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