Multiparticle Work

System kinetic energy and external work

For particles of total mass $M$ , write each velocity as a centre-of-mass part plus a relative part,

v_{i} = V_{cm} + v_{i}^{'} .

The total kinetic energy separates as

K = i \sum \frac{1}{2} m_{i} v_{i}^{2} = \frac{1}{2} M V_{cm}^{2} + i \sum \frac{1}{2} m_{i} v_{i}^{'}^{2} = K_{cm} + K^{'} .

The cross term vanishes because $\sum_{i} m_{i} v_{i}^{'} = 0$ . The first term is translation of the system centre of mass. The second is kinetic energy relative to the centre-of-mass frame: rotation, vibration, and other internal motion. A rigid body translating without rotation has $K^{'} = 0$ ; a spinning body at rest at its centre of mass has $K_{cm} = 0$ but can have large $K^{'}$ .

External force determines centre-of-mass motion:

M A_{cm} = \sum F_{ext} .

The work associated with the net external force and centre-of-mass displacement is

W_{cm} = \int (\sum F_{ext}) \cdot d R_{cm} = Δ K_{cm} .

Centre-of-mass work measures translation only. Total external work uses the application-point displacement of every force and can change internal kinetic energy as well as centre-of-mass motion.

External resultant force changes center-of-mass translation, while forces acting at different points can also create internal rotation or vibration.

Work of individual external forces.

The total work of external forces is the sum of work at their individual points of application,

W_{ext} = a \sum \int F_{a}^{ext} \cdot d r_{a} .

Application points generally do not share the centre-of-mass displacement. A force on a rotating wheel can do work even if the centre of mass is fixed; a force through the centre of mass can translate the body without changing its rotation. For a rigid body, the displacement of a point can be separated into translational and rotational parts. The external work then contains a centre-of-mass contribution and a torque contribution,

P_{ext} = (\sum F_{ext}) \cdot V_{cm} + τ_{cm} \cdot ω .

Net force and net torque govern distinct power terms. A pure couple has zero resultant force but nonzero torque and can increase rotational kinetic energy. A force whose line of action passes through the centre of mass can have zero torque but nonzero translational power.

Force and velocity must be read at the same instant and material point. An off-centre actuator, a rotating shaft, and a translating support can share a force magnitude yet deliver different power, because their application-point velocities differ. Force magnitude alone does not fix the energy transfer without the matching kinematics.

Internal work, rolling, and center-of-mass scope

A pair of particles has equal-and-opposite internal forces, but their works need not cancel because the particles generally have different displacements. A stretched spring pulls both ends; its forces sum to zero, yet its potential energy can become kinetic energy of the attached masses. Internal force cancellation governs the momentum sum; work requires the individual application-point displacements.

For conservative internal interactions, potential energy accounts for the work:

W_{int} = - Δ U .

For dissipative interactions, internal work can raise thermal energy or permanent deformation. A complete system-energy statement has the form

Δ K_{cm} + Δ K^{'} + Δ U + Δ E_{th} = W_{ext} .

The division into internal and external terms depends on the system boundary. For a crate-only system, a person's push is external work. For a person-crate system, the contact force is internal, while chemical energy in the person decreases.

A spring transfers internal potential energy into relative kinetic energy. The center of mass remains fixed when no external force acts.

Rolling and contact work.

In ideal rolling on a fixed surface, static friction does no work on the rolling body: the contact point is instantaneously at rest. It still supplies torque and changes angular speed. Kinetic friction does work because the contact point slides, and a moving conveyor or accelerating support shifts the contact-point displacement so that even static friction can transfer energy.

Derivation of centre-of-mass work.

From $R = M^{- 1} \sum_{i} m_{i} r_{i}$ , two differentiations give

M A = i \sum m_{i} a_{i} .

Newton's second law expresses each particle acceleration through external and internal forces. Equal-and-opposite internal pairs cancel in the sum, leaving

M A = \sum F_{ext} .

Dot multiplication by $d R = V d t$ gives

\int \sum F_{ext} \cdot d R = \int M A \cdot V d t = Δ (\frac{1}{2} M V^{2}) .

