Work against gravity and force on pedals cycling up a hill

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
5 replies · 6K views
Triad
Messages
3
Reaction score
0

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
 
Physics news on Phys.org
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 ?