# Friction/Inclined plane question, help would be appreciated

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 dont get the answer which is supposedly 0.8m

Thanks

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I rearange the equations using F=ma, using trig etc, and i dont 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
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 :)

ah thank you, was for some reason doing a^2 :)
You're Welcome !!