1.54PH1 · Section 1: Forces and motion

Work, energy and power

Key learnings

Work done: W = Fs (in joules)
Work done = force × distance moved in the direction of the force; $W = Fs$ ; unit joule (J). Two elements: a force acts AND there is displacement along the force. At an angle $\theta$ , use the component along motion: $W = Fs\cos\theta$ . 1 J = the work when 1 N moves its point of application 1 m in the force's direction. E.g. a 40 N push over 3 m does $W = 40 \times 3 = 120$ J; at 30° it is $40 \times 3 \times \cos 30° \approx 104$ J.
KE, GPE and conservation of energy
$\text{KE} = \tfrac{1}{2}mv^2$ ( $m$ in kg, $v$ in m/s); $\text{GPE} = mgh$ ( $g = 9.8$ N/kg, $h$ in m); both in joules. Conservation: energy is only transferred between stores, never created or destroyed. Ideally GPE lost = KE gained: $\tfrac{1}{2}mv^2 = mgh$ ; with friction some goes to the thermal store, so less KE is gained. The eight stores: kinetic, gravitational potential, elastic potential, thermal, chemical, nuclear, magnetic, electrostatic.
Power and efficiency
Power = rate of energy transfer (rate of doing work); $P = E/t = W/t$ ; unit watt (W); 1 W = 1 J/s. When a force $F$ moves at constant velocity $v$ , $P = Fv$ . Efficiency = useful energy output ÷ total energy input, as a fraction or ×100%. It always lies between 0 and 100% — over 100% means the ratio is inverted. E.g. 600 J useful from 1000 J: $\tfrac{600}{1000} \times 100 = 60\%$ .

Common mistakes

Drawn from real examiner reports.

Not squaring the speed in KE = ½mv²

In $\text{KE} = \tfrac{1}{2}mv^2$ the speed must be squared before multiplying. Writing $\tfrac{1}{2}mv$ (no square) or $(\tfrac{1}{2}mv)^2$ is wrong. For 2 kg at 6 m/s: $\tfrac{1}{2} \times 2 \times 6^2 = \tfrac{1}{2} \times 2 \times 36 = 36$ J. Forgetting the square gives $\tfrac{1}{2} \times 2 \times 6 = 6$ J — six times too small. Square $v$ first, then multiply.

November 2024 Paper 1P Q2(a)(ii): the examiner explicitly noted that a significant number of candidates did not square the speed when calculating kinetic energy, giving an incorrect evaluation.

Treating a power (W) as an energy (J)

"Power is energy" is wrong — it omits time. The mark-scheme form is "rate of energy transfer"; $P = E/t$ , unit W. In calculations, do not feed a heater's power rating straight into $E = mc\Delta\theta$ : find the energy first, $E = Pt$ . A 2 kW heater for 5 minutes transfers $E = 2000 \times 300 = 600\,000$ J, NOT 2000 J. Always check: is the value in W or J?

November 2024 Paper 2P Q5(b): many candidates used the power of the heater without multiplying by time at all, scoring a maximum of one mark because they treated a power value as an energy value.

W = Fs with distance left in cm

The joule is N·m, so the distance must be in metres. Leave it in cm and the answer comes out in N·cm — not a valid unit, so the unit mark is lost. Convert first: 80 cm = 0.80 m, so $W = 50 \times 0.80 = 40$ J. Using 80 gives $50 \times 80 = 4000$ — 100 times too large and wrongly-unitted. The same applies to $W = Fs\cos\theta$ .

June 2024 Paper 2P Q2(c)(i)-(ii): a sizable minority of candidates forgot to convert the distance from cm to m before calculating work done (W = Fs), so the answer was not in joules. The examiner explicitly noted: "Check what unit is on the answer line."

Efficiency over 100% (ratio inverted)

Efficiency = useful output ÷ total input, so it can never exceed 100% — a machine cannot give out more useful energy than it takes in. An answer like 130% means total input was put on top; swap numerator and denominator. E.g. useful 500 J from 650 J input is $\tfrac{500}{650} \times 100 \approx 77\%$ , not 130%.

Using sin θ instead of cos θ for W

For a force at angle $\theta$ to the motion, the work is $W = Fs\cos\theta$ — the component ALONG the displacement. Using $\sin\theta$ takes the perpendicular component, which does no work. Check: a purely sideways force ( $\theta = 90°$ ) gives $\cos 90° = 0$ , i.e. zero work. Draw the force to see which component lies along the motion.

Height in GPE measured along the slope

In $\text{GPE} = mgh$ , $h$ is the vertical height gained, not the distance travelled along a ramp or slope. Using the slope length overstates the GPE. On an incline, find the vertical rise (often slope length × sin of the incline angle) before substituting into $mgh$ .

Exam tips

Energy scaffold: name stores, then calculate
Multi-step energy questions score per stage: name the store that falls and the one that rises, state energy is conserved (or lost to thermal), then compute. For efficiency, state which output is "useful" before writing the formula, or the mark cannot be awarded.
GPE lost = KE gained — carry the value
On a frictionless fall or ramp, take the KE straight from the GPE value instead of recomputing it from other data. The mass usually cancels ( $gh = \tfrac{1}{2}v^2$ ), so the speed at the bottom is independent of mass — a quick check on your answer.
Check the unit on the answer line
Before writing the final answer, check its unit: power in W (multiply P by t for energy in J), work in J (distance in m, not cm), efficiency as a %. A wrong or missing unit loses the mark even when the number is correct.

Revision notes