use the concept of efficiency to solve problems

Energy Conservation, Work, Power & Efficiency (Cambridge International AS & A Level Physics 9702)

1 Quantities, Units & Uncertainties

• Work (W) – scalar; \(W = F\,s\) (force × displacement, when the force is parallel to the motion).
  SI unit: joule (J) = N·m = kg m² s⁻².
• Energy (E) – the ability to do work; for a closed system the total energy is conserved.
  SI unit: joule (J).
• Power (P) – rate of energy transfer.
  \[ P = \frac{E}{t}= \frac{W}{t}=F\,v \] SI unit: watt (W) = J s⁻¹.
• Efficiency (η) – ratio of useful output to input (dimensionless). Expressed as a decimal in calculations or as a percentage (×100 %).
• All measured quantities must be quoted to the correct number of significant figures (SF).
  • When multiplying or dividing, the result is limited to the smallest number of SF in the data.
  • When adding or subtracting, the result is limited to the smallest number of decimal places.
• For experimental efficiency work, propagate relative uncertainties: \[ \frac{\Delta \eta}{\eta}= \sqrt{\left(\frac{\Delta E_{\text{out}}}{E_{\text{out}}}\right)^{2} +\left(\frac{\Delta E_{\text{in}}}{E_{\text{in}}}\right)^{2}} \] (or the equivalent expression using powers).

2 Work, Energy & the Conservation Law

• Work–energy theorem: the net work done on a particle equals the change in its kinetic energy, \[ W_{\text{net}} = \Delta E_{k}. \]
• Law of Conservation of Energy: In an isolated system the total energy remains constant; energy can change form but cannot be created or destroyed.

  

• Energy‑transfer diagrams are a useful way of visualising where energy enters, leaves, and is stored in a system.

3 Kinetic & Gravitational Potential Energy

Both are forms of mechanical energy. In the absence of non‑conservative forces (friction, air resistance) the sum \(E_{k}+E_{p}\) is constant.

4 Power

• Two equivalent forms: \[ P = \frac{E}{t}\qquad\text{and}\qquad P = F\,v \] Use the first when energy and time are known; use the second when a constant force moves an object at constant speed.
• Rated power of a device (e.g. a 500 W motor) is the maximum continuous power it can deliver. The actual average power may be lower depending on load and operating conditions.

5 Efficiency – Definition & General Formulae

Because power is the rate of energy transfer, the same efficiency can be written either way:

6 Using Efficiency in Problem Solving (AO2)

1. Identify the type of quantity supplied – energy (E_in) or power (P_in).
2. Determine the required useful output. This may involve:
   • Mechanical work: \(W = F\,s\) or \(W = mgh\) or \(W = \tfrac12 mv^{2}\).
   • Thermal energy: \(Q = mc\Delta T\) or \(Q = mL\) (phase change).
   • Electrical energy: \(E = VIt\) or \(P = VI\).
3. Read or calculate the efficiency (convert % to a decimal).
4. Apply the appropriate relation:
   • Energy form: \(E_{\text{out}} = \eta\,E_{\text{in}}\) or \(E_{\text{in}} = \dfrac{E_{\text{out}}}{\eta}\).
   • Power form: \(P_{\text{out}} = \eta\,P_{\text{in}}\) or \(P_{\text{in}} = \dfrac{P_{\text{out}}}{\eta}\).
5. If time is required, connect power and energy with \(P = E/t\).
6. Check the answer against the conservation of energy and evaluate whether the calculated loss is realistic.
7. Report the final result with the correct number of significant figures and appropriate units.

7 Worked Examples

7.1 Electrical Device – Electric Kettle

Given: Power rating 1500 W, efficiency 80 %, mass of water 0.500 kg, temperature rise 20 °C → 100 °C (ΔT = 80 K), specific heat capacity of water \(c = 4186\;\text{J kg}^{-1}\text{K}^{-1}\).

1. Useful energy needed: \[ E_{\text{useful}} = mc\Delta T = (0.500)(4186)(80) = 1.67\times10^{5}\;\text{J}\;(3\;\text{SF}) \]
2. Input energy (η = 0.80): \[ E_{\text{in}} = \frac{E_{\text{useful}}}{\eta} = \frac{1.67\times10^{5}}{0.80}=2.09\times10^{5}\;\text{J} \]
3. Time from rated power: \[ t = \frac{E_{\text{in}}}{P} = \frac{2.09\times10^{5}}{1500}=1.40\times10^{2}\;\text{s} \approx 140\;\text{s}\;(2\;\text{SF}) \]

7.2 Mechanical Device – Motor‑Driven Lift

Given: Motor rated 500 W, efficiency 70 %, load 200 kg, lift height 5 m.

