coefficient of kinetic friction URGENT PLS. HELP!!!

[brunettegurl]

1. The problem statement, all variables and given/known data

A box of mass 35 kg is sliding down a patch of frictionless snow inclined at 25deg to the horizontal with a speed of 6.0m/s when its a bare patch of dirt that's 1.50 m accross.If the box's speed is reduced by 0.5m/s by the time it reaches the far side of the bare patch, whtas the coefficient of kinetic friction between box and the dirt?

2. Relevant equations

vf^2=vi^2+2ad
\sumF=ma

3. The attempt at a solution
ok first i used the equation vf^2=vi^2+2ad and rearranged it for a by using the speeds and not their components(vocotheta).. then w/the a= -1.917 i used the
\sumF=ma
Ff-Fg=ma
mu*mg*costheta-mgcostheta=ma
mue= \frac{ma+mgcostheta}{mgcostheta}

and solved i'm getting an answer of 0.78 while i should be getting an answer 0.68 pls. help

[LowlyPion]

The component of gravity down the incline is sin25.

[brunettegurl]

so then this would be the correct equation
\sumF=ma
Ff-Fg=ma
mu*mg*costheta-mgcostheta=ma
mue= \frac{ma+mgsintheta}{mgcostheta}

also does it make sense when finding the velocity to not do vcostheta ..??

[LowlyPion]

also does it make sense when finding the velocity to not do vcostheta ..??

The V2 equation is for a uniformly accelerating object.

So what you are calculating there is the deceleration over the distance given. I guess I don't understand why you would have an interest in v*cos25 unless they were giving you only the horizontal component of velocity. But in this problem they are giving you the distance along the slope. So, ... use the velocity along the slope as given.

Your equation above for μ should yield the desired result.

[brunettegurl]

ok i got the answer but i have one question.. when i plug in the value of a why does it work without the negative sign?? ..thanks :))

[LowlyPion]

ok i got the answer but i have one question.. when i plug in the value of a why does it work without the negative sign?? ..thanks :))

Because it slows down.