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IB Physics SL/Notes/A.3 Work, energy and power

IB Physics SLA.3 Work, energy and powerNotes

Track Energy Through the System

Energy conservation starts with the chosen system. If the system is isolated, the total energy inside it stays constant even though energy can move between kinetic, gravitational potential, elastic, thermal, chemical, or internal stores. If the system is not isolated, energy may enter or leave by work done, heating, radiation, or electrical transfer. IB answers should avoid saying energy is lost. A better phrase is that energy is transferred or dissipated to the surroundings, often becoming less useful for the intended process.

Energy cannot be created or destroyed; it is transferred between stores or between systems.
For an isolated system, total energy remains constant.
Individual energy stores such as kinetic or gravitational potential energy can increase or decrease.
If energy crosses the system boundary by work, heating, radiation, or electrical transfer, include it in the accounting.
Dissipated energy is still conserved, but it is spread into less useful stores such as thermal energy of the surroundings.

Choose the conserved quantity after checking the system boundary.

Decision
pendulum with negligible air resistance, system = pendulum + Earth
pendulum with air resistance, system = pendulum + Earth + surrounding air
pendulum with air resistance, system = pendulum only

For a described process, define the system and explain how energy is conserved while changing stores or crossing the boundary.

define the system boundary
state total energy of an isolated system is conserved
identify energy stores involved
identify transfers across the boundary when system is not isolated
use dissipated/transferred rather than lost/destroyed

IB questions often ask why energy is not “lost” when mechanical energy decreases; the expected answer names the transfer to other stores or surroundings.

Energy is conserved overall. Within an isolated system, total energy remains constant while energy transfers between stores. If the chosen system is not isolated, energy may cross the boundary by work, heating, radiation, or electrical transfer, so include those transfers in the accounting.

Claiming that energy is lost, or saying a single store such as kinetic energy is conserved without checking the process.

For a described process, define the system and explain how energy is conserved while changing stores or crossing the boundary.

Choose

Make Work an Energy Transfer

Work is not a new energy store; it is a way energy is transferred. A force does work only when the object has a displacement and the force has a component along that displacement. For a constant force, W = Fs cosθ. Positive work transfers energy into the object or chosen system; negative work transfers energy out of it or reduces a mechanical energy store. A perpendicular force, such as ideal centripetal force in uniform circular motion, does no work because cos90° = 0.

Work done by a force is a transfer of energy.
Only the component of force parallel to the displacement does work.
For a constant force, W = Fs cosθ where θ is the angle between force and displacement.
Work is measured in joules; 1 J = 1 N m.
Work is positive when the force component is along displacement, negative when opposite, and zero when perpendicular or when displacement is zero.

Build the work-done relation and decide the sign from the angle.

Formula
Target formula W = Fs cosθ
W
work done or energy transferred
J
F
constant force magnitude
N
s
displacement magnitude
m
θ
angle between force and displacement
degree or rad
1Identify the displacement of the object/system.s
2Use the force component along displacement.F_parallel = F cosθ
3Multiply component by displacement.W = Fs cosθ
4Interpret the sign of cosθ.positive, zero, or negative work
5Link work to energy transfer.work done = energy transferred

Explain or calculate the work done by a force, including the direction of energy transfer.

state work is energy transfer by a force
use the component of force along displacement
write W = Fs cosθ for constant force
give units J or N m
interpret positive/negative/zero work physically

IB questions often test the angle between force and displacement, sign of work, and the phrase “energy transfer”.

Work done by a constant force is W = Fs cosθ, where θ is the angle between force and displacement. It represents energy transferred by the force. Work is positive if the force component is along the displacement, negative if opposite, and zero if perpendicular.

Using F s without resolving the force, or saying work is done when there is no displacement in the force direction.

Explain or calculate the work done by a force, including the direction of energy transfer.

