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brunettegurl
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coefficient of kinetic friction URGENT PLS. HELP!
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 across.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?
vf^2=vi^2+2ad
[tex]\sum[/tex]F=ma
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
[tex]\sum[/tex]F=ma
Ff-Fg=ma
mu*mg*costheta-mgcostheta=ma
mue= [tex]\frac{ma+mgcostheta}{mgcostheta}[/tex]
and solved I'm getting an answer of 0.78 while i should be getting an answer 0.68 pls. help
Homework Statement
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 across.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?
Homework Equations
vf^2=vi^2+2ad
[tex]\sum[/tex]F=ma
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
[tex]\sum[/tex]F=ma
Ff-Fg=ma
mu*mg*costheta-mgcostheta=ma
mue= [tex]\frac{ma+mgcostheta}{mgcostheta}[/tex]
and solved I'm getting an answer of 0.78 while i should be getting an answer 0.68 pls. help