The Work-Energy Theorem

Derivation

The Work-Energy Theorem

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Statement

For a particle of constant mass \(m\) moving under a net force \(\vec{F}\), the work done by that net force as the particle travels from point \(A\) to point \(B\) along its actual trajectory equals the change in its kinetic energy: \(\displaystyle \int_{A}^{B}\vec{F}\cdot d\vec{r} = \tfrac{1}{2}mv_B^{2}-\tfrac{1}{2}mv_A^{2}\), where \(K=\tfrac12 mv^2\) and \(v=|\vec v|\) is the speed.

Why it matters

The work-energy theorem converts a vector equation of motion, \(\vec{F}=m\vec{a}\), into a single scalar statement about speed. When the detailed geometry of the path is awkward but the endpoints and forces are known, energy bookkeeping often reaches the answer with no need to solve the trajectory in time.

It is also the bridge to the entire language of energy. Kinetic energy, potential energy, and eventually the conservation of mechanical energy all grow from this one identity, which holds for any net force, conservative or not.

Assumptions

The mass \(m\) is constant.If \(m\) varies (a rocket shedding fuel, a relativistic particle), \(\vec{F}=m\vec{a}\) is no longer the correct form of Newton's law and the manipulation \(\vec{F}\cdot d\vec{r}=m\,\vec{v}\cdot d\vec{v}\) fails; one must return to \(\vec{F}=d\vec{p}/dt\).

\(\vec{F}\) is the net (total) force on the particle.If \(\vec{F}\) is only one of several forces acting, the left side is the work of that single force alone and no longer equals the total change in kinetic energy; the theorem in the stated form breaks.

Newton's second law holds in the chosen frame — the frame is inertial.In a non-inertial (accelerating or rotating) frame, real forces alone do not obey \(\vec{F}=m\vec{a}\); fictitious forces (centrifugal, Coriolis) must be included in \(\vec{F}\) for the theorem to remain valid.

The particle is a point mass — no internal structure or rotation.For an extended or spinning body the kinetic energy acquires a rotational term \(\tfrac{1}{2}I\omega^{2}\), and \(\vec{v}\) must be read as the centre-of-mass velocity with a separate accounting for internal energy.

Derivation

\[ \vec{F} = m\vec{a} = m\frac{d\vec{v}}{dt} \]

Newton's second law for a constant-mass particle; acceleration is the time derivative of velocity. A

\[ W = \int_{A}^{B}\vec{F}\cdot d\vec{r} \]

Definition of the work done by \(\vec{F}\) along the path from \(A\) to \(B\); the line integral of force through displacement. A

\[ d\vec{r} = \frac{d\vec{r}}{dt}\,dt = \vec{v}\,dt \]

Parametrise the path by time; the infinitesimal displacement is velocity times \(dt\), converting the line integral into an integral over time. A

\[ W = \int_{t_A}^{t_B} m\frac{d\vec{v}}{dt}\cdot \vec{v}\,dt = \int_{t_A}^{t_B} m\,\vec{v}\cdot\frac{d\vec{v}}{dt}\,dt \]

Substitute Steps 1 and 3 into Step 2. The dot product is commutative, so \(\tfrac{d\vec v}{dt}\cdot\vec v = \vec v\cdot\tfrac{d\vec v}{dt}\). B

\[ \vec{v}\cdot\frac{d\vec{v}}{dt} = \frac{1}{2}\frac{d}{dt}\left(\vec{v}\cdot\vec{v}\right) = \frac{1}{2}\frac{d}{dt}\left(v^{2}\right) \]

Product rule on \(\vec v\cdot\vec v\): \(\tfrac{d}{dt}(\vec v\cdot\vec v)=2\,\vec v\cdot\tfrac{d\vec v}{dt}\). Here \(v^{2}\equiv\vec v\cdot\vec v\) is the squared speed. C

\[ W = \int_{t_A}^{t_B} \frac{d}{dt}\left(\frac{1}{2}mv^{2}\right)dt \]

