A.3 Work, energy and power

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.

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.

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

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

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.

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.

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.

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.

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.

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.

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.

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.

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.

Mechanical shortcut

For a constant force parallel to velocity,

$$P=Fv$$

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.

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.

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.

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