t{Total Power Input}}$ (or equivalent using energy) B1\n- Multiplies input power by efficiency: $120 \\times 0.40$ OR divides useful energy by time in seconds: $\\frac{2880}{60}$ M1\n- Obtains the final answer of $48\\text{ kW}$ A1","marks":4,"order":0,"markingRubric":null,"explanation":{"methods":[{"id":"energy_and_efficiency","label":"Energy and efficiency","steps":[{"type":"text","order":1,"title":"Convert to input power","content":"Power is the rate at which energy is transferred. Here the energy is given in $\text{kJ}$ each minute, so divide by the $60\text{ s}$ in one minute:\n\n$$\\text{input power}=\\frac{7200\\text{ kJ}}{60\\text{ s}}=120\\text{ kW}$$\n\nThis earns the first **M1**. The units work directly because $\text{kJ s}^{-1}=\text{kW}$.","highlights":[{"text":"The fuel burned by the generator's engine supplies a total of $7200\\text{ kJ}$ of energy every minute.","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\htmlData{anchor=relation}{\\eta=\\frac{P_{\\mathrm{useful}}}{P_{\\mathrm{input}}}}"},{"text":"Using efficiency = useful power output / total power input earns the B1."},{"text":"Multiply both sides by the input power to rearrange:"},{"latex":"\\htmlData{anchor=rearranged}{P_{\\mathrm{useful}}=\\eta\\times P_{\\mathrm{input}}}"},{"latex":"P_{\\mathrm{useful}}=0.40\\times120\\ \\mathrm{kW}"}],"connections":[{"to":"rearranged","from":"relation","color":"green","label":"rearrange"}]},"order":2,"title":"Apply the efficiency","description":"Efficiency is the fraction of input power that becomes useful output."},{"type":"hyper_latex","block":{"nodes":[{"latex":"P_{\\mathrm{useful}}=0.40\\times120\\ \\mathrm{kW}=\\boxed{48\\ \\mathrm{kW}}"},{"text":"The power is already in kilowatts, so no further unit conversion is needed."}]},"order":3,"title":"Calculate the output","description":"Carrying out 120 x 0.40 earns the second M1. The correct final answer, 48 kW, earns the A1."}]}],"generatedAt":"2026-07-22T09:13:22.861Z","modelVersion":"gpt-5.6-terra+diagrams-1","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null},{"id":"88321c12-4ea4-45cf-8177-b6218879fb17","revisionId":"a5593ed8-a99b-4e70-95a4-22e1e9dc69f2","questionNumber":null,"questionType":"LA","specification":"","difficulty":4,"difficultyBand":"medium","totalMarks":3,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"daf9c320-ce18-47c4-811f-76c8896306fe","label":null,"content":"A crate of building supplies weighing $18\\,000\\text{ N}$ is lifted vertically upwards through a height of $14\\text{ m}$ by a construction crane.\n\nThe useful power developed by the crane's electric motor is $1200\\text{ W}$.\n\nCalculate the time taken to lift the crate.","markscheme":"- Calculates the work done as $W = 18\\,000 \\times 14 = 252\\,000\\text{ J}$ M1\n- Uses or rearranges $\\text{Power} = \\frac{\\text{Work done}}{\\text{time}}$ to express $\\text{time} = \\frac{\\text{Work done}}{\\text{Power}}$ M1\n- Evaluates $t = \\frac{252\\,000}{1200} = 210\\text{ s}$ A1","marks":3,"order":0,"markingRubric":null,"explanation":{"methods":[{"id":"work_and_power","label":"Work done and power","steps":[{"type":"text","order":1,"title":"Choose the work equation","content":"For a vertical lift, the work done against gravity is the crate's weight multiplied by its vertical height: $$W=Fd$$ Here $F$ is the weight and $d$ is the vertical height."},{"type":"hyper_latex","block":{"nodes":[{"latex":"W=Fd"},{"latex":"W=18\\,000\\times14"},{"latex":"W=\\htmlData{anchor=work}{252\\,000\\ \\mathrm{J}}"}],"highlights":[{"color":"green","style":"box","anchor":"work"}]},"order":2,"title":"Calculate the work done","description":"Substitute the given weight and height. M1: This substitution gives 252 000 J."},{"type":"hyper_latex","block":{"nodes":[{"latex":"P=\\frac{W}{t}"},{"latex":"Pt=W"},{"latex":"t=\\frac{W}{P}"}]},"order":4,"title":"Rearrange the power equation","description":"Power is work done divided by tim

Power and efficiency

Question 3

Power and efficiency Questions

Power and efficiency

Question 3

Power and efficiency Questions