Hence $W_{cm} = Δ K_{cm}$ . The theorem uses center-of-mass displacement, not the displacement of an arbitrary material point. That distinction is what excludes rotational and deformational energy from $K_{cm}$ .

Total particle work.

Applying the work-energy theorem to every particle gives

Δ (K_{cm} + K^{'}) = W_{ext} + W_{int} .

The external work is evaluated at each force's point of application. For conservative internal forces, $W_{int} = - Δ U$ . For dissipative internal forces, it can increase thermal energy and permanent deformation. The recoverable system-energy form is

Δ K_{cm} + Δ K^{'} + Δ U + Δ E_{th} = W_{ext} .

Internal forces cancelling in the net-force equation does not require their work to cancel. The two ends of a spring have different displacements, so spring forces can convert elastic energy into relative kinetic energy while producing no centre-of-mass acceleration.

Deformation, rotation, and accelerating frames

Two carts connected by a compressed spring provide a direct example. The system is initially at rest and has no external force. After release, the carts have opposite momenta, so $K_{cm} = 0$ remains unchanged. The spring potential energy decreases and relative kinetic energy increases:

Δ K^{'} = - Δ U_{s} .

An inelastic collision reverses this energy path. The centre-of-mass velocity is unchanged when external impulse is negligible, but relative kinetic energy becomes internal deformation and thermal energy. Centre-of-mass work alone cannot predict that heating because it contains no relative-motion term.

A rigid body has external power with translational and rotational pieces,

P_{ext} = F_{net} \cdot V_{cm} + τ_{cm} \cdot ω .

The first term changes centre-of-mass translation and the second changes rotation. A pure couple has zero net force but nonzero torque and can supply rotational power. A force through the centre of mass can supply translational power with zero torque.

Checks and boundaries

All energy terms have units of joules. An isolated system has $W_{cm} = 0$ and constant centre-of-mass speed, but it can have changing total kinetic energy. System boundaries determine whether a force is external or internal. A rope force is external to a block, internal to a block-rope system, and changes category again when a person pulling the rope is included. The force itself is unchanged; only the energy accounting changes.

Translational and rotational work for rigid bodies.

The velocity of point $a$ on a rigid body is

v_{a} = V_{cm} + ω \times r_{a}^{'},

where $r_{a}^{'}$ runs from the centre of mass to the point. Substitution into the sum of force powers gives

a \sum F_{a} \cdot v_{a} = (a \sum F_{a}) \cdot V_{cm} + (a \sum r_{a}^{'} \times F_{a}) \cdot ω .

The second parenthesis is torque about the centre of mass. The common formula $P = F v$ holds only at the particular point where the force acts. When a body rotates, its points have different velocities, so a force may do positive work at one point while the centre of mass has no displacement along it. Power accounting begins with the point of application.

Description	Power expression	Required velocity
One applied force	$P_{a} = F_{a} \cdot v_{a}$	velocity at point $a$
Net translation	$F_{net} \cdot V_{cm}$	centre-of-mass velocity
Rotation about CM	$τ_{cm} \cdot ω$	angular velocity and moment arm

An ideal hinge is a limiting case. A force at a fixed hinge point does zero work because that point has zero displacement, yet other forces can rotate the body and change its kinetic energy. The hinge force can be large without supplying energy. A motor shaft is different: it exerts torque through an angle and supplies work $W = \int τ d θ$ .

Work in accelerating frames.

The centre-of-mass theorem is most direct in an inertial frame. In a uniformly accelerating frame, an inertial force $- M a_{frame}$ is included in the effective external-force sum. The centre of mass then obeys the frame's component equation. Apparent-work terms can be represented by an effective potential only under restricted conditions. Mixing a centre-of-mass velocity measured in one frame with force work measured in another produces an inconsistent energy calculation.

Detailed deformable-body example.

