Friction/Inclined plane question, help would be appreciated

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Homework Help Overview

The discussion revolves around a physics problem involving friction and an inclined plane. The scenario describes a box sliding up a plane inclined at 25.0° with an initial speed and a given coefficient of kinetic friction. Participants are attempting to determine how far the box travels before coming to rest.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are discussing the application of Newton's second law and the role of friction in the motion of the box. There are attempts to rearrange equations and apply trigonometric functions to solve for distance. Some participants express confusion regarding their calculations and seek clarification on the results.

Discussion Status

The discussion is ongoing, with participants sharing their calculations and questioning each other's approaches. There is an acknowledgment of potential errors in the calculations, and some guidance has been offered to revisit the steps taken.

Contextual Notes

Participants are working within the constraints of the problem as posed, including the initial conditions and the coefficient of friction. There is a noted discrepancy between calculated results and an expected answer, prompting further exploration of the assumptions and methods used.

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Friction/Inclined plane question, help would be appreciated :)

A box is sent sliding up a plane, inclined to the horizontal at 25.0◦ with an
initial speed of 3.0 m/s.
The coefficient of kinetic friction between the box
and the plane is 0.17. How far up the plane does the box go before coming to
rest



f=F[itex]\mu[/itex]



I rearange the equations using F=ma, using trig etc, and i don't get the answer which is supposedly 0.8m

Thanks
 
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jami8337 said:
I rearange the equations using F=ma, using trig etc, and i don't get the answer which is supposedly 0.8m

Please explain how you got a result, and what the result was ...
 


I got, F=mgcos25
and fk(kinetic friction) = 0.17mgcos25

then
f=ma
hence
fk+mgsin25=f
ma=0.17mgcos25+mgsin25
a=0.17gcos25+gsin25
=5.66m/s^2

i then put that into v^2=u^2+2ax

and got x=0.14
 


You did something wrong here
jami8337 said:
i then put that into v^2=u^2+2ax

and got x=0.14
Try it again ...
 


ah thank you, was for some reason doing a^2 :)
 


jami8337 said:
ah thank you, was for some reason doing a^2 :)

You're Welcome !
 

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