Problem
Determine the formula to calculate potential energy if given weight, and show how to compute the gravitational potential energy change for an object raised (or lowered) by a vertical height difference.
Physical meaning and assumptions
The phrase “potential energy” in classical mechanics commonly refers to gravitational potential energy near Earth, where the gravitational field is approximately uniform and the object’s weight is essentially constant over the height change.
Under this uniform-field assumption, \(g\) is treated as constant and the weight is \(W = m g\).
Step 1: Start from the standard gravitational potential energy formula
Near Earth’s surface (uniform \(g\)), gravitational potential energy relative to a chosen reference level is:
\[ U = m g h \]
For a height change from \(h_1\) to \(h_2\), the change in potential energy is:
\[ \Delta U = U_2 - U_1 = m g (h_2 - h_1) = m g\,\Delta h \]
Step 2: Replace \(mg\) with weight
Weight is the gravitational force on the object:
\[ W = m g \]
Substituting into the expression for \(\Delta U\) gives the requested relationship:
\[ \Delta U = W\,\Delta h \]
Step 3: Unit check (why the formula makes sense)
A newton is \(1\ \text{N} = 1\ \text{kg}\cdot\text{m}/\text{s}^2\). A joule is \(1\ \text{J} = 1\ \text{N}\cdot\text{m}\). Therefore:
\[ W\,\Delta h \;\; \text{has units} \;\; (\text{N})\cdot(\text{m}) = \text{J} \]
The product of weight (force) and vertical displacement (distance) matches energy units, consistent with the idea that energy change equals work done against gravity.
How to use the formula in practice
When given the weight \(W\) directly (in newtons) and a vertical height change \(\Delta h\) (in meters), compute:
\[ \Delta U = W\,\Delta h \]
If the object is raised, \(\Delta h > 0\) and \(\Delta U > 0\). If the object is lowered, \(\Delta h < 0\) and \(\Delta U < 0\).
Worked example
An object has weight \(W = 500\ \text{N}\) and is lifted upward by \(\Delta h = 2.0\ \text{m}\). The change in gravitational potential energy is:
\[ \Delta U = (500\ \text{N})\cdot(2.0\ \text{m}) = 1.0\times 10^3\ \text{J} \]
Quick reference table
| Given | Compute | Result units | Notes |
|---|---|---|---|
| Weight \(W\) and height change \(\Delta h\) | \(\Delta U = W\,\Delta h\) | J | Valid when weight is approximately constant over the motion |
| Mass \(m\) and height change \(\Delta h\) | \(\Delta U = m g\,\Delta h\) | J | Use \(g \approx 9.81\ \text{m}/\text{s}^2\) near Earth if \(W\) is not provided |
Visualization: \(\Delta U\) grows linearly with height when weight is constant
Final formula
\[ \Delta U = W\,\Delta h \qquad \text{(uniform gravitational field; weight approximately constant)} \]