Standard +0.3 This is a straightforward work-energy problem requiring application of the work-energy principle with multiple energy terms (KE change, PE change, work against resistance). The calculation involves standard formulas and arithmetic with clearly stated values, making it slightly above average difficulty due to the multiple components but requiring no novel insight or complex problem-solving.
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]
Use of work-energy equation. 5 dimensionally correct terms.
F=18.5
A1
5
Answer
Marks
Guidance
Question
Answer
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Question 2:
2 | Initial KE = 1×75×102
2
Final KE = 1×75×52
2 | B1 | Either correct
PE gained =75g×700sin1.5 [=13 743] | B1
WD by F = F × 700 | B1 | For WD by F=F×d
WD by F + Initial KE = FinalKE+PEgain+2000 | M1 | Use of work-energy equation. 5 dimensionally correct terms.
F=18.5 | A1
5
Question | Answer | Marks | Guidance
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]
\hfill \mbox{\textit{CAIE M1 2019 Q2 [5]}}