Conservation of mechanical energy: Cyclist up a hill

AI Thread Summary
The discussion focuses on calculating the work done by a cyclist ascending an 8.00-degree hill with a vertical height of 115 meters, given a total mass of 80 kg. The first part of the problem is solved, yielding a work of 90,160 Joules against gravity. For the second part, participants suggest calculating the average force exerted on the pedals by determining the potential energy increase per pedal revolution. The conversation emphasizes the importance of using correct units and understanding the relationship between work, force, and distance in the context of mechanical energy conservation. The discussion concludes with a clarification that the average force can be derived from the increase in potential energy divided by the distance moved per revolution.
Triad
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Homework Statement



A cyclist intends to cycle up a 8.00 hill whose vertical height is 115 . The pedals turn in a circle of diameter 36.0 .

1:Assuming the mass of bicycle plus person is 80.0 , calculate how much work must be done against gravity.

2: If each complete revolution of the pedals moves the bike 5.90 along its path, calculate the average force that must be exerted on the pedals tangent to their circular path. Neglect work done by friction and other losses.


Homework Equations


1: W = F*d

2: I don't know. Somehow I am sure it involves 1/2mv^2 + mgy


The Attempt at a Solution



1: W= (80kg*9.8m/s)*115m = 90160 N*m = 9.02*10^4 J (this one I solved)

2: shrug
 
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Triad said:

Homework Statement



A cyclist intends to cycle up a 8.00 hill whose vertical height is 115 . The pedals turn in a circle of diameter 36.0 .

1:Assuming the mass of bicycle plus person is 80.0 , calculate how much work must be done against gravity.

2: If each complete revolution of the pedals moves the bike 5.90 along its path, calculate the average force that must be exerted on the pedals tangent to their circular path. Neglect work done by friction and other losses.


Homework Equations


1: W = F*d

2: I don't know. Somehow I am sure it involves 1/2mv^2 + mgy


The Attempt at a Solution



1: W= (80kg*9.8m/s)*115m = 90160 N*m = 9.02*10^4 J (this one I solved)

2: shrug

Welcome to PF.

So what are your units?
You can't get the right answer without the right units.
 
Ack!. The first attempt to post I had all the units. I hurridly relaid it out.

8.00 Degrees
115m
36.0 cm Breaks down into .36 m with a radii of .16 m
80.0 kg
5.90 m


and the onyl equation I can assume for 2 is 1/2 mv^2+mgy
 
Perhaps you can approach 2) by identifying how much increase in Potential Energy for each revolution. Then knowing that amount of work to do that and the distance over which you had to do it ...
 
W = (80kg*9.8m/s)*(5.90m/2pi.16) = 1160 J ??

I am not understanding at all. I can believe I am being stumped by this when I could get the complete total for #1
 
Triad said:
W = (80kg*9.8m/s)*(5.90m/2pi.16) = 1160 J ??

I am not understanding at all. I can believe I am being stumped by this when I could get the complete total for #1

Well what's a joule? A N-m

And the increase in Y is what determines your increase in PE.

So doesn't that mean that your increase in PE/5.9 = Favg ?
 

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