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Physics · HL · Unit 1: Space, Time and Motion

Unit 1Topic A.3Key Concepts144 min read

Work, energy and power

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Work, energy, and power

Energy is a scalar quantity that makes it possible to predict change without following every force and acceleration at every instant. A force can transfer energy by doing work; the transferred energy can appear in kinetic, gravitational, elastic, thermal, or other stores.

The same situation can be described at two levels. Energy conservation accounts for every energy store and transfer; the total energy of an isolated system remains constant. In the mechanical model, mechanical-energy conservation means that the sum of kinetic and potential energies remains constant: the net work done by non-conservative forces is zero. Friction and drag can transfer mechanical energy into non-mechanical stores.

Power answers a different question: how quickly is energy transferred? Efficiency compares the useful output with the total input. A complete model therefore follows the chain: identify the system, identify the transfers and stores, conserve total energy, and then calculate the rate and useful fraction.

Key terms

Energy
A scalar quantity that is conserved in an isolated system and can be transferred between stores.
Work
Energy transferred by a force acting through a displacement.
Power
The rate at which work is done or energy is transferred.
Efficiency
The useful energy or power output divided by the total input.

Worked examples

Lift, work, and power

A weightlifter raises a 25 kg25\,\mathrm{kg} mass vertically by 1.8 m1.8\,\mathrm{m} in 1.5 s1.5\,\mathrm{s}. Find the useful work and average useful power. Use g=9.81 m s−2g=9.81\,\mathrm{m\,s^{-2}}.

  1. 1The useful transfer is into gravitational potential energy: W=mghW=mgh.
  2. 2W=25×9.81×1.8=441 JW=25\times 9.81\times 1.8=441\,\mathrm{J} to three significant figures.
  3. 3P=Wt=4411.5=294 WP=\frac{W}{t}=\frac{441}{1.5}=294\,\mathrm{W}.

Answer: The useful work is 441 J441\,\mathrm{J} and the average useful power is 294 W294\,\mathrm{W}.

An energy-transfer snapshot

A device receives 90 kJ90\,\mathrm{kJ} in 60 s60\,\mathrm{s} and transfers 72 kJ72\,\mathrm{kJ} usefully. Find its input power and efficiency.

  1. 1Input power is the input energy per time: Pin=90 00060=1500 WP_{\mathrm{in}}=\frac{90\,000}{60}=1500\,\mathrm{W}.
  2. 2Efficiency is η=EusefulEin=7290=0.80\eta=\frac{E_{\mathrm{useful}}}{E_{\mathrm{in}}}=\frac{72}{90}=0.80.
  3. 3The remaining 18 kJ18\,\mathrm{kJ}, or 20% of the input, is not included in the stated useful transfer. These figures alone do not establish whether it is stored in the device or transferred elsewhere; describe it as dissipated only if that is specified.

Answer: The input power is 1.5 kW1.5\,\mathrm{kW} and the efficiency is 80%80\%.

Rules and formulas

Conservation of total energy

Einitial=EfinalE_{\mathrm{initial}}=E_{\mathrm{final}}
EE
total energy of the chosen isolated system (JJ)

Average power

Pavg=ΔEΔt=WΔtP_{\mathrm{avg}}=\frac{\Delta E}{\Delta t}=\frac{W}{\Delta t}
PavgP_{\mathrm{avg}}
average power (WW)
ΔE\Delta E
energy transferred (JJ)
Δt\Delta t
time interval (ss)

Efficiency

η=EusefulEinput=PusefulPinput\eta=\frac{E_{\mathrm{useful}}}{E_{\mathrm{input}}}=\frac{P_{\mathrm{useful}}}{P_{\mathrm{input}}}
η\eta
efficiency as a decimal (nonenone)

Work and force–displacement graphs

For a constant force, work depends on the component parallel to the displacement: W=Fscos⁡θW=Fs\cos\theta, where θ\theta is the angle between the force and displacement vectors. Work is a scalar and its SI unit is the joule: 1 J=1 N m1\,\mathrm{J}=1\,\mathrm{N\,m}.

Work is positive when the parallel force component points with the displacement, negative when it points against the displacement, and zero when it is perpendicular. A normal force on an object sliding along a fixed surface therefore does no work because the force and displacement are at 90∘90^\circ.

If force changes with position, split the motion into narrow displacement intervals. The signed area under a force–displacement graph gives the work: area above the displacement axis is positive and area below it is negative. Rectangles, triangles, and trapezia often provide the required area without calculus.

Key terms

Joule
The SI unit of work and energy; one joule equals one newton metre.
Positive work
Energy transfer associated with a force component in the direction of displacement.
Negative work
Energy transfer associated with a force component opposite to displacement.
Variable force
A force whose magnitude or direction changes with position.

