Kinetic & Potential Energy Calculator

Use mgh for near-surface calculations when height h is small relative to planet radius R (typically h < 0.01R or ~60 km for Earth). This approximation assumes gravitational acceleration g is constant. Example: Lifting an object to 10 m height on Earth uses mgh = m(9.81)(10). Use the two-body formula U = -GMm/r for high altitudes, orbital mechanics, or any scenario where g varies significantly with distance. Example: Satellites in low Earth orbit (~400 km altitude) require the two-body model because g decreases by ~10% from surface value. The two-body formula is always more accurate but requires knowing the central body mass M and using separation distance r from the center (not height above surface). For Earth: r = R_Earth + h, where R_Earth ≈ 6.371×10⁶ m.

Gravitational potential energy is defined with zero at infinite separation (r → ∞). Bringing two masses together from infinity requires negative work relative to this reference, resulting in negative potential energy: U(r) = -GMm/r < 0. The negative sign indicates the system is gravitationally bound—energy must be added to separate the masses to infinity (escape). Physical interpretation: At r = ∞, objects are free and U = 0. As they approach (r decreases), U becomes more negative, representing energy released as kinetic energy or lost to other processes. Only changes in potential energy ΔU = U_final - U_initial matter for predictions, not the absolute value. The choice of reference point (r = ∞) is conventional but convenient for orbital mechanics and astrophysics.

If air resistance and friction are negligible (ideal free fall), use mechanical energy conservation: initial energy equals final energy. At release (height h, velocity 0): E_i = mgh + 0. At bottom (height 0, velocity v): E_f = 0 + ½mv². Setting E_i = E_f: mgh = ½mv². Cancel mass m and solve for velocity: v = √(2gh). This is independent of mass—all objects fall at the same rate in vacuum. Example: Dropping from h = 5 m on Earth (g = 9.81 m/s²): v = √(2 × 9.81 × 5) ≈ 9.9 m/s. For large heights where g varies, use two-body conservation: -GMm/r_initial + 0 = -GMm/r_final + ½mv². Solve for v. For realistic scenarios with air drag, energy is dissipated: E_final < E_initial, so v < √(2gh). Drag force increases with speed squared, eventually balancing gravity to produce terminal velocity.

Potential energy is defined relative to an arbitrary reference point (zero level) because only changes in potential energy ΔU have physical meaning. For gravitational potential near Earth's surface (U = mgh), the reference is typically ground level (h = 0, U = 0), but you could choose any convenient height. Example: If you define the second floor as h = 0, then ground level has h = -3 m and U = -mg(3) (negative potential). Predictions about motion don't change—only ΔU matters. For two-body gravitational potential (U = -GMm/r), the reference is conventionally r = ∞ (U = 0 at infinite separation). This makes bound orbits have negative total energy. For springs (U = ½kx²), the reference is equilibrium position (x = 0, U = 0). Choosing different references shifts all potential energies by a constant but doesn't affect force (F = -dU/dx) or energy conservation (ΔK = -ΔU). Always be consistent within a single problem.

Yes, total kinetic energy is the sum of translational (linear) and rotational components: K_total = K_trans + K_rot = ½mv² + ½Iω². This applies to rolling objects like wheels, balls, or cylinders that simultaneously translate and rotate. Example: A solid sphere (mass m, radius r) rolling without slipping down a ramp has both: (1) Translational KE from center-of-mass velocity v: K_trans = ½mv². (2) Rotational KE about the center of mass: K_rot = ½Iω² where I = (2/5)mr² for a solid sphere. For rolling without slipping: v = rω (linear speed equals tangential speed at contact point). Total: K_total = ½mv² + ½(2/5)mr²(v/r)² = ½mv²(1 + 2/5) = (7/10)mv². This shows rolling objects move slower than sliding objects for the same initial potential energy because some energy goes into rotation. In this calculator, select modes separately to calculate each component, then add manually if both are present.

Friction and air drag are non-conservative forces that perform negative work, removing mechanical energy from the system and converting it to thermal energy (heat). Conservation of mechanical energy (K + U = constant) only holds when no non-conservative forces act. With friction/drag: E_final = E_initial - W_non-conservative, where W_non-conservative > 0 is energy dissipated. Example: Sliding a block down a ramp with friction. Initial: E_i = mgh (at top, v = 0). Final: E_f = ½mv² (at bottom, h = 0). Energy lost to friction: W_friction = μmg cos(θ) × d, where μ is friction coefficient, θ is ramp angle, d is distance traveled. Result: mgh = ½mv² + μmgd cos(θ), so final speed is less than frictionless case. Air drag (F_drag ∝ v²) increases with speed, eventually balancing gravity to produce constant terminal velocity where K no longer increases despite losing height (U decreasing). To model accurately, you must separately calculate work done by friction/drag and subtract from initial mechanical energy.

Use SI units consistently: kg for mass, m for distance, s for time, J (joules) for energy. Common mistakes: (1) Mixing unit systems: Don't use kg with ft or lbm with meters—stick to one system. (2) Angular velocity units: ω must be in rad/s, not rpm. Convert: ω (rad/s) = rpm × 2π/60. Example: 3000 rpm = 3000 × 2π/60 ≈ 314 rad/s. (3) Gravitational units: g must match your mass/distance units. Earth: g ≈ 9.81 m/s² (SI) or 32.2 ft/s² (Imperial). (4) Energy units: 1 J = 1 N·m = 1 kg·m²/s². Note that 1 N·m as torque is conceptually different from 1 J as energy, though dimensionally equivalent. (5) Spring constant: k is in N/m (SI) or lbf/ft (Imperial). Verify by checking dimensional analysis: [½kx²] = [N/m][m²] = [N·m] = [J]. (6) Moment of inertia: I has units kg·m² (SI) or slug·ft² (Imperial). For geometric shapes, formulas give I in terms of mass and size (e.g., solid disk: I = ½mr²). Always verify final answer has correct energy units (J or ft·lbf) before interpreting results.