Question

The record time for a Tour de France cyclist to ascend the 1100-m-high Alpe d'Huez is 37.5 min . The rider and his bike had a mass of 65 kg. At the average speed of the ride, the metabolic power to ride on level ground would be 700 W. To this, we must add the power to make the climb.

What was his total metabolic power?

Answer 1

Increase in the gravitational potential energy of the cyclist to rise the top of the mountain is given as

$U = mgh$

given that

m = 65 kg

g = 9.8 m/s^2

h = 1100 m

now from above equation we will have

$U = 65 \times 9.8 \times 1100$

$U = 700700 J$

now it will take total time t = 37.5 min

so power consumed in this process is given as

$P = \frac{U}{t}$

$P = \frac{700700}{37.5 \times 60}$

$P = 311.4 W$

now total metabolic power required for this is given as

$P = 700 + 311.4 = 1011.4 W$

Answer 2

Answer:

The total metabolic power of the cyclist is $\boxed{1011.4\,{\text{W}}}$.

Further Explanation:

Given:

The mass of the cyclist is $65\,{\text{kg}}$.

The height that the cyclist climbs is $1100\,{\text{m}}$.

The acceleration due to gravity is $9.8\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}$ .

Concept:

The metabolic power of the cyclist required to climb the Alpe d’Huez is the sum of the power required to climb and the power on the level ground.

The potential energy of the cyclist at the top is given by:

${E_P} = mgh$              …… (1)                                            

The amount of energy required to climb up is equal to the potential energy of the cyclist.

 $\begin{aligned}{E_P} &= 65 \times 9.8 \times 1100\\&= 700700\,{\text{J}}\\\end{aligned}$

The power consumed by the cyclist in climbing up:

$P = \dfrac{{{E_P}}}{T}$       …… (2)                                                      

Substitute the value of the potential energy and the time taken.

 $\begin{aligned}P &= \frac{{700700\,{\text{J}}}}{{\left( {37.5 \times 60} \right)\,{\text{s}}}}\\&= 311.4\,{\text{W}}\\\end{aligned}$

Since the total metabolic power consumed in climbing up is the sum of the power on the level ground and the power consumed in climbing.

 $\begin{aligned}{P_T} &= {P_{level}} + P\\&= 700 + 311.4\\&= 1011.4\,{\text{W}}\\\end{aligned}$

Therefore, the total metabolic power consumed in climbing is  $\boxed{1011.4\,{\text{W}}}$.

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Answer Details:

Grade: High School

Chapter: Energy and Power

Subject: Physics

Keywords:Tour de France, cyclist, 1100 m, 37.5 min, rider and his bike, potential energy, climbing up, metabolic power, level ground, 700 W.