5.1 Energy conservation
- Syllabus
- 9702–2028–2029
- Topic
- 5.1
- Level
- AS
Learning objectives
5.1.1The concept of work, and recall and use work done = force × displacement in• understand the concept of work, and recall and use work done = force × displacement in the direction of the force5.1.2The principle of conservation of energy• recall and apply the principle of conservation of energy5.1.3That the efficiency of a system is the ratio of useful energy output• recall and understand that the efficiency of a system is the ratio of useful energy output from the system to the total energy input5.1.4The concept of efficiency to solve problems• use the concept of efficiency to solve problems5.1.5Power as work done per unit time• define power as work done per unit time5.1.6P = W / t• solve problems using P = W / t5.1.7P = Fv and use it to solve problems• derive P = Fv and use it to solve problems
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/totalinputenergy
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=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.