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Read the Sankey Diagram

Sankey diagrams are visual energy accounts. The width of each arrow represents the amount of energy transferred, or sometimes the power flow if the diagram is per unit time. The total input must equal the total output when all pathways are included. Useful output is the energy transferred into the intended store or pathway; dissipated output is energy transferred to less useful stores, often thermal energy of the surroundings. Read the diagram by comparing widths or labelled values, not by the physical length of an arrow.

A Sankey diagram shows energy transfers or power flows through a system.
Arrow width is proportional to the amount of energy or power represented.
The input arrow equals the sum of useful output and dissipated output arrows.
Dissipated energy is usually shown branching away to the surroundings.
Efficiency can be read from the useful output divided by the total input.

Label the Sankey arrows and use their widths as the energy account.

Label
Labels
4

Use a Sankey diagram to identify input, useful output, dissipated output, and the energy or power represented by arrow widths.

state arrow width is proportional to energy or power
identify input, useful output, and dissipated pathways
check input equals sum of all outputs
describe dissipated energy as transferred to surroundings
use useful output/input for efficiency if asked

IB questions may ask for useful output, wasted/dissipated transfer, efficiency, or whether the diagram satisfies conservation of energy.

Read the Sankey diagram by arrow width. The input equals the sum of all output arrows. Useful output is the intended pathway; dissipated energy is transferred to less useful stores such as thermal energy of the surroundings.

Reading arrow length instead of width, or ignoring a dissipated branch when calculating output totals.

Use a Sankey diagram to identify input, useful output, dissipated output, and the energy or power represented by arrow widths.

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Work by constant force

This calculation card is about choosing the correct component and area. If force is constant and parallel to displacement, work is simply force times displacement. If the force is angled, first resolve it along the displacement, F_parallel = F cosθ, then multiply by s. A constant-force graph against displacement gives a rectangular area, so W is the area under the F_parallel-s graph. When the force opposes displacement, the work done by that force is negative.

For a constant force parallel to displacement, W = Fs.
For a force at angle θ to displacement, use W = Fs cosθ.
Equivalently, W = F_parallel s where F_parallel is the component along displacement.
On a force-displacement graph, work is the area under the graph; for constant force this is a rectangle.
Use signs: work by friction or a force opposite motion is negative in the chosen displacement direction.

Build the constant-force work calculation from component and displacement.

Formula
Target formula W = F_parallel s = Fs cosθ = area under F-s graph
W
work done by the force
J
F_parallel
force component along displacement
N
F
force magnitude
N
s
displacement
m
θ
angle between force and displacement
degree or rad
1Identify the displacement direction and distance.s
2Find the force component along the displacement.F_parallel = F cosθ
3Multiply by displacement for constant force.W = F_parallel s
4On a force-displacement graph, use rectangular area.W = area = F_parallel × s
5Apply sign from direction.opposite displacement -> negative work

A constant force moves an object through a displacement. Calculate the work done and interpret it from a force-displacement graph.

identify displacement direction
use force component parallel to displacement
write W = Fs cosθ or W = F_parallel s
use signed area under force-displacement graph
give work in joules

IB questions may include angled forces, friction, or force-displacement graphs where the area represents work done.

Resolve the constant force along the displacement, then multiply by displacement: W = F_parallel s = Fs cosθ. On a force-displacement graph, this is the signed area under the graph.

Using the full force instead of the parallel component, or missing negative work for a force opposite displacement.

A constant force moves an object through a displacement. Calculate the work done and interpret it from a force-displacement graph.

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Work-energy change

The work-energy idea turns a force-and-displacement problem into an energy-change problem. For one object, the net or resultant work is equal to the change in kinetic energy, W_net = ΔE_k = E_k,final - E_k,initial. For a broader system, work done across the boundary transfers energy into or out of the system and may change gravitational, elastic, thermal, or kinetic stores. This is why the setup matters: identify the system, decide whether you are using resultant work or work by a named force, then name the energy store being changed.

Work done on a system is an energy transfer into or out of that system.
For a single object, the work done by the resultant force equals the change in kinetic energy: W_net = ΔE_k.
Work done by individual forces can increase or decrease different energy stores depending on the chosen system.
Before applying a work-energy equation, define the system and identify the energy store that changes.
Use signs consistently: negative net work means the kinetic energy decreases.

