Introduction
Raising an object requires work against gravity. For a rigid object lifted vertically without changing its orientation, every part of the object rises through the same distance. This makes the calculation depend on the object's total mass and the vertical displacement, rather than on its shape.
There is an important distinction to keep clear. We can calculate the work done against gravity by summing gravity's effect over the object. To conclude that this is also the work done by the lifting force, we need an additional condition about the motion and any other forces doing work.
Weight and Work Against Gravity
Near Earth's surface, we often treat gravitational acceleration \(g\) as constant. A small portion of an object with mass \(dm\) then has a downward gravitational force of magnitude \(g\,dm\). The object's total weight has magnitude \(mg\), where \(m\) is its total mass.
Suppose the whole object is raised vertically by a distance \(d\). Each mass portion moves upward by that distance. Gravity acts downward, opposite to the displacement, so the work done by gravity on a mass portion is negative. The work required against gravity is positive.
For a uniform object, its mass is distributed uniformly through its volume or along its length. This can make it convenient to express \(dm\) using a constant density. But the final result will depend only on the total mass: every portion has the same displacement, so the contributions add to \(g d\) times the total mass.
Deriving the Lifting Formula
We can justify the formula by partitioning the object into small mass portions. For each portion, its contribution to work against gravity is its gravitational force magnitude multiplied by its upward distance. Adding these contributions gives the integral over the object's mass.
A portion with mass \(dm\) has gravitational force magnitude \(g\,dm\), and it rises by distance \(d\). Its work against gravity is \(g\,dm\,d\).
Integrating the contributions over all mass portions gives \(W_{\text{against gravity}}=\int g d\,dm\).
Because both are the same for every portion, \(W_{\text{against gravity}}=gd\int dm=mgd\), since the total mass is \(m\).
This derivation calculates work against gravity. It does not assert that the lifting force on each small portion equals that portion's weight. In a rigid object, forces transmitted between parts can affect the forces on individual portions.
To identify the total work done by the lifting force, use the work–energy theorem: the net work on an object equals its change in kinetic energy. If the object moves at constant speed, that change is zero. Gravity does work \(-mgd\) during the upward lift. If the lifting force and gravity are the only forces doing work, their total work must be zero, so the lifting force's total work is \(mgd\).
Worked Examples
Worked Example: Raising an Object With Known Mass
A \(12\text{ kg}\) object is raised \(2.5\text{ m}\) vertically near Earth's surface. Take \(g=9.8\text{ m/s}^2\). Find the work against gravity. If the object is raised at constant speed and no other forces do work, find the lifting force's total work as well.
The gravitational force magnitude on the whole object is
The work against gravity is force magnitude times vertical distance:
Under the stated constant-speed and no-other-work conditions, the lifting force's total work is also \(294\text{ J}\). As a check, \(mgd=(12)(9.8)(2.5)=294\), and the units reduce to \(\text{kg}\cdot\text{m}^2/\text{s}^2=\text{J}\).
Worked Example: Lifting a Uniform Bar
A uniform bar is \(3\text{ m}\) long and has mass \(6\text{ kg}\). It is lifted vertically, without rotating, through \(4\text{ m}\). Find the work against gravity. Use \(g=9.8\text{ m/s}^2\).
The bar's mass per unit length is constant:
Let \(x\) measure distance along the bar, from \(0\text{ m}\) to \(3\text{ m}\). A short piece of length \(dx\) has mass \(dm=\lambda\,dx=(2\text{ kg/m})\,dx\). It rises \(4\text{ m}\), just like every other piece. Therefore,
The factors \(2\text{ kg/m}\) and \(3\text{ m}\) give the total mass \(6\text{ kg}\); thus the result is \(9.8\cdot4\cdot6=235.2\text{ J}\), matching \(mgd\). This check confirms that distributing the calculation along the uniform bar gives the same answer as using its total mass.
Worked Example: Finding Mass From Uniform Density
A uniform rectangular object has dimensions \(0.4\text{ m}\), \(0.5\text{ m}\), and \(0.8\text{ m}\), and density \(750\text{ kg/m}^3\). It is raised vertically by \(1.2\text{ m}\). Find the work against gravity, using \(g=9.8\text{ m/s}^2\).
First find its volume and mass:
The density units cancel with the volume units to leave kilograms. The object's weight magnitude is
Therefore the work against gravity is
Checking directly with \(mgd\) gives \((120)(9.8)(1.2)=1411.2\). The units are newton-metres, or joules, as required.
Units and Interpretation
Work is force multiplied by distance in the direction of displacement. In SI units, force is measured in newtons and distance in metres, so work is measured in joules: \(1\text{ J}=1\text{ N}\cdot\text{m}\). When using mass in kilograms, use \(g\) in \(\text{m/s}^2\) and distance in metres; then \(mgd\) is in joules.
In other unit systems, keep force and distance units compatible. For example, a weight given in pounds-force multiplied by a height in feet gives work in foot-pounds-force. Do not treat a mass in pounds as if it were a force in pounds-force.
A Common Mistake
A frequent error is to multiply mass by distance and call the result work. For example, for a \(10\text{ kg}\) object raised \(3\text{ m}\), the calculation \(10\cdot3=30\text{ J}\) is incorrect. It omits the gravitational acceleration, and its units are \(\text{kg}\cdot\text{m}\), not joules.
The correction is to first convert mass to weight using \(g\):
A second conceptual error is to claim that each piece of a rigid object experiences an upward lifting force equal to its own weight. That equality need not hold for individual pieces. Instead, sum the work against gravity over the mass, and use the work–energy theorem separately if you need to identify the lifting force's total work.
Why the Result Is Useful
The formula \(mgd\) allows a lifting problem to be solved without tracking the object's detailed shape or how the lifting force is distributed across it. Uniformity can help determine mass from a density, but once the total mass is known, the work against gravity depends only on the vertical rise.
It also gives a reliable way to check a model: the work against gravity is positive for an upward lift, increases in proportion to mass and height, and has units of energy. The work done by gravity itself has the opposite sign, because gravity points downward while the object moves upward.
Check Your Understanding
Use \(g=9.8\text{ m/s}^2\) where needed. State units and distinguish work against gravity from the lifting force's work when the conditions matter.
- What is the work against gravity when a \(5\text{ kg}\) object is raised \(2\text{ m}\)?
- A uniform bar has mass \(8\text{ kg}\) and is raised \(1.5\text{ m}\). Explain why its shape does not change the work against gravity.
- Why does work against gravity equal \(mgd\) when every mass portion rises by the same distance?
- What additional conditions allow \(mgd\) to be identified with the lifting force's total work?
- For a \(4\text{ kg}\) object raised \(3\text{ m}\), identify the unit error in \(4\cdot3=12\text{ J}\), and give the corrected work.
