How Does Friction Impact Block Velocity in a Multi-Height Path?

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Homework Help Overview

The problem involves a block moving through a multi-height path with friction affecting its velocity. The context includes kinetic and potential energy considerations, as well as the impact of friction on the block's speed at different points along its path.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between kinetic and potential energy at different points, questioning how to account for friction and initial velocity. There are attempts to clarify the energy conservation principles involved.

Discussion Status

Participants are exploring the energy relationships at various points in the problem, with some guidance provided on how to approach the calculations. There is an acknowledgment of the initial kinetic energy and potential energy at point A, but no consensus has been reached on the specific calculations for points B and C.

Contextual Notes

There is mention of a coefficient of kinetic friction and specific heights that may influence the calculations. Some participants have noted that certain aspects of the problem have already been resolved, while others are still seeking clarity on the energy transitions involved.

dominus96
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Kinetic Energy with diagram!

Homework Statement



In the figure below, a small block is sent through point A with a speed of 7.5 m/s. Its path is without friction until it reaches the section of length L = 12 m, where the coefficient of kinetic friction is 0.70. The indicated heights are h1 = 6.5 m and h2 = 1.6 m.

(a) What is the speed of the block at point B?
(b) What is the speed of the block at point C?

W0156A-N.jpg


Homework Equations



K=.5(mv^2)
U=mgh
F=ma
Friction force=mu*force normal

The Attempt at a Solution



First off, ignore the friction stuff because I already got those. I just need help with the 2 specific questions I listed. For a I did K=U so I set .5v^2=gh since the masses cancel out. But I got velocity and it was wrong...
 
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Ooo! I did this one a while back.

Ok the reason you missed the first one is because U @ point a does NOT equal K @ point B. U+K @ a does though. Don't forget it's initially moving.
 
So U+K(point A)=K(point B)?
 
what's point d in there for then, lol
 
It's for another part of the problem I already figured out so don't worry about it lol.
 
cooolio
 
indeed. the second question is similar. Energy total @ point b = energy total @ c

Ergo... K(@b)=U+K(@c)
 
Thanks!
 
Sure
 

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