Spot and repair the setup errors before using the work-energy relation.

Spot Errors

Use the work-energy theorem to relate work done by the resultant force to the change in kinetic energy, or explain energy changes for a chosen system.

define the system or selected object
distinguish resultant work from work by a named force
write W_net = ΔE_k for a single object
use ΔE_k = E_k,final - E_k,initial
include negative work terms when forces oppose displacement

IB questions often ask students to find speed from work done, or to explain why a named force’s work is not the same as net work.

For a single object, the net work done by all forces equals the change in kinetic energy: W_net = ΔE_k. If several forces do work, add their signed work terms. For a wider system, work crossing the boundary changes the system’s energy stores.

Equating one force’s work with ΔE_k when other forces also do work, or reversing final and initial energy.

Use the work-energy theorem to relate work done by the resultant force to the change in kinetic energy, or explain energy changes for a chosen system.

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Mechanical energy

Mechanical energy is a useful subset of total energy. It includes kinetic energy plus potential energy stores that can be converted mechanically, mainly gravitational potential and elastic potential in A.3. Use E_k = 1/2mv^2 for motion, E_p = mgh for height changes near Earth with constant g, and E_el = 1/2kx^2 for Hooke’s-law springs. If friction, drag, heating, or deformation is important, some mechanical energy is transferred into non-mechanical stores, so total mechanical energy may not be conserved.

Mechanical energy is the sum of kinetic energy and mechanical potential energy stores.
Kinetic energy is associated with motion: E_k = 1/2 mv^2.
Gravitational potential energy near Earth is associated with height in a gravitational field: E_p = mgh.
Elastic potential energy is stored in a stretched or compressed spring: E_el = 1/2 kx^2.
Mechanical energy is not the same as total energy when thermal, internal, chemical, or other stores are involved.

Match each mechanical energy store to its expression and situation.

Match
Reasons
0/5

Identify the mechanical energy stores in a situation and write the expression for total mechanical energy.

include kinetic energy when object moves
include gravitational potential energy for height changes near Earth
include elastic potential energy for springs in Hooke’s-law region
exclude dissipated thermal/internal stores from mechanical energy unless accounting separately
state mechanical energy may not be conserved when dissipative forces act

IB questions may ask students to write an energy equation, so the first mark is often choosing the correct stores.

The total mechanical energy for this A.3 system is the sum of kinetic, gravitational potential, and elastic potential stores: E_mech = E_k + E_p + E_el, using 1/2mv^2, mgh, and 1/2kx^2 where the conditions apply.

Treating mechanical energy as total energy and ignoring thermal/internal energy produced by friction.

Identify the mechanical energy stores in a situation and write the expression for total mechanical energy.

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Mechanical energy conservation

Practice

Use mechanical energy conservation only after checking the model conditions. If gravity and ideal springs are the only forces doing work, energy can transfer between kinetic, gravitational potential, and elastic potential stores without changing their total. If friction, air resistance, an external motor, or deformation is significant, mechanical energy changes because energy transfers into or out of non-mechanical stores. Then write an energy account with an added dissipated or external work term instead of forcing mechanical energy to be constant.

Mechanical energy is conserved when energy transfers only between kinetic, gravitational potential, and elastic potential stores.
Mechanical energy is not conserved when friction, drag, deformation, or other dissipative processes transfer energy to non-mechanical stores.
If non-conservative work is present, include it as a work or energy-transfer term.
A common mechanical-energy equation is E_k,i + E_p,i + E_el,i = E_k,f + E_p,f + E_el,f when losses are negligible.
Total energy is still conserved even when mechanical energy is not.

Decide whether mechanical energy is conserved in each situation.

