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Derivation

Conservation of Mechanical Energy

D-024 Home PU-101 Threads energy · symmetry Depends on The Work-Energy Theorem, Conservative Forces and Potential Energy
Statement

For a system of particles acted upon only by conservative forces, the total mechanical energy \(E = T + U\) — the sum of kinetic energy \(T\) and potential energy \(U\) — is constant along any physically realized trajectory: \(\frac{dE}{dt} = 0\), equivalently \(E(t_2) = E(t_1)\) for all times \(t_1, t_2\).

Why it matters

Conservation of mechanical energy converts a second-order differential problem (Newton's law) into a first-order algebraic relation between speed and position. Without ever integrating the equation of motion, one can read off the speed of a body at any point of its path from geometry alone — the workhorse behind pendulums, orbits, roller coasters, and escape velocity.

Deeper still, it is the mechanical face of a universal principle. By Noether's theorem the constancy of \(E\) is the direct consequence of time-translation symmetry: the laws governing the system do not change from one instant to the next. Energy conservation is thus not an accident of particular forces but a signature of a symmetry of nature.

Assumptions
All forces doing work are conservative.If a non-conservative force (friction, drag, a time-varying applied push) acts, it does work not accounted for by any potential and \(E\) is not conserved: \(\frac{dE}{dt}=P_{\text{nc}}\neq 0\).
Each conservative force derives from a potential, \(\vec{F}_i = -\nabla_i U\).If a force cannot be written as the gradient of a scalar function of position, no potential energy exists to store the work and the bookkeeping fails.
The potential has no explicit time dependence, \(\frac{\partial U}{\partial t}=0\).If \(U=U(\vec{r},t)\) depends explicitly on time (e.g. a moving wall, a ramping field), then \(\frac{dE}{dt}=\frac{\partial U}{\partial t}\neq 0\); energy leaks in or out through the explicit time channel even though the force is "conservative" in space.
Constraints, if any, do no work.Rigid rods, frictionless tracks, and normal forces are permitted because \(\vec{F}\cdot\vec{v}=0\); a constraint that does work (a sliding contact with friction) breaks conservation.
Derivation
1
\[ W_{1\to 2} = T_2 - T_1 \]
The work–energy theorem: the net work done on the body over its path equals its change in kinetic energy. This is the assumed prior result. A
2
\[ W_{1\to 2} = \int_{1}^{2} \vec{F}\cdot d\vec{r} \]
Definition of work as the line integral of the net force along the actual trajectory from configuration 1 to configuration 2. A
3
\[ \vec{F} = -\nabla U \]
By assumption every force is conservative, so it is the gradient of a scalar potential (prior result: conservative force ⇔ potential energy). B
4
\[ W_{1\to 2} = -\int_{1}^{2} \nabla U \cdot d\vec{r} = -\int_{1}^{2} dU \]
Substitute step 3 into step 2. Along the path \(\nabla U\cdot d\vec{r}\) is exactly the total differential \(dU\), because \(dU = \frac{\partial U}{\partial x}dx + \frac{\partial U}{\partial y}dy + \frac{\partial U}{\partial z}dz = \nabla U\cdot d\vec{r}\). This is where path-independence enters: the integrand is an exact differential. C
5
\[ W_{1\to 2} = -\left( U_2 - U_1 \right) = U_1 - U_2 \]
Fundamental theorem of calculus applied to the exact differential: the integral of \(dU\) depends only on the endpoints, giving \(U_2-U_1\). B
6
\[ T_2 - T_1 = U_1 - U_2 \]
Equate the two expressions for the same work, step 1 and step 5. A
7
\[ T_1 + U_1 = T_2 + U_2 \]
Rearrange, collecting quantities at each configuration. Since 1 and 2 were arbitrary, the combination \(T+U\) takes the same value at every point of the motion. A
8
\[ \frac{dE}{dt} = \frac{d}{dt}(T+U) = \vec{F}\cdot\vec{v} + \nabla U\cdot\vec{v} = \left(\vec{F}+\nabla U\right)\cdot\vec{v} = 0 \]
Differential form: \(\frac{dT}{dt}=\vec{F}\cdot\vec{v}\) (from \(\frac{d}{dt}\tfrac12 m v^2 = m\vec{a}\cdot\vec{v}=\vec{F}\cdot\vec{v}\)) and \(\frac{dU}{dt}=\nabla U\cdot\vec{v}+\frac{\partial U}{\partial t}\). With \(\vec{F}=-\nabla U\) and \(\frac{\partial U}{\partial t}=0\) the bracket vanishes identically. C
Result
\[ E = T + U = \tfrac{1}{2}mv^2 + U(\vec{r}) = \text{const}, \qquad \frac{dE}{dt}=0 \]

