iciency/questions","formulaUrl":null,"initialIndex":19,"initialQuestionWindow":[{"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 time. M1: Rearranging for time earns this mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"t=\\frac{W}{P}"},{"latex":"t=\\frac{252\\,000}{1200}"},{"latex":"t=\\htmlData{anchor=time}{210\\ \\mathrm{s}}"}],"highlights":[{"color":"green","style":"box","anchor":"time"}]},"order":6,"title":"Find the time","description":"Now divide the work done by the useful power. A1: The final answer is 210 s."}]}],"generatedAt":"2026-07-22T11:56:33.268Z","modelVersion":"gpt-5.6-terra+diagrams-1","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null},{"id":"90f56a2e-9fa3-48b0-aa52-14f7854ac453","revisionId":"c7407668-dba8-49a3-bf45-5ebf2379f7b9","questionNumber":null,"questionType":"LA","specification":"","difficulty":7,"difficultyBand":"medium","totalMarks":3,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"832bb99e-64a7-4f9f-97c6-19d7678870bc","label":null,"content":"A solar-powered pump is used to lift $1200\\text{ kg}$ of water vertically upwards through a height of $35\\text{ m}$ from an underground source.\n\nThe electrical power supplied to the pump's motor is $1600\\text{ W}$, and the motor operates at an efficiency of $75\\%$.\n\nTaking $g = 9.8\\text{ m s}^{-2}$, calculate the time taken to pump this quantity of water.","markscheme":"- Calculates the work done (or gravitational potential energy gained) in lifting the water: $\\text{Work Done} = mgh = 1200 \\times 9.8 \\times 35 = 411\\,600\\text{ J}$ M1\n- Determines the useful power output using efficiency: $\\text{Useful Power} = 1600 \\times 0.75 = 1200\\text{ W}$ M1\n- Calculates the time taken: $\\text{time} = \\frac{411\\,600}{1200} = 343\\text{ s}$ A1","marks":3,"order":0,"markingRubric":null,"explanation":{"methods":[{"id":"energy_useful_power","label":"Energy and useful power","steps":[{"type":"text","order":1,"title":"Identify the energy gain","content":"Because the water is lifted vertically, it gains gravitational potential energy. Use the mass, vertical height and gravitational field strength given in the question.",

Power and efficiency

Question 5

Power and efficiency Questions

Power and efficiency

Question 5

Power and efficiency Questions