A-LevelMathematicsMechanics9709/41 · Oct/Nov 2019 · Q25 Marks
The total mass of a cyclist and her bicycle is 75 kg. The cyclist ascends a straight hill of length 0.7 km inclined at 1.5° to the horizontal. Her speed at the bottom of the hill is 10 m s⁻¹ and at the top it is 5 m s⁻¹. There is a resistance to motion, and the work done against this resistance as the cyclist ascends the hill is 2000 J. The cyclist exerts a constant force of magnitude F N in the direction of motion. Find F.
- The total mass of a cyclist and her bicycle is 75 kg. The cyclist ascends a straight hill of length 0.7 km inclined at 1.5° to the horizontal. Her speed at the bottom of the hill is 10 m s⁻¹ and at the top it is 5 m s⁻¹. There is a resistance to motion, and the work done against this resistance as the cyclist ascends the hill is 2000 J. The cyclist exerts a constant force of magnitude F N in the direction of motion. Find F.[5]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.
- [B1] Initial KE = ½ × 75 × 10² or Final KE = ½ × 75 × 5² (Either correct)
- [B1] PE gained = 75g × 700 sin 1.5 [=13 743]
- [B1] WD by F = F × 700
- [M1] Use of work-energy equation. 5 dimensionally correct terms. (WD by F + Initial KE = Final KE + PE gain + 2000)
- [A1] F = 18.5
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