What Is the Cyclist's Power Output to Climb a 7.0 Degree Hill at 5.0 m/s?

In summary, the conversation discusses the power output required for a cyclist to climb a 7.0 degree hill at a steady speed of 5.0 m/s, assuming a total mass of 75kg. The formula for power, work per unit time, is discussed and two methods are presented to calculate the power required. The final answer is determined to be 900W, with the explanation that the cyclist must exert a force equal to the horizontal component of the force of gravity, as well as the force of friction, in order to maintain a constant speed while going uphill.
  • #1
ammu
6
0
A bicyclist coasts down a 7.0 degree hill at a steady speed of 5.0 m/s. Assuming a total mass of 75kg (bicycle plus rider), what must be the cyclist's power output to climb the same hill at the same speed?

I got 90N * 5.0 m/s = 450W but that is not the answer

The book says 9.0 * 10^2 W
i am sorry for the trouble because this question has already been asked.
please can somebody help me coz i didnt understand the one explained b4?
 
Last edited:
Physics news on Phys.org
  • #2
First start of with the definition of power. Power is defined as work per unit time. In other words, how much work does this guy have to do to go up this incline every single second?
 
  • #3
yea, i first used the formula and got 450 W, but the answer is 900 W, i don't know why i can't get 900W as the answer.
 
  • #4
You should say what formula you used and what it means. There are two ways that you can do this. Both lead to your answer, which I believe is correct if you have been given the right information.

One way:
[tex]Power = \frac{dW}{dt} = \frac{d}{dt}(F*x) = F \frac{dx}{dt}[/tex]
This should be familiar
[tex]P = Fv[/tex]
Power, given a constant force, is simply the product of force and velocity.
[tex]P = mgsin(\theta)*v[/tex]

Work more directly
[tex]P = \frac{dW}{dt}[/tex]
Find how much work happens in one second
[tex]W/sec = F \cdot ds/sec[/tex]
The force and displacement are in the same direction, so the dot prod gives
[tex]W/sec = F*distance/sec[/tex]
At this point it is clear the two are really the same
[tex]dist = speed * time = 5m/s * 1 sec = 5m[/tex]
Then plug in and get the same answer
[tex]W/sec = mgsin\theta * 5m/s = P[/tex]
 
Last edited:
  • #5
thanks for the reply,
when i use that formula i get the answer as 450W but the actual answer of the book is 900W. why is it like that? please help
 
  • #6
The book is correct. You are missing something.

First, he coasts downhill with a constant speed.
Normally he should be accelerating downhill because the horizontal component of the force of gravity acts on him. But he is not accelerating, he is going with a constant speed. Therefore, there must be no net force acting on him:
Fn=Fgy
and
Ff=Fgx
So there is force of friction acting on him going downhill. It is equal to the horizontal component of the force of gravity. Ff = Fgx = mg * sin 7 = 89.57 N.
To go up at the same constant speed, he has to exert a force equal to Ff + Fgx. Again, there should be no net force acting on him, as the speed is constant.
So force to go up is F = Ff + Fgx = Ff + mg * sin 7 = 179.15 N.
The power is F * v = 896 = 900 W.
 
  • #7
thankyou very much for the help. i understood my mistake. it was a great help.
 

Related to What Is the Cyclist's Power Output to Climb a 7.0 Degree Hill at 5.0 m/s?

1. What is cyclist's power output problem?

The cyclist's power output problem refers to the challenge of accurately measuring and tracking the power output of a cyclist while they are riding. This can be important for training, performance analysis, and equipment testing.

2. How is cyclist's power output measured?

Cyclist's power output is typically measured using a power meter, which is a device attached to the bike that measures the force being applied to the pedals and calculates the power output in watts. Other methods include using a stationary bike with a known resistance and measuring heart rate and speed to estimate power output.

3. What factors affect cyclist's power output?

Several factors can affect a cyclist's power output, including their fitness level, weight, bike setup, road conditions, and wind speed. The type of terrain, such as flat or hilly, can also have a significant impact on power output.

4. Can cyclist's power output be improved?

Yes, cyclist's power output can be improved through training and proper technique. This can include strength training, interval training, and optimizing pedaling efficiency. Improving overall fitness and reducing weight can also lead to an increase in power output.

5. How can cyclist's power output data be used?

Cyclist's power output data can be used for a variety of purposes, including tracking progress, setting training goals, and analyzing performance. It can also help with bike fit and equipment selection, as well as identifying areas for improvement in technique and strategy.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
4K
  • Introductory Physics Homework Help
Replies
7
Views
10K
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
3K
Replies
3
Views
4K
  • Introductory Physics Homework Help
Replies
3
Views
2K
Back
Top