Decision
cart moves on a frictionless track under gravity only
cart moves down a rough track and warms the wheels
ideal spring launches a glider with negligible friction
motor pulls an object upward at constant speed
falling object experiences significant air resistance

State whether mechanical energy is conserved in a described system and write the appropriate energy equation.

identify mechanical energy stores
state conservation only if dissipative/external transfers are negligible
write initial mechanical energy equals final mechanical energy when valid
include dissipated energy or external work when not valid
distinguish total energy conservation from mechanical energy conservation

IB questions often hide the condition in words like “frictionless”, “negligible air resistance”, “rough”, or “constant speed motor”.

If friction and other dissipative transfers are negligible, E_k + E_p + E_el is conserved between initial and final states. If friction, drag, or external work is significant, include a dissipated-energy or work term rather than setting mechanical energies equal.

Using mechanical energy conservation despite friction or drag, or saying energy is lost instead of dissipated.

State whether mechanical energy is conserved in a described system and write the appropriate energy equation.

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Mechanical energy transformations

Mechanical energy transformations describe which stores increase and decrease during a process. A falling object loses gravitational potential energy and gains kinetic energy. A rising object does the reverse. An ideal spring can store energy elastically and release it as kinetic energy. With friction or drag, the transformation is not purely mechanical: some mechanical energy is transferred to thermal/internal stores. When writing an energy equation, name the stores before choosing formulas.

Mechanical energy can transform between kinetic, gravitational potential, and elastic potential stores.
Falling without air resistance transfers gravitational potential energy into kinetic energy.
Moving upward transfers kinetic energy into gravitational potential energy.
A spring can transfer elastic potential energy into kinetic energy, or kinetic energy into elastic potential energy.
If friction or drag is present, some mechanical energy transfers to thermal/internal stores.

Sort each situation by the main energy transformation.

Sort
Unsorted
6
GPE to KE
0
KE to GPE
0
elastic PE to KE
0
KE to elastic PE
0
mechanical energy to thermal/internal
0

Describe the energy transformations in a mechanical process and state whether mechanical energy remains in mechanical stores.

identify initial and final energy stores
state which store decreases and which increases
include elastic potential energy for springs
include thermal/internal energy when friction or drag is present
avoid saying energy is lost

IB questions often ask for energy-store descriptions before or after a calculation, especially with springs, falling objects, or friction.

A correct transformation sentence names both stores. For example, a falling object transfers gravitational potential energy to kinetic energy; a compressed spring transfers elastic potential energy to kinetic energy; friction transfers mechanical energy to thermal/internal energy.

Listing formulas without naming stores, or omitting the dissipated store when friction is present.

Describe the energy transformations in a mechanical process and state whether mechanical energy remains in mechanical stores.

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Kinetic energy

Kinetic energy depends on mass and speed, not velocity direction. Use speed in E_k = 1/2mv^2, so a negative velocity does not make negative kinetic energy. Because speed is squared, changes in speed have a large effect: twice the speed gives four times the kinetic energy. The alternative form E_k = p^2/(2m) is useful when momentum and mass are known.

Translational kinetic energy is the energy store associated with motion.
For a mass m moving at speed v, E_k = 1/2 mv^2.
Kinetic energy is a scalar and is never negative, even if velocity is assigned a negative sign.
Doubling speed quadruples kinetic energy.
Using p = mv, kinetic energy can also be written as E_k = p^2/(2m).

Build the kinetic energy formula and check the speed-squared dependence.

Formula
Target formula E_k = 1/2 mv^2 = p^2/(2m)
E_k
kinetic energy
J
m
mass
kg
v
speed
m s^-1
p
momentum magnitude
kg m s^-1
1Use speed, the magnitude of velocity.v ≥ 0
2Square the speed.v^2
3Multiply by half the mass.E_k = 1/2 mv^2
4If momentum is given, use the equivalent form.E_k = p^2/(2m)
5Check proportionality.E_k ∝ v^2

Calculate kinetic energy from mass and speed, or compare kinetic energies when speed or momentum changes.

write E_k = 1/2mv^2
use speed in m s^-1 and mass in kg
give joules
treat kinetic energy as scalar
use p^2/(2m) when momentum is the given quantity

IB questions often test proportional reasoning: speed doubled means kinetic energy quadrupled, and direction signs do not make kinetic energy negative.