Worked examples

Pulling at an angle

A student pulls a crate 12 m12\,\mathrm{m} along a horizontal floor with a constant 80 N80\,\mathrm{N} force directed 30∘30^\circ above the horizontal. Find the work done by the pulling force.

  1. 1The angle between the force and horizontal displacement is 30∘30^\circ.
  2. 2Use W=Fscos⁡θW=Fs\cos\theta: W=80×12×cos⁡30∘W=80\times12\times\cos30^\circ.
  3. 3W=831 JW=831\,\mathrm{J} to three significant figures.

Answer: The pulling force does 8.31×102 J8.31\times10^2\,\mathrm{J} of positive work.

Triangular force–displacement graph

The component of force along an object's straight-line displacement increases linearly with displacement from 00 to 40 N40\,\mathrm{N} as the object moves 0.30 m0.30\,\mathrm{m}. Find the work done by this force from its force–displacement graph.

  1. 1The area under the graph is a triangle.
  2. 2W=12×base×height=12×0.30×40W=\tfrac12\times\text{base}\times\text{height}=\tfrac12\times0.30\times40.
  3. 3The area is 6.0 N m=6.0 J6.0\,\mathrm{N\,m}=6.0\,\mathrm{J}.

Answer: The work done is 6.0 J6.0\,\mathrm{J}.

Rules and formulas

Work by a constant force

W=Fscos⁡θW=Fs\cos\theta
WW
work done by the force (JJ)
FF
force magnitude (NN)
ss
displacement magnitude (mm)
θ\theta
angle between force and displacement (degreeorraddegree or rad)

Work by the parallel component

W=F∥sW=F_{\parallel}s
F∥F_{\parallel}
signed force component parallel to displacement (NN)

Work from a variable force

W=signed area under the F∥ against s graphW=\text{signed area under the }F_{\parallel}\text{ against }s\text{ graph}
F∥F_{\parallel}
parallel force plotted against displacement (NN)

Kinetic energy and the work–energy theorem

Translational kinetic energy is Ek=12mv2=p22mE_k=\tfrac12mv^2=\frac{p^2}{2m}, where p=mvp=mv is the magnitude of momentum. It is scalar, cannot be negative, and depends on speed rather than velocity direction. Doubling speed multiplies kinetic energy by four; doubling mass at fixed speed only doubles it.

The work–energy theorem states that the net work done by all external forces equals the change in kinetic energy: Wnet=ΔEkW_{\mathrm{net}}=\Delta E_k. Positive net work increases speed, while negative net work decreases speed.

Net work is not generally the work done by just one convenient force. If a driving force, friction, gravity, and a normal force act, add the work done by every force or first find the resultant force component along the displacement.

Key terms

Kinetic energy
Energy associated with the translational motion of an object.
Work–energy theorem
The net work on an object equals its change in kinetic energy.
Net work
The sum of the work done by all forces on the object.
Speed dependence
Kinetic energy is proportional to the square of speed.

Worked examples

Car accelerating

A 1200 kg1200\,\mathrm{kg} car speeds up from 20 m s−120\,\mathrm{m\,s^{-1}} to 25 m s−125\,\mathrm{m\,s^{-1}}. Find the net work done on it.

  1. 1Apply Wnet=ΔEk=12m(v2−u2)W_{\mathrm{net}}=\Delta E_k=\tfrac12m(v^2-u^2).
  2. 2Wnet=0.5×1200×(252−202)W_{\mathrm{net}}=0.5\times1200\times(25^2-20^2).
  3. 3Wnet=135 000 JW_{\mathrm{net}}=135\,000\,\mathrm{J}, which is positive because the speed increases.

Answer: The net work is +1.35×105 J+1.35\times10^5\,\mathrm{J}.

Finding speed without time

A 15 kg15\,\mathrm{kg} box starts from rest. A constant resultant force does 480 J480\,\mathrm{J} of work. Find the final speed.

  1. 1The work–energy theorem gives 480=12mv2−0480=\tfrac12mv^2-0.
  2. 2Substitute the mass: 480=0.5×15×v2480=0.5\times15\times v^2, so v2=64v^2=64.
  3. 3Take the positive square root because speed is a magnitude: v=8.0 m s−1v=8.0\,\mathrm{m\,s^{-1}}.

Answer: The final speed is 8.0 m s−18.0\,\mathrm{m\,s^{-1}}.

Rules and formulas

Translational kinetic energy

Ek=12mv2=p22mE_k=\frac12mv^2=\frac{p^2}{2m}
EkE_k
kinetic energy (JJ)
mm
mass (kgkg)
vv
speed (ms−1m s⁻¹)
pp
magnitude of momentum (kgms−1kg m s⁻¹)

Work–energy theorem

Wnet=ΔEk=12m(v2−u2)W_{\mathrm{net}}=\Delta E_k=\frac12m(v^2-u^2)
uu
initial speed (ms−1m s⁻¹)
vv
final speed (ms−1m s⁻¹)

Gravitational potential energy near Earth's surface

Near Earth's surface, a change in height Δh\Delta h changes gravitational potential energy by ΔEg=mgΔh\Delta E_g=mg\Delta h. The result depends only on the initial and final heights, not on the path between them, because gravity is a conservative force.

