A.3 Work, energy and power

Energy is conserved

Energy cannot be created or destroyed. In a defined system, energy is transferred between stores or across the system boundary, so the total energy accounting remains balanced.

Define the system first

Name the objects included and identify transfers by work, heating, radiation or electrical means. A falling object may transfer gravitational potential energy to kinetic energy, internal energy or sound.

Follow the chain

Write the initial store, the useful output store and any dissipated or transferred energy. A Sankey diagram or energy-flow statement should account for all significant branches.

Common trap

Energy “lost” from a useful store has been transferred elsewhere; it has not disappeared.

Work transfers energy

Work done by a force is the energy transferred by that force. For a constant force,

$$W=Fs\cos\theta$$

where $\theta$ is the angle between force and displacement.

Use the sign

Positive work transfers energy into the object’s relevant store; negative work transfers energy out of it. A force perpendicular to displacement does zero work.

Follow the physical process

Wind can transfer kinetic energy to a turbine through work, while resistive forces can transfer mechanical energy to internal energy of the surroundings.

Common trap

Do not call every force an energy transfer. Check whether the force has a component along the displacement.

Read the width as energy

A Sankey diagram shows an input energy flowing into useful output and other transfers. Arrow width is proportional to energy, so the branches must account for the whole input.

Identify useful output

Label the useful branch before calculating efficiency. Other branches may represent heating, sound or unwanted mechanical transfers.

Connect to efficiency

The useful fraction of the input is

$$\eta=\frac{E_{useful}}{E_{input}}=\frac{P_{useful}}{P_{input}}$$

Common trap

Do not compare branch widths without checking whether the diagram uses the same scale and whether the requested quantity is energy or power.

Constant-force work

For a force $F$ acting through displacement $s$,

$$W=Fs\cos\theta$$

Only the component parallel to displacement transfers energy by work.

Area under a force–distance graph

For a variable force, the area under an $F$-against-$s$ graph gives work. A negative area represents work against the chosen displacement direction.

Check the angle

Use the angle between force and displacement, not the angle between the force and an unrelated axis unless the component has first been resolved.

Worked example from local Question Bank row 39177

A kite pulls a ship with force $2.50\times10^5\,\mathrm{N}$ at $39^\circ$ to its $1.00\,\mathrm{km}$ displacement. Convert $1.00\,\mathrm{km}=1.00\times10^3\,\mathrm{m}$, then

$$W=Fs\cos\theta=(2.50\times10^5)(1.00\times10^3)\cos39^\circ=1.94\times10^8\,\mathrm{J}\approx1.9\times10^8\,\mathrm{J}$$

Only the force component along the ship's displacement transfers energy.

Work–energy theorem

The net work done by the resultant force on a system equals its change in kinetic energy:

$$W_{net}=\Delta E_k$$

Use force–distance area

For a variable resultant force, the signed area under the force–distance graph gives the work and therefore the kinetic-energy change.

Include all resultant forces

Friction, applied forces and gravity may each do work. Add their signed contributions before relating the result to the final kinetic energy.

Worked example from local Question Bank row 31356

A constant net force of $100\,\mathrm{N}$ moves an object from rest through $2.0\,\mathrm{m}$ until its speed is $10\,\mathrm{m\,s^{-1}}$.

$$W_{net}=Fs=(100)(2.0)=200\,\mathrm{J}$$
$$200=\Delta E_k=\frac12m(10)^2-0$$
$$m=4.0\,\mathrm{kg}$$

The positive net work is exactly the object's kinetic-energy gain.

Common trap

Do not use the work of one force as the net work unless all other force contributions are zero or already included.

Mechanical energy stores

Mechanical energy is the sum of translational kinetic energy, gravitational potential energy and elastic potential energy:

$$E_{mech}=E_k+E_{p,g}+E_{p,elastic}$$

Use the chosen system

Mechanical energy describes these stores within the system. Internal energy, chemical energy and sound may also be present in the full energy account but are not mechanical energy.

Common trap

Do not call all conserved energy mechanical energy; classify the store before applying a mechanical-energy equation.

Condition for conservation

Mechanical energy is conserved when only conservative forces transfer energy within the system and friction or other resistive transfers are absent or negligible.

Write the balance

$$E_{k,i}+E_{p,i}=E_{k,f}+E_{p,f}$$

Choose a convenient zero for potential energy and keep the same reference throughout.

When it is not conserved

Friction, drag or deformation transfer mechanical energy to internal energy. Total energy is still conserved, but the mechanical-energy equation needs an additional transfer term.

Conservative transformations

When mechanical energy is conserved, energy can move between translational kinetic, gravitational potential and elastic potential stores without changing their sum.

Use the endpoints

For a car descending a frictionless track, gravitational potential energy decreases while kinetic energy increases. For a spring system, elastic potential energy can become kinetic energy and then return.

Add non-conservative transfers

If friction or drag acts, part of the mechanical energy transfers to internal energy. The endpoint equation must include that loss from the mechanical stores.

Kinetic-energy forms

Translational kinetic energy is

$$E_k=\frac12mv^2=\frac{p^2}{2m}$$

Choose the known quantity

Use $\frac12mv^2$ when mass and speed are given, or $p^2/(2m)$ when momentum is given. Kinetic energy is scalar and cannot be negative.