Kinetic energy is E_k = 1/2mv^2. Since v is speed, the value is non-negative and measured in joules. If momentum is known instead, E_k = p^2/(2m).

Using velocity sign to make kinetic energy negative, or treating E_k as proportional to v rather than v^2.

Calculate kinetic energy from mass and speed, or compare kinetic energies when speed or momentum changes.

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Map Gravitational potential energy

Gravitational potential energy near Earth is a height-based energy store. Because the zero level is chosen by the solver, absolute E_p values are less important than changes. Use ΔE_p = mgΔh when g is approximately constant over the height change. Lifting an object increases E_p; lowering or falling decreases E_p and can transfer energy to kinetic or other stores. Be careful to use vertical height change, not distance travelled along a slope.

Near Earth, where g is approximately constant, ΔE_p = mgΔh.
Only changes in gravitational potential energy matter; the zero height/reference level can be chosen for convenience.
If height increases, Δh is positive and gravitational potential energy increases.
If height decreases, Δh is negative and gravitational potential energy decreases.
Use g in N kg^-1 and height change in metres to get joules.

Build the gravitational potential energy change formula using vertical height change.

Formula
Target formula ΔE_p = mgΔh
ΔE_p
change in gravitational potential energy
J
m
mass
kg
g
gravitational field strength
N kg^-1
Δh
vertical height change
m
1Choose or note the zero height reference.reference level
2Use vertical height change, final minus initial.Δh = h_f - h_i
3Multiply mass, g, and height change.ΔE_p = mgΔh
4Interpret sign from upward or downward motion.up -> positive; down -> negative
5Check g is approximately constant.near Earth / small height range

Calculate the change in gravitational potential energy for an object moving between two heights near Earth.

use ΔE_p = mgΔh
use vertical height change final minus initial
state or imply near-Earth constant g
give units J
interpret sign or energy transfer if object rises/falls

IB questions often hide the height change inside a ramp, vertical circle, or graph, so using vertical Δh is the key mark.

Near Earth, ΔE_p = mgΔh, where Δh is the vertical height change relative to a chosen reference level. Rising gives positive ΔE_p; falling gives negative ΔE_p.

Using path length instead of vertical height change, or treating the zero reference as physically fixed when only changes matter.

Calculate the change in gravitational potential energy for an object moving between two heights near Earth.

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Elastic potential energy

A spring stores elastic potential energy when it is stretched or compressed from its natural length. In the Hooke’s-law region, force increases linearly with extension, F = kx, so the energy stored is the triangular area under the force-extension graph: E_el = 1/2kx^2. Use x as the displacement from natural length, not the full length of the spring. Because x is squared, stretching and compressing by the same amount store the same energy, as long as the spring remains elastic.

Elastic potential energy in a Hooke’s-law spring is E_el = 1/2kx^2.
x is extension or compression from the natural length, measured in metres.
The formula is valid in the linear Hooke’s-law region, before the proportional limit is exceeded.
Elastic potential energy is the area under the force-extension graph.
Doubling extension or compression quadruples the stored elastic potential energy.

Build the elastic potential energy expression from Hooke’s law and graph area.

Formula
Target formula E_el = 1/2 kx^2
E_el
elastic potential energy
J
k
spring constant
N m^-1
x
extension or compression from natural length
m
1Measure extension/compression from natural length.x
2Use the Hooke’s-law force relation in the linear region.F = kx
3Take the triangular area under the F-x graph.area = 1/2 × x × kx
4Simplify the energy expression.E_el = 1/2kx^2
5Check the spring remains elastic/linear.before proportional limit

Calculate the elastic potential energy stored in a spring and state the condition for using the formula.

use x as extension/compression from natural length
write E_el = 1/2kx^2
state valid in Hooke’s-law/linear region
connect formula to area under force-extension graph
give joules

IB questions often combine Hooke’s law with energy by using the area under a force-extension graph.

