04 // Interactive Formula Compendium // PHYSICSE_total = K + U = ½ m v² + m g h + ½ k x² = constant
Conservation of Mechanical Energy — Kinetic, Potential & Spring Sheet
Conservation of Mechanical Energy states that when only conservative forces act, ΔK + ΔU = 0, so ½mv₁² + mgh₁ + ½kx₁² = ½mv₂² + mgh₂ + ½kx₂².
Formula Reference
Main Formula
E_total = K + U = ½ m v² + m g h + ½ k x² = constant
Rearranged Forms
K = ½ m v² = p² / (2m) (Translational Kinetic Energy)
U_g = m g h, U_s = ½ k x² (Gravitational & Hookean Spring Potential)
v_bottom = √(v₀² + 2 g Δh) (Frictionless Descent Speed)
Variables & Units
| Symbol | Name | Unit |
|---|---|---|
| E_total | Total Mechanical Energy (½mv² + mgh) | J |
| m | Object Mass (m) | kg |
| v | Speed (v) | m/s |
| h | Elevation Height (h) | m |
| g | Gravitational Acceleration (g) | m/s² |
Inputs & Parameters
[kg]
[m/s]
[m]
[m/s²]
Calculated Result (Total Mechanical Energy (½mv² + mgh))
E_total =12,306.65J
Formula Notes & Background
- By the Work-Energy Theorem, net work done on a particle equals its change in kinetic energy: W_net = ΔK.
- For a conservative force F = −∇U, the work done moving from point 1 to point 2 is W_cons = −(U₂ − U₁) = −ΔU.
- Equating ΔK = −ΔU gives K₁ + U₁ = K₂ + U₂ = E_total.
Step-by-Step Examples
#1: A 50 kg rollercoaster car passes a 20 m crest at v₁ = 10 m/s. Find its total mechanical energy and speed at ground level (h₂ = 0).
m = 50 kgv₁ = 10 m/sh₁ = 20 mg = 9.80665 m/s²
E = ½(50)(100) + 50(9.80665)(20) = 2,500 J + 9,806.65 J = 12,306.65 J
→ E_total = 12,306.65 J; speed at bottom v₂ = √(2E/m) = 22.19 m/s
#2: Notice that mass m cancels out when solving for v₂ at the bottom of the drop.
v₁ = 10 m/sΔh = 20 m
v₂ = √(10² + 2 × 9.80665 × 20) = √(100 + 392.266)
→ v₂ = 22.19 m/s for any mass m
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Frequently Asked Questions
What happens when friction or air resistance is present?+
Non-conservative work W_nc = −f_k · d converts mechanical energy into thermal internal energy ΔE_thermal, so K₁ + U₁ + W_nc = K₂ + U₂.
Does the choice of zero reference height (h = 0) affect physical predictions?+
No, because only potential energy differences ΔU = mg(h₂ − h₁) govern velocity changes and forces.
How is rotational kinetic energy included for a rolling sphere or wheel?+
Add rotational kinetic energy K_rot = ½ I ω² to translational ½ m v², so a rolling solid sphere (I = ⅖mr²) has total kinetic energy (7/10)mv².