Work, energy and power

• Work counts only the displacement along the force; energy is never lost, only dissipated; and \(F\) in \(P = Fv\) is the driving force, not the resultant.

• 5.1.1 I can understand — understand the concept of work, and recall and use work done = force × displacement in the direction of the force
• 5.1.2 I can recall — recall and apply the principle of conservation of energy
• 5.1.3 I can recall — recall and understand that the efficiency of a system is the ratio of useful energy output from the system to the total energy input
• 5.1.4 I can use — use the concept of efficiency to solve problems
• 5.1.5 I can define — define power as work done per unit time
• 5.1.6 I can — solve problems using P = W / t
• 5.1.7 I can — derive P = Fv and use it to solve problems
• 5.2.1 I can — derive, using W = Fs, the formula ΔEP = mgΔh for gravitational potential energy changes in a uniform gravitational field
• 5.2.2 I can recall — recall and use the formula ΔEP = mgΔh for gravitational potential energy changes in a uniform gravitational field
• 5.2.3 I can — derive, using the equations of motion, the formula for kinetic energy EK = ½mv²
• 5.2.4 I can recall — recall and use EK = ½mv²

Units, significant figures and working are part of the physics. Give a calculated answer to the same number of significant figures as the least precise data, or one more; keep full precision in the working and round only at the end; write the unit with every final answer. A fifth of the qualification is experimental: Papers 3 and 5 test AO3 only, and their questions may be set in contexts outside the syllabus content, so the practical work in this chapter is set out as method, recording, graphs and uncertainties rather than as theory.

• “Work done = force × distance moved.” Correct Work done = force × displacement in the direction of the force. For a force at angle \(\theta\) to the displacement, only the component \(F\cos\theta\) does work, so \(W = Fs\cos\theta\). A force at right angles to the motion does no work. A force opposing the motion does negative work.
• “Energy is lost to friction.” Correct Energy cannot be destroyed. Friction and air resistance transfer it, or dissipate it, to thermal energy of the surfaces and the surroundings, where it is no longer useful. Write "dissipated" or "transferred to thermal energy", never "lost".
• “\(\Delta E_P = mg \times\) the distance travelled along the slope.” Correct \(\Delta h\) is the change in vertical height. The weight acts vertically, so only the vertical part of the displacement is in the direction of the force. The slope length is used only for the work done against a resistive force that acts along the slope.
• “In \(P = Fv\), \(F\) is the resultant force, so a car at constant speed needs no power.” Correct \(F\) is the driving force that does the work. At constant velocity the driving force equals the total resistive force, so \(P = (\text{resistive force}) \times v\), even though the resultant force is zero.
• “Efficiency = 0.45 J” or “the efficiency is 115%.” Correct Efficiency is a ratio of two energies, or of two powers, so it has no unit. The useful output can never exceed the input, so the efficiency is never greater than 1 (100%).
• “Doubling the speed doubles the kinetic energy.” Correct \(E_K \propto v^2\), so doubling the speed quadruples the kinetic energy, and it quadruples the braking distance too.
• “The three energy formulas are on the Data sheet.” Correct None of \(W = Fs\), \(P = W/t\), \(P = Fv\), efficiency, \(\Delta E_P = mg\Delta h\) or \(E_K = \tfrac{1}{2}mv^2\) is printed there. They are recall, and three of them you must also be able to derive.
• “Carrying a heavy bag across a level room does a lot of work on it.” Repair Work is force × displacement in the direction of the force. The upward force on the bag is perpendicular to its horizontal displacement, so it does no work on the bag. Your muscles do transfer energy internally, but not to the bag.
• “Work is always positive.” Repair A force opposing the displacement does negative work, \(W = Fs\cos 180^\circ = -Fs\). Friction on a sliding box takes kinetic energy out of it.
• “\(W = Fs\sin\theta\) for a force at \(\theta\) to the motion.” Repair The component along the displacement is \(F\cos\theta\), with \(\theta\) measured between the force and the displacement. Draw the angle, then choose: adjacent is cos.
• “Energy was lost to friction.” Repair Energy cannot be destroyed. It was transferred, or dissipated, to thermal energy of the surfaces and the surroundings.
• “The efficiency is 115%.” Repair The useful output cannot exceed the input (conservation of energy). An answer above 1 means the useful output and the input have been swapped or misidentified.
• “Efficiency = 0.45 J.” Repair A ratio of two energies has no unit. Write 0.45, or 45%.
• “A kilowatt-hour is a unit of power.” Repair It is a power multiplied by a time, so it is an energy: \(1\,\mathrm{kWh} = 3.6 \times 10^6\,\mathrm{J}\).
• “In \(P = Fv\), \(F\) is the resultant force, so a car at constant speed needs no power.” Repair \(F\) is the driving force. At constant speed it equals the resistive force, and \(P = (\text{resistive force}) \times v\).
• “To lift a load at constant speed, the lifting force must be bigger than its weight.” Repair At constant speed the acceleration is zero, so the resultant force is zero and the lifting force equals the weight. That is exactly the step the \(mg\Delta h\) derivation uses.
• “\(\Delta E_P = mg \times\) (distance along the slope).” Repair \(\Delta h\) is the vertical height change. The slope length is used only for the work done against a resistive force that acts along the slope.
• “Doubling the speed doubles the kinetic energy.” Repair \(E_K \propto v^2\), so doubling the speed quadruples the kinetic energy.
• “\(\Delta E_K = \tfrac{1}{2}m(v - u)^2\).” Repair Square each speed, then subtract: \(\Delta E_K = \tfrac{1}{2}m(v^2 - u^2)\). From 2 m s−1 to 14 m s−1 the bracket is 192 m2 s−2, not 144 m2 s−2.
• “\(E_K = \tfrac{1}{2}mv^2\) only holds for uniform acceleration, because that is how it was derived.” Repair The constant force was a convenience for the derivation. The result depends only on \(m\) and \(v\).
• “The main source of error was human error in timing.” Repair Name the quantity and the reason: the time was measured by hand over a short interval, so reaction time is a large fraction of it. Then improve it: use light gates connected to a timer, or a larger height.

