# Mechanics unknown angle

A cyclist and bike have a mass of 80kg. The resistance to motion is proportional to the speed. On level ground they can travel 10m/s, they can free-wheel down an incline of angle θ at a steady speed of 14m/s. What is the maximum speed at which they can go up the same incline. Am I correct in thinking I should be using power = driving force * velocity What is the force to go 10m/s?

• I don't know why you are getting close votes it's a perfectly good question and you ask an appropriate question to solve it. The force to go at $10$m/s is $\frac{10}{14}$ of the force to go at $14$ m/s which is $80\cdot 9.8\sin(\theta)$N. – Phil H Aug 30 at 0:58

Power of rider = Force times Velocity = $$80\cdot 9.8\sin(\theta) \cdot \frac{10}{14}\cdot 10 = 5600\sin(\theta)$$ watts
Going up the incline $$5600\sin(\theta) = 80\cdot9.8\sin(\theta)v + 80\cdot9.8\sin(\theta)\cdot \frac{10}{14}v$$
$$5600 = 80\cdot 9.8\cdot \frac{24}{14}v$$
$$v = \frac{5600\cdot 14}{80\cdot 9.8\cdot 24}$$
$$v = 4.167$$ m/s