Insert Step 5 and pull the constant \(m\) inside; the integrand is now a total time derivative of the scalar \(\tfrac12 m v^2\). B

\[ W = \left[\frac{1}{2}mv^{2}\right]_{t_A}^{t_B} = \frac{1}{2}mv_B^{2}-\frac{1}{2}mv_A^{2} \]

Fundamental theorem of calculus: the integral of a total derivative is the change in its antiderivative between the endpoints. A

\[ K \equiv \frac{1}{2}mv^{2} \quad\Longrightarrow\quad W = K_B - K_A = \Delta K \]

Define kinetic energy \(K\); the net work equals the change in kinetic energy. A

Result

\[ W_{\text{net}} = \int_{A}^{B}\vec{F}\cdot d\vec{r} = \frac{1}{2}mv_B^{2}-\frac{1}{2}mv_A^{2} = \Delta K \]

Reading. The scalar quantity accumulated by projecting the net force onto each step of the path — the net work — is stored entirely as a change in the particle's kinetic energy. Positive net work speeds the particle up; negative net work (force opposing motion) slows it down. Only the endpoint speeds enter, not the intermediate detail.

Units check. \([\vec F\cdot d\vec r] = \mathrm{N\cdot m = J}\). On the right, \([mv^{2}] = \mathrm{kg\,(m/s)^{2} = kg\,m^{2}/s^{2} = J}\). Both sides are joules; \(\tfrac12\) is dimensionless. Consistent.

Limiting cases

Force perpendicular to motion (\(\vec F\perp\vec v\), e.g. uniform circular motion or a magnetic force): \(\vec F\cdot d\vec r=0\) everywhere, so \(W=0\) and speed is constant even though velocity is not.
Constant force along a straight line: \(W=Fd\) recovers the elementary \(Fd=\tfrac12 mv_B^2-\tfrac12 mv_A^2\), which with kinematics reproduces \(v_B^2=v_A^2+2ad\).
Zero net force: \(W=0\Rightarrow \Delta K=0\), so speed is constant — Newton's first law as an energy statement.
Closed path with a conservative net force: \(\oint\vec F\cdot d\vec r=0\), so the particle returns to its starting point with its original speed.

Breaks when

Variable mass. For a body gaining or losing mass (a rocket, a raindrop accreting vapour), \(\vec F=d\vec p/dt\neq m\vec a\), so Step 1 is invalid and the simple \(\Delta K\) form fails; the ejected mass carries away momentum and energy.
Relativistic speeds. When \(v\) approaches \(c\), kinetic energy is \(K=(\gamma-1)mc^{2}\) with \(\gamma=1/\sqrt{1-v^2/c^2}\), not \(\tfrac12 mv^2\); the low-speed identity in Step 5 is only the leading term of the expansion.
Non-inertial frame without fictitious forces. If \(\vec F\) omits centrifugal or Coriolis terms in a rotating frame, \(\vec F=m\vec a\) fails and the computed work no longer matches \(\Delta K\).
Extended or deformable bodies. For a real object, part of the work goes into internal (thermal, elastic, rotational) energy, so \(\int\vec F\cdot d\vec r\) over the centre of mass need not equal the total kinetic-energy change of the parts — one needs the centre-of-mass (pseudowork) theorem separately.

Failure modes

Using a single force instead of the net force. Setting the work of gravity alone equal to \(\Delta K\) when tension or friction also act. Only the total work equals \(\Delta K\).
Sign errors on the dot product. Forgetting \(\vec F\cdot d\vec r = F\,dr\cos\theta\), so a retarding force (\(\theta=180^\circ\)) does negative work; students often add its magnitude.
Squaring the speed difference. Writing \(\tfrac12 m(v_B-v_A)^2\) instead of \(\tfrac12 m v_B^2-\tfrac12 m v_A^2\); kinetic energy is not linear in the velocity difference.
Treating \(v\) as a vector inside the square. \(v^2=\vec v\cdot\vec v\) is the squared speed; only the magnitude matters, so pure direction changes contribute no work.
Assuming \(W=Fd\) always. The product form holds only for a constant force along a straight path; for varying force or a curved path the integral is mandatory.
Confusing work with power or impulse. Work is \(\int\vec F\cdot d\vec r\) (joules), not \(\vec F\cdot\vec v\) (watts) nor \(\int\vec F\,dt\) (newton-seconds).