A bullet embeds in a suspended block. During the short collision, external impulse is negligible in the horizontal direction, so horizontal centre-of-mass velocity is approximately constant. The block and bullet deform, heat, and acquire internal vibration. Their total kinetic energy decreases even though $K_{cm}$ is unchanged. After the collision, the combined body rises; during that separate stage, the centre-of-mass kinetic energy converts to gravitational potential energy. The collision equation and the subsequent energy equation cannot be merged because the internal energy transfer occurs only during impact.

The same bookkeeping applies to a falling deformable object striking the ground. An object-ground system classifies contact forces as internal. Gravitational potential energy can become thermal energy of both surfaces, sound, and residual deformation. An object-only system classifies the ground's contact force as external work. Both system boundaries are valid if their energy terms are classified consistently.

Two failure modes.

An isolated internal explosion accelerating the centre of mass. This violates momentum conservation; a zero external resultant keeps $V_{cm}$ fixed.
Internal force pairs always doing zero total work. This confuses equal-and-opposite forces with equal displacements, since spring ends move by different amounts.

The translational theorem and the full system-energy balance answer different questions and draw on different displacement data.

Requested quantity	Relation	Required kinematic data
Centre-of-mass translation	$W_{ext, cm} = Δ K_{cm}$	centre-of-mass displacement and net external force
Total power	$P = \sum_{a} F_{a} \cdot v_{a}$	velocity of every force application point
Rigid-body power	$P = F_{net} \cdot V_{cm} + τ_{cm} \cdot ω$	resultant force, torque, and angular speed
Internal storage	change in relative, elastic, thermal, or deformation energy	declared system boundary

State the approximation explicitly. If external impulse is appreciable, $V_{cm}$ changes and the matching external work stays in the balance. Internal energy terms may be dropped only once the model has shown that they are unchanged.

External and internal work at distinct points

The external-work sum uses the displacement of the point where each external force acts. Substituting the centre-of-mass displacement everywhere holds only for special cases, such as a rigid body under a resultant force with no torque contribution. An off-centre force can translate the system and change its rotational or vibrational energy at the same time.

Internal forces need the same point-by-point care. Equal and opposite forces need not do equal and opposite work, because their application points can move through different displacements: a relaxing spring can do positive work on both attached masses while its stored potential energy decreases. The internal-force sum is zero at each instant, yet the internal work still converts energy among the system's relative degrees of freedom. The system boundary sets the classification, not the physical identity of the force. A hand's push is external to a cart-only system; for a hand-cart system the same contact force is internal, and the chemical-energy change in the hand joins the account.

Boundary sets the bookkeeping.

A person pushes a crate $2.0 m$ with a $40 N$ horizontal force, supplying $+ 80 J$ of external work to the crate-only system. In the person-crate system the push is internal, so an $80 J$ chemical-energy decrease and any thermal change carry the account instead. Entering both the external push work and the internal chemical-energy decrease in one equation double-counts the same $80 J$ .

Center-of-mass energy versus relative energy.

$K_{cm} = \frac{1}{2} M V_{cm}^{2}$ carries translation of the centre of mass; the remaining $K^{'}$ , measured in the centre-of-mass frame, holds the opposite motions of two separating fragments, rotation about the centre, and vibration. The external resultant controls only $K_{cm}$ : a zero resultant fixes $V_{cm}$ yet leaves $K^{'}$ free to change, as in an explosion at rest ( $K_{cm} = 0$ , chemical energy feeding $K^{'}$ ) or an inelastic collision ( $K_{cm}$ fixed, $K^{'}$ falling into heat).

Mass weighting shapes the split. The centre of mass sits nearer the heavier particle, and for equal and opposite momenta the lighter particle moves faster and takes the larger share of $K^{'}$ , because at fixed momentum magnitude the kinetic energy $p^{2} /2 m$ grows as $m$ shrinks.

Two particles with opposite centre-of-mass-frame momenta have a fixed center of mass but nonzero relative kinetic energy. The lighter particle moves faster, so the internal kinetic-energy partition is mass-weighted even though the system center of mass remains at rest.

Two-body spring release.

