# Cyclist's power output problem

1. May 11, 2007

### ammu

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: May 11, 2007
2. May 11, 2007

### Mindscrape

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. May 12, 2007

### ammu

yea, i first used the formula and got 450 W, but the answer is 900 W, i dont know why i cant get 900W as the answer.

4. May 12, 2007

### Mindscrape

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

Work more directly
$$P = \frac{dW}{dt}$$
Find how much work happens in one second
$$W/sec = F \cdot ds/sec$$
The force and displacement are in the same direction, so the dot prod gives
$$W/sec = F*distance/sec$$
At this point it is clear the two are really the same
$$dist = speed * time = 5m/s * 1 sec = 5m$$
Then plug in and get the same answer
$$W/sec = mgsin\theta * 5m/s = P$$

Last edited: May 12, 2007
5. May 12, 2007

### ammu

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? plz help

6. May 12, 2007

### husky88

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. May 13, 2007

### ammu

thankyou very much for the help. i understood my mistake. it was a great help.