CourseCourseQuestionQuestionDocumentDocument

5. Work, energy and power

Syllabus
9702–2028–2029
Section
5
Level
AS

5.1 Energy conservation

Syllabus
9702–2028–2029
Topic
5.1
Level
AS

Work transfers energy when a force causes displacement in its direction

For a constant force, work done is W=Fs when the displacement s is along the force. More generally W=Fs cosθ for angle θ.

Work is energy transferred, so choose the force and displacement pair and use the component of force along the displacement.

A 40 N force moving a box 3.0 m in its direction transfers 120 J; a perpendicular force transfers zero work.

A force can act without doing work if there is no displacement or if it is perpendicular to the motion.

Energy is conserved while it is transferred between stores and pathways

The principle of conservation of energy says total energy is constant: energy is transferred between stores, not created or destroyed.

Define the system boundary and identify where energy enters, leaves or changes store. Include useful and dissipated pathways in a complete account.

A falling object transfers gravitational potential energy mainly to kinetic energy, then to thermal energy through drag.

“Wasted” energy has not vanished; it has usually spread into less useful thermal stores.

Efficiency compares useful output energy with total input energy

efficiency=usefuloutputenergy/totalinputenergyefficiency = useful output energy / total input energy

Useful output is the energy transferred in the intended way. Other output is not destroyed; it is transferred to less useful stores, often heating the system or surroundings.

A motor receives 500 J and transfers 350 J usefully to mechanical energy. Efficiency = 350/500 = 0.70 = 70%. The remaining 150 J is transferred by non-useful pathways.

Use useful output, not total output, in the numerator. Efficiency is a ratio with no unit and cannot exceed 1, or 100%, for the complete energy account.

Rearrange efficiency to find useful output, input or dissipation

Write efficiency as a decimal, then use η = Euseful/Einput. Hence Euseful = ηEinput and Einput = Euseful/η. Non-useful energy = input − useful output.

A machine is 80% efficient and must deliver 2.0 kJ of useful energy. Einput = 2.0/0.80 = 2.5 kJ, so 0.5 kJ is transferred by non-useful pathways.

When efficiency is below 100%, total input must be greater than useful output. Do not multiply by 100 twice, and do not put dissipated energy in the denominator.

Power measures the rate at which work is done or energy is transferred

Power is the rate of energy transfer: P=W/t=∆E/t, measured in watts, where 1 W=1 J s⁻¹.

Use the energy transferred over the relevant time interval; average power need not equal instantaneous power when the rate changes.

A device transferring 900 J in 30 s has average power 30 W.

Power is not the same as total work: two machines can do the same work in different times and therefore have different powers.

Use P=W/t when work is transferred over a known time

For a process with work W completed in time t, average power is P=W/t. Rearranging gives W=Pt or t=W/P.

Keep time in seconds for watts and identify whether W is useful work or total input work before substituting.

A 1.2 kW lift doing 36 kJ of work takes 30 s if its power is constant.

A larger power does not mean more total work unless the operating time is also considered.

Derive power as force times velocity along the force

For a constant force F acting along displacement s: work W = Fs. Power P = W/t = Fs/t. Since velocity v = s/t, P = Fv.

P=FvP = Fv

F must be the force component parallel to the velocity. The relation gives the instantaneous mechanical power transferred by that force; if force and velocity are perpendicular, that force transfers zero power.

A 400 N driving force parallel to a vehicle moving at 15 m s⁻¹ supplies P = 400 × 15 = 6000 W = 6.0 kW.

P = Fv is an energy-transfer-rate relation, not P = ma. Use the specified driving force or parallel component, not automatically the resultant of unrelated forces.

5.2 Gravitational potential energy and kinetic energy

Syllabus
9702–2028–2029
Topic
5.2
Level
AS

Derive gravitational potential energy change from work

Raise a mass m slowly through vertical height Δh in a uniform gravitational field, so the upward applied force equals its weight mg and the displacement is along that force.

Work done W = Fs. Substitute F = mg and s = Δh: W = mgΔh. This work transfers energy to the gravitational potential store, so ΔEP = W = mgΔh.

ΔEP=mgΔhΔEP = mgΔh

The derivation uses vertical height change, not path length, and assumes g is uniform. Raising gives positive ΔEP; lowering gives negative ΔEP when Δh is signed.

Use ∆E_P=mg∆h for a uniform gravitational field near Earth

In a uniform field, the change in gravitational potential energy is ∆E_P=mg∆h, with g treated as constant over the height interval.

Define the reference level and keep the sign of ∆h consistent. Only differences in potential energy affect energy conservation calculations.

A 0.50 kg mass lowered 4.0 m has ∆E_P=−19.6 J relative to its starting level; that energy can become kinetic or thermal.

Zero potential at the floor is a choice, not a physical claim that the object has no energy anywhere else.

Derive kinetic energy from resultant work and motion

Let a constant resultant force F accelerate a constant mass m through displacement s, changing its speed from u to v. The resultant work is W = Fs and F = ma, so W = mas.

From v² = u² + 2as, as = (v² − u²)/2. Substitute into W = mas: W = ½m(v² − u²) = ½mv² − ½mu².

Resultant work equals the change in kinetic energy, so EK = ½mv² relative to rest, and ΔEK = ½mv² − ½mu² for a speed change.

Kinetic energy uses speed squared and is scalar. The derivation assumes constant mass; signs of velocity disappear only after the vector dynamics have established the speed change.

Kinetic energy is the energy of motion, E_K=½mv²

The kinetic-energy store of a mass m moving at speed v is E_K=½mv². It is a scalar and is never negative.

Use speed magnitude and consistent units; doubling speed quadruples kinetic energy, while doubling mass doubles it.

A 4.0 kg trolley moving at 3.0 m s⁻¹ has kinetic energy 18 J.

Kinetic energy does not carry the direction sign of momentum, and stopping does not destroy it—it transfers it to other stores.