A compressed spring between two free masses gives a clean centre-of-mass picture. On a frictionless surface starting from rest, external force and impulse vanish, so the centre of mass stays fixed. Release drops the elastic potential energy and gives the masses opposite momenta and positive relative kinetic energy, drawn from internal spring deformation rather than external work.

Momentum conservation sets the ratio of final speeds; the spring-energy decrease sets their scale. With final momenta of equal magnitude $p$ and opposite direction,

K^{'} = \frac{p ^{2}}{2 m _{1}} + \frac{p ^{2}}{2 m _{2}},

and equating $K^{'}$ to the released spring energy fixes $p$ . The lighter mass leaves faster and with the larger share of kinetic energy, while the centre of mass holds its original position.

The centre-of-mass theorem and the full system-energy balance are complementary: the first tracks translation, the second the internal changes translation cannot resolve. Both require one consistent system boundary and inertial frame.

Deformation and relative energy

External work on a deformable body is generally not the resultant force times centre-of-mass displacement, since material points move by different amounts; the work sum uses each external force with its own application-point displacement. Equal and opposite external forces can have zero resultant and zero centre-of-mass acceleration yet still do positive work by stretching the body, storing energy as elastic deformation, vibration, or heat rather than translation.

A bar pulled at both ends illustrates the distinction. If the end forces pull outward while the ends separate, each force does positive work on the bar. Their resultant is zero, so $W_{cm} = 0$ , but the bar's internal energy increases. A centre-of-mass work calculation alone would report no translational kinetic-energy change and would miss the energy stored in strain. Releasing the bar can reverse the transfer and launch waves or masses attached to its ends.

An ideal elastic deformation stores external work reversibly as elastic potential energy. Plastic deformation or internal damping converts part to thermal energy that is not recovered on unloading. The same external-force pattern can therefore produce different final energy partitions depending on the material model. The system boundary and constitutive assumption must be stated before assigning a single “internal energy” term.

Equal and opposite external pulls can deform a bar without moving its center of mass. Each force acts through an outward endpoint displacement and does positive work; the zero external resultant leaves CM translation unchanged while the work appears as internal strain or thermal energy.

Energy in the center-of-mass frame.

The centre-of-mass frame removes the translational kinetic-energy term by definition: $V_{cm} = 0$ , so the total kinetic energy measured in that frame is $K^{'}$ . This makes the centre-of-mass frame well suited to collisions, explosions, and two-body interactions. It isolates the energy available for relative motion from the energy associated with the system moving as a whole past an external observer.

For two particles, relative kinetic energy can be expressed with the reduced mass $μ = m_{1} m_{2} / (m_{1} + m_{2})$ and relative speed $v_{rel}$ :

K^{'} = \frac{1}{2} μ v_{rel}^{2} .

The reduced-mass form replaces the two-body motion with one relative coordinate and gives the same value as summing individual centre-of-mass-frame kinetic energies, but it makes clear that $K^{'}$ depends on how rapidly the particles approach or separate. The reduced mass is smaller than either total mass and is weighted toward the lighter particle, reflecting the unequal shares of motion in the centre-of-mass frame.

Frame choice changes the numerical total kinetic energy but not the relative speed or the internal kinetic energy $K^{'}$ . Two particles observed from a train and from the ground have different centre-of-mass velocities, yet their separation rate and reduced-mass energy are the same under a constant-velocity frame transformation. $K^{'}$ is therefore the frame-independent energy quantity for internal interactions.

Energy transfer through an internal force.

Internal forces transfer energy between parts of a system even when their vector sum is zero. For an ideal spring the two internal works sum to the negative change in spring potential energy, and each end's work depends on that end's own motion. Momentum conservation follows from the cancellation of internal force pairs, but energy transfer needs the velocities of the application points. A damper is the contrasting case: its forces also sum to zero, yet the work it removes from relative motion becomes heat rather than recoverable elastic energy.

The total internal power is the sum of the two force-velocity dot products: negative while spring potential energy rises, positive while it falls. The centre-of-mass velocity can stay constant throughout, so this internal power changes $K^{'}$ and $U$ without touching $K_{cm}$ .

