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Potential Energy

Reference data and engineering information about potential energy for dynamics applications.

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Overview

Engineering reference data for Potential Energy in dynamics.

Key Formulas

Newton's Second Law

F=maF = ma

Force = mass × acceleration.

Kinetic Energy

Ek=12mv2E_k = \frac{1}{2}mv^2

Energy of motion.

Momentum

p=mvp = mv

Mass × velocity.

Work

W=FdcosθW = Fd\cos\theta

Force × displacement × cos(angle).

Variables

SymbolDescriptionUnit
FFForceN
mmMasskg
aaAccelerationm/s²
vvVelocitym/s
EkE_kKinetic energyJ

Practical Examples

The fundamental relationship Ep=mgΔhE_p = m \cdot g \cdot \Delta h can be applied directly to engineering problems using either mass or force inputs.

SI Units Example: A mass of 1000kg1000 \, \text{kg} is elevated by Δh=10m\Delta h = 10 \, \text{m}. Ep=(1000kg)(9.81m/s2)(10m)=98,100J=98kJ0.027kWhE_p = (1000 \, \text{kg}) \cdot (9.81 \, \text{m/s}^2) \cdot (10 \, \text{m}) = 98{,}100 \, \text{J} = 98 \, \text{kJ} \approx 0.027 \, \text{kWh}

Imperial Units Example (Force): A body with a weight of 500lbf500 \, \text{lbf} is elevated by Δh=30ft\Delta h = 30 \, \text{ft}. Ep=(500lbf)(30ft)=15,000ft⋅lbE_p = (500 \, \text{lbf}) \cdot (30 \, \text{ft}) = 15{,}000 \, \text{ft·lb}

Imperial Units Example (Mass): A body with a mass of 15slugs15 \, \text{slugs} is elevated by Δh=30ft\Delta h = 30 \, \text{ft}. Ep=(15slugs)(32.174ft/s2)(30ft)14,478ft⋅lbE_p = (15 \, \text{slugs}) \cdot (32.174 \, \text{ft/s}^2) \cdot (30 \, \text{ft}) \approx 14{,}478 \, \text{ft·lb}

Hydropower Potential Calculation

The potential energy stored in a reservoir or tank can be estimated for hydropower applications. The total energy is the integral of the energy contained in horizontal slices of water, which depends on the elevation of each slice above the outlet.

For a slice of water at elevation zz with volume dVdV, the potential energy is dEp=ρgzdVdE_p = \rho \cdot g \cdot z \cdot dV, where ρ\rho is water density. The total energy is: Ep=ρg0HzA(z)dzE_p = \rho \cdot g \cdot \int_{0}^{H} z \cdot A(z) \, dz where A(z)A(z) is the cross-sectional area at height zz, and HH is the total water height.

Engineering Approach: This integral is commonly solved numerically by dividing the reservoir into horizontal slices and summing the contributions. A spreadsheet calculator can automate this for various reservoir shapes (e.g., cylindrical, conical, irregular). You can adapt such a calculator by inputting the reservoir geometry and water level data.

Note on Hydraulic Head: The change in elevation Δh\Delta h is often referred to as the hydraulic head in hydropower contexts. The available power is then approximated as P=ηρgQΔhP = \eta \cdot \rho \cdot g \cdot Q \cdot \Delta h, where η\eta is system efficiency and QQ is volumetric flow rate.

References