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10.4 Work, Energy, Work-Energy Theorem, Conservative Forces & Power

Key Takeaways

  • Mechanical work is defined as the scalar product W=F⋅d=Fdcos⁡θW = \mathbf{F} \cdot \mathbf{d} = Fd\cos\theta, vanishing when the force is perpendicular to displacement (θ=90∘\theta = 90^\circ), such as centripetal forces and ideal normal contact reactions.

  • A force field is conservative if work done around any arbitrary closed loop vanishes (∮F⋅dr=0\oint \mathbf{F} \cdot d\mathbf{r} = 0), which permits the definition of a scalar potential energy function related to force by F=−∇U\mathbf{F} = -\nabla U (or F=−dUdxF = -\frac{dU}{dx}).

  • The Work-Energy Theorem proves that net work performed by all combined forces acting on a body equals its net variation in kinetic energy (Wnet=ΔK=12mvf2−12mvi2W_{\text{net}} = \Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2).

  • Kinetic energy is quadratic in linear momentum (K=p22mK = \frac{p^2}{2m}), implying that percentage changes in momentum produce amplified percentage alterations in kinetic energy.

  • Instantaneous power is evaluated as P=F⋅vP = \mathbf{F} \cdot \mathbf{v} in watts, commercial electrical energy is quantified in kilowatt-hours (1 kWh=3.6×106 J1\text{ kWh} = 3.6 \times 10^6\text{ J}), and efficiency η=PoutPin×100%\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% remains strictly below 100%100\% in physical systems due to non-conservative dissipation.

Last updated: October 2026

Work, Energy, Work-Energy Theorem, Conservative Forces & Power

Work and energy principles form the primary alternative to Newtonian vectorial dynamics, providing scalar methods to determine final velocities, displacements, and force interactions without requiring time-dependent acceleration integrals. In the Intermediate Physics Part I syllabus and the UET ECAT entry test, mechanical energy problems emphasize work by constant and variable forces, the work-energy theorem, conservative vector fields, gravitational and elastic potential energies, instantaneous power, and mechanical efficiency.


1. Mechanical Work Done by Constant and Variable Forces

Work Done by a Constant Force

The mechanical work WW performed by a constant force F\mathbf{F} acting on a body that undergoes displacement d\mathbf{d} is defined as the scalar (dot) product of force and displacement:

W=F⋅d=Fdcos⁡θW = \mathbf{F} \cdot \mathbf{d} = F d \cos\theta

where θ\theta is the angle between the force vector and the displacement vector.

  • Positive Work (0∘≤θ<90∘0^\circ \le \theta < 90^\circ): The force possesses a component along the direction of motion, adding mechanical energy to the system (e.g., an engine pulling a carriage, θ=0∘  ⟹  W=Fd\theta = 0^\circ \implies W = Fd).
  • Zero Work (θ=90∘\theta = 90^\circ): The force acts perpendicular to displacement, contributing no work: W=Fdcos⁡(90∘)=0W = F d \cos(90^\circ) = 0 Examples of zero mechanical work include:
    • Centripetal force acting on a satellite in circular orbit (Fc⊥v\mathbf{F}_c \perp \mathbf{v}).
    • The tension in a simple pendulum string (T⊥v\mathbf{T} \perp \mathbf{v}).
    • Normal reaction force on a crate sliding across a flat horizontal floor.
    • A porter carrying a load on his head while walking horizontally at constant speed (mg⊥dmg \perp d).
  • Negative Work (90∘<θ≤180∘90^\circ < \theta \le 180^\circ): The force opposes the direction of motion, removing mechanical energy from the system (e.g., kinetic friction acting at θ=180∘  ⟹  W=−fkd\theta = 180^\circ \implies W = -f_k d).