1. Useful work (increase in GPE): \[ W_{\text{useful}} = mgh = (200)(9.81)(5)=9.81\times10^{3}\;\text{J} \]
2. Input energy: \[ E_{\text{in}} = \frac{W_{\text{useful}}}{0.70}=1.40\times10^{4}\;\text{J} \]
3. Time: \[ t = \frac{E_{\text{in}}}{P_{\text{in}}} = \frac{1.40\times10^{4}}{500}=2.8\times10^{1}\;\text{s} \approx 28\;\text{s} \]

7.3 Thermal Device – Simple Heat Engine (Carnot‑like)

Given: Hot reservoir at \(T_{\text{h}} = 500\;\text{K}\), cold reservoir at \(T_{\text{c}} = 300\;\text{K}\). The engine absorbs 5.0 kJ of heat from the hot reservoir each cycle.

1. Maximum (Carnot) efficiency: \[ \eta_{\text{Carnot}} = 1-\frac{T_{\text{c}}}{T_{\text{h}}} = 1-\frac{300}{500}=0.40\;(40\%) \]
2. Maximum useful work per cycle: \[ W_{\text{max}} = \eta_{\text{Carnot}}\,Q_{\text{in}} =0.40\times5.0\times10^{3}=2.0\times10^{3}\;\text{J} \]
3. Heat rejected to the cold reservoir: \[ Q_{\text{out}} = Q_{\text{in}}-W_{\text{max}} = 5.0\times10^{3}-2.0\times10^{3}=3.0\times10^{3}\;\text{J} \]

This example links the efficiency concept to the thermodynamics part of the syllabus (heat engines, Carnot efficiency).

8 Efficiency of Non‑Mechanical Devices (AO1)

• Transformer: Efficiency = \(\dfrac{P_{\text{out}}}{P_{\text{in}}}\times100\%\). Typical values 95–99 % because losses are mainly core hysteresis and copper resistance.
• Heat‑engine (e.g., internal‑combustion engine): η = useful mechanical work ÷ chemical energy of fuel. Real engines achieve 20–35 %.
• Refrigerator / heat‑pump: Use the coefficient of performance (COP) rather than η, but the idea of useful output (heat removed) versus electrical input is analogous.

9 Practical Activity – Measuring the Efficiency of a Small DC Motor (AO3)

1. Objective: Determine the efficiency of a 12 V, 2.0 A DC motor under three different loads.
2. Apparatus: DC motor, variable resistive load (e.g. a set of known masses lifted by a pulley), digital voltmeter, ammeter, stopwatch, ruler, balance.
3. Method:
   1. Measure the supply voltage (V) and current (I) while the motor lifts each load at constant speed; calculate input power \(P_{\text{in}} = VI\).
   2. Record the time (t) taken to raise the load through a known height (h). Compute useful work \(W_{\text{out}} = mgh\) and output power \(P_{\text{out}} = W_{\text{out}}/t\).
   3. Calculate efficiency for each load: \(\eta = P_{\text{out}}/P_{\text{in}}\times100\%\).
   4. Estimate uncertainties in V, I, m, h, and t, then propagate them to obtain \(\Delta\eta\) using the formula given in Section 1.
4. Data‑handling:
   • Tabulate all measured quantities with uncertainties.
   • Plot η (%) against load (kg) and comment on the trend.
5. Typical observation: η rises with load because frictional and electrical losses are roughly constant, so they form a smaller fraction of the total input power at higher useful output.

10 Data‑Handling Task (AO2) – Example Table

Students should:

• Calculate η for each row (already shown).
• Draw a graph of η (%) vs. load (kg).
• Explain why efficiency improves with increasing load, referencing constant frictional/electrical losses.

11 Common Pitfalls

• Forgetting to convert a percentage to a decimal before using it in a calculation.
• Assuming 100 % efficiency for real devices; always check the specification or experimental value.
• Mixing up energy (J) and power (W); remember \(P = E/t\).
• Neglecting phase‑change energy when heating/cooling (e.g., boiling water requires latent heat of vaporisation).
• Using the rated power of a motor as the average power without confirming the operating condition.
• Incorrect handling of significant figures or uncertainties.

12 Summary

• Work, energy, and power are fundamental scalar quantities; the work–energy theorem links force, displacement and kinetic energy.
• Energy is conserved; it can appear as kinetic, gravitational potential, thermal, electrical, etc.
• Efficiency quantifies how well a real device converts input energy (or power) into the desired useful form. It is always < 100 %.
• Problem solving follows a clear sequence: identify inputs, determine the required useful output, apply the efficiency relation, use \(P = E/t\) if needed, and check the result against physical expectations.
• Experimental work requires careful measurement, uncertainty propagation, and data presentation (tables, graphs, trend analysis).
• Understanding efficiency links the idealised conservation law to the performance of everyday devices—electric kettles, lifts, transformers, heat engines, and more.