For a Hooke’s-law spring, force increases linearly with extension, so the stored elastic potential energy is the area under the force-extension graph: E_el = 1/2kx^2. The extension x is measured from natural length.

Using full spring length for x, omitting the 1/2, or using the formula beyond the proportional limit.

Calculate the elastic potential energy stored in a spring and state the condition for using the formula.

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Use Energy per Time

Power tells how quickly energy is transferred. If 500 J is transferred in 10 s, the average power is 50 W. The same energy transferred in a shorter time has greater power. In mechanical problems, P = Fv is valid when F is the component of force along the motion; more generally use P = F_parallel v. Distinguish average power over a time interval from instantaneous power at a moment.

Power is the rate of energy transfer or the rate of doing work.
Average power over an interval is P = ΔE/Δt = W/Δt.
The watt is the unit of power: 1 W = 1 J s^-1.
For a force moving an object at speed v, P = F_parallel v, where F_parallel is the force component along the velocity.
High power means the same energy is transferred in less time, not necessarily more total energy.

Build the power relation from energy transfer and time.

Formula
Target formula P_avg = ΔE/Δt = W/Δt; P = F_parallel v
P
power
W
ΔE
energy transferred
J
W
work done
J
Δt
time interval
s
F_parallel
force component along velocity
N
v
speed
m s^-1
1Start with energy transferred per time.P_avg = ΔE/Δt
2If energy transfer is work, use work per time.P_avg = W/Δt
3State the unit relation.1 W = 1 J s^-1
4For mechanical power, use the force component along motion.P = F_parallel v
5Interpret rate versus amount.same energy in less time -> greater power

Calculate average power from energy transferred or work done over time, and apply P = Fv when a force moves at constant speed.

write P = ΔE/Δt or W/Δt
use seconds and joules for watts
distinguish power from total energy
use P = F_parallel v only with force along motion
state average power over interval if appropriate

IB questions often compare devices doing the same work in different times, or use P = Fv for vehicles/lifts at steady speed.

Power is the rate of energy transfer: P_avg = ΔE/Δt = W/Δt. The unit is W = J s^-1. For mechanical motion, P = F_parallel v, where only the force component along the velocity counts.

Comparing power by energy alone without time, or using P = Fv when the force is not parallel to motion.

Calculate average power from energy transferred or work done over time, and apply P = Fv when a force moves at constant speed.

Choose

Compare Useful and Input Energy

Efficiency measures how much of the supplied energy or power becomes the useful output. The denominator is total input, not wasted output. The numerator is the useful output for the intended purpose. If powers are used, they must refer to the same process/time interval, so the time cancels. A realistic device has efficiency less than 1 or less than 100% because some energy is dissipated to less useful stores.

Efficiency is the fraction of input energy or power transferred usefully.
η = useful output energy / total input energy.
The same ratio can be calculated using power: η = useful output power / total input power.
Use matching quantities over the same time interval; do not mix energy with power.
Efficiency may be left as a fraction or multiplied by 100% for a percentage.

Build the efficiency ratio from useful output and total input.

Formula
Target formula η = E_useful/E_input = P_useful/P_input; percentage = η × 100%
η
efficiency as a fraction
E_useful
useful output energy
J
E_input
total input energy
J
P_useful
useful output power
W
P_input
total input power
W
1Identify the useful output for the intended purpose.E_useful or P_useful
2Identify total input, not wasted output.E_input or P_input
3Use energy with energy or power with power.same quantity type
4Form useful divided by total input.η = useful/input
5Convert to percent only if requested.η × 100%

Calculate the efficiency of a device from useful output and total input energy or power.

identify useful output for the stated purpose
identify total input
use matching energy or power units
calculate useful/input
convert to percentage only at the end if required

IB questions commonly give Sankey diagrams or input/output powers and require a fraction or percentage efficiency.