• Read the command word before you decide how much to write. This syllabus has fifteen of them: calculate, comment, compare, define, describe, determine, explain, give, identify, justify, predict, show (that), sketch, state and suggest. Define wants a precise meaning — for a physical quantity, usually an equation in words with every quantity named. State and give want a fact and nothing more. Describe wants the points or the features. Explain wants the reasons and the relationships — a describe-level answer to an explain question is incomplete however well written it is. Show (that) gives you the result and asks for the structured evidence that leads to it, so every step must appear — and a final value worked to one more significant figure than the one printed makes it plain that you calculated it rather than copied it. Sketch wants a freehand graph with its key features — intercepts, asymptotes, the shape — correct, but no plotted scale.
• Say which force and which body. "The work done" is ambiguous. Write "the work done by the tension on the sledge". A question can ask about the work done by one force (the tension), by another (friction, which is negative), or by the resultant force, and the three numbers differ.
• Interleave with the chapters that use this one. Topic 6 (Deformation of solids) finds elastic potential energy as a work done: re-answer "define work done" there. Topic 9 (Electricity) uses power as a rate of energy transfer: re-do the motor efficiency item. Topic 13 (Gravitational fields) takes up the case where \(g\) is not uniform: re-answer "state the condition for \(\Delta E_P = mg\Delta h\)". Recalling a topic inside a new context is worth more than another pass over this chapter on its own; at A Level, Paper 4 assumes the whole of the AS content, so nothing here is ever finished with.

• Cambridge International AS & A Level Physics 9702 has five components. Topic 5 is AS Level content, so it is examined in Papers 1, 2 and 3. AS Level content: examined in Paper 1 (multiple choice), Paper 2 (AS structured) and, as practical context, Paper 3. Assumed knowledge for Papers 4 and 5. AS Level candidates take Papers 1, 2 and 3; A Level candidates take all five, either staged over two years (Papers 1–3 in year one, Papers 4 and 5 in year two) or together in one series. Examinations are available in the June and November series, and in March in India.
• Across both the AS Level and the A Level the assessment objectives are weighted AO1 40% (knowledge and understanding), AO2 40% (handling, applying and evaluating information) and AO3 20% (experimental skills and investigations). AS candidates are graded a–e; A Level candidates A*–E. The Data and formulas sheet is printed as page 2 of Papers 1 and 2 and as pages 2 and 3 of Paper 4: it gives the constants and a short list of formulas. Every other equation in this chapter is one the syllabus says you must recall, and this chapter says which is which.
• A Paper 1 item on this topic can turn on one choice: the component of a force along the displacement (\(\cos\theta\), not \(\sin\theta\)), the vertical height rather than the slope length, the driving force rather than the resultant in \(P = Fv\), or \(v^2\) rather than \(v\). A Paper 2 question may ask you to define work or power, to show that \(P = Fv\), \(\Delta E_P = mg\Delta h\) or \(E_K = \tfrac{1}{2}mv^2\) follows from \(W = Fs\), to state the principle of conservation of energy, or to explain where the energy went.
• The calculations are energy accounts: energy lost as \(mg\Delta h\) or \(\tfrac{1}{2}mv^2\), energy gained, energy dissipated, and a resistive force from the energy dissipated divided by the path length. They also include efficiency forwards and backwards, and \(P = Fv\) combined with \(F = ma\) for a vehicle. None of \(W = Fs\), \(P = W/t\), \(P = Fv\), efficiency, \(\Delta E_P = mg\Delta h\) or \(E_K = \tfrac{1}{2}mv^2\) is on the Data and formulas sheet; \(g\) and \(v^2 = u^2 + 2as\) are.
• Topic 5 has no describe-an-experiment outcome, but Papers 3 and 5 may set their skills in any context, including energy measurements. One Paper 3-style context is a motor lifting a load: vary the mass, time the rise between two marks with a stopwatch, read the current and p.d., and calculate the efficiency \(mgh/(VIt)\). Hand timing over a few seconds is the largest uncertainty. A Paper 5-style context, used in this chapter, is a block sliding down a ramp, with \(v^2\) plotted against \(h\) to find a resistive force.
• Read the command word before you decide how much to write. This syllabus has fifteen of them: calculate, comment, compare, define, describe, determine, explain, give, identify, justify, predict, show (that), sketch, state and suggest. Define wants a precise meaning — for a physical quantity, usually an equation in words with every quantity named. State and give want a fact and nothing more. Describe wants the points or the features. Explain wants the reasons and the relationships — a describe-level answer to an explain question is incomplete however well written it is. Show (that) gives you the result and asks for the structured evidence that leads to it, so every step must appear — and a final value worked to one more significant figure than the one printed makes it plain that you calculated it rather than copied it. Sketch wants a freehand graph with its key features — intercepts, asymptotes, the shape — correct, but no plotted scale.