Introduction
When only conservative forces act, the total mechanical energy E = K + U remains constant. This powerful conservation law simplifies problem-solving by relating positions and velocities without time dependence. It provides deep insight into the interplay between motion and configuration in mechanical systems.
Statement of Conservation
If only conservative forces do work, total mechanical energy is conserved: E = K + U = constant. This means K1 + U1 = K2 + U2 at any two points. Equivalently: ΔK + ΔU = 0, or ΔK = -ΔU. Any decrease in potential energy appears as increase in kinetic energy, and vice versa. Energy transforms between kinetic and potential forms but total remains unchanged.
Conditions for Conservation
Mechanical energy is conserved when: (1) Only conservative forces act; (2) Non-conservative forces (like friction) do no work; (3) External forces are absent or do no work. If non-conservative forces do work, mechanical energy changes: W_nc = ΔE = ΔK + ΔU. Friction converts mechanical energy to thermal energy, decreasing total mechanical energy.
\nIntroduction
When only conservative forces act, the total mechanical energy E = K + U remains constant. This powerful conservation law simplifies problem-solving by relating positions and velocities without time dependence. It provides deep insight into the interplay between motion and configuration in mechanical systems.
Statement of Conservation
If only conservative forces do work, total mechanical energy is conserved: E = K + U = constant. This means K1 + U1 = K2 + U2 at any two points. Equivalently: ΔK + ΔU = 0, or ΔK = -ΔU. Any decrease in potential energy appears as increase in kinetic energy, and vice versa. Energy transforms between kinetic and potential forms but total remains unchanged.
Conditions for Conservation
Mechanical energy is conserved when: (1) Only conservative forces act; (2) Non-conservative forces (like friction) do no work; (3) External forces are absent or do no work. If non-conservative forces do work, mechanical energy changes: W_nc = ΔE = ΔK + ΔU. Friction converts mechanical energy to thermal energy, decreasing total mechanical energy.
\nProblem-Solving Using Conservation
Strategy: (1) Identify initial and final states; (2) Write K1 + U1 = K2 + U2; (3) Substitute expressions for K and appropriate U; (4) Solve for unknown. Particularly useful when: time is not given or needed; position-velocity relations are required; forces are complicated but conservative. Avoid when non-conservative forces (friction) are significant.
Examples
Simple pendulum: At top (height h), v = 0, so E = mgh; at bottom (h = 0), E = 1/2mv2, giving v = √(2gh). Roller coaster: Maximum speed at lowest point, converts to height at top of next hill. Spring-mass: 1/2kx_max2 = 1/2mv_max2, giving v_max = x_max√(k/m). Orbit: Total energy E = -GMm/(2a) is constant for elliptical orbits.
Energy Diagrams
Energy diagrams plot K, U, and E versus position. Where U < E, motion is allowed (K > 0). Turning points occur where U = E (K = 0, v = 0). Regions where U > E are forbidden (classically). For bound systems, E < U(∞), creating potential wells. The shape of U(x) determines possible motions and stability.
\nSolved Example: Roller Coaster Loop
A roller coaster cart starts from rest at height h = 25 m. Find: (a) Speed at bottom, (b) Minimum height to complete vertical loop of radius R = 7 m. Solution: (a) Using conservation: mgh = 1/2mv2 → v = √(2gh) = √(2×9.8×25) = 22.1 m/s at bottom. (b) To complete loop, cart needs minimum speed at top: v_top ≥ √(gR) = √(9.8×7) = 8.28 m/s (so normal force stays positive). Energy at top: E = 1/2mv_top2 + mg(2R) = 1/2m(68.7) + m(137.2) = 171.6m J. This equals initial PE: mgh_min = 171.6m → h_min = 17.5 m. Cart must start at least 17.5 m high to complete loop. Actual coasters start higher for safety margin.
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