the whole cable weighs \(100\text{ N}\), and its slices have lifting distances ranging uniformly from \(0\) to \(10\text{ m}\), with average distance \(5\text{ m}\). Thus the work is \(100\text{ N}\times5\text{ m}=500\text{ J}\), agreeing with the integral. The integrand has units \((\text{N/m})(\text{m})(\text{m})=\text{J}\).
Pumping Liquid: Separate Slice Weight from Lift Distance
For a liquid of constant weight density \(\rho g\), a slice with volume \(dV\) weighs \(\rho g\,dV\). Its work contribution is this weight times the vertical distance from the slice to the outlet. The slice's height coordinate and its lifting distance are generally not the same: the latter depends on where the outlet is.
Worked Example: Pumping Water from a Cylindrical Tank
A cylindrical tank has horizontal cross-sectional area \(0.5\text{ m}^2\), height \(3\text{ m}\), and water filling it to a depth of \(2\text{ m}\). Water is pumped to an outlet at the top of the tank. Use water's weight density \(9800\text{ N/m}^3\). Find the work required, neglecting losses.
Measure \(y\) upward from the tank bottom. The water occupies \(0\le y\le2\), and each thin horizontal slice has volume \(dV=(0.5\text{ m}^2)\,dy\). The outlet is at height \(3\text{ m}\), so a slice at height \(y\) is lifted \(3-y\) metres. Its weight is \((9800\text{ N/m}^3)(0.5\text{ m}^2)\,dy=4900\,dy\text{ N}\). Therefore,
Evaluating the integral,
Check independently: the water volume is \((0.5\text{ m}^2)(2\text{ m})=1\text{ m}^3\), so its weight is \(9800\text{ N}\). Its slices range from \(3\text{ m}\) of lift at the bottom to \(1\text{ m}\) at the water surface, so the average lift is \(2\text{ m}\). The work is \(9800\text{ N}\times2\text{ m}=19600\text{ J}\). The integral's units are \((\text{N/m}^3)(\text{m}^2)(\text{m})(\text{m})=\text{J}\).
Piecewise Forces: Account for Every Part of the Motion
A piecewise force law assigns different formulas or values to different position intervals. The work over the whole motion is the sum of the integrals over those intervals. A single formula must not be extended beyond the interval where it applies, and an interval must not be omitted simply because the force changes there. The value at one isolated boundary does not change the integral.
Worked Example: Work Under a Two-Part Force Law
An object moves in the positive \(x\)-direction from \(x=0\text{ m}\) to \(x=4\text{ m}\). The force component is \(3\text{ N}\) for \(0\le x<2\text{ m}\) and \(5\text{ N}\) for \(2\le x\le4\text{ m}\). Find the work done by this force.
Both pieces contribute. Split the integral at the point where the force changes:
The first interval contributes \(3(2-0)=6\text{ J}\). The second contributes \(5(4-2)=10\text{ J}\). Hence
Check by adding force-times-distance on each constant-force interval: \(3\text{ N}\times2\text{ m}+5\text{ N}\times2\text{ m}=6\text{ J}+10\text{ J}=16\text{ J}\). Omitting the second interval would give only \(6\text{ J}\), the work on the first half of the motion, not the total.
Common Error Checks
Before accepting a work integral, check its setup rather than relying on whether the final number seems plausible. These checks catch many errors early:
- Force sign: State the positive coordinate direction, then assign the force component. Do not label a force component negative merely because the force resists motion.
- Coordinate match: Use the same variable in the force law, differential, and bounds. If a coordinate is measured from the bottom but the outlet is at the top, write the lift distance as outlet height minus slice height.
- Travel distance: For a cable or liquid, determine how far each slice moves. Do not substitute the whole cable length or tank height for every slice without checking the geometry.
- Intervals: For a piecewise law, split at each change and include every portion of the path.
- Units: A work integrand must reduce to force times distance, or \(\text{N}\cdot\text{m}=\text{J}\).
- Independent check: When possible, compare with a simple force-times-distance estimate, a symmetry or average-distance argument, or the expected sign.
Check Your Understanding
For each question, identify the modeling choice that matters before calculating.
- An object moves from \(x=5\text{ m}\) to \(x=2\text{ m}\) under a constant force pointing in the positive \(x\)-direction. What are the signs of the force component and the work?
- A spring is stretched from \(0.1\text{ m}\) to \(0.3\text{ m}\) beyond its natural length. Which coordinate should supply the bounds in the spring-force integral?
- A uniform cable hangs below a platform. Why does a slice near the platform require less work to lift than a slice near the bottom?
- Water occupies the bottom of a tank and is pumped to an outlet above it. If \(y\) is measured from the bottom, how can the slice's lift distance be expressed using the outlet height?
- A force law changes at \(x=2\text{ m}\), while the object moves from \(x=0\text{ m}\) to \(x=5\text{ m}\). How should the work integral reflect that change?