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

Gravitational Potential Energy (GPE) and Kinetic Energy

Learning Objective

Derive, using the work‑energy principle \$W = F s\$, the expression

\$\Delta EP = mg\,\Delta h\$

for the change in gravitational potential energy of an object moving in a uniform gravitational field.

Key Concepts

• Work: \$W = \mathbf{F}\cdot\mathbf{s}=Fs\cos\theta\$ where \$F\$ is the magnitude of a constant force, \$s\$ the displacement, and \$\theta\$ the angle between them.
• Uniform gravitational field: Near the Earth’s surface the acceleration due to gravity \$g\$ is approximately constant (≈ 9.81 m s⁻²).
• Gravitational force: \$F_g = mg\$, directed vertically downward.
• Potential energy change: The work done by a conservative force (gravity) is stored as potential energy.

Derivation Using \$W = Fs\$

1. Consider an object of mass \$m\$ moving vertically a distance \$\Delta h\$ in a uniform gravitational field.
2. The weight of the object is the constant force \$F_g = mg\$ acting downward.
3. If the object is displaced upward, the angle between the force (downward) and the displacement (upward) is \$\theta = 180^\circ\$, so \$\cos\theta = -1\$.
4. Apply the definition of work:

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

5. For a conservative force, the work done by the force equals the negative change in the associated potential energy:

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

6. Combine the two expressions for \$W_{\text{gravity}}\$:

   \$- \Delta EP = - mg\,\Delta h \quad\Longrightarrow\quad \Delta EP = mg\,\Delta h.\$

Assumptions Made

• The gravitational field is uniform (constant \$g\$) over the height interval \$\Delta h\$.
• Air resistance and other non‑conservative forces are neglected.
• The displacement is strictly vertical; horizontal components do not affect the work done by gravity.

Table of Symbols

Example Problem

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. Identify the given values: \$m = 2.0\;\text{kg}\$, \$\Delta h = 0.50\;\text{m}\$, \$g = 9.81\;\text{m s}^{-2}\$.
2. Apply the derived formula:

   \$\Delta EP = mg\,\Delta h = (2.0\;\text{kg})(9.81\;\text{m s}^{-2})(0.50\;\text{m}) = 9.81\;\text{J}.\$

3. Therefore, the block’s gravitational potential energy increases by \$9.8\;\text{J}\$ (to two significant figures).

Connection to Kinetic Energy

When an object moves under gravity without non‑conservative forces, the total mechanical energy is conserved:

\$\Delta EP + \Delta KE = 0 \quad\Longrightarrow\quad \Delta KE = -\Delta EP.\$

This relationship explains why an object falling a height \$\Delta h\$ gains kinetic energy \$KE = mg\,\Delta h\$.

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\$, leading directly to \$\Delta EP = mg\,\Delta h\$.
• This expression is valid for any motion in a uniform gravitational field, provided non‑conservative forces are negligible.