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Nov 2017 p42 q5
3433
A cyclist is riding up a straight hill inclined at an angle \(\alpha\) to the horizontal, where \(\sin \alpha = 0.04\). The total mass of the bicycle and rider is 80 kg. The cyclist is riding at a constant speed of 4 m s\(^{-1}\). There is a force resisting the motion. The work done by the cyclist against this resistance force over a distance of 25 m is 600 J.
(i) Find the power output of the cyclist.
The cyclist reaches the top of the hill, where the road becomes horizontal, with speed 4 m s\(^{-1}\). The cyclist continues to work at the same rate on the horizontal part of the road.
(ii) Find the speed of the cyclist 10 seconds after reaching the top of the hill, given that the work done by the cyclist during this period against the resistance force is 1200 J.
Solution
(i) To find the power output of the cyclist, we first calculate the resistance force: \(F_{\text{resistance}} = \frac{600}{25} = 24 \text{ N}\).
The weight component along the incline is \(80g \sin \alpha = 80 \times 9.8 \times 0.04 = 32 \text{ N}\).
The total force exerted by the cyclist is \(24 + 32 = 56 \text{ N}\).
Power is given by \(P = Fv\), where \(F\) is the total force and \(v\) is the velocity. Thus, \(P = 56 \times 4 = 224 \text{ W}\).
(ii) The work done by the cyclist in 10 seconds is \(224 \times 10 = 2240 \text{ J}\).
The initial kinetic energy (KE) at the top of the hill is \(\frac{1}{2} \times 80 \times 4^2 = 640 \text{ J}\).
Using the work-energy principle: \(\frac{1}{2} \times 80 \times v^2 = 640 + 2240 - 1200\).
Solving for \(v\), we get \(v^2 = 42\), so \(v = \sqrt{42} \approx 6.48 \text{ m s}^{-1}\).