Worked example from local Question Bank row 35674

For $m=0.14\,\mathrm{g}=1.4\times10^{-4}\,\mathrm{kg}$ and $v=3.1\,\mathrm{m\,s^{-1}}$,

$$E_k=\frac12(1.4\times10^{-4})(3.1)^2=6.7\times10^{-4}\,\mathrm{J}=0.67\,\mathrm{mJ}$$

Converting grams to kilograms before substitution keeps the energy unit in joules.

Common trap

Doubling speed quadruples kinetic energy; do not scale it linearly with speed.

Near-Earth gravitational potential energy

For a height change $\Delta h$ in a uniform gravitational field,

$$\Delta E_{p,g}=mg\Delta h$$

Use the height change

Raising an object gives positive change in gravitational potential energy; lowering it gives negative change relative to the chosen reference.

Link to power

If height changes at constant speed, the rate of gravitational potential-energy gain is $mgv$, before accounting for efficiency or other transfers.

Worked example from local Question Bank row 37039

An object's weight is $6.10\times10^2\,\mathrm{N}$ and it rises vertically by $8.0\,\mathrm{m}$. Since $mg$ is its weight,

$$\Delta E_{p,g}=(6.10\times10^2)(8.0)=4.88\times10^3\,\mathrm{J}\approx4.9\,\mathrm{kJ}$$

The positive result means the gravitational potential-energy store increases.

Common trap

Use the local value of $g$ and the vertical height change, not the distance along a slope.

Elastic store

For a spring within its linear range,

$$E_{p,elastic}=\frac12k(\Delta x)^2$$

where $\Delta x$ is extension or compression from the natural length.

Area under the graph

The elastic potential energy equals the work done in stretching or compressing the spring. On a force–extension graph it is the area under the graph.

Worked example from local Question Bank row 31357

A spring with $k=100\,\mathrm{N\,m^{-1}}$ is compressed by $0.10\,\mathrm{m}$.

$$E_{p,elastic}=\frac12(100)(0.10)^2=0.50\,\mathrm{J}$$

This is the energy available for transfer when the ideal spring is released.

Common trap

Do not use the total spring length as $\Delta x$, and remember that doubling extension quadruples the stored energy in the ideal model.

Power is rate

Power is the rate of work or energy transfer:

$$P=\frac{\Delta W}{\Delta t}=\frac{\Delta E}{\Delta t}$$

Mechanical shortcut

For a constant force parallel to velocity,

$$P=Fv$$

Keep energy and power distinct

Energy is measured in joules; power is measured in watts, or joules per second. Multiply power by time to recover transferred energy.

Worked example from local Question Bank row 29322

A student of weight $600\,\mathrm{N}$ climbs $6.0\,\mathrm{m}$ vertically in $8.0\,\mathrm{s}$.

$$\Delta W=(600)(6.0)=3.6\times10^3\,\mathrm{J}$$
$$P=\frac{3.6\times10^3}{8.0}=4.5\times10^2\,\mathrm{W}=450\,\mathrm{W}$$

The result is the average rate of energy transfer against gravity.

Useful fraction

Efficiency is the ratio of useful output to total input:

$$\eta=\frac{E_{useful}}{E_{input}}=\frac{P_{useful}}{P_{input}}$$

Choose matching quantities

Use energy ratios for the same process and time interval, or power ratios when input and output are rates. Efficiency is dimensionless and is often reported as a percentage.

Worked example from local Question Bank row 29709

Solar intensity is $240\,\mathrm{W\,m^{-2}}$ over $2.50\times10^4\,\mathrm{m^2}$, so input power is

$$P_{in}=(240)(2.50\times10^4)=6.0\times10^6\,\mathrm{W}=6.0\,\mathrm{MW}$$

For a useful output of $1.6\,\mathrm{MW}$,

$$\eta=\frac{1.6}{6.0}=0.27=27\%$$

The remaining input is transferred through non-useful pathways.

Common trap

Do not invert the ratio or use the total output, including unwanted transfers, as the useful output.

Energy per volume

For the current IB Physics definition, fuel energy density $u$ is the transferable energy per unit volume:

$$u=\frac{E}{V}$$

Its SI unit is $\mathrm{J\,m^{-3}}$. This lets fuels be compared when storage volume is the constraint.

Connect it to a fuel flow

If fuel flows at volume rate $\dot V$, its input power is $P_{in}=u\dot V$. Apply efficiency only after finding the input energy or power.

Worked example from local Question Bank row 36970

An engine produces $20\,\mathrm{kW}$ useful power at $50\%$ efficiency while consuming $1.0\times10^{-5}\,\mathrm{m^3\,s^{-1}}$ of fuel.

$$P_{in}=\frac{20\,\mathrm{kW}}{0.50}=40\,\mathrm{kW}$$
$$u=\frac{P_{in}}{\dot V}=\frac{4.0\times10^4}{1.0\times10^{-5}}=4.0\times10^9\,\mathrm{J\,m^{-3}}=4.0\,\mathrm{GJ\,m^{-3}}$$

Common trap

Specific energy is energy per unit mass, measured in $\mathrm{J\,kg^{-1}}$. Some sources use the words loosely, so let the stated definition and units determine whether to divide by volume or mass.