Conservation of Energy/Angular Momentum

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Homework Help Overview

The problem involves a wooden block sliding up an incline, where the concepts of conservation of energy and the effects of friction are to be analyzed. The goal is to determine the maximum height the block reaches in terms of a previously defined height.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of conservation of energy, questioning how to express the distance traveled up the ramp in terms of height and angle. There is also exploration of how to relate the energy equations to the height found in a previous part of the problem.

Discussion Status

Some participants have provided guidance on expressing distance as a function of height and angle. Others are clarifying the distinction between different height variables and exploring the implications of rotational motion for a different scenario involving a hoop.

Contextual Notes

Participants note constraints regarding the variables that can be used in the final answer and the requirement to express the solution in terms of a previously defined height. There is also mention of the roughness of the incline affecting the energy calculations.

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Homework Statement


Hi.
The problem is "A wooden block of mass m slides with an initial speed of vi up a straight incline that makes an angle of theta with the horizontal. If the incline is rough with a coefficient of kinetic friction uk, determine the maximum height it rises in terms of h found in part a (which was H = vi^2/2g)"


Homework Equations


conservation of energy
You can only use m (mass of block), vi (initial velocity), theta, and g (gravity)


The Attempt at a Solution


KE at the bottom = PE at the top + Work.
mgh + Fd = 1/2mv^2
h = [mvi^2 - ukmgcos(theta)d]/mg
Problem is, I don't know what d is and it can't be part of my final answer. I know it's the distance that the block travels up the ramp, but how do I figure that out?
Also, since it says in terms of h found in part a, does that mean I set everything equal to vi^2/2g?

How would I do this with a hoop with radius R?
 
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So, you correctly applied the conservation of mechanical energy. :smile:
Could you right d as a function of h and [itex]\theta[/itex]? :rolleyes:

The hoop will have in the end not only linear velocity but also angular velocity. Apply again conservation of mechanical energy, but this time consider also the kinetic rotational energy. :-p
 
Okay this is going to sound really confusing but am I on the right path for the block one?

I got to the point where gh = 1/2(vi)^2 - ukg(cot(theta))h
At this point, do I plug in what I got for H in part (a) since it wants it says "in terms of h found in part a"? What exactly are they asking me to do?
 
Nevermind. I was being stupid and forgot they are two different H's
 

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