Internal-power interpretation.

When two masses move apart and stretch their spring, each spring force opposes its endpoint velocity, so the spring does negative work on both: relative kinetic energy falls and spring potential energy rises. On release the velocities reverse relative to the forces, the spring does positive work on both, and relative kinetic energy returns. An isolated system holds the same centre-of-mass velocity throughout, since the internal forces have no resultant.

Constraints, collisions, and energy audits

An ideal constraint does not automatically imply zero work on every member of a system. A fixed, smooth guide exerts a normal force perpendicular to its contact point displacement and therefore does no work on the guided particle. A moving guide can exert a normal force through a nonzero displacement and transfer energy. A taut string of fixed length can transfer energy between its endpoints even when the net work of tension on the complete ideal string-and-mass system is zero. The relevant question is always which point moves under which force.

Constraint geometry often links velocities and therefore links power. In a simple massless rope over an ideal pulley, tension acts along each rope segment. The power on one attached mass is tension times that endpoint's velocity component along the rope; the power on the other mass has the opposite sign when the endpoints move in opposite directions with equal speeds. Tension can remove kinetic energy from one object while delivering it to the other without changing the total energy of the ideal constrained pair.

Nonideal constraints add internal energy channels. Pulley axle friction, rope stretch, sliding contact, or deformation can convert part of the transferred mechanical energy to thermal energy. The common-tension and zero-net-constraint-work assumptions then need revision. A diagram that labels an ideal constraint is making a quantitative energy statement about the constraint.

Collision energy and center-of-mass motion.

During a short collision with negligible external impulse, the centre-of-mass velocity is unchanged. That does not mean total kinetic energy is unchanged. The centre-of-mass kinetic term is fixed by the conserved total momentum, while the relative kinetic term can decrease as the objects deform, heat, or vibrate. In a perfectly elastic collision, relative kinetic energy is preserved. In an inelastic collision, some or all of it becomes internal energy, even though the centre of mass continues with the same velocity.

The centre-of-mass frame makes this contrast direct. Before a head-on collision, the particles approach with opposite momenta. After a perfectly inelastic collision, the combined object is at rest in the centre-of-mass frame and the relative kinetic energy is zero. In a laboratory frame, that same combined object may continue to move with substantial kinetic energy because the centre of mass is translating. The lab-frame final kinetic energy is then precisely the centre-of-mass translational term of the combined mass.

Energy lost from relative motion is not negative; it is added deformation, thermal, and sometimes sound energy inside the larger system. Momentum conservation alone sets the common post-collision centre-of-mass velocity; an energy statement is needed to find how much relative kinetic energy remains and how much turns internal.

In a short inelastic collision, the center of mass continues uniformly when external impulse is negligible, while relative kinetic energy collapses into deformation and thermal energy. The laboratory-frame motion after sticking can remain nonzero because it contains the unchanged center-of-mass translation term.

Complete two-object system energy audit.

For two carts joined by a spring and pulled by an external agent, the external work can change centre-of-mass translation, spring deformation, and relative motion at once. The internal spring's own work is not entered again as an external term; its transfer shows up through the change in elastic potential and relative kinetic energy. Adding both the spring work and the spring-potential change to the same total-system balance double-counts it.

A two-object energy audit separates center-of-mass translation from relative motion and internal storage. External work crosses the system boundary; spring deformation and thermal energy are internal terms whose changes explain energy transfer even when the external resultant and CM motion are zero.

Variable mass and coupled bodies

An energy statement for a system whose mass changes requires special care because matter can carry kinetic energy, internal energy, and momentum across the boundary. A rocket-only system, for example, loses mass through its nozzle. The expelled fuel crosses the boundary with velocity relative to the chosen frame and carries kinetic energy with it. A simple fixed-mass equation for the rocket alone omits this energy flux unless it is written explicitly.

