What Speed Can a Cyclist Climb a Hill Using the Same Power as Descending?

[skiboka33]

A bicyclist of mass 80kg (including the bike) can coast down a 4 degree hill at a steady speed of 6.0 km/h. Pumping hard, the cyclist can descend the hill at 30 km/h. Using the same ower, at what speed can the cyclist climb the hill? Assume the force of friction is proportional to the square of the speed; that is, Ff = bv^2, where b is constant.

thanks...

 

[Janitor]

Check me on this, but can you say that when coasting downhill at 6 km/hr, the frictional force b * 6^2 is exactly equal (and opposite) to the effective force of gravity, which would be m * g * sin(4 deg)? If so, that would allow you to solve for b. If that is correct, then it would get you started on attacking the remainder of the problem.

 

[Jez]

that is incorrect. Frictional Force is the static friction coefficient*Force Normal which is perpendicular to the surfuace. so when on an incline it is not equal to m*g.

 

[Janitor]

Jez,

I was thinking of it as a drag force, due mostly to wind resistance, but if you are correct in your post, then skiboka can throw my idea out the window, I guess.

I was thinking once skiboka had b figured, the next thing would be to calculate b * 30^2 to get the drag force at that speed, then subtract from it m * g * sin(4 deg) to get the force the rider is providing with his legs, and then multiply by the speed of 30 to get power.

Then for the uphill part of the problem, it would be a matter of finding speed v for which power required to go uphill at that speed plus power to overcome drag was equal to the power calculated previously.

 

[Jez]

i was thinking of friction between the tires and the surface. hmmmm... now you got me thinking again.

 

[mmwave]

that is incorrect. Frictional Force is the static friction coefficient*Force Normal which is perpendicular to the surfuace. so when on an incline it is not equal to m*g.

I agree with Janitor. Since the wheels are rolling not sliding, the force of gravity projected along the incline equals the friction force from wind resistance. It's a safe bet to ignore friction in the bearings for this problem.

 

[skiboka33]

ok i found b to be 1.52

then i found Fapplied = bv^2 - mgsin@ = .0309

then i found power = F * v = .927

so F required = Fg + Ff right? which = 54.7 + 1.52v^2
P = (54.7 + 1.52v^2)(v) = ?

now what? did i mess up somewhere?

 

[Janitor]

If you are going to work in MKS units, you need to convert 6 km/hr to m/s, which makes it 1.666 m/s. Thus, in MKS units, the value of b is actually 19.7. Then convert 30 km/ hr to m/s, which is 8.33 m/s. The force that the rider is applying to the bicycle going downhill is then 19.7 * 8.33^2 - 80 * 9.8 * 0.0697 = 1367 - 55 = 1312 Newtons. Power provided by the rider is 1312 * 8.33 = 10935 watts.

So the remainder of the problem is to find a value of v for which 10935 = b v^3 + v m g sin(4 deg). I get 8.1 m/s or 29 km/hr.

Note that I am approximating g as 9.8 m/sec^2. Maybe you have a more precise value of g to use.

 

[Janitor]

Warning!

I calculated that the rider is capable of putting out 11 kilowatts. If you multiply kilowatts by 1.3 to get horsepower, that is 14 horsepower. That seems way too high, even for a short burst by human legs. So either the problem is not realistic, or I have goofed up. Check me carefully.

 

[Doc Al]

bicyclist problem

Please see my analysis of this problem here: https://www.physicsforums.com/showthread.php?p=183868#post183868

 

[Janitor]

I just now saw another thread where Doc Al has solved the same problem in a more straightforward way, without having to bother with calculating power:

https://www.physicsforums.com/showthread.php?t=18959

 

[Janitor]

Oops!

As I was posting the link, Doc Al was also doing that.