Discussion

The theorem is remarkable for what it discards. Newton's second law is three coupled vector equations tracking the full time history of position; the work-energy theorem collapses this to one scalar relation between two speeds. The price is that the path integral \(\int\vec F\cdot d\vec r\) must be evaluated, but when the force depends only on position and the path is simple, this is far easier than integrating the equations of motion. Energy methods trade dynamical detail for scalar economy.

Crucially, the theorem holds for every net force — friction, drag, tension, a time-dependent push. It does not require the force to be conservative. Conservativeness is an extra property that lets us replace the path integral with a potential-energy difference, \(\int_A^B\vec F\cdot d\vec r = -(U_B-U_A)\). Combining that special case with the theorem yields conservation of mechanical energy, \(K_A+U_A=K_B+U_B\). Thus the work-energy theorem is the general parent; energy conservation is its restriction to conservative forces.

The heart of the proof, \(\vec v\cdot d\vec v=\tfrac12 d(v^2)\), is why speed and not velocity appears in the answer. Only the component of force along the velocity changes the speed; the perpendicular component turns the velocity without doing work. This is the same reason the magnetic force \(q\vec v\times\vec B\), always perpendicular to \(\vec v\), can never change a charged particle's kinetic energy — it steers but never speeds up.

At a deeper level the theorem is the particle-mechanics shadow of a symmetry-linked conservation law. Its differential form \(\frac{dK}{dt}=\vec F\cdot\vec v = P\) states that power is the rate of kinetic-energy change. In Lagrangian mechanics, when the Lagrangian has no explicit time dependence, Noether's theorem guarantees a conserved energy; the work-energy theorem is what remains when that time-symmetry is broken by an explicit external force feeding power \(P=\vec F\cdot\vec v\) into the system. Seen this way, work is precisely the energy transferred across the system boundary by a force, and \(\Delta K\) is the ledger entry recording where it went.

Common misconceptions. Work is not force, and not force times time (that is impulse) — it is force times displacement along the motion. A force can be large yet do zero work (holding a weight still, or any centripetal force). And negative work is not an error: it simply means the force is draining kinetic energy, as friction does. Kinetic energy is always \(\ge 0\), but its change may be of either sign.

Worked examples

Example 1 — Block pushed against friction.

\[ W_{\text{net}} = (F_{\text{push}} - f)\,d = \Delta K = \tfrac12 m v_f^2 - \tfrac12 m v_i^2 \]

Net force is the applied push minus kinetic friction, both along the straight motion; set its work equal to \(\Delta K\). A

\[ v_f = \sqrt{v_i^2 + \frac{2(F_{\text{push}}-f)\,d}{m}} \]

Solve symbolically for the final speed, \(v_i\) the initial speed. B

\[ m=4.0\,\mathrm{kg},\; F_{\text{push}}=20\,\mathrm{N},\; f=6.0\,\mathrm{N},\; d=3.0\,\mathrm{m},\; v_i=0 \]

Insert numbers with units after the symbolic result is fixed. A

\[ v_f = \sqrt{0 + \frac{2(20-6.0)(3.0)}{4.0}} = \sqrt{\frac{84}{4.0}} = \sqrt{21}\ \mathrm{m/s} \]

Arithmetic; net work \(=(14\,\mathrm N)(3.0\,\mathrm m)=42\,\mathrm J\), so \(\tfrac12 m v_f^2 = 42\,\mathrm J\). A

\[ v_f = \sqrt{21}\approx 4.6\ \mathrm{m/s},\qquad W_{\text{net}} = 42\ \mathrm{J} \]

Reading. The 42 J of net work all became kinetic energy; friction reduced the push's 60 J of work by 18 J.

Units check. \(2Fd/m\) has units \(\mathrm{(N\,m)/kg = (kg\,m/s^2\cdot m)/kg = m^2/s^2}\), whose square root is m/s.