The zero of gravitational potential energy is a chosen reference level. Changing the reference changes the numerical value of EgE_g, but it does not change ΔEg\Delta E_g or any prediction. State the reference when an absolute value is shown.

Gravity does positive work when an object moves downward and negative work when it moves upward: Wg=−ΔEgW_g=-\Delta E_g. When gravity is the only force doing work, a decrease in EgE_g appears as an equal increase in kinetic energy.

Key terms

Gravitational potential energy
Energy associated with the position of masses in a gravitational field.
Reference level
The chosen height at which gravitational potential energy is defined to be zero.
Conservative force
A force whose work depends only on the initial and final positions, not on the path.
Gravitational field strength
Force per unit mass; near Earth its magnitude is approximately 9.81 N kg−1kg^{-1}.

Worked examples

Climber gaining height

A 65 kg65\,\mathrm{kg} climber rises through a vertical height of 18 m18\,\mathrm{m}. Find the increase in gravitational potential energy.

  1. 1Use the vertical height change, not the distance along the climbing route.
  2. 2ΔEg=mgΔh=65×9.81×18\Delta E_g=mg\Delta h=65\times9.81\times18.
  3. 3ΔEg=11 478 J≈11.5 kJ\Delta E_g=11\,478\,\mathrm{J}\approx11.5\,\mathrm{kJ}.

Answer: The gravitational potential energy increases by 11.5 kJ11.5\,\mathrm{kJ}.

Speed after a vertical drop

A cart is released from rest 12 m12\,\mathrm{m} above the bottom of a frictionless track. Find its speed at the bottom.

  1. 1The loss in gravitational potential energy equals the gain in kinetic energy: mgh=12mv2mgh=\tfrac12mv^2.
  2. 2Mass cancels, so v=2gh=2×9.81×12v=\sqrt{2gh}=\sqrt{2\times9.81\times12}.
  3. 3v=15.3 m s−1v=15.3\,\mathrm{m\,s^{-1}}.

Answer: The cart reaches approximately 15.3 m s−115.3\,\mathrm{m\,s^{-1}}.

Rules and formulas

Change in gravitational potential energy

ΔEg=mgΔh\Delta E_g=mg\Delta h
ΔEg\Delta E_g
change in gravitational potential energy (JJ)
gg
gravitational field strength (Nkg−1N kg⁻¹)
Δh\Delta h
final height minus initial height (mm)

Gravitational potential energy from a chosen zero

Eg=mghE_g=mgh
hh
height above the chosen reference (mm)

Work done by gravity

Wg=−ΔEgW_g=-\Delta E_g
WgW_g
work done by gravity (JJ)

Elastic potential energy and springs

Within its linear elastic range, a spring obeys Hooke's law: the restoring force is opposite to displacement and has magnitude kxkx. The extension or compression xx is measured from the spring's natural, unstrained length.

Because spring force grows from zero to kxkx, the work required to deform it is the triangular area under its force–extension graph. The elastic potential energy stored is therefore Es=12kx2E_s=\tfrac12kx^2, not kxkx or kx2kx^2.

When the spring returns toward its natural length, it can transfer elastic potential energy into kinetic energy or another store. The quadratic energy model requires a linear force–deformation relationship. An object can still be elastic beyond its limit of proportionality; beyond its elastic limit, it may not return fully to its original shape.

Key terms

Elastic potential energy
Energy stored when an elastic object is stretched or compressed.
Spring constant
The force per unit extension in the linear region; it measures stiffness.
Natural length
The unstrained length of a spring, from which extension or compression is measured.
Elastic limit
The deformation limit beyond which an object may not return fully to its original shape.

Worked examples

Spring launcher

A spring of constant 250 N m−1250\,\mathrm{N\,m^{-1}} is compressed by 0.080 m0.080\,\mathrm{m} and launches a 0.20 kg0.20\,\mathrm{kg} cart on a level frictionless track. Find the stored energy and launch speed.

  1. 1Es=12kx2=0.5×250×0.0802=0.800 JE_s=\tfrac12kx^2=0.5\times250\times0.080^2=0.800\,\mathrm{J}.
  2. 2Conservation gives Es=EkE_s=E_k: 0.800=12×0.20×v20.800=\tfrac12\times0.20\times v^2.
  3. 3v=8.00=2.83 m s−1v=\sqrt{8.00}=2.83\,\mathrm{m\,s^{-1}}.

