derive, using W = Fs, the formula ∆EP = mg∆h for gravitational potential energy changes in a uniform gravitational field

Gravitational Potential Energy (GPE) – Derivation and Application (AS 5.2)

Learning Objective (AO1, AO2, AO3)

By the end of this topic you will be able to:

  • Derive, from the work‑energy principle \(W=\mathbf{F}\!\cdot\!\mathbf{s}\), the expression \(\displaystyle \Delta EP = mg\,\Delta h\) for a uniform gravitational field (AO1).

  • State and use the relationship \(\Delta EP + \Delta KE = 0\) for a closed system where only gravity does work (AO2).

  • Apply the formula to a range of Cambridge A‑Level style problems, interpreting sign conventions and reference levels correctly (AO3).

Key Concepts

  • Work (scalar): \(W = \mathbf{F}\!\cdot\!\mathbf{s}=Fs\cos\theta\) for a constant force \(\mathbf{F}\) acting over a straight‑line displacement \(\mathbf{s}\).

  • Work‑energy theorem: The net work done on a body equals the change in its kinetic energy,

    \[

    W_{\text{net}} = \Delta KE .

    \]

  • Uniform gravitational field (near Earth): \(g\approx9.81\ \text{m s}^{-2}\) is taken as constant over the height interval considered.

  • Weight (gravitational force): \(\mathbf{F}_g = mg\) directed vertically downward.

  • Conservative force: For a conservative force the work done can be stored as potential energy,

    \[

    W_{\text{gravity}} = -\Delta EP .

    \]

Sign Convention & Reference Level

  • Take upward displacement as positive (\(\Delta h>0\)).

  • The zero of potential energy may be chosen arbitrarily – e.g. the floor, a table, or the centre of the Earth (for large‑scale problems).

  • If the chosen reference level is changed, the *numerical* value of \(EP\) changes, but the *difference* \(\Delta EP\) remains the same.

Two possible reference levels: (a) ground as zero, (b) table top as zero. The vertical displacement \(\Delta h\) is measured from the initial to the final position, independent of the chosen zero.

Derivation of \(\displaystyle \Delta EP = mg\,\Delta h\)

  1. Consider a body of mass \(m\) displaced vertically by \(\Delta h\) in a uniform field.

  2. The only external force that does work is the weight \(\mathbf{F}_g = mg\) (downward).

  3. For an upward displacement the angle between \(\mathbf{F}_g\) (down) and \(\mathbf{s}\) (up) is \(\theta = 180^{\circ}\); thus \(\cos\theta = -1\).

  4. Work done by gravity:

    \[

    W{\text{gravity}} = Fg\,\Delta h\cos\theta = mg\,\Delta h(-1) = -mg\,\Delta h .

    \]

  5. Because gravity is conservative,

    \[

    W_{\text{gravity}} = -\Delta EP .

    \]

  6. Equating the two expressions gives

    \[

    -\Delta EP = -mg\,\Delta h \;\Longrightarrow\; \boxed{\Delta EP = mg\,\Delta h}.

    \]

Energy‑Conservation Statement (ΔEP + ΔKE = 0)

  • When only gravity does work (no friction, air resistance, or applied forces), the net work equals the work of gravity:

    \[

    W{\text{net}} = W{\text{gravity}} .

    \]

  • From the work‑energy theorem and the result above,

    \[

    \Delta KE = W_{\text{gravity}} = -\Delta EP .

    \]

  • Re‑arranging,

    \[

    \boxed{\Delta EP + \Delta KE = 0}.

    \]

  • This relation holds for a *closed* system in a uniform field; it is the basis of the mechanical‑energy‑conservation method used in many A‑Level questions.

Assumptions (and When They Break Down)

AssumptionWhy It Is NeededWhat Happens If It Fails?
Uniform gravitational field (\(g\) constant)Allows \(mg\) to be taken outside the integral of work.For large height changes (e.g. satellite orbits) \(g\) varies; the simple formula must be replaced by \( \Delta EP = GMm\left(\frac1{r1}-\frac1{r2}\right)\).
Displacement is strictly verticalOnly the vertical component of \(\mathbf{s}\) contributes to work by gravity.If the path has a horizontal component, the work by gravity is still \(-mg\Delta h\); the extra horizontal motion affects kinetic energy but not \(\Delta EP\).
Negligible non‑conservative forces (air resistance, friction)Ensures \(W{\text{net}} = W{\text{gravity}}\) and the simple conservation law holds.Presence of dissipative forces adds a negative term \(-W{\text{nc}}\) to the energy balance: \(\Delta EP + \Delta KE = -W{\text{nc}}\).