Efficiency is useful output divided by total input. It may be calculated from energies or powers: η = E_useful/E_input = P_useful/P_input. As a percentage, multiply the fraction by 100%.

Using wasted output as the numerator, using total output rather than useful output, or mixing energy and power in one ratio.

Calculate the efficiency of a device from useful output and total input energy or power.

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Compare Energy per Fuel Mass

This card compares how much energy different fuels release for the same mass. If a fuel has a value in J kg^-1, multiply by the fuel mass in kg to find total energy released. Convert MJ kg^-1 to J kg^-1 when needed. A high energy-per-mass fuel is useful when mass matters, but it does not automatically mean the device using it is efficient or powerful; efficiency depends on useful output divided by input, and power depends on time.

Energy per fuel mass is often quoted as fuel energy density or specific energy, with unit J kg^-1 or MJ kg^-1.
energy per mass = energy released / mass of fuel.
Energy released can be calculated from E = (energy per mass) × mass.
Compare fuels on the same mass basis and with consistent units, especially kg versus g and J versus MJ.
Energy per mass is not the same as efficiency or power.

Build the energy-per-mass comparison for a fuel.

Formula
Target formula specific energy = E/m; E = (specific energy)m
specific energy
energy released per unit mass of fuel
J kg^-1
E
energy released
J
m
mass of fuel
kg
1Put energy and mass into consistent units.J and kg, or MJ and kg
2For comparison, divide energy by fuel mass.specific energy = E/m
3For a known fuel mass, multiply by mass.E = specific energy × m
4Compare fuels on the same mass basis.same kg basis
5Apply efficiency only if useful output is asked.E_useful = ηE_input

Compare two fuels using energy released per unit mass, or calculate energy released from a given fuel mass.

use E/m for energy per mass
convert MJ to J or g to kg when needed
compare fuels on the same mass basis
distinguish energy per mass from efficiency and power
apply useful output = η input only if efficiency is given

IB questions may give a table of fuel values in MJ kg^-1 and ask for comparison, total energy, or useful output after efficiency.

Energy per unit mass is E/m with units J kg^-1. For a fuel mass m, E = (energy per mass)m. This compares fuels by mass; efficiency and power are separate quantities unless the question includes them.

Using grams with J kg^-1 without conversion, or assuming the fuel with higher energy per mass has higher efficiency.

Compare two fuels using energy released per unit mass, or calculate energy released from a given fuel mass.

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Retrieve the A.3 Work, energy and power Model

Review

A.3 is an energy-accounting topic. Start by choosing the system and the energy stores. If the question gives a force and displacement, use work. If it gives before/after speeds, heights, or springs, choose the relevant energy stores. If it gives time, use power. If it gives useful and input pathways, use efficiency. If it compares fuels, use energy per unit mass. The strongest IB answers state the model condition before the arithmetic.

Define the system boundary before applying conservation of energy.
Use work as energy transfer by a force component through displacement.
Track mechanical energy stores and check whether dissipative transfers are negligible before conserving mechanical energy.
Use power as energy transfer per time, and efficiency as useful output divided by total input.
Keep fuel energy per mass separate from efficiency and power.

Match each A.3 trigger to the model move it should trigger before calculation.

Match
Reasons
0/10

A mixed A.3 problem includes a device, motion, or fuel data. State the model choice and condition before calculating.

define the system or device boundary
identify stores and transfers
choose work, mechanical energy, power, efficiency, or fuel-energy model
state validity conditions such as negligible dissipation or same time interval
use consistent units and interpret the result

A.3 marks usually reward choosing the right energy account before substituting: store changes, work terms, time rates, useful ratios, or fuel mass basis.

Begin with the system boundary and the requested quantity. Use work for force through displacement, mechanical energy when stores change, power for rate of energy transfer, efficiency for useful over input, and fuel energy per mass for fuel comparisons.

Using a memorized formula without checking whether the question asks for energy, rate, useful fraction, or energy per mass.

A mixed A.3 problem includes a device, motion, or fuel data. State the model choice and condition before calculating.

Choose