Answer: The spring stores 0.800 J0.800\,\mathrm{J} and the cart leaves at 2.83 m s−12.83\,\mathrm{m\,s^{-1}}.

Extra work over an extension interval

A spring obeys Hooke's law with k=160 N m−1k=160\,\mathrm{N\,m^{-1}}. How much additional work must an external force do to stretch it slowly, without energy losses, from an extension of 0.050 m0.050\,\mathrm{m} to 0.150 m0.150\,\mathrm{m}?

  1. 1The required work equals the increase in elastic potential energy.
  2. 2ΔEs=12k(x22−x12)\Delta E_s=\tfrac12k(x_2^2-x_1^2).
  3. 3ΔEs=0.5×160×(0.1502−0.0502)=1.60 J\Delta E_s=0.5\times160\times(0.150^2-0.050^2)=1.60\,\mathrm{J}.

Answer: The additional work is 1.60 J1.60\,\mathrm{J}.

Rules and formulas

Hooke's law

Fs=−kxF_s=-kx
FsF_s
spring restoring force (NN)
kk
spring constant (Nm−1N m⁻¹)
xx
extension or compression from natural length (mm)

Elastic potential energy

Es=12kx2E_s=\frac12kx^2
EsE_s
elastic potential energy (JJ)

Change in elastic potential energy

ΔEs=12k(x22−x12)\Delta E_s=\frac12k(x_2^2-x_1^2)
x1x_1
initial deformation (mm)
x2x_2
final deformation (mm)

Conservation of mechanical energy and energy transfers

Mechanical energy is the combined kinetic and potential energy of the chosen system: Emech=Ek+Eg+EsE_{\mathrm{mech}}=E_k+E_g+E_s. If only conservative forces transfer energy within an isolated mechanical system, this total remains constant.

An energy equation compares states rather than describing the route between them. List every relevant store at the initial and final states, include any transfer across the system boundary, and cancel terms that are zero or unchanged. Mass often cancels in gravity-only motion, explaining why the predicted speed is independent of mass.

Schematic ideal and frictional cart-track snapshots. Full energy bars remain equal in height; friction increases thermal energy while mechanical energy decreases. All illustrated states retain kinetic energy.Schematic snapshots, not numerical scale drawings. On the ideal track, kinetic and gravitational potential energy exchange while their sum remains constant. With friction, some mechanical energy becomes thermal energy. Each full bar represents the same total for the cart–track–Earth system, assuming negligible energy transfer to its surroundings. The first and last snapshots still have kinetic energy: they are not release-from-rest or turning points. The right-hand cart is shown at one lower uphill position, not at a calculated maximum height. The gravitational potential-energy reference is arbitrary, so its valley value need not be zero.

Open graph full size (opens in a new tab)

Key terms

Mechanical energy
The total of kinetic and relevant potential-energy stores in a system.
Isolated system
A system that exchanges neither matter nor energy with its surroundings during the process considered.
System boundary
The conceptual surface separating the selected system from its surroundings.
Energy transfer
A movement of energy between stores or across a system boundary.

Worked examples

Ideal roller-coaster valley

A cart starts from rest and drops 12 m12\,\mathrm{m} on an ideal track. Explain why its mass does not affect its speed at the bottom and find that speed.

  1. 1Conservation gives mgh=12mv2mgh=\tfrac12mv^2.
  2. 2The same factor mm appears on both sides, so it cancels: v=2ghv=\sqrt{2gh}.
  3. 3v=2×9.81×12=15.3 m s−1v=\sqrt{2\times9.81\times12}=15.3\,\mathrm{m\,s^{-1}}.

Answer: The ideal speed is 15.3 m s−115.3\,\mathrm{m\,s^{-1}}, independent of the cart's mass.

Pendulum through equilibrium

A pendulum bob is released from rest 0.45 m0.45\,\mathrm{m} above its lowest point. Ignore air resistance. Find its speed at the lowest point.

  1. 1Choose the lowest point as Eg=0E_g=0. Initially the energy is mghmgh; finally it is 12mv2\tfrac12mv^2.
  2. 2Set the stores equal and cancel mass: v=2ghv=\sqrt{2gh}.
  3. 3v=2×9.81×0.45=2.97 m s−1v=\sqrt{2\times9.81\times0.45}=2.97\,\mathrm{m\,s^{-1}}.

Answer: The speed at the lowest point is 2.97 m s−12.97\,\mathrm{m\,s^{-1}}.