Work Done by a Variable Force

When the applied force changes in magnitude or direction along the displacement path xx, the total work is obtained by integrating over infinitesimal displacements dxdx:

W=∫xixfF(x)cos⁡θ dxW = \int_{x_i}^{x_f} F(x) \cos\theta \, dx

For one-dimensional motion along the xx-axis:

W=∫xixfFx(x) dxW = \int_{x_i}^{x_f} F_x(x) \, dx

Graphical Work Evaluation (F−xF-x Graph)

On a Force versus Position (F−xF-x) graph, the definite integral equals the signed geometric area under the curve:

W=Area under the F−x curve between xi and xfW = \text{Area under the } F-x \text{ curve between } x_i \text{ and } x_f


2. Conservative vs. Non-Conservative Forces

Forces in nature are fundamentally divided into conservative and non-conservative forces based on their path dependence.

Conservative Forces

A force is conservative if it satisfies any of the following mutually equivalent physical definitions:

  1. The work done by the force in moving a particle between any two points depends solely on the initial and final endpoints and is completely independent of the path taken: WA→B(Path 1)=WA→B(Path 2)=WA→B(Path 3)W_{A \to B}^{(\text{Path 1})} = W_{A \to B}^{(\text{Path 2})} = W_{A \to B}^{(\text{Path 3})}
  2. The work done by the force around any closed loop is identically zero: ∮F⋅dr=0\oint \mathbf{F} \cdot d\mathbf{r} = 0
  3. The force can be expressed as the negative spatial gradient of a scalar potential energy function UU: F=−∇U  ⟹  Fx=−∂U∂x\mathbf{F} = -\nabla U \implies F_x = -\frac{\partial U}{\partial x}

Examples of Conservative Forces:

  • Gravitational force: Fg=−GMmr2r^\mathbf{F}_g = -\frac{G M m}{r^2}\hat{\mathbf{r}}
  • Electrostatic Coulomb force: Fe=14πε0q1q2r2r^\mathbf{F}_e = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}\hat{\mathbf{r}}
  • Elastic restoring force of an ideal Hookean spring: Fs=−kxF_s = -kx

Non-Conservative (Dissipative) Forces

A force is non-conservative if the work it performs depends on the actual path traversed. Around a closed path, non-conservative forces do non-zero work, dissipating mechanical energy into heat, sound, or internal deformation:

∮Fnc⋅dr≠0\oint \mathbf{F}_{\text{nc}} \cdot d\mathbf{r} \ne 0

Examples of Non-Conservative Forces:

  • Friction (fkf_k): always opposes motion, doing negative work along every path segment.
  • Viscous drag in fluids (Fd=6πηrvF_d = 6\pi\eta r v).
  • Air resistance.
  • Tension in a flexible cord, normal contact forces with deformation, and motor propulsion forces.

3. Kinetic and Potential Energy Formulations

Kinetic Energy (KK)

Kinetic energy is the capacity of a body to do work by virtue of its motion:

K=12mv2K = \frac{1}{2}mv^2

Relating kinetic energy to linear momentum p=mvp = mv:

K=p22m  ⟺  p=2mKK = \frac{p^2}{2m} \iff p = \sqrt{2mK}

ECAT Percentage Change Rule for Momentum and Energy

  • If momentum increases by a factor of (1+x)(1 + x), kinetic energy increases by a factor of (1+x)2(1 + x)^2: ΔK%=[(1+x)2−1]×100%\Delta K\% = [(1 + x)^2 - 1] \times 100\% For example, if momentum increases by 50%50\% (x=0.5x = 0.5): ΔK%=[(1.5)2−1]×100%=[2.25−1]×100%=125%\Delta K\% = [(1.5)^2 - 1] \times 100\% = [2.25 - 1] \times 100\% = 125\%
  • If kinetic energy increases by 300%300\%, the new energy is 4K4K, meaning momentum increases by 4−1=1=100%\sqrt{4} - 1 = 1 = 100\%.

Gravitational Potential Energy (UgU_g)

Gravitational potential energy is the energy stored in a body by virtue of its position in a gravitational field. Near Earth's surface (where gg is uniform):

Ug=mghU_g = mgh

where hh is measured relative to an arbitrary reference datum where Ug=0U_g = 0.