Reading. As the body moves, kinetic and potential energy trade back and forth, but their sum never changes. A gain in speed is paid for exactly by a drop in potential, and vice versa. The constant value is fixed once and for all by the initial conditions.

Units check. \([T]=\mathrm{kg}\cdot(\mathrm{m/s})^2=\mathrm{kg\,m^2\,s^{-2}}=\mathrm{J}\); \([U]=\mathrm{J}\) (work of a force, \(\mathrm{N\cdot m}=\mathrm{kg\,m\,s^{-2}}\cdot\mathrm{m}\)). Both terms are joules, so \(E\) is in joules and \(\frac{dE}{dt}\) in watts, here zero.

Limiting cases
  • Uniform gravity near Earth. \(U=mgh\Rightarrow \tfrac12 mv^2+mgh=\text{const}\); speed at the bottom of a drop \(h\) is \(v=\sqrt{2gh}\), independent of mass and path.
  • Simple harmonic oscillator. \(U=\tfrac12 kx^2\Rightarrow \tfrac12 mv^2+\tfrac12 kx^2=\tfrac12 kA^2\); energy sloshes fully between kinetic and potential twice per period.
  • Kepler / gravitational orbit. \(U=-\frac{GMm}{r}\Rightarrow \tfrac12 mv^2-\frac{GMm}{r}=E\); bound orbits have \(E<0\), the escape threshold is \(E=0\).
  • Free particle. \(U=\text{const}\Rightarrow T=\text{const}\); constant speed, recovering Newton's first law.
Breaks when
  • Dissipative forces act. Friction, air drag, viscous damping, or inelastic collisions do work with no associated potential; mechanical energy decreases monotonically, \(\frac{dE}{dt}=\vec{F}_{\text{nc}}\cdot\vec{v}<0\), and reappears as heat. Conservation of total energy survives, but mechanical energy alone does not.
  • The potential is explicitly time-dependent. A moving piston, a charging capacitor, a rocket losing mass, or any externally driven field gives \(\frac{\partial U}{\partial t}\neq 0\); then \(\frac{dE}{dt}=\frac{\partial U}{\partial t}\) and energy is pumped in or out. Time-translation symmetry is broken, so by Noether there is no conserved energy.
  • Open systems and constraint forces that do work. If mass or momentum crosses the system boundary, or a constraint slides with friction, the accounting is incomplete and \(T+U\) drifts.
Failure modes
  • Double-counting the drop. Writing \(mgh\) for potential and separately adding the work done by gravity — gravity's work is already encoded in \(-\Delta U\). Use one or the other, never both.
  • Sign of the potential. Taking \(\vec{F}=+\nabla U\), which flips the direction of every force; the minus sign in \(\vec{F}=-\nabla U\) is not optional.
  • Applying it through friction. Setting \(\tfrac12 mv_1^2+mgh_1=\tfrac12 mv_2^2+mgh_2\) across a rough surface, ignoring the \(-f d\) dissipation term.
  • Confusing height with path length. In \(U=mgh\), \(h\) is the vertical drop only; the distance travelled along an incline is irrelevant to the potential.
  • Forgetting rotational kinetic energy. For a rolling body \(T=\tfrac12 mv^2+\tfrac12 I\omega^2\); omitting the spin term overestimates translational speed.
  • Inconsistent reference level. Measuring \(h\) from two different origins on the two sides of the equation.
Discussion

Energy conservation is best understood as a first integral of Newton's equations. The equation of motion \(m\ddot{\vec{r}}=-\nabla U\) is second order; taking the dot product with \(\vec{v}\) and integrating once produces \(\tfrac12 mv^2+U=\text{const}\). We have traded a differential equation for an algebraic constraint, at the cost of losing directional information — energy alone gives speed, not the direction of travel. This is why one scalar conservation law can replace only part of the three-component vector equation.

