Determining the Acceleration (Inclined Plane Problem)

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Homework Help Overview

The problem involves a crate sliding down an inclined plane at an angle of 22.0° with a given coefficient of kinetic friction. Participants are tasked with determining the acceleration of the crate and its speed upon reaching the bottom after starting from rest.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply force equations to find the acceleration and speed of the crate. Some participants question the accuracy of the calculations and suggest clarifying the definitions of forces involved, particularly the normal force.

Discussion Status

Participants are actively engaging in clarifying the setup of the problem and the definitions of forces. Some guidance has been offered regarding the normal force and the direction of friction, but there is no explicit consensus on the correct approach to calculating acceleration.

Contextual Notes

There is a noted confusion regarding the terminology used for forces, and participants are addressing potential misunderstandings in the original poster's equations. The problem context includes assumptions about the initial conditions and the effects of friction.

yandereni
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Homework Statement


A crate lies on a plane tilted at an angle θ = 22.0° to the horizontal, with μkin = 0.12.
a.) determine the acceleration of the crate as it slides down the plane
b.) if the carton starts from rest 9.30m up the plane from its base, what will be the carton's speed when it reaches the bottom?

this is the most confusing so far...

Homework Equations


ΣFhorizontal = Ffrict+Fgrav(sin22) = ma
ΣFvertical = Fnorm + Fgrav(cos22) = ma
Vfinal2= VInitial2 + 2a(displacement)
ples.jpg

The Attempt at a Solution


ΣFhorizontal = (0.12)(Fnorm)+Fgrav(sin22) = ma
= 0.12Fn + 0.375Fg=ma
= 0.12(0.279Fg + ma) = ma
= 0.111Fg + 0.12ma = ma
= 0.111Fg = 0.88ma
=Fg = 7.928ma
ΣFvertical = Fnorm + Fgrav(cos22) = ma
= Fn - 0.927Fg = ma
=Fn=0.927Fg+ma
= Fn= 0.927(7.928ma)
=Fn = 7.349ma

since I didnt really know what to do next..
7.928ma/7.349ma= 1.079
so i assumed that was the acceleration

then...

Vfinal2 = (0m/s)2 + (2)(1.079m/s2)(9.3m)
Vfinal2=(20.07m/s)
(Vfinal2)1/2=(20.07m/s)1/2
Vfinal=4.480m/sThanks in Advance!
 
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Your approach to part b) is correct, but the value for the acceleration is wrong.

To be honest, I couldn't really understand what you were doing in part a).

The first step is to find the normal force acting on the block. If we call this ##F_n##, can you work out what ##F_n## is?
 
You have written subscript "horizontal" where you should call it "down the slope". Similarly, you wrote "vertical" where you should have "normal to the slope".

Don't forget that friction acts opposite to the direction of motion. So with the block sliding down the slope here, friction will be acting up the slope.
 
yandereni said:
ΣFvertical = Fnorm + Fgrav(cos22) = ma
As NascentOxygen posted, you mean Fnormal, not Fvertical, but what is a in that equation? Yes, I know it's an acceleration, but what acceleration exactly?
 

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