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Nov 2011 p42 q1
3470
A racing cyclist, whose mass with his cycle is 75 kg, works at a rate of 720 W while moving on a straight horizontal road. The resistance to the cyclist’s motion is constant and equal to \(R N\).
Given that the cyclist is accelerating at \(0.16 \, \text{m/s}^2\) at an instant when his speed is \(12 \, \text{m/s}\), find the value of \(R\).
Given that the cyclist’s acceleration is positive, show that his speed is less than \(15 \, \text{m/s}\).
Solution
(i) The power \(P\) is given by \(P = Fv\), where \(F\) is the force and \(v\) is the velocity. Thus, \(F = \frac{P}{v} = \frac{720}{12} = 60 \, \text{N}\).
Using Newton's second law, \(F - R = ma\), where \(m = 75 \, \text{kg}\) and \(a = 0.16 \, \text{m/s}^2\). Therefore, \(60 - R = 75 \times 0.16\).
Solving for \(R\), we get \(R = 60 - 12 = 48\).
(ii) Given \(a > 0\), we have \(\frac{P}{v} - R = ma\). Since \(a > 0\), it implies \(\frac{P}{v} > R\).
Substituting \(R = 48\), we have \(\frac{720}{v} > 48\).
Solving for \(v\), \(v < \frac{720}{48} = 15\). Thus, the speed is less than \(15 \, \text{m/s}\).