What is the speed of a block sliding down an inclined plane with friction?

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Homework Help Overview

The problem involves a block sliding down an inclined plane with friction, specifically examining the speed of the block after sliding a certain distance. The incline angle is 17.5 degrees, and the coefficient of friction is 0.085.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the forces acting on the block, including friction, normal force, gravitational force, and the parallel component of gravity. There are attempts to calculate acceleration and speed using kinematic equations. Some participants question the accuracy of calculations and mention potential typos in the reasoning.

Discussion Status

The discussion includes various attempts to analyze the problem using different approaches, including force analysis and kinematic equations. While some calculations have been presented, there is no explicit consensus on the final speed, and alternative methods, such as energy considerations, have been suggested.

Contextual Notes

Participants note that the problem involves specific values for mass, angle, and friction, which may influence the calculations. There is mention of potential errors in the calculations, indicating a need for careful review of the assumptions made.

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Homework Statement



A board is inclined at an angle of 17.5 degrees. The coefficient of friction between a 15.0 kilogram block and the board is .085. what will be the speed of the block if it slides down the board a distance of 4.0 meters, starting from rest?
 
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Hi sskicker23,

sskicker23 said:

Homework Statement



A board is inclined at an angle of 17.5 degrees. The coefficient of friction between a 15.0 kilogram block and the board is .085. what will be the speed of the block if it slides down the board a distance of 4.0 meters, starting from rest?


What have you tried so far?
 
what have it tried so far:

Friction
mu* Fnormal
.085*15*9.8*Cos(17.5)=11.91669

Fnormal
M*G*Cos(17.5)= 140.1964N

Fperp
M*G*Cos(17.5)= 140.1964N

Fgravitational
m*g
15*9.8=147N

Fparallel= m*g*sin(17.5) =140.1964N

axmax= -gsin(17.5)+Mu*G* Cos(17.5)
A=.9.8*sin(17.5)+9.80*Cos(17.5)
A= 2.1525

v^2=Vi^2+2AX
V^2=0+2*2.1525*4
v=4.15m/s
 
sskicker23 said:
what have it tried so far:

Friction
mu* Fnormal
.085*15*9.8*Cos(17.5)=11.91669

Fnormal
M*G*Cos(17.5)= 140.1964N

Fperp
M*G*Cos(17.5)= 140.1964N

Fgravitational
m*g
15*9.8=147N

Fparallel= m*g*sin(17.5) =140.1964N

axmax= -gsin(17.5)+Mu*G* Cos(17.5)
A=.9.8*sin(17.5)+9.80*Cos(17.5)

This line has a few problems, but I think they are just typos.

A= 2.1525

v^2=Vi^2+2AX
V^2=0+2*2.1525*4
v=4.15m/s

That looks like the right answer to me.


(An alternative approach to this problem would be to use the energy equation.)
 
Thank you for your help!
 

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