Work–Energy Theorem · aiwiki.page

The work–energy theorem is a result in classical mechanics relating the work performed by forces to a change in kinetic energy. For a particle, the total work done by all forces during a displacement equals its final kinetic energy minus its initial kinetic energy: $W_{\mathrm{net}}=\Delta K$ . It provides a way to connect forces, displacement, and speed without necessarily determining the complete motion as a function of time. (openstax.org)

Mathematical statement

For a nonrelativistic particle of constant mass $m$ , moving between positions $A$ and $B$ ,

W_{\mathrm{net}} =\int_A^B \mathbf F_{\mathrm{net}}\cdot d\mathbf r =K_B-K_A =\frac12 mv_B^2-\frac12 mv_A^2.

Here $\mathbf F_{\mathrm{net}}$ is the sum of all forces acting on the particle, $d\mathbf r$ is an infinitesimal displacement along its actual trajectory, and $v_A$ and $v_B$ are its initial and final speeds. Every force acting on the particle must be included, whether or not it is conservative. (openstax.org)

The line integral defines work along a path. The dot product, an example of an inner product, selects the component of force parallel to displacement:

dW=\mathbf F\cdot d\mathbf r.

For a constant force and displacement,

W=Fs\cos\theta,

where $\theta$ is their angle. Work can be positive, negative, or zero; a force perpendicular to displacement does no work. Work is measured in joules in the International System of Units, with $1\,\mathrm J=1\,\mathrm{N\,m}$ . (openstax.org)

Kinetic energy has the same units as work. Its classical expression, $K=\tfrac12 mv^2$ , applies when speeds are much smaller than the speed of light. (openstax.org)

Derivation from Newton’s second law

The theorem follows from Newton’s second law. For constant mass,

\mathbf F_{\mathrm{net}}=m\frac{d\mathbf v}{dt}.

Since $d\mathbf r=\mathbf v\,dt$ ,

\begin{aligned} dW_{\mathrm{net}} &=\mathbf F_{\mathrm{net}}\cdot d\mathbf r\\ &=m\frac{d\mathbf v}{dt}\cdot\mathbf v\,dt\\ &=m\mathbf v\cdot d\mathbf v\\ &=d\left(\frac12m\mathbf v\cdot\mathbf v\right). \end{aligned}

Taking the integral between the initial and final states gives $W_{\mathrm{net}}=K_B-K_A$ . Neither constant acceleration nor a straight trajectory is required. The derivation concerns the particle’s actual displacement, not simply the distance between its endpoints. (live.ocw.mit.edu)

Dividing the differential relation by time gives the instantaneous rate of work:

\frac{dW_{\mathrm{net}}}{dt} =\mathbf F_{\mathrm{net}}\cdot\mathbf v =\frac{dK}{dt}.

Thus the net mechanical power delivered to a particle equals the rate at which its kinetic energy changes. (openstax.org)

Physical interpretation and applications

Positive net work increases a particle’s speed, negative net work decreases it, and zero net work leaves its initial and final speeds equal. Equal speeds do not imply equal velocity vectors: a force can change the direction of motion while remaining perpendicular to the velocity and doing no work. The theorem therefore constrains speed rather than fully determining the trajectory. (openstax.org)

Motion under gravity. In a uniform gravitational field, the work done by gravity during a downward height change $h$ is $mgh$ . If other forces do no work,

\frac12mv_B^2-\frac12mv_A^2=mgh, \qquad v_B^2=v_A^2+2gh.

This relates the speed to the vertical drop without requiring the travel time. (ocw.mit.edu)

Stopping by friction. As an illustrative application, consider a body modeled as a particle, initially moving at speed $v_0$ , and stopped by a constant opposing force of magnitude $f$ . Over a stopping distance $s$ ,

-fs=0-\frac12mv_0^2, \qquad s=\frac{mv_0^2}{2f}.

