What Is the Coefficient of Kinetic Friction Between the Box and the Dirt?

AI Thread Summary
The discussion focuses on calculating the coefficient of kinetic friction between a box and dirt after it slides down an incline. A box with a mass of 35 kg experiences a speed reduction from 6.0 m/s to 5.5 m/s over a distance of 1.5 m on dirt. The participant initially calculated the acceleration and friction using the wrong equations but later corrected their approach. The correct equation for the coefficient of kinetic friction was discussed, emphasizing the importance of using the velocity along the slope. The participant successfully found the answer but questioned the use of a negative sign in their calculations, which was clarified as unnecessary since the deceleration is accounted for in the equations.
brunettegurl
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coefficient of kinetic friction URGENT 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
\sumF=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
\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
 
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The component of gravity down the incline is sin25.
 


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 ..??
 


brunettegurl said:
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.
 


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 :))
 


brunettegurl said:
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.
 
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