limetres to metres: $15\\text{ mm} = 0.015\\text{ m}$ B1\n- Recalls and rearranges the equation: $k = \\frac{F}{x}$ M1\n- Calculates the spring constant: $k = \\frac{1.2}{0.015} = 80\\text{ N/m}$ A1","marks":3,"order":0,"markingRubric":null,"explanation":{"methods":[{"id":"hookes_law","label":"Use Hooke's law","steps":[{"type":"text","order":0,"title":"Convert the extension","content":"The force is given in newtons, so use the extension in metres because the answer must be in $\\text{N/m}$. Since $1\\text{ mm}=0.001\\text{ m}$, divide by $1000$:\n\n$$15\\text{ mm}=0.015\\text{ m}$$\n\nThis correct unit conversion earns the **B1** mark.","highlights":[{"text":"15\\text{ mm}","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"F = k\\htmlData{anchor=x}{x}"},{"latex":"\\htmlData{anchor=k}{k} = \\frac{F}{\\htmlData{anchor=x-used}{x}}"}],"highlights":[{"color":"green","style":"box","anchor":"k"}],"connections":[{"to":"x-used","from":"x","color":"sky","label":"divide by x"}]},"order":1,"title":"Rearrange the equation","description":"Start with Hooke's law, then divide both sides by x to make k the subject. This correct rearrangement earns the M1 mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"k = \\htmlData{anchor=fraction}{\\frac{1.2\\,\\mathrm{N}}{0.015\\,\\mathrm{m}}}"},{"latex":"k = \\htmlData{anchor=answer}{80\\,\\mathrm{N/m}}"}],"highlights":[{"color":"green","style":"box","anchor":"answer"}],"connections":[{"to":"answer","from":"fraction","color":"orange","label":"calculate"}]},"order":2,"title":"Substitute and calculate","description":"Use the converted extension in k = F divided by x. The correct calculation earns the A1 mark."}]}],"generatedAt":"2026-07-22T11:43:17.489Z","modelVersion":"gpt-5.6-terra+diagrams-1","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null},{"id":"78544f05-b3b4-40a2-8480-a07082283bdf","revisionId":"8a7296b5-62d1-455d-98e1-2b0540df2c80","questionNumber":null,"questionType":"LA","specification":"An engineer is testing a heavy-duty spring designed for a maritime mooring system. The tension $T$ applied to the spring (measured in kilonewtons, $\\text{kN}$) and the corresponding extension $x$ (measured in millimetres, $\\text{mm}$) are recorded in the table below.\n\n| Tension $T$ ($\\text{kN}$) | Extension $x$ ($\\text{mm}$) |\n| :--- | :--- |\n| 1.2 | 4.2 |\n| 2.4 | 8.4 |\n| 3.6 | 12.6 |\n| 4.8 | 16.8 |\n| 6.0 | 21.0 |\n| 7.2 | 25.2 |","difficulty":5,"difficultyBand":"medium","totalMarks":3,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"1a1a18b0-76e9-4377-87c3-6777395f60d0","label":null,"content":"An apprentice reviews the data and writes this conclusion:\n\n*\"The extension of the mooring spring is directly proportional to the tensile force applied to it.\"*\n\nComment on the apprentice's conclusion.","markscheme":"- States that the apprentice's conclusion is correct or fully supported by the data B1\n- Provides quantitative evidence from the table, e.g. showing that doubling the tension from $1.2\\text{ kN}$ to $2.4\\text{ kN}$ exactly doubles the extension from $4.2\\text{ mm}$ to $8.4\\text{ mm}$ (or tripling from $1.2\\text{ kN}$ to $3.6\\text{ kN}$ triples the extension from $4.2\\text{ mm}$ to $12.6\\text{ mm}$) M1\n- Explains the mathematical relationship by showing that the ratio of extension to tension is constant, i.e. $\\frac{x}{T} = 3.5\\text{ mm/kN}$ for all data points, confirming a straight-line relationship that passes through the origin $(0,0)$ A1","marks":3,"order":0,"markingRubric":null,"explanation":{"methods":[{"id":"constant_ratio","label":"Check the constant ratio","steps":[{"type":"text","order":1,"title":"Compare a doubled force","content":"Compare two rows of the table. When the tension doubles from $1.2\\,\\text{kN}$ to $2.4\\,\\text{kN}$, the extension also doubles from $4.2\\,\\text{mm}$ to $8.4\\,\\text{mm}$.\n\nThis quantitative comparison earns the **M1** mark: directly proportion

Forces and springs

Question 17

Forces and springs Questions

Forces and springs

Question 17

Forces and springs Questions