The potential energy is defined only up to an additive constant, since only differences \(U_2-U_1\) — and gradients \(\nabla U\) — carry physical meaning. Choosing the zero of \(U\) (ground level, or infinity for gravity) is a matter of convenience and never affects the dynamics. This gauge freedom in the potential is the mechanical shadow of a much larger structure in field theory.

The sharpest statement is Noether's: to every continuous symmetry of the action corresponds a conserved quantity. Invariance of the Lagrangian \(L=T-U\) under time translation, \(\frac{\partial L}{\partial t}=0\), yields a conserved Hamiltonian \(H=\sum_i \dot{q}_i \frac{\partial L}{\partial \dot{q}_i} - L\); for velocity-independent potentials and time-independent constraints this Hamiltonian is precisely \(E=T+U\). Thus mechanical energy conservation and the homogeneity of time are two descriptions of one fact. When the potential depends explicitly on time the symmetry is gone and so is the conservation law — the "breaks when" cases are not exceptions but direct corollaries.

Common misconceptions. Energy conservation does not say kinetic energy is conserved (it is not — it converts to potential), nor that "energy is always conserved so friction is impossible." Mechanical energy is a subset of total energy; friction moves energy out of the mechanical ledger into thermal microscopic motion, which the mechanical bookkeeping simply does not track. The total energy of the universe is conserved; the mechanical energy of a chosen system is conserved only under the stated assumptions.

Worked examples
1
Roller-coaster drop. A car of mass \(m=500\ \mathrm{kg}\) starts from rest at the top of a frictionless track of height \(h=25\ \mathrm{m}\). Find its speed at the bottom. A
2
\[ \tfrac12 m v_{\text{top}}^2 + m g h = \tfrac12 m v_{\text{bot}}^2 + 0 \]
Conservation of mechanical energy between top and bottom; take \(U=0\) at the bottom.
3
\[ v_{\text{bot}} = \sqrt{2gh}\quad(\text{since }v_{\text{top}}=0) \]
Symbols first: mass cancels, giving a path- and mass-independent result.
4
\[ v_{\text{bot}} = \sqrt{2(9.81\ \mathrm{m/s^2})(25\ \mathrm{m})} \]
Insert numbers with units.
\[ v_{\text{bot}} = \sqrt{490.5}\ \mathrm{m/s} \approx 22.1\ \mathrm{m/s} \]

Reading. About \(80\ \mathrm{km/h}\), independent of the car's mass and of the track's shape — only the vertical drop matters.

1
Pendulum swing. A bob of mass \(m=0.30\ \mathrm{kg}\) on a string of length \(L=1.2\ \mathrm{m}\) is released from rest at angle \(\theta_0=40^\circ\) from vertical. Find its speed at the lowest point. B
2
\[ \Delta h = L\left(1-\cos\theta_0\right) \]
Geometry: the height risen above the lowest point when the string makes angle \(\theta_0\) with the vertical.
3
\[ m g L(1-\cos\theta_0) = \tfrac12 m v^2 \implies v = \sqrt{2gL(1-\cos\theta_0)} \]
Energy conservation; the string tension does no work (perpendicular to motion), so it is a legitimate workless constraint. Mass cancels.
4
\[ v = \sqrt{2(9.81)(1.2)(1-\cos 40^\circ)}\ \mathrm{m/s},\quad \cos 40^\circ = 0.766 \]
Insert numbers; \(1-0.766=0.234\).
\[ v = \sqrt{2(9.81)(1.2)(0.234)}\ \mathrm{m/s} = \sqrt{5.51}\ \mathrm{m/s} \approx 2.35\ \mathrm{m/s} \]

Reading. The bob is fastest at the bottom where all the released potential has become kinetic; the string tension never enters the energy balance because it does no work.

