Friction/Inclined plane question, help would be appreciated

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A box slides up an inclined plane at 25.0° with an initial speed of 3.0 m/s and a coefficient of kinetic friction of 0.17. The calculations involve using forces, including gravitational and frictional forces, to determine the box's deceleration. One participant initially miscalculated the distance traveled before coming to rest, arriving at 0.14 m instead of the correct 0.8 m. After clarification and corrections, the importance of accurately applying the equations of motion was emphasized. The discussion highlights the need for careful calculations in physics problems involving friction and inclined planes.
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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\mu



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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