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4.5 Energy, work and power

Syllabus
9709–2028–2029
Topic
4.5
Level
AS

Learning objectives

4.5.1Scalar product• understand the concept of the work done by a force, and calculate the work done by a constant force when its point of application undergoes a displacement not necessarily parallel to the force cosWF d i=, Use of the scalar product is not required.4.5.2Gravitational potential energy• understand the concepts of gravitational potential energy and kinetic energy, and use appropriate formulae4.5.3Forces• understand and use the relationship between the change in energy of a system and the work done by the external forces, and use in appropriate cases the principle of conservation of energy Including cases where the motion may not be linear (e.g. a child on a smooth curved 'slide'), where only overall energy changes need to be considered.4.5.4Complex loci• use the definition of power as the rate at which a force does work, and use the relationship between power, force and velocity for a force acting in the direction of motion Including calculation of (average) power as Time taken Work done. P = Fv.4.5.5Instantaneous acceleration• solve problems involving, for example, the instantaneous acceleration of a car moving on a hill against a resistance.

Calculate work from the force component along displacement

For constant force magnitude $F$ and displacement $d$ of its point of application, with included angle $\theta$:W=Fd\cos\theta.Unitsarejoules(J).Units are joules (J).

Angle Work by the force Meaning
0≤θ<90∘0\le\theta<90^\circ positive transfers energy to motion/system
θ=90∘\theta=90^\circ zero no component along displacement
90∘<θ≤180∘90^\circ<\theta\le180^\circ negative removes mechanical energy

A $20$ N force acts through $5$ m at $60^\circ$ to the motion:W=20\times5\times\cos60^\circ=50\text{ J}.

Use displacement of the force’s point of application and the angle between the two directions. Resolve first if that makes the along-motion component clearer.

Use of the scalar product is not required. W=FdW=Fd applies only when force and displacement are parallel in the same direction.

Track kinetic and gravitational potential energy

KE=\frac12mv^2,\qquad \Delta GPE=mg(h_2-h_1).In Mechanics use $g=10$ m s$^{-2}$ unless stated otherwise.

Energy Depends on Does not depend on
kinetic mass and speed squared velocity direction
gravitational potential change mass, gg, vertical height change path shape/length

A $2$ kg particle rises $3$ m and changes speed from $4$ to $1$ m s$^{-1}$:\Delta GPE=60\text{ J},\qquad \Delta KE=\tfrac12(2)(1^2-4^2)=-15\text{ J}.

Potential-energy zero is chosen conveniently; only differences matter. Keep joules and use vertical height, including on a curved path.

KE is never negative and uses speed. Do not use 9.89.8 by default in this component or replace vertical height by travelled distance.

Balance initial energy, external work and final energy

ChoosethesystemandwriteChoose the system and writeE_{\text{initial}}+W_{\text{external}}=E_{\text{final}},usingkineticandgravitationalpotentialenergies.Equivalently,externalworkequalsthechangeinsystemenergy.using kinetic and gravitational potential energies. Equivalently, external work equals the change in system energy.

If no external non-conservative work changes mechanical energy, KE+GPE=constant.KE+GPE=\text{constant}. Smooth contact reactions often do no work along the motion.

Work by resistance/friction is negative and must be included, or represented as energy dissipated on the loss side. Driving work is positive when along motion.

Energy depends on initial/final speed and height, not path shape. A child moving on a smooth curved slide can be solved from overall height change without resolving forces along every segment.

Do not conserve KE+GPEKE+GPE when friction or another external force does non-zero work unless that work is included in the ledger.

Distinguish average power from instantaneous $Fv$

Averagepower:Average power:P_{\text{avg}}=\frac{W}{t}.Foraforceactinginthedirectionofinstantaneousvelocity:For a force acting in the direction of instantaneous velocity:P=Fv.Units are watts, $1\text{ W}=1\text{ J s}^{-1}$.

If force is not parallel to motion, only its along-velocity component contributes: P=Fvcos⁡θP=Fv\cos\theta. The syllabus P=FvP=Fv form assumes force acts in the direction of motion.

An engine does $120$ kJ in $20$ s, so average power is $6.0$ kW. If it exerts $1500$ N along motion at $8$ m s$^{-1}$, instantaneous power is $12$ kW.

At non-zero speed, driving force from known power is F=P/vF=P/v; this feeds the next objective’s force equation.

Power is a rate, not energy or force. Do not use peak force with total time to claim average power, or divide by zero speed in F=P/vF=P/v.

Convert engine power into instantaneous acceleration on a hill

At speed $v>0$, driving force along motion isD=\frac Pv.ThenresolvealongthehillandapplyThen resolve along the hill and apply\sum F=ma.

For a car of mass $m$ moving uphill at angle $\theta$ against resistance $R$:\frac Pv-R-mg\sin\theta=ma,sosoa=\frac{P/v-R-mg\sin\theta}{m}.

Convert power units to watts, identify the instantaneous speed, compute D=P/vD=P/v, draw/resolve slope forces with one positive direction, then calculate aa and interpret its sign as velocity change—not automatically speed change.

If motion is downhill, reassign signs: the component of weight along the slope aids motion while resistance opposes it. Re-derive rather than memorising one sign pattern.

PP is not a force. The formula P/vP/v is not usable at v=0v=0, and resistance plus the weight component must not be omitted.