# Power output of cyclist up slope at costant speed

1. Sep 25, 2011

### mrsteve

1. The problem statement, all variables and given/known data

A bicyclist coasts down a 7.0 degree slope at a steady speed of 5.0 m/s. Assuming a total mass of 75 kg (bicycle plus rider), what must the cyclist's power output be to pedal up the same slope at the same speed?

2. Relevant equations
$$P=Fvcos\theta$$
$$K=\frac{1}{2}mv^2$$
$$F_g=mg$$

3. The attempt at a solution

I need to find the net force but since there is presumably no acceleration (constant velocity of 5.0 m/s) the only force is the force to overcome gravity and maintain a constant speed of 5.0 m/s. Correct?

I'm missing a formula somewhere I think. How do I find the force? Once I have that I can determine the Power using the first formula.

2. Sep 25, 2011

### Staff: Mentor

Don't worry about force. When he climbs his potential energy changes.

3. Sep 25, 2011

### mrsteve

Potential energy isn't covered until the next chapter. Also, I don't know the height, so how would I find PE?

4. Sep 25, 2011

### Staff: Mentor

Speed and angle give you height change per second.

5. Sep 25, 2011

### mrsteve

$$F_biker=mg\sin\theta => (75)(9.8)(\sin7)=9.0*10^1$$

$$P=(90)(5.0)(\cos7)=4.47*10^2 => 4.5*10^2 = 450W$$

6. Sep 25, 2011

### Spinnor

Why do you have cos(7)? It doesn't change the answer much but I don't think it should be there.

7. Sep 25, 2011

### mrsteve

I thought you needed it because theta is the angle between the force vector and the velocity vector.

No?

It changes the answer a lot without it. 90*50 = 4.5*10^3 I got 4.5*10^2

8. Sep 26, 2011

### Staff: Mentor

Check your math - cos(7) is 0.993, it can't change the answer tenfold. You have already taken care of the angle calculating vertical component of the force, assuming no friction vertical component is the only one doing work.