Conservation of Mechanical Energy

Start from a result you already trust, the work–kinetic energy theorem: Wₙₑₜ = ΔK. Sort the work by the forces that did it, Wₙₑₜ = W_c + W_nc, where W_c comes from conservative forces such as gravity and ideal springs. For each of those, the work done over any displacement equals −ΔU, which is how its potential energy was defined in the first place. Substituting gives ΔK = −ΔU + W_nc, which rearranges to ΔK + ΔU = W_nc. So K + U is constant exactly when W_nc = 0. Two readings of that line matter. The law is conditional: before using it, check that no nonconservative force does work, rather than assuming it. And it is an accounting identity, not new physics: every joule the conservative forces hand to K is a joule taken out of U, so a spring that has lost 3.0 J of stored energy has given the block on its end exactly 3.0 J of kinetic energy.

In one dimension a conservative force is F = −dU/dx, so Newton's second law reads m dv/dt = −dU/dx. Multiply both sides by v = dx/dt. On the left, mv dv/dt = d(½mv²)/dt by the chain rule; on the right, −(dU/dx)(dx/dt) = −dU/dt, again by the chain rule. Therefore d(½mv² + U)/dt = 0 at every instant, not just between a start and an end. This is what energy buys. Newton's law is second order and needs two integrations before it yields positions; E = ½mv² + U(x) is first order and gives the speed at any position at once, v(x) = √(2[E − U(x)]/m). A 0.40 kg cart on a spring with k = 250 N m⁻¹, released from rest at x = 0.12 m, has E = ½ × 250 × 0.12² = 1.8 J. At x = 0.050 m, U = 0.31 J, so v = √(2 × 1.49 ÷ 0.40) = 2.7 m s⁻¹, found without ever knowing when the cart gets there.

Mechanical energy belongs to a system, so name the system before writing any energy. Gravitational potential energy is shared by an object and Earth, and elastic energy is stored in a spring; a U term belongs in the equation only when both partners of that interaction sit inside the boundary. Then test each force with its power, P = F⋅v, the rate at which it transfers energy. A normal force from a fixed surface and the tension from a fixed pivot stay perpendicular to the velocity, so F⋅v = 0 at every instant and they transfer nothing, however large they are. A force with a component along v does transfer energy: kinetic friction and drag inside the system, or a hand, rope or motor acting from outside, each break conservation. Zero net force is not the test. A 25 kg crate lowered 3.0 m at a steady 0.40 m s⁻¹ keeps its kinetic energy while mgy falls by 25 × 9.81 × 3.0 = 736 J, because the rope does −736 J of work on the crate–Earth system from outside.

Write E₁ and E₂ as sums of the same terms, one per form: ½mv², mgy, ½kx². The two potential energies take their zeros differently. For gravity the zero is yours: only differences in mgy enter ΔU, so measure y from the floor, the table or the release point, provided every snapshot uses the same level. For a spring it is not: ½kx² needs x measured from the natural length, and the spring's energy is zero only there. Take a 0.20 kg block pressed 0.15 m into a spring with k = 400 N m⁻¹ on a frictionless table 0.90 m high. Measuring y from the table top gives E = 4.5 J; measuring from the floor gives E = 4.5 + 1.77 = 6.27 J. The totals differ, yet both predict 6.7 m s⁻¹ as the block leaves the spring, because both contain the same differences. An energy bar chart keeps the ledger honest: draw K, U_g and Uₛ for each snapshot and check that the stacks reach the same total before any algebra.

With gravity the only force doing work, ½mv₁² + mgy₁ = ½mv₂² + mgy₂. Divide by m: v₂² = v₁² + 2g(y₁ − y₂). Mass has gone, and so has every detail of the route. A crate released from rest at the top of a frictionless 20° ramp 3.2 m high reaches the bottom at √(2 × 9.81 × 3.2) = 7.9 m s⁻¹, and so does a crate on a 45° ramp or a curved chute of the same height, whatever either weighs. Now notice what cancelled. Along the 20° ramp the acceleration is g sin 20° = 3.36 m s⁻² over 9.36 m, a trip of 2.36 s; the 45° ramp takes 1.14 s. The final velocities also point in different directions. Energy fixes the speed at each height and nothing else, so questions about when, or which way, still need kinematics or Newton's second law.

Conservation never returns a force by itself, but it supplies the speed that Newton's second law needs to find one. Take a small cart sliding without friction round a vertical loop of radius R after release from rest at height h. At the top, 2R above the bottom, energy gives v² = 2g(h − 2R). There the weight and the track's normal force both point down, toward the centre, so N + mg = mv²/R. The cart keeps contact only while N ≥ 0, which needs v² ≥ gR, so 2g(h − 2R) ≥ gR, or h ≥ 5R/2. A loop with R = 4.0 m therefore needs a release point at least 10 m up. Carry that same h to the bottom of the loop: v² = 2gh = 5gR, and N − mg = mv²/R gives N = 6mg. A 60 kg rider is pressed into the seat with 3.5 kN, six times their weight, which is one reason real coaster loops are tighter at the top than at the bottom.