Elastic Potential Energy of a Hookean Spring (UsU_s)

According to Hooke's Law, the restoring force exerted by a spring stretched or compressed by distance xx from its equilibrium position is:

Fs=−kxF_s = -kx

where kk is the spring stiffness constant (N/m\text{N/m}). The external force required to stretch the spring without acceleration is Fext=+kxF_{\text{ext}} = +kx. The work performed in extending the spring from x=0x = 0 to xx is stored as elastic potential energy:

Us=∫0xFextdx=∫0xkxdx=12kx2U_s = \int_0^x F_{\text{ext}} dx = \int_0^x kx dx = \frac{1}{2}kx^2


4. The Work-Energy Theorem

The Work-Energy Theorem states that the net work done by all forces (conservative, non-conservative, internal, and external) acting on a particle equals the resulting change in the particle's kinetic energy:

Wnet=ΔK=Kf−Ki=12mvf2−12mvi2W_{\text{net}} = \Delta K = K_f - K_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

Derivation from Newton's Second Law

Using Fnet=ma=m(vdvdx)F_{\text{net}} = ma = m\left(v\frac{dv}{dx}\right):

Wnet=∫xixfFnetdx=∫xixfm(vdvdx)dx=∫vivfmvdv=[12mv2]vivf=12mvf2−12mvi2W_{\text{net}} = \int_{x_i}^{x_f} F_{\text{net}} dx = \int_{x_i}^{x_f} m\left(v\frac{dv}{dx}\right) dx = \int_{v_i}^{v_f} mv dv = \left[\frac{1}{2}mv^2\right]_{v_i}^{v_f} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

Generalized Work-Energy Principle

Separating net work into conservative work WcW_c and non-conservative work WncW_{\text{nc}}:

Wnet=Wc+Wnc=ΔKW_{\text{net}} = W_c + W_{\text{nc}} = \Delta K

Because work done by conservative forces equals the negative change in potential energy (Wc=−ΔUW_c = -\Delta U):

−ΔU+Wnc=ΔK  ⟹  Wnc=ΔK+ΔU=Δ(K+U)=ΔEmech-\Delta U + W_{\text{nc}} = \Delta K \implies W_{\text{nc}} = \Delta K + \Delta U = \Delta(K + U) = \Delta E_{\text{mech}}

The work done by all non-conservative forces equals the change in the total mechanical energy of the system.


5. Law of Conservation of Mechanical Energy

If only conservative forces perform work on an isolated system (Wnc=0W_{\text{nc}} = 0):

ΔEmech=0  ⟹  Ki+Ui=Kf+Uf=constant\Delta E_{\text{mech}} = 0 \implies K_i + U_i = K_f + U_f = \text{constant}

Total mechanical energy remains constant throughout the motion.

Energy Interconversions for a Freely Falling Body

Consider a body of mass mm dropped from rest at height hh:

  1. At top release point (y=hy = h): v=0  ⟹  K1=0,U1=mgh  ⟹  E1=mghv = 0 \implies K_1 = 0, \quad U_1 = mgh \implies E_1 = mgh
  2. At intermediate elevation yy (0<y<h0 < y < h): After falling a distance (h−y)(h - y), velocity is v=2g(h−y)v = \sqrt{2g(h - y)}: K2=12mv2=12m[2g(h−y)]=mg(h−y)K_2 = \frac{1}{2}mv^2 = \frac{1}{2}m[2g(h - y)] = mg(h - y) U2=mgyU_2 = mgy E2=K2+U2=mg(h−y)+mgy=mghE_2 = K_2 + U_2 = mg(h - y) + mgy = mgh
  3. Just before ground contact (y=0y = 0): Velocity is v=2ghv = \sqrt{2gh}: K3=12m(2gh)=mgh,U3=0  ⟹  E3=mghK_3 = \frac{1}{2}m(2gh) = mgh, \quad U_3 = 0 \implies E_3 = mgh

At every point in the descent, the sum of kinetic and gravitational potential energy equals the initial potential energy mghmgh.

Falling in the Presence of Air Friction (ff)

When a non-conservative retarding drag force ff opposes motion over vertical height hh:

Loss in Potential Energy=Gain in Kinetic Energy+Work against Friction\text{Loss in Potential Energy} = \text{Gain in Kinetic Energy} + \text{Work against Friction}

mgh=12mv2+fh  ⟹  v=2gh−2fhmmgh = \frac{1}{2}mv^2 + fh \implies v = \sqrt{2gh - \frac{2fh}{m}}


6. Power and Mechanical Efficiency

Power (PP)

Power is the rate at which work is performed or energy is transferred.