One remedy is to choose a larger system that includes the rocket and all exhaust under consideration. Then the chemical-energy decrease, kinetic energy of the rocket, and kinetic energy of the exhaust are internal terms. Another is to retain a smaller system and account for energy carried across its boundary by the entering or leaving mass. Neither choice is automatically simpler; the correct choice is the one whose boundary terms can be identified without mixing frames or velocities.

The same caution applies to a conveyor dropping material into a cart, leaking fluid, or a sandbag landing on a moving wagon. The system mass changes and the incoming or outgoing material has a velocity not generally equal to the centre-of-mass velocity of the retained system. Centre-of-mass work and fixed-mass kinetic-energy formulas must be applied only after the system definition has been made explicit.

A variable-mass system boundary must account for energy carried by matter crossing it. Exhaust leaving a rocket has its own kinetic energy in the selected frame; enlarging the boundary to include rocket and exhaust converts that boundary flux into internal kinetic and chemical-energy changes.

Exhaust energy is not negligible.

If $0.20 kg$ of exhaust leaves a vehicle at high speed, its kinetic energy matters even though the remaining vehicle is far heavier. A vehicle-only equation needs an outflow-energy term; a vehicle-plus-exhaust equation makes that term internal but must include the chemical-energy decrease that drove both motions. The two accounts agree only for the same frame and the same material crossing.

Coupled-body energy calculation.

Two bodies linked by an ideal string offer a compact system-energy calculation when their motions are constrained. The string enforces a relation between their speeds, and tension is internal to the two-body-string system. Its individual works can be nonzero on each mass, yet they cancel in the total ideal-system energy balance when the string is massless and taut. Gravitational potential changes, kinetic energy of both masses, and any friction on a contact surface remain as the relevant terms.

The system approach is efficient for finding the common speed after a displacement. It does not immediately supply tension, because tension has been removed as an internal interaction. To obtain tension, return to an individual-body force equation after the energy calculation has established the speed or acceleration. This division of labour avoids treating the internal tension as both cancelled and externally available in the same line of reasoning.

Energy methods determine reachability. If the decrease in gravitational potential of a descending mass is smaller than the increase required by the rising mass plus friction work, the assumed displacement cannot occur from rest without an external energy source. A negative result for speed squared indicates a physically inaccessible configuration, not a negative speed.

For two masses coupled by an ideal string, tension is internal to the enlarged two-body system while gravity and surface friction supply the net energy change. The linked displacements give a common speed magnitude, allowing one energy equation to determine motion without first solving for tension.

Linkages and pulley work

A pulley or rigid linkage can redirect force and motion while transferring mechanical energy between components. For an ideal fixed pulley, the two rope endpoints move equal distances, so equal tension magnitudes produce equal and opposite tension works on the attached bodies. The pulley changes direction but does not create energy. A movable pulley has endpoint distances differing by a geometric factor. The force ratio and displacement ratio compensate so that ideal input and output work match.

The work balance avoids treating mechanical advantage as an energy advantage. A linkage that lets a person lift a load with half the force generally requires twice the input displacement. Friction, pulley rotation, and rope stretch reduce output energy relative to input, but no ideal arrangement can supply more mechanical energy than it receives. A system-energy diagram should show where any losses are assigned: axle heating, rope deformation, or contact friction.

Power has the same conservation structure. In an ideal linkage, input force times input speed equals output force times output speed at each instant, with signs set by the motion directions. A slow, high-force load motion can therefore be driven by a fast, lower-force input motion. This is an instantaneous constraint relation, not an additional energy source.

A movable pulley trades force for displacement while preserving ideal work. Two supporting rope segments lift the load, so the input end moves twice as far as the load; the doubled load-side tension force is balanced by the doubled input displacement in the ideal energy transfer.

Boundary and losses.

State whether the string and pulley are inside the system. When they are ideal and included, tension is internal and cancels from the total energy balance; a hand pulling the free end does work across the boundary; modelled axle friction or rope stretch adds a thermal or elastic term. The displacement ratio must come from the actual string or linkage geometry. Using a force advantage without its matching motion constraint invents energy and signals an inconsistent model, and the same geometry must hold for power at every instant.