How Does Friction Affect the Maximum Height of a Block on an Incline?

  • Thread starter Thread starter Ineedphysicshelp
  • Start date Start date
  • Tags Tags
    Formula
Click For Summary

Homework Help Overview

The discussion revolves around the effects of friction on the maximum height achieved by a block sliding down a frictionless track and then ascending an inclined plane. The problem involves energy methods and the coefficient of kinetic friction.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between gravitational potential energy and work done against friction. Some explore the implications of the cotangent function in the context of the problem, while others express confusion over their calculations and the setup of the energy conservation equation.

Discussion Status

Some participants have offered guidance on applying conservation of energy principles, suggesting a focus on the initial potential energy and the work done against friction. There is recognition of misunderstandings in the mathematical representation of the problem, with attempts to clarify the correct formulation of equations.

Contextual Notes

Participants express challenges in understanding the physics behind the equations and the implications of friction on the maximum height. There is mention of the need to consider gravitational forces and the role of kinetic friction in the energy balance.

Ineedphysicshelp
Messages
7
Reaction score
0
A block slides down a curved frictionless track and then up an inclined plane. the coefficient of kinetic friction between the block and the incline is Uk. Use energy methods to show that the maximum height reached by the block is:

Ymax=

h
-----------------
1 + Uk cot(theta)
 
Physics news on Phys.org
I don't know how to delete the other post. Sorry about that. Ok I know that because it's cot(theta) that is equal to cos(theta)/sin(theta). I've been trying to figure it out in reverse. So I multiplied the right side by Fg/Fg so that I got something that looks like: Fgcos(theta)h / [Fg+UkFgsin(theta)] which looks like Fgyh / (Fg + FgxUk) I don't know where to go from here. I've filled up pages of calculations and it has just stopped making sense. Sorry again about the double posting.
 
I don't know how to delete the other post. Sorry about that. Ok I know that because it's cot(theta) that is equal to cos(theta)/sin(theta). I've been trying to figure it out in reverse. So I multiplied the right side by Fg/Fg so that I got something that looks like: Fgcos(theta)h / [Fg+UkFgsin(theta)] which looks like Fgyh / (Fg + FgxUk) I don't know where to go from here. I've filled up pages of calculations and it has just stopped making sense. Sorry again about the double posting.

It seems like you are making it way too hard for yourself. This is physics, not math. If you understand work and energy, the math will follow. Working backward from the equation will not help you to understand the physics of the situation.
 
Don't bother "reverse engineering" the equation...just apply conservation of energy.

How much energy does the block start with? (Consider gravitational PE.)

Call the height reached Ymax. In reaching that height up the incline: How much work was done against friction? What's the final PE?

Apply conservation of energy: The initial PE = final PE + work done against friction.
 
Ok, thank you, I realize what I did wrong. I had it as hcos(theta)/[1+Uksin(theta)] which is upside down to what it's supposed to be: hsin(theta)/[1+Uksin(theta)] that's why it wasn't making any sense. :P . Thank you.
 
I mean hsin(theta)/[1+Ukcos(theta)]
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 33 ·
2
Replies
33
Views
4K
Replies
19
Views
5K
Replies
6
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
Replies
5
Views
2K
Replies
12
Views
2K
Replies
4
Views
4K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K