1.70625 · Section 1: Motion, Forces and Energy

Energy, work and power

Key learnings

Work done: W = Fd, in joules (J)
Work done = force × distance moved in the direction of the force: $W = Fd$ ; unit joule (J). Two elements: a force acts, and displacement is in the direction of that force — a perpendicular force does no work. 1 J is the work done when a force of 1 N moves its point of application 1 m. Cambridge accepts $W = Fs$ . Work done equals the energy transferred between stores. Mini-example: 25 N pushing a box 4 m does $W = 25 \times 4 = 100$ J.
GPE = mgh and KE = ½mv²; energy is conserved
GPE (gravitational potential energy) $= mgh$ , with $g = 10$ N/kg (Cambridge); unit J. KE (kinetic energy) $= \tfrac{1}{2}mv^2$ — the speed $v$ must be squared. Conservation of energy: energy is never created or destroyed, only transferred between stores. As an object falls, the gravitational store transfers to the kinetic store (plus thermal if friction acts); with no friction $mgh = \tfrac{1}{2}mv^2$ , so $v = \sqrt{2gh}$ and mass cancels.
Power P = E/t (W); efficiency is a ratio
Power = rate of energy transfer: $P = E/t$ (or $P = W/t$ ); unit watt (W); 1 W = 1 J/s. Two elements: it is a rate, of energy transfer. (E) $P = Fv$ when a constant force acts at constant speed. Efficiency $= \dfrac{\text{useful energy output}}{\text{total energy input}} \times 100\%$ ; always between 0% and 100% — a value above 100% is impossible and means the ratio is inverted. Mini-example: a 500 W motor running 4 s transfers $E = Pt = 2000$ J.

Common mistakes

Drawn from real examiner reports.

Work done is W = Fd — never divide

Work done has two parts: a force AND the distance moved in the direction of the force. Wrong forms in Cambridge mark schemes — distance ÷ force, force ÷ distance, distance ÷ time — all score zero. It is always $W = Fd$ (multiply). Second trap: going up stairs, use the vertical height gained for the GPE change, not the slope length.

June 2024 examiner report (0625 s24_er): "Many candidates struggled to recall the quantities needed to calculate the work done on the box. Most candidates incorrectly believed that the width of the stairs was required in the calculation." Also: most common wrong attempts were distance ÷ force, work done = force ÷ distance, or work done = distance ÷ time (Paper 2 Q1(b)(ii)).

Forgetting to square v in KE = ½mv²

In $\text{KE} = \tfrac{1}{2}mv^2$ the speed must be squared. Writing $\tfrac{1}{2}mv$ (no square) gives an answer $v$ times too small: for $m = 2$ kg at $v = 6$ m/s the correct value is $\tfrac{1}{2} \times 2 \times 36 = 36$ J, not 6 J. Squaring the whole bracket, $(\tfrac{1}{2}mv)^2$ , is also wrong. Always write the unit J — omitting it loses the final mark.

November 2023 examiner report (0625 w23_er): "Most candidates remembered the formula for kinetic energy correctly. When substituting values into the equation, a significant number of candidates did not square the velocity … Weaker candidates either omitted the unit, gave an incorrect unit or were unable to recall the correct formula."

Name the store in full, not just "potential"

Cambridge names its energy stores precisely. "Potential energy" alone is ambiguous (elastic? electrostatic?) and is not credited — write "gravitational potential energy" or GPE. Use "thermal (internal) energy", not "heat"; a compressed spring stores "elastic (strain) potential energy", not "spring energy".

November 2024 mark scheme (0625_w24_ms_41): "from kinetic (energy store) … to internal / thermal (energy store as final store)" — the word "store" is required or implied. June 2024 examiner report: "Weaker candidates only referred to potential energy instead of gravitational potential energy and there were incomplete or incorrect statements about the energy transfer when the ball hit the ground."

Power (W) and energy (J) are not the same

Power (watt, W) is the rate of energy transfer; energy (joule, J) is the total amount transferred. A 500 W motor running for 4 s transfers $E = Pt = 500 \times 4 = 2000$ J. Substituting a power value where an energy value is needed (or the reverse) is a common multi-mark slip. From $P = E/t$ : $E = Pt$ and $t = E/P$ .

Use g = 10 N/kg, not 9.8

Cambridge 0625 uses $g = 10$ N/kg as standard. Using 9.8 (the Edexcel value) makes GPE answers slightly wrong: a 5.0 kg mass raised 2.0 m gains $mgh = 5.0 \times 10 \times 2.0 = 100$ J, not 98 J. Only use a different value if the question states one. The unit of $g$ is N/kg (equivalent to m/s²).

Convert time to seconds before P = E/t

The formula $P = E/t$ needs the time in seconds. A crane transferring 45 000 J in 3.0 minutes has $t = 3.0 \times 60 = 180$ s, so $P = 45\,000 / 180 = 250$ W. Substituting 3.0 directly gives 15 000 W — 60 times too large. Always convert minutes to seconds before dividing.

Exam tips

Name the store, the direction, then calculate
Marks come step by step. To "describe the energy transfers": name the initial store, then the final store(s), say which increases or decreases, invoke conservation, then calculate. Write each formula in symbols, substitute with units, and state a unit.
Show working — method marks even if wrong
"Show that" and "Calculate" credit the method: write the equation in symbols, substitute with units, then evaluate. Correct working earns marks even when the final answer is wrong, but a bare wrong answer scores zero. To change an answer, cross the old one out.
Check efficiency is between 0% and 100%
Write efficiency as $\dfrac{\text{useful output}}{\text{total input}} \times 100\%$ and check it is between 0% and 100%. Above 100% means the fraction is inverted — swap numerator and denominator. Keep the "×100%" and write "%" for a percentage.

Revision notes