Rules and formulas

Mechanical energy

Emech=Ek+Eg+EsE_{\mathrm{mech}}=E_k+E_g+E_s
EmechE_{\mathrm{mech}}
mechanical energy (JJ)

Ideal mechanical-energy conservation

Ek,i+Eg,i+Es,i=Ek,f+Eg,f+Es,fE_{k,i}+E_{g,i}+E_{s,i}=E_{k,f}+E_{g,f}+E_{s,f}
ii
initial state (nonenone)
ff
final state (nonenone)

System energy change when work is the only boundary energy transfer

Wext=ΔEsystemW_{\mathrm{ext}}=\Delta E_{\mathrm{system}}
WextW_{\mathrm{ext}}
net work transferring energy across the system boundary (JJ)

Dissipation, friction, drag, and braking

Friction and drag can do negative work on a moving object. When they reduce its mechanical energy, that energy has not disappeared: it is transferred mainly into internal or thermal energy of the object and surroundings, and sometimes sound. Friction from a moving belt or drag from moving air can instead do positive work on the chosen object; identify the system and the direction of each transfer.

For kinetic friction of approximately constant magnitude, the thermal transfer is ΔEth=fks\Delta E_{\mathrm{th}}=f_k s. More generally, the work done by non-conservative forces equals the change in mechanical energy of the selected system.

Braking is a direct work–energy application. With the same mass and braking force, Fd=12mv2Fd=\tfrac12mv^2, so braking distance is proportional to speed squared. Doubling speed therefore requires about four times the braking distance, before adding the driver's reaction distance.

Key terms

Dissipation
Transfer of energy into stores that are less available for the intended useful process.
Thermal energy
Energy associated with the microscopic motion and interactions of particles.
Non-conservative force
A force whose work depends on the path and can change mechanical energy.
Braking distance
Distance travelled while braking forces reduce a vehicle's kinetic energy.

Worked examples

Stopping a sliding block

An 8.0 kg8.0\,\mathrm{kg} block moves at 10 m s−110\,\mathrm{m\,s^{-1}} on a horizontal surface. A constant 20 N20\,\mathrm{N} friction force stops it. Find the stopping distance.

  1. 1Initial kinetic energy is Ek=0.5×8.0×102=400 JE_k=0.5\times8.0\times10^2=400\,\mathrm{J}.
  2. 2Friction transfers this energy thermally: fkd=400f_kd=400.
  3. 3d=40020=20 md=\frac{400}{20}=20\,\mathrm{m}.

Answer: The block stops after 20 m20\,\mathrm{m}.

Effect of doubling speed

A car's braking distance on a level road is 18 m18\,\mathrm{m} at speed vv. Estimate the braking distance at 2v2v if the mass is unchanged and the same constant resultant braking force acts in both cases.

  1. 1Braking distance satisfies d∝v2d\propto v^2 for constant mass and braking force.
  2. 2Replacing vv by 2v2v multiplies v2v^2 by 22=42^2=4.
  3. 3The new distance is 4×18=72 m4\times18=72\,\mathrm{m}.

Answer: The estimated braking distance is 72 m72\,\mathrm{m}, not 36 m36\,\mathrm{m}.

Rules and formulas

Thermal transfer by constant kinetic friction

ΔEth=fks\Delta E_{\mathrm{th}}=f_k s
fkf_k
kinetic-friction magnitude (NN)
ss
sliding distance (mm)

Non-conservative work

Wnc=ΔEmechW_{\mathrm{nc}}=\Delta E_{\mathrm{mech}}
WncW_{\mathrm{nc}}
net work done by non-conservative forces in the mechanical energy balance, including internal dissipative work (JJ)

Idealized braking distance

d=mv22Fbd=\frac{mv^2}{2F_b}
dd
braking distance (mm)
FbF_b
constant braking-force magnitude, or its average over the braking distance (NN)

Power and rate of energy transfer

Average power is energy transferred per time interval: Pavg=ΔEΔtP_{\mathrm{avg}}=\frac{\Delta E}{\Delta t}. One watt is one joule per second. Two machines can perform the same work but the machine that completes it sooner has greater average power.

For a force acting on an object moving with instantaneous velocity, mechanical power is P=F⃗⋅v⃗=Fvcos⁡θP=\vec F\cdot\vec v=Fv\cos\theta. Only the force component parallel to velocity transfers energy at that instant.

At constant speed on a level road, the driving force balances the total resistance, so engine output power at the wheels is P=FresvP=F_{\mathrm{res}}v. At changing speed, distinguish the driving force from the resultant force because some engine power changes kinetic energy while some is dissipated.

Key terms

Watt
The SI unit of power; one watt equals one joule per second.
Average power
Energy transferred divided by the duration of a process.
Instantaneous power
The rate of energy transfer at a particular instant.
Mechanical power
Power delivered by a force, equal to the scalar product of force and velocity.

Worked examples

Car cruising at constant speed

A car travels at a constant 25 m s−125\,\mathrm{m\,s^{-1}} against total resistive forces of 900 N900\,\mathrm{N}. Find the mechanical power delivered at the wheels.