For sliding on a horizontal stationary surface with kinetic friction $f=\mu_kmg$ , this becomes $s=v_0^2/(2\mu_kg)$ . Under these idealized assumptions, stopping distance grows with the square of initial speed. This is a direct calculation using the theorem, not an empirical braking model. (openstax.org)

Relationship to potential energy and energy conservation

The theorem does not require that mechanical energy be conserved. Where a time-independent conservative force can be represented by potential energy $U$ , its work satisfies

W_{\mathrm c}=-\Delta U.

Separating conservative and nonconservative contributions gives

\Delta K=-\Delta U+W_{\mathrm{nc}},

\Delta(K+U)=W_{\mathrm{nc}}.

Mechanical energy $K+U$ is constant when the remaining forces do zero net work. Conservative forces may still change kinetic energy by exchanging it with potential energy. (openstax.org)

A loss of mechanical energy is not necessarily a loss of total energy. For a system containing interacting bodies, energy may be redistributed through internal interactions. It is consequently important to distinguish work on a single particle from the energy balance of a larger system. (arxiv.org)

Particle systems and rigid bodies

For a fixed collection of constant-mass particles, summing the individual work–energy equations yields

\Delta K_{\mathrm{total}} =W_{\mathrm{external}}+W_{\mathrm{internal}}, \qquad K_{\mathrm{total}}=\sum_i\frac12m_iv_i^2.

Internal forces cannot generally be omitted. Equal and opposite interaction forces can do nonzero combined work because their points of application may undergo different displacements. Thus external work alone need not equal the change in the total kinetic energy of a deformable system. (arxiv.org)

For an ideal rigid body, internal constraint forces do no net work. External forces must nevertheless be evaluated using the displacements of their actual points of application. A pair of forces with zero resultant can still do work by producing rotation. (live.ocw.mit.edu)

For rotation about a fixed axis,

W_{\mathrm{net}} =\int_{\theta_A}^{\theta_B}\tau_{\mathrm{net}}\,d\theta =\frac12I\omega_B^2-\frac12I\omega_A^2,

where $\tau_{\mathrm{net}}$ is the net torque about that axis, $I$ is the constant moment of inertia, and $\omega$ is angular velocity. This is the rotational form of the theorem. (openstax.org)

Reference frames and limits of applicability

Work and kinetic energy depend on the reference frame. The theorem holds in every inertial reference frame, provided forces, displacements, and velocities are evaluated consistently in that frame. A force that does zero work in one frame may do nonzero work in another. In an accelerating, nonrotating frame, the work of the corresponding inertial force must be included in the kinetic-energy balance. (arxiv.org)

The elementary derivation assumes constant mass. A body gaining or losing material cannot generally be analyzed merely by replacing $m$ with $m(t)$ ; the particles entering or leaving the chosen system carry momentum and energy that must be accounted for. (arxiv.org)

In special relativity, the work–energy relation remains applicable to a particle of constant rest mass, but kinetic energy becomes

K=(\gamma-1)mc^2, \qquad \gamma=\frac{1}{\sqrt{1-v^2/c^2}}.

With force defined through momentum as $\mathbf F=d\mathbf p/dt$ , net work equals the change in this relativistic kinetic energy. The expression $\tfrac12mv^2$ is recovered in the low-speed limit. (openstax.org)

References

3 Work-Energy Theorem — University Physics Volume 1openstax.org
1 Work — University Physics Volume 1openstax.org
2 Kinetic Energy — University Physics Volume 1openstax.org
Work and Energy — MIT OpenCourseWarelive.ocw.mit.edu
Conservative Internal Forces and Potential Energy — MIT OpenCourseWareocw.mit.edu
3 Conservation of Energy — University Physics Volume 1openstax.org
Remarks on the conservation of mechanical energy in introductory mechanicsarxiv.org
8 Work and Power for Rotational Motion — University Physics Volume 1openstax.org
Work and energy in inertial and non inertial reference framesarxiv.org
9 Relativistic Energy — University Physics Volume 3openstax.org