Example 2 — Stopping distance of a braking car.

\[ W_{\text{net}} = -f\,d = 0 - \tfrac12 m v_i^2 \]

The only horizontal force is friction \(f=\mu m g\) opposing motion (negative work); final kinetic energy is zero at rest. A

\[ \mu m g\, d = \tfrac12 m v_i^2 \;\Longrightarrow\; d = \frac{v_i^2}{2\mu g} \]

Substitute \(f=\mu m g\); mass cancels, giving the stopping distance symbolically. B

\[ v_i = 30\ \mathrm{m/s},\quad \mu = 0.80,\quad g = 9.8\ \mathrm{m/s^2} \]

Insert numbers with units into the symbolic result. A

\[ d = \frac{(30)^2}{2(0.80)(9.8)} = \frac{900}{15.68}\ \mathrm{m} \]

Arithmetic. A

\[ d = \frac{900}{15.68} \approx 57\ \mathrm{m} \]

Reading. Stopping distance grows as \(v_i^2\): doubling the speed quadruples the distance, a direct consequence of kinetic energy scaling with speed squared.

Units check. \([v^2/(\mu g)] = \mathrm{(m^2/s^2)/(m/s^2) = m}\). Correct.

Problems

A 0.50 kg ball is thrown straight up at 12 m/s. Using only the work-energy theorem (take \(g=9.8\,\mathrm{m/s^2}\), ignore air), find the height at which its speed has dropped to 6.0 m/s.
Solution
Net force is gravity; \(W=-mg\,h=\tfrac12 m v_f^2-\tfrac12 m v_i^2\). Mass cancels: \(-g h=\tfrac12(v_f^2-v_i^2)\Rightarrow h=\frac{v_i^2-v_f^2}{2g}=\frac{12^2-6.0^2}{2(9.8)}=\frac{144-36}{19.6}=\frac{108}{19.6}\approx 5.5\ \mathrm{m}\).
A 1200 kg car accelerates from 10 m/s to 25 m/s. What net work was done on it, and what constant net force would achieve this over 150 m?
Solution
\(W=\Delta K=\tfrac12 m(v_f^2-v_i^2)=\tfrac12(1200)(625-100)=600\times525=3.15\times10^5\ \mathrm J\). Constant force: \(F=W/d=3.15\times10^5/150=2.1\times10^3\ \mathrm N\).
A 2.0 kg puck moves in a horizontal circle of radius 0.75 m at constant speed 4.0 m/s, held by a string. How much work does the string tension do over one full revolution? Explain.
Solution
The tension is centripetal, always perpendicular to the velocity, so \(\vec F\cdot d\vec r=0\) at every instant. Over one revolution \(W=\oint\vec F\cdot d\vec r=0\ \mathrm J\). Consistently, the speed (and hence \(K\)) is unchanged, so \(\Delta K=0\). The radius and mass are irrelevant.
A 0.20 kg arrow is fired into a bale of hay and stops in 0.15 m. If it entered at 60 m/s, find the average resistive force the hay exerts.
Solution
\(W_{\text{net}}=-F d=0-\tfrac12 m v_i^2\Rightarrow F=\frac{\tfrac12 m v_i^2}{d}=\frac{\tfrac12(0.20)(60)^2}{0.15}=\frac{0.10\times3600}{0.15}=\frac{360}{0.15}=2.4\times10^3\ \mathrm N\) opposing motion.
A 8.0 kg block is pulled 5.0 m along the floor by a rope at \(30^\circ\) above horizontal with tension 40 N; kinetic friction has coefficient \(\mu=0.25\). Starting from rest, find the final speed. (\(g=9.8\,\mathrm{m/s^2}\).)
Solution
Horizontal pull: \(T\cos30^\circ=40(0.866)=34.6\,\mathrm N\). Normal force: \(N=mg-T\sin30^\circ=8.0(9.8)-40(0.5)=78.4-20=58.4\,\mathrm N\); friction \(f=\mu N=0.25(58.4)=14.6\,\mathrm N\). Net horizontal force \(=34.6-14.6=20.0\,\mathrm N\). Net work \(=20.0\times5.0=100\,\mathrm J=\tfrac12 m v_f^2\Rightarrow v_f=\sqrt{2(100)/8.0}=\sqrt{25}=5.0\ \mathrm{m/s}\).