  1. 1At constant velocity, the driving force equals the resistance in magnitude.
  2. 2The force and velocity are parallel, so P=Fv=900×25P=Fv=900\times25.
  3. 3P=22 500 W=22.5 kWP=22\,500\,\mathrm{W}=22.5\,\mathrm{kW}.

Answer: The required mechanical power is 22.5 kW22.5\,\mathrm{kW}.

Elevator's useful power

An elevator of total mass 750 kg750\,\mathrm{kg} rises 12 m12\,\mathrm{m} in 18 s18\,\mathrm{s} at steady speed. Find its average useful power.

  1. 1Useful energy transferred is ΔEg=mgh=750×9.81×12=88 290 J\Delta E_g=mgh=750\times9.81\times12=88\,290\,\mathrm{J}.
  2. 2P=ΔEΔt=88 29018P=\frac{\Delta E}{\Delta t}=\frac{88\,290}{18}.
  3. 3P=4905 W≈4.91 kWP=4905\,\mathrm{W}\approx4.91\,\mathrm{kW}.

Answer: The average useful power is 4.91 kW4.91\,\mathrm{kW}.

Rules and formulas

Average power

Pavg=WΔt=ΔEΔtP_{\mathrm{avg}}=\frac{W}{\Delta t}=\frac{\Delta E}{\Delta t}
PavgP_{\mathrm{avg}}
average power (WW)

Instantaneous mechanical power

P=F⃗⋅v⃗=Fvcos⁡θP=\vec F\cdot\vec v=Fv\cos\theta
v⃗\vec v
instantaneous velocity (ms−1m s⁻¹)
θ\theta
angle between force and velocity (degreeorraddegree or rad)

Parallel force form

P=F∥vP=F_{\parallel}v
F∥F_{\parallel}
force component parallel to velocity (NN)

Efficiency and Sankey energy-flow diagrams

Efficiency is the fraction of input energy or power delivered in the useful output pathway. It is dimensionless and can be written as a decimal or multiplied by 100%100\%. The numerator and denominator must describe the same time interval and use matching energy or power quantities.

What counts as useful depends on the stated purpose. Light from a lamp may be useful while thermal transfer is dissipated; for a heater, the thermal transfer is the intended useful output. Always identify the function before labelling pathways.

In a Sankey diagram, arrow width represents the amount of energy or power in each pathway. For a device with no net energy storage change, the input arrow splits into useful and dissipated outputs whose combined widths equal the input width. If energy is being stored or released, include that change in the energy balance rather than assuming the output alone equals the input.

Key terms

Useful output
The energy or power transferred in the pathway intended for the device's purpose.
Dissipated output
Energy transferred into pathways not useful for the stated purpose.
Sankey diagram
A flow diagram in which arrow widths represent energy or power transfers.
Percentage efficiency
The efficiency fraction multiplied by 100%.

Worked examples

Efficiency from energy data

A machine receives 500 J500\,\mathrm{J} and transfers 360 J360\,\mathrm{J} usefully. Assume that no energy accumulates in the machine and that all non-useful output is dissipated. Find its efficiency and the dissipated energy.

  1. 1η=EusefulEinput=360500=0.72\eta=\frac{E_{\mathrm{useful}}}{E_{\mathrm{input}}}=\frac{360}{500}=0.72.
  2. 2Percentage efficiency is 0.72×100%=72%0.72\times100\%=72\%.
  3. 3Dissipated energy is 500−360=140 J500-360=140\,\mathrm{J}.

Answer: The efficiency is 72%72\% and 140 J140\,\mathrm{J} is dissipated.

Input power for a pump

A pump increases water's gravitational potential energy by 60 kJ60\,\mathrm{kJ} in 25 s25\,\mathrm{s}. Its efficiency is 75%75\%. Find the input power.

  1. 1Useful power is Puseful=60 00025=2400 WP_{\mathrm{useful}}=\frac{60\,000}{25}=2400\,\mathrm{W}.
  2. 2Use η=PusefulPinput\eta=\frac{P_{\mathrm{useful}}}{P_{\mathrm{input}}}.
  3. 3Pinput=24000.75=3200 W=3.2 kWP_{\mathrm{input}}=\frac{2400}{0.75}=3200\,\mathrm{W}=3.2\,\mathrm{kW}.

Answer: The required input power is 3.2 kW3.2\,\mathrm{kW}.