Table of Symbols

SymbolQuantityUnits
\(m\)Mass of the objectkg
\(g\)Acceleration due to gravity (uniform)m s\(^{-2}\)
\(\Delta h\)Vertical displacement (positive upward)m
\(\Delta EP\)Change in gravitational potential energyJ
\(W_{\text{gravity}}\)Work done by the weight of the objectJ
\(v\)Speed of the objectm s\(^{-1}\)
\(KE\)Kinetic energyJ

Worked Examples

Example 1 – Lifting a Block (Positive Δh)

Question: A 2.0 kg block is lifted vertically by 0.50 m from the floor. Calculate the increase in its gravitational potential energy.

Solution:

  1. Given: \(m=2.0\ \text{kg}\), \(\Delta h = +0.50\ \text{m}\), \(g=9.81\ \text{m s}^{-2}\).

  2. Apply \(\Delta EP = mg\,\Delta h\):

    \[

    \Delta EP = (2.0)(9.81)(0.50) = 9.81\ \text{J}.

    \]

  3. Result: The block’s GPE increases by \(\boxed{9.8\ \text{J}}\) (to two significant figures).

Example 2 – Falling Object (Negative Δh)

Question: A 1.5 kg stone falls from a height of 3.0 m to the ground. Determine the change in its GPE and the gain in kinetic energy, assuming air resistance is negligible.

Solution:

  1. Take upward as positive, so \(\Delta h = -3.0\ \text{m}\).

  2. Change in GPE:

    \[

    \Delta EP = mg\,\Delta h = (1.5)(9.81)(-3.0) = -44.1\ \text{J}.

    \]

  3. Since only gravity does work, \(\Delta KE = -\Delta EP = +44.1\ \text{J}\).

  4. Result: GPE decreases by 44 J and kinetic energy increases by the same amount.

Example 3 – Energy‑Conservation Problem (Throwing Upwards)

Question: A ball of mass 0.20 kg is thrown straight up with an initial speed of 5.0 m s\(^{-1}\). Ignoring air resistance, find the maximum height above the launch point.

Solution:

  1. Initial kinetic energy: \(KE_i = \tfrac12 mv^2 = \tfrac12(0.20)(5.0)^2 = 2.5\ \text{J}\).

  2. At the highest point \(KE_f = 0\); all the kinetic energy has been converted into GPE: \(\Delta EP = -\Delta KE = -2.5\ \text{J}\).

  3. Using \(\Delta EP = mg\,\Delta h\):

    \[

    -2.5 = (0.20)(9.81)\,\Delta h \;\Longrightarrow\; \Delta h = \frac{2.5}{0.20\times9.81} \approx 1.27\ \text{m}.

    \]

  4. Result: The ball rises about \(\boxed{1.3\ \text{m}}\) above the launch point.

Quick‑Recall Box

  • ΔEP (GPE change): \(\displaystyle \Delta EP = mg\,\Delta h\)

  • KE: \(\displaystyle KE = \tfrac12 mv^{2}\)

  • Work‑energy theorem: \(W_{\text{net}} = \Delta KE\)

  • Mechanical‑energy conservation (gravity only): \(\displaystyle \Delta EP + \Delta KE = 0\)

  • Reference level: Arbitrary; only \(\Delta EP\) is physically meaningful.

Suggested Diagram for Classroom Use

Block of mass \(m\) being lifted a height \(\Delta h\). Weight \(mg\) acts downward, displacement \(\mathbf{s}\) is upward; the angle between them is \(180^{\circ}\). Include a second diagram showing the same block falling a distance \(|\Delta h|\) to illustrate the sign change.

Summary

  • The work done by a constant gravitational force over a vertical displacement \(\Delta h\) is \(W_{\text{gravity}} = -mg\,\Delta h\).

  • Because gravity is conservative, \(W_{\text{gravity}} = -\Delta EP\), giving the fundamental relation \(\Delta EP = mg\,\Delta h\).

  • In the absence of non‑conservative forces, mechanical energy is conserved: \(\Delta EP + \Delta KE = 0\). This provides a powerful shortcut for many A‑Level problems.

  • Always state the chosen reference level, apply the correct sign convention, and check that the assumptions (uniform \(g\), no friction/air resistance, vertical motion) are satisfied.