Energy, work and power

• Energy for “how fast, how far”; power for “how quickly”. Count the weight once — as potential energy or as work, never both — and measure \(h\) vertically.
• From MF19 you may still need \(v^2 = u^2 + 2as\) and its three companions for a constant-force check — but never for a vehicle at constant power.
• If three things come back instantly on day 30, the chapter has stuck: \(W = Fd\cos\theta\); work by the applied forces = gain in KE + gain in PE + work against resistances; and \(D = P/v\) into \(F = ma\), for an acceleration at one instant only.

• 4.5.1 I can understand — 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
• 4.5.2 I can understand — understand the concepts of gravitational potential energy and kinetic energy, and use appropriate formulae
• 4.5.3 I can understand — 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
• 4.5.4 I can use — 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
• 4.5.5 I can — solve problems involving, for example, the instantaneous acceleration of a car moving on a hill against a resistance

Why a second method at all? \(F = ma\) needs the forces along one straight line and gives the acceleration; the constant-acceleration formulae then need that acceleration to be constant. Energy needs neither: it compares the start and the end, so it works on a curved path where the forces change direction all the way down, and it never asks for the time.

• “The slide is 6 m long, so \(h = 6\).” Correct \(h\) is the vertical drop between the start and the end, never the length of the slope or the slide. On a straight slope of length \(d\) at \(\alpha\), \(h = d\sin\alpha\); on a curved slide, \(h\) is given as a height (section C, figure 2).
• “The acceleration at \(15\ \mathrm{m\,s^{-1}}\) is \(0.433\ \mathrm{m\,s^{-2}}\), so after 10 s the speed is \(15 + 4.33\).” Correct An acceleration found from \(D = P/v\) is instantaneous: as the speed rises, \(D\) falls, so the acceleration falls too. The constant-acceleration formulae must never be used for a vehicle working at constant power (section E).
• “Work done by the rope = \(80 \times 50\).” Correct Only the component along the displacement does work: \(W = Fd\cos\theta = 80 \times 50 \times \cos 25^\circ\) (section A).
• “Loss of PE \(mgh\), plus the work done by gravity \(mgh\)…” Correct The weight is counted once: either as a change in potential energy or as the work done by gravity, never both. This chapter uses potential energy throughout (section C).
• “The normal force pushes the child along the slide, so it does work.” Correct A normal force is perpendicular to the motion at every instant, so \(\cos 90^\circ = 0\) and it does no work, however the slide curves (section A).
• “\(P = 24\) kW, so \(D = 24/12 = 2\) N.” Correct \(P = Fv\) needs watts: 24 kW = 24 000 W, so \(D = 24\,000/12 = 2000\) N (section D).
• “Work done by the rope \(= Fd = 80 \times 50\)”, for a rope at \(25^\circ\) to the motion. Repair Only the component along the displacement does work: \(W = Fd\cos\theta = 80 \times 50 \times \cos 25^\circ = 3630\) J.
• Using the length of a curved slide, or of a slope, as \(h\). Repair \(h\) is the vertical drop between the start and the end. On a straight slope of length \(d\) at \(\alpha\), \(h = d\sin\alpha\).
• Including the work done by the normal force in an energy equation. Repair The normal force is perpendicular to the motion at every instant and does no work.
• Counting gravity twice: \(mgh\) as a loss of PE and as work done by the weight. Repair Use one or the other. This chapter always uses potential energy and leaves the weight out of the work terms.
• “\(\text{KE} = mv^2\)”, or a KE change of \(\tfrac12 m(v - u)^2\). Repair \(\text{KE} = \tfrac12 mv^2\), and the change is \(\tfrac12 m(v^2 - u^2)\).
• “Work done against friction \(= \mu mg \times d\)” on a slope. Repair On a slope \(R = mg\cos\alpha\) (when nothing else pushes into it), so the friction is \(\mu mg\cos\alpha\) and the work against it is \(\mu mg\cos\alpha \times d\).
• Adding a loss of kinetic energy as if it were more work the cyclist must supply. Repair A loss of kinetic energy supplies energy: its “gain” is negative, so it is subtracted. In worked example 8, \(16\,000 - 1560 + 12\,000\).
• “\(P = Fv\), so \(D = 800 \times 30\)”, when asked for the driving force at \(12\ \mathrm{m\,s^{-1}}\). Repair The driving force at a stated speed is \(D = P/v\) at that speed: \(24\,000/12 = 2000\) N.
• Using \(v = u + at\) to find how long a car at constant power takes to reach a speed. Repair \(D = P/v\) changes with \(v\), so \(a\) is not constant; \(F = ma\) gives \(a\) at one instant only, and no constant-acceleration formula applies.
• Forgetting the weight component on a hill: \(D - \text{resistance} = ma\). Repair Up the hill, \(D - mg\sin\alpha - \text{resistance} = ma\); down the hill, \(D + mg\sin\alpha - \text{resistance} = ma\). The normal contact force \(R\) is perpendicular to the hill and never appears in this equation.
• Power in kilowatts substituted as 24 into \(P = Fv\). Repair 24 kW = 24 000 W. Convert before substituting.
• Using \(g = 9.8\). Repair The syllabus says questions use \(g = 10\ \mathrm{m\,s^{-2}}\), and so does every answer in this chapter.
• “The normal force on the smooth slide slows the child down.” Repair On a smooth slide only the weight does work, so KE + PE is conserved. The normal force changes the direction of motion, never the speed.

• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.
• Say which forces do work before you calculate. A one-line list — “\(R\) is perpendicular to the motion, so does no work; friction does work \(-Fd\)” — is part of the method, and it guards against two errors: including the normal force, and forgetting friction.
• Two numbers, two different jobs. The stated speed goes into \(D = P/v\) for the acceleration; the greatest speed is found from \(a = 0\) and has nothing to do with the stated speed. Writing “at this instant” beside the acceleration shows that you know it is not constant.
• Interleave with the chapters that use this one. This is the last Paper 4 chapter, so interleave backwards: redo chapter 21’s rough-plane example by energy (it is worked example 4 here), and re-answer chapter 19’s constant-acceleration drills asking which of them an energy equation would have solved faster. Recalling a method inside a new problem is worth more than another pass over this chapter on its own: the syllabus says an individual examination question may involve ideas and methods from more than one section of the content for that paper, so nothing here is ever finished with.

• Cambridge International AS & A Level Mathematics 9709 has six components, and a candidate takes two of them for the AS Level and four for the A Level. This chapter is Mechanics content, examined in Paper 4. Paper 4 (Mechanics) is 40% of an AS Level that includes it and 20% of an A Level that includes it. It assumes the Paper 1 content. An A Level route that includes Paper 4 takes Papers 1, 3, 4 and 5; Paper 4 cannot be combined with Paper 6. Every paper is a written examination of compulsory structured questions, answered on the question paper, with MF19 (the list of formulae and statistical tables) supplied. Examinations are available in the June and November series, and in March in India.
• Across the whole qualification the assessment objectives are weighted AO1 55% (knowledge and understanding: concepts, terminology, notation and accurate manipulative technique) and AO2 45% (application and communication: choosing the procedure, combining techniques to solve problems, and presenting the work clearly and logically) at AS Level, and AO1 52%, AO2 48% at A Level. AS candidates are graded a–e; A Level candidates A*–E.
• A work–energy question: a particle on a slope or a curved path, with the speed, distance or work against a resistance to find; a part may be set as a show that for one work term, as in this chapter's worked example 8. A power question: average power from work ÷ time, or \(P = Fv\) at a stated speed. A vehicle question: the acceleration at one instant and the greatest speed, on the level or on a hill. The reasoning sits in the energy equation, or the equation of motion, written before its numbers.
• MF19's Mechanics entry is the four constant-acceleration formulae of chapter 19. \(W = Fd\cos\theta\), \(\tfrac12 mv^2\), \(mgh\), the work–energy principle, conservation of energy and \(P = Fv\) must all be known, as must \(g = 10\ \mathrm{m\,s^{-2}}\). The MF19 card lists them.
• Energies in J (or kJ), power in W: convert kW first. Keep four or more significant figures through the working and round once to 3 s.f.; leave an exact answer exact (\(200/9\), \(\sqrt6\)). Name the zero level and the forces that do work, and write the energy equation with its terms before substituting: an unsupported number earns nothing.
• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.