Rules and formulas

Energy efficiency

η=EusefulEinput\eta=\frac{E_{\mathrm{useful}}}{E_{\mathrm{input}}}
η\eta
efficiency (nonenone)

Power efficiency

η=PusefulPinput\eta=\frac{P_{\mathrm{useful}}}{P_{\mathrm{input}}}
PusefulP_{\mathrm{useful}}
useful output power (WW)
PinputP_{\mathrm{input}}
total input power (WW)

Percentage efficiency

η%=100η\eta_{\%}=100\eta
η%\eta_{\%}
efficiency expressed as a percentage (%\%)

Applications and multistage energy systems

Real systems often combine several stores and transfers: a skier loses gravitational potential energy while gaining kinetic and thermal energy; a hydroelectric station turns a gravitational transfer into electrical output; a motor may lift a load while warming its surroundings. For a fuel, volumetric energy density is u=EfuelVu=\frac{E_{\mathrm{fuel}}}{V}, in J m−3\mathrm{J\,m^{-3}}; specific energy is e=Efuelme=\frac{E_{\mathrm{fuel}}}{m}, in J kg−1\mathrm{J\,kg^{-1}}. Hence Efuel=uV=emE_{\mathrm{fuel}}=uV=em and u=ρeu=\rho e, where VV is fuel volume, mm its mass, and ρ=mV\rho=\frac{m}{V} its mass density. Useful output is ηEfuel\eta E_{\mathrm{fuel}}. Compare fuels on the same mass or volume basis: greater energy density reduces the volume needed for a given energy, but does not by itself imply greater efficiency.

For a multistage problem, compare the first and last states whenever intermediate details are unnecessary. Then add any known dissipated transfer. This energy route is often shorter than calculating acceleration and time separately, but it does not directly provide direction or trajectory.

For a simple cascade of devices, suppose each stage receives exactly the useful output of the previous stage, with no extra input or bypass pathway. The overall efficiency is then the product of the stage efficiencies as decimals, which explains why many moderately efficient stages can produce a much lower total efficiency.

Key terms

Energy pathway
A sequence of transfers and stores connecting an input to an output.
Multistage system
A system in which energy passes through two or more conversion stages.
Overall efficiency
The useful final output divided by the original input for the complete transfer chain.
Mass flow rate
Mass passing a point per unit time, used in continuous-flow power calculations.

Worked examples

Skier with dissipative forces

A 70 kg70\,\mathrm{kg} skier descends 25 m25\,\mathrm{m} vertically, speeding up from 4.04.0 to 18 m s−118\,\mathrm{m\,s^{-1}}. Estimate the energy transferred thermally.

  1. 1Gravitational energy decrease: mgΔh=70×9.81×25=17 168 Jmg\Delta h=70\times9.81\times25=17\,168\,\mathrm{J}.
  2. 2Kinetic energy increase: ΔEk=0.5×70×(182−42)=10 780 J\Delta E_k=0.5\times70\times(18^2-4^2)=10\,780\,\mathrm{J}.
  3. 3The difference becomes thermal energy: 17 168−10 780=6388 J17\,168-10\,780=6388\,\mathrm{J}.

Answer: Approximately 6.39 kJ6.39\,\mathrm{kJ} is transferred thermally.

Hydroelectric output

Water flows through a turbine at 250 kg s−1250\,\mathrm{kg\,s^{-1}} with a vertical drop of 18 m18\,\mathrm{m}. The plant is 82%82\% efficient. Estimate its useful output power.

  1. 1The available gravitational power is Pg=m˙ghP_g=\dot m gh.
  2. 2Pg=250×9.81×18=44 145 WP_g=250\times9.81\times18=44\,145\,\mathrm{W}.
  3. 3Useful output is 0.82×44 145=36 200 W0.82\times44\,145=36\,200\,\mathrm{W} to three significant figures.

Answer: The useful output power is approximately 36.2 kW36.2\,\mathrm{kW}.

Rules and formulas

Overall efficiency of a simple useful-output cascade

ηoverall=η1η2η3⋯\eta_{\mathrm{overall}}=\eta_1\eta_2\eta_3\cdots
ηn\eta_n
efficiency of one stage as a decimal (nonenone)

Gravitational power of a mass flow

Pg=m˙ghP_g=\dot m gh
m˙\dot m
mass flow rate (kgs−1kg s⁻¹)
hh
vertical height drop of the flowing material (mm)

Mechanical-to-thermal balance with no external energy transfer

Emech,i=Emech,f+ΔEthE_{\mathrm{mech},i}=E_{\mathrm{mech},f}+\Delta E_{\mathrm{th}}
ΔEth\Delta E_{\mathrm{th}}
increase in thermal energy (JJ)

Experiments, data analysis, and exam workflow

A spring investigation measures the applied stretching-force magnitude for several extensions, plots that force on the vertical axis against extension on the horizontal axis, and uses the best-fit gradient for the spring constant. The area under this graph gives the work stored elastically, provided the loading remains within the linear elastic region. A graph of the signed restoring force instead has gradient −k.

A human-power investigation measures mass, vertical height, and climbing time, then estimates useful power from the gravitational energy gained per time. Repeated timing reduces random variation; a clearly defined start and finish and a calibrated height measurement reduce systematic error.