Problems
  1. (A) A \(2.0\ \mathrm{kg}\) block slides down a frictionless \(30^\circ\) incline from rest through a slope distance of \(4.0\ \mathrm{m}\). Find its speed at the bottom.
    Solution Vertical drop \(h=4.0\sin30^\circ=2.0\ \mathrm{m}\). Energy: \(mgh=\tfrac12 mv^2\Rightarrow v=\sqrt{2gh}=\sqrt{2(9.81)(2.0)}=\sqrt{39.24}=6.26\ \mathrm{m/s}\). Mass is irrelevant.
  2. (A) A ball is thrown straight up at \(15\ \mathrm{m/s}\). Using energy conservation, find its maximum height.
    Solution \(\tfrac12 mv_0^2=mgh_{\max}\Rightarrow h_{\max}=\frac{v_0^2}{2g}=\frac{15^2}{2(9.81)}=\frac{225}{19.62}=11.5\ \mathrm{m}\).
  3. (B) A \(0.50\ \mathrm{kg}\) mass on a spring (\(k=200\ \mathrm{N/m}\)) is pulled \(0.10\ \mathrm{m}\) from equilibrium and released on a frictionless surface. Find its maximum speed.
    Solution Maximum speed occurs at equilibrium where all spring PE has become kinetic: \(\tfrac12 kA^2=\tfrac12 mv_{\max}^2\Rightarrow v_{\max}=A\sqrt{k/m}=0.10\sqrt{200/0.50}=0.10\sqrt{400}=0.10(20)=2.0\ \mathrm{m/s}\).
  4. (B) A bead slides on a frictionless wire from height \(h_1=3.0\ \mathrm{m}\) (speed \(4.0\ \mathrm{m/s}\)) to height \(h_2=1.0\ \mathrm{m}\). Find its speed at the lower point.
    Solution \(\tfrac12 v_1^2+gh_1=\tfrac12 v_2^2+gh_2\Rightarrow v_2=\sqrt{v_1^2+2g(h_1-h_2)}=\sqrt{16+2(9.81)(2.0)}=\sqrt{16+39.24}=\sqrt{55.24}=7.43\ \mathrm{m/s}\).
  5. (C) A projectile of mass \(m\) is launched radially from Earth's surface (radius \(R=6.37\times10^6\ \mathrm{m}\), \(GM=3.986\times10^{14}\ \mathrm{m^3/s^2}\)) at \(8.0\ \mathrm{km/s}\). Using \(U=-GMm/r\), find the maximum distance \(r_{\max}\) from Earth's centre (ignore atmosphere and rotation).
    Solution At \(r_{\max}\) the radial speed is zero. Energy conservation per unit mass: \(\tfrac12 v_0^2-\frac{GM}{R}=-\frac{GM}{r_{\max}}\). Compute \(\tfrac12(8000)^2=3.20\times10^7\ \mathrm{J/kg}\) and \(\frac{GM}{R}=\frac{3.986\times10^{14}}{6.37\times10^6}=6.258\times10^7\). The constant is \(3.20\times10^7-6.258\times10^7=-3.058\times10^7\ \mathrm{J/kg}<0\) (bound). Then \(\frac{GM}{r_{\max}}=3.058\times10^7\Rightarrow r_{\max}=\frac{3.986\times10^{14}}{3.058\times10^7}=1.303\times10^7\ \mathrm{m}\), about \(2.0R\) (roughly \(6.7\times10^6\ \mathrm{m}\) altitude). Since \(E<0\) it does not escape — consistent with escape speed \(\sqrt{2GM/R}=11.2\ \mathrm{km/s}>8.0\ \mathrm{km/s}\).