Chapter 6, Problem 71P

To determine

The cyclist power output to climb the same hill at the same speed.

Answer to Problem 71P

Solution:$614.63\text{ watt}$

Explanation of Solution

Given:

Angle made by hill with horizontal $=6.0°$

Steady speed of cyclist in downward motion $\text{=}\text{4}\text{.0}\text{m/s}$

Total mass of bicycle plus rider $=75\text{kg}$

Formula used:

  $\begin{array}{c}P=\frac{W}{t}\\ =FV \cos \theta \end{array}$

W : Work

t : time

F: force

V: speed

θ : angle between force and velocity.

Newton’s second law

  $\sum F_x=m{a}_x$

  $\sum F_x=m{a}_y$

Calculation:

When cyclist moves down ward with steady speed.

  

N : normal reaction by hill on bicycle

Mg : weight of cyclist plus rider.

F : friction force acting upwards along the hill.

  $\sum F_x=0$

Considering x -axis along the inclined plane.

  $mg \sin \theta -f=0$

  $f=mg \sin \theta \to \left(1\right)$

When cyclist moves up ward with steady speed $\left(4\text{m}/\text{s}\right)$

  

Friction force acts down ward, because cycle moves upward.

N : normal reaction by hill on bicycle

Mg : weight of cyclist plus rider

F : friction force

  $F_p$: pull on bicycle in upward direction

Normal, and contact surface is same so friction on the cyclist remain same only direction will be reversed. So

Friction force $\left(f\right)=mg \sin \theta$

Write Newton’s second law along the plane.

  $F_p=mg \sin \theta +f$

  $F_p=\left(mg \sin \theta +mg \sin \theta \right)$

  $F_p=2mg \sin \theta$

Calculation of power

Here θ = $0$

  $\begin{array}{l}P=FV\\ =\left(2mg \sin \theta \text{N}\right)\left(\text{4}\text{m/s}\right)\\ =8mg \sin \theta \text{ watt}\end{array}$

Substitute the values and solve:

  $\begin{array}{l}m=75\text{kg}\\ g=9.8\text{m}/{\text{s}}^2\\ \theta =6.0°\end{array}$

  $\begin{array}{c}P=8\left(75\right)\left(9.8\right)\left(\text{sin}6.0°\right)\\ =614.63\text{ watt}\end{array}$

Conclusion: Power output of bicyclist $614.63\text{ watt}.$