For exam calculations, sketch the initial and final states, choose the system and zero level, list energy stores and transfers, write a symbolic conservation equation, substitute SI values, and finish with units and a reasonableness check. Do not start by hunting for a formula without an energy model.

Key terms

Best-fit gradient
The slope representing the overall trend of experimental graph data.
Random uncertainty
Unpredictable variation that causes repeated measurements to scatter.
Systematic error
A consistent bias that shifts measurements in the same direction.
Control variable
A quantity held constant so that its effect does not confound the relationship tested.

Worked examples

Spring graph data

A best-fit force–extension line passes through (0,0)(0,0) and (0.024 m,3.6 N)(0.024\,\mathrm{m},3.6\,\mathrm{N}). Find kk and the elastic energy at 0.024 m0.024\,\mathrm{m}.

  1. 1The gradient is k=ΔFΔx=3.60.024=150 N m−1k=\frac{\Delta F}{\Delta x}=\frac{3.6}{0.024}=150\,\mathrm{N\,m^{-1}}.
  2. 2Stored energy is the triangular graph area: Es=12FxE_s=\tfrac12Fx.
  3. 3Es=0.5×3.6×0.024=0.0432 JE_s=0.5\times3.6\times0.024=0.0432\,\mathrm{J}.

Answer: k=150 N m−1k=150\,\mathrm{N\,m^{-1}} and Es=0.0432 JE_s=0.0432\,\mathrm{J}.

Stair-climbing power

A 55 kg55\,\mathrm{kg} student gains 3.2 m3.2\,\mathrm{m} of vertical height in 5.8 s5.8\,\mathrm{s}. Estimate average useful power.

  1. 1ΔEg=mgh=55×9.81×3.2=1727 J\Delta E_g=mgh=55\times9.81\times3.2=1727\,\mathrm{J}.
  2. 2P=ΔEgΔt=17275.8P=\frac{\Delta E_g}{\Delta t}=\frac{1727}{5.8}.
  3. 3P=298 WP=298\,\mathrm{W} to three significant figures.

Answer: The estimated average useful power is 298 W298\,\mathrm{W}.

Rules and formulas

Spring constant from a graph

k=ΔFΔxk=\frac{\Delta F}{\Delta x}
kk
gradient of the linear applied stretching-force versus extension graph (Nm−1N m⁻¹)

Elastic work from graph area

W=12Fx=12kx2W=\frac12Fx=\frac12kx^2
FF
final applied stretching-force magnitude in the linear elastic region (NN)

Useful climbing power

P=mgΔhΔtP=\frac{mg\Delta h}{\Delta t}
Δt\Delta t
measured climbing time (ss)

Topic summary

  • Work is energy transferred by a force through a displacement; for a constant force, W=Fscos⁡θW=Fs\cos\theta.
  • Only the force component parallel to the displacement does work.
  • Positive net work increases an object's kinetic energy; negative net work reduces it.
  • The signed area under a force–displacement graph equals the work done.
  • Kinetic energy is Ek=12mv2=p2(2m)E_k=\tfrac12mv^2=\frac{p^2}{(2m)}, where pp is momentum magnitude; at fixed mass it is proportional to speed squared.
  • The work–energy theorem is Wnet=ΔEkW_{\mathrm{net}}=\Delta E_k.
  • Near Earth's surface, ΔEg=mgΔh\Delta E_g=mg\Delta h; only vertical height change matters.
  • The gravitational potential-energy reference level is arbitrary, but energy changes are not.
  • A linear spring stores Es=12kx2E_s=\tfrac12kx^2 within its elastic range.
  • Total energy remains constant in an isolated system.
  • In the mechanical model, mechanical energy is conserved when the net work done by non-conservative forces is zero.
  • Friction and drag can reduce mechanical energy through dissipation, but a moving surface or fluid can also supply energy to the chosen object; total energy does not disappear.
  • With a constant braking force, braking distance is proportional to speed squared.
  • Average power is P=ΔEΔtP=\frac{\Delta E}{\Delta t}, and 1 W=1 J s−11\,\mathrm{W}=1\,\mathrm{J\,s^{-1}}.
  • Instantaneous mechanical power is P=F⃗⋅v⃗=Fvcos⁡θP=\vec F\cdot\vec v=Fv\cos\theta.
  • Power describes the rate of transfer, not the total amount of energy transferred.
  • Efficiency is useful output divided by total input and cannot exceed 100% for a real passive device.
  • In a Sankey diagram, arrow width represents energy or power flow.
  • For a simple cascade in which each stage's useful output is the next stage's entire input, overall efficiency is the product of the decimal stage efficiencies.
  • A reliable solution identifies the system, states initial and final stores, includes every transfer, uses SI units, and checks the result.