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

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In summary, 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. Using energy methods, it can be shown that the maximum height reached by the block is equal to h/(1+Uk cot(theta)), where h is the initial height and theta is the angle of the incline. To solve this problem, conservation of energy is applied, considering the initial potential energy and the work done against friction as the block reaches its maximum height. It is important to note that the equation hsin(theta)/[1+Ukcos(theta)] should be used instead of hcos(theta)/[1+Uksin(theta)] for the correct solution.
  • #1
Ineedphysicshelp
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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)
 
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  • #3
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.
 
  • #4
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.
 
  • #5
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.
 
  • #6
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.
 
  • #7
I mean hsin(theta)/[1+Ukcos(theta)]
 

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