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?

  • 1
    $\begingroup$ 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. $\endgroup$ – 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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.