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
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 ?
 
Thread 'Struggling to make relation between elastic force and height'

Thread 'Struggling to make relation between elastic force and height'

Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
View full post »
Thread 'Voltmeter readings for this circuit with switches'

Thread 'Voltmeter readings for this circuit with switches'

TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
View full post »

Similar threads

Conservation of Mechanical Energy - Which Equation To Use?
Replies
9
Views
1K
Cycling Up a Hill: Calculating Work Against Gravity
Replies
8
Views
3K
Conservation of Energy when lifting a box up off the floor
Replies
5
Views
2K
Find mechanical energy lost when a cyclist goes down a hill
Replies
3
Views
2K
Is potential energy defined only for internal conservative forces?
Replies
15
Views
1K
Proving that the total mechanical energy is conserved with time
Replies
2
Views
2K
Cyclist coasting down a hill cons. of energy
Replies
4
Views
3K
Sledding Down a Hill: Calculating Friction and Velocity
Replies
31
Views
3K
Conservation of mechanical energy
Replies
8
Views
2K
A question regarding energy and momentum conservation
Replies
9
Views
3K

Hot Threads

Recent Insights

Back
Top