Average Power=Pavg=ΔWΔt\text{Average Power} = P_{\text{avg}} = \frac{\Delta W}{\Delta t}

Instantaneous Power=P=lim⁡Δt→0ΔWΔt=dWdt=F⋅drdt=F⋅v=Fvcos⁡θ\text{Instantaneous Power} = P = \lim_{\Delta t \to 0}\frac{\Delta W}{\Delta t} = \frac{dW}{dt} = \frac{\mathbf{F} \cdot d\mathbf{r}}{dt} = \mathbf{F} \cdot \mathbf{v} = F v \cos\theta

If a body moves with instantaneous velocity v\mathbf{v} under applied force F\mathbf{F}, instantaneous power is the dot product of force and velocity.

Units of Work, Energy, and Power

QuantityUnit NameSymbolEquivalence in SI Base Units
Work / EnergyJouleJ\text{J}1 J=1 N⋅m=1 kg⋅m2⋅s−21\text{ J} = 1\text{ N}\cdot\text{m} = 1\text{ kg}\cdot\text{m}^2\cdot\text{s}^{-2}
PowerWattW\text{W}1 W=1 J/s=1 kg⋅m2⋅s−31\text{ W} = 1\text{ J/s} = 1\text{ kg}\cdot\text{m}^2\cdot\text{s}^{-3}
HorsepowerHorsepowerhp\text{hp}1 hp=746 W≈0.75 kW1\text{ hp} = 746\text{ W} \approx 0.75\text{ kW}
Electrical EnergyKilowatt-hourkWh\text{kWh}1 kWh=(1000 W)×(3600 s)=3.6×106 J=3.6 MJ1\text{ kWh} = (1000\text{ W}) \times (3600\text{ s}) = 3.6 \times 10^6\text{ J} = 3.6\text{ MJ}

ECAT Unit Distinction: The kilowatt-hour (kWh\text{kWh}) has units of power multiplied by time; therefore, it is a unit of energy, not power.

Mechanical Efficiency (η\eta)

Efficiency is the ratio of useful work output to total energy input, expressed as a percentage:

η=Useful Work OutputTotal Energy Input×100%=Power OutputPower Input×100%\eta = \frac{\text{Useful Work Output}}{\text{Total Energy Input}} \times 100\% = \frac{\text{Power Output}}{\text{Power Input}} \times 100\%

Because non-conservative forces like friction and viscous drag dissipate mechanical energy as heat, the efficiency of any real machine is strictly less than 100%100\% (η<1\eta < 1).


Work, Energy, and Power Relationships Reference Table

Physical Theorem / FormulaMathematical ExpressionApplicable Physical ConditionsKey ECAT Conceptual Implication
Work by Constant ForceW=F⋅d=Fdcos⁡θW = \mathbf{F} \cdot \mathbf{d} = F d \cos\thetaUniform force vector F\mathbf{F}Zero work when F⊥d\mathbf{F} \perp \mathbf{d} (centripetal force, normal force)
Work by Variable ForceW=∫F(x)dxW = \int F(x) dx1D spatially variable forceArea under the Force-Displacement (F−xF-x) curve
Conservative Force Field∮F⋅dr=0,F=−dUdx\oint \mathbf{F} \cdot d\mathbf{r} = 0, \quad F = -\frac{dU}{dx}Conservative interaction (gravity, spring)Path-independent work; potential energy uniquely definable
Work-Energy TheoremWnet=ΔK=12mvf2−12mvi2W_{\text{net}} = \Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2Net resultant of all forces actingConverts mechanical work directly to kinetic energy change
Mechanical Energy ConservationKi+Ui=Kf+UfK_i + U_i = K_f + U_fStrictly conservative forces (Wnc=0W_{\text{nc}} = 0)Total mechanical energy is constant throughout trajectory
Energy Dissipation by FrictionWnc=Δ(K+U)=−fkdW_{\text{nc}} = \Delta(K + U) = -f_k dNon-conservative friction or drag presentMechanical energy converts irreversibly into thermal energy
Instantaneous PowerP=F⋅v=Fvcos⁡θP = \mathbf{F} \cdot \mathbf{v} = F v \cos\thetaInstantaneous force and velocityScalar product; maximum when force is aligned with velocity
Kinetic Energy vs MomentumK=p22m  ⟺  p=2mKK = \frac{p^2}{2m} \iff p = \sqrt{2mK}Any translational mass mmpp changes by Δp%  ⟹  K\Delta p\% \implies K changes quadratically by [(1+Δp)2−1][(1+\Delta p)^2 - 1]

7. Non-Conventional Energy Sources

The work–energy chapter of Intermediate physics also surveys energy sources other than fossil fuels. Each converts one form of energy into another, so the same conservation reasoning applies:

SourceEnergy conversionKey fact or formula
Solar (photovoltaic)Light → electricalSunlight delivers roughly 1.4 kW/m21.4\text{ kW/m}^2 above the atmosphere and about 1 kW/m21\text{ kW/m}^2 at the ground on a clear day; cell efficiencies are typically 15–25%
Solar (thermal)Light → heatCollectors and concentrators heat water or a working fluid
WindKinetic → electricalPower in the wind P=12ρAv3P = \frac{1}{2}\rho A v^3, so doubling wind speed gives 8×8\times the power
HydroelectricGravitational PE → electricalP=ηρgQhP = \eta \rho g Q h for flow rate QQ and head hh
TidalKinetic/PE of tides → electricalUses the rise and fall of sea level twice a day
GeothermalInternal heat of Earth → heat/electricalHot rocks heat underground water to steam
Biomass and wasteChemical → heat/electricalBurning or fermenting plant and animal waste (biogas is mostly methane)

Worked example (wind): air density ρ=1.2 kg/m3\rho = 1.2\text{ kg/m}^3, swept area A=100 m2A = 100\text{ m}^2, wind speed v=10 m/sv = 10\text{ m/s}: P=12(1.2)(100)(103)=60 kWP = \frac{1}{2}(1.2)(100)(10^3) = 60\text{ kW} of kinetic energy flux (a real turbine extracts only part of this).

Worked example (hydro): Q=10 m3/sQ = 10\text{ m}^3/\text{s}, h=50 mh = 50\text{ m}, η=0.8\eta = 0.8: P=0.8×1000×9.8×10×50≈3.9 MWP = 0.8 \times 1000 \times 9.8 \times 10 \times 50 \approx 3.9\text{ MW}.

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Decision Flowchart for Work-Energy Theorem and Power
Test Your Knowledge

If the linear momentum of a moving particle is increased by 50%50\%, by what percentage does its kinetic energy increase?

A

50%50\%

B

125%125\%

C

100%100\%

D

225%225\%

Test Your Knowledge

A vehicle of mass 1000 kg1000\text{ kg} accelerates uniformly from rest to a speed of 20 m/s20\text{ m/s} in a time duration of 10 s10\text{ s} along a horizontal level highway. Neglecting frictional losses, what is the average power delivered by the vehicle's engine during this acceleration interval?

A

20 kW20\text{ kW}

B

10 kW10\text{ kW}

C

40 kW40\text{ kW}

D

200 kW200\text{ kW}

Test Your Knowledge

A two-dimensional force field is described by F(x,y)=(2xy)i^+(x2)j^\mathbf{F}(x, y) = (2xy)\hat{\mathbf{i}} + (x^2)\hat{\mathbf{j}}. Which fundamental physical and mathematical criterion confirms that this force field is conservative?

A

The work done in moving a particle between two points depends on the specific path taken.

B

The force vector always acts perpendicular to the instantaneous velocity of the particle.

C

The force produces a constant acceleration on any object of mass mm.

D

The line integral of the force along any arbitrary closed loop is identically zero (∮F⋅dr=0\oint \mathbf{F} \cdot d\mathbf{r} = 0).

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