Physics theoremStatement: AI-adaptedProof: AI-adaptedprecheck pass

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Power of the total force and the work-energy theorem: $K^{'} = P$ and $K (t_{1}) - K (t_{0}) = W ([t_{0}, t_{1}])$ along a motion

Physical scope

The work-energy identity of the force branch: along every motion the time derivative of the kinetic energy equals the power of the total force, and the change of kinetic energy over a time interval equals the work of the total force. The result is a conditional deduction from the adopted equation of motion and uses no potential, no pair structure and no conservation hypothesis; it asserts no frame-independent value and no measured quantity.

Statement

Adopt the force model and one inertial frame of Total mass, total momentum, centre of mass, angular momentum, torque, kinetic energy, power and work of a particle model in an inertial frame: masses $m_{1}, \dots, m_{N} > 0$ , an open configuration domain $Q$ , the force map $F$ with the standing hypotheses of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance clause (iii), and the quantities $w_{i}$ , $K = \frac{1}{2} \sum_{i} m_{i} ∥ w_{i} ∥_{2}^{2}, P = \sum_{i} ⟨ F_{i} (t, x, w), w_{i} ⟩, W ([t_{0}, t_{1}]) = \int_{t_{0}}^{t_{1}} P (t) d t$ along a curve family. Let $x : J \to Q$ be a motion of the model on an interval $J$ with nonempty interior. Then:

$K$ is of class $C^{1}$ on $J$ and $K^{'} (t) = P (t) (t \in J) :$ the rate of change of the kinetic energy equals the power of the total force;
for all $t_{0} < t_{1}$ in $J$ , $K (t_{1}) - K (t_{0}) = \int_{t_{0}}^{t_{1}} P (t) d t = W ([t_{0}, t_{1}]) :$ the change of the kinetic energy equals the work of the total force.

No potential, no pair decomposition and no conservation hypothesis is used: the identity is a consequence of the equation of motion alone, in the fixed frame. The power and the work are frame-dependent (see clause 4 of the frame-transformation item of this page); the mechanical energy statement of the potential branch is the energy balance of this page.

Facts & Assumptions

Given: A force model of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance with a motion $x : J \to Q$ on an interval $J$ with nonempty interior, and the quantities $K$ , $P$ , $W$ of Total mass, total momentum, centre of mass, angular momentum, torque, kinetic energy, power and work of a particle model in an inertial frame.

[L1]

Notation and equation of motion (Total mass, total momentum, centre of mass, angular momentum, torque, kinetic energy, power and work of a particle model in an inertial frame, The associated first-order system, motions, maximal motions and the Newtonian flow of a force model): $w_{i} = x_{i}^{'}$ is $C^{1}$ because $x_{i}$ is $C^{2}$ ; $p_{i} = m_{i} w_{i}$ ; and $m_{i} {\ddot{x}}_{i} (t) = F_{i} (t, x (t), w (t))$ for all $i$ , $t \in J$ . The Euclidean inner product satisfies $∥ u ∥_{2}^{2} = ⟨ u, u ⟩$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ) and finite sums are linear (Laws of finite sums and finite products).

[L2]

Derivative of the squared norm (Differentiation of sums, scalar multiples, inner products, cross products and squared norms of differentiable curves in $R^{3}$ clause 4): for a differentiable curve $u$ , the function $t \mapsto ∥ u (t) ∥_{2}^{2}$ is differentiable with derivative $2 ⟨ u^{'} (t), u (t) ⟩$ .

[L3]

Newton–Leibniz (Newton–Leibniz needs only continuity on $[a, b]$ , differentiability on $(a, b)$ , and a Riemann-integrable extension of the interior derivative): if $G : [a, b] \to R$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , and $f$ is Riemann integrable with $f = G^{'}$ on $(a, b)$ , then $\int_{a}^{b} f = G (b) - G (a)$ . The interval $J$ has nonempty interior (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Proof

technique · direct; differentiate the kinetic energy using the equation of motion and integrate the resulting identity

1.1givenL1L2algebra

For each $i$ , $w_{i}$ is differentiable with derivative $w_{i}^{'} = {\ddot{x}}_{i}$ ; [L2] gives $\frac{d}{d t} ∥ w_{i} (t) ∥_{2}^{2} = 2 ⟨ {\ddot{x}}_{i} (t), w_{i} (t) ⟩ .$ Multiplying by $\frac{1}{2} m_{i}$ and summing over $i$ with the linearity of finite sums, $K^{'} (t) = \frac{1}{2} \sum_{i} m_{i} \cdot 2 ⟨ {\ddot{x}}_{i} (t), w_{i} (t) ⟩ = \sum_{i} ⟨ m_{i} {\ddot{x}}_{i} (t), w_{i} (t) ⟩ = \sum_{i} ⟨ F_{i} (t, x (t), w (t)), w_{i} (t) ⟩ = P (t),$ where the third equality is the equation of motion of [L1]. Each term is continuous in $t$ (the $F_{i}$ and $w_{i}$ are continuous), so $K^{'}$ is continuous and $K$ is $C^{1}$ ; this is clause 1.

2.1step 1.1L1L3∎

Clause 2: fix $t_{0} < t_{1}$ in $J$ . The function $K$ is continuous on $[t_{0}, t_{1}]$ and differentiable on $(t_{0}, t_{1})$ with $K^{'} = P$ by step 1.1, and $P$ is continuous on $[t_{0}, t_{1}]$ , hence Riemann integrable there; [L3] gives $\int_{t_{0}}^{t_{1}} P = K (t_{1}) - K (t_{0}),$ which is the first equality of clause 2, and the second equality is the definition of $W ([t_{0}, t_{1}])$ .

Remarks

No structure beyond the equation of motion. The power of the total force is the sum of the powers of whatever contributions the model declares, but the identity does not require the sum to split over pairs, and it holds for velocity-dependent and time-dependent forces alike.
Work depends on the frame and on the path. For the boost velocity $v$ in Transformation of the state quantities under a change of inertial frame: centre of mass, total momentum, kinetic energy, angular momentum, torque, power, work and mechanical energy clause 4, the power gains $⟨ R Σ, v ⟩$ , and the work gains the integral of this term. For a motion with the antisymmetric pair presentation, the momentum balance replaces that integral by $⟨ R (P (t_{1}) - P (t_{0})), v ⟩$ ; without those hypotheses the force-integral form applies. The work-energy identity holds in each frame separately.
Not yet an energy conservation statement. Clause 2 relates the kinetic energy to the work of the total force; a conserved mechanical energy requires a potential, which is the subject of Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets and Mechanical energy balance for a potential presentation: $E^{'} = \partial_{t} U$ plus the residual power, conservation for a static potential with no residual forces, and the integrated work form.

Power of the total force and the work-energy theorem: $K^{'} = P$ and $K (t_{1}) - K (t_{0}) = W ([t_{0}, t_{1}])$ along a motion

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Power of the total force and the work-energy theorem: K′=PK′=P and K(t1)−K(t0)=W([t0,t1])K(t1​)−K(t0​)=W([t0​,t1​]) along a motion

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Power of the total force and the work-energy theorem: $K^{'} = P$ and $K (t_{1}) - K (t_{0}) = W ([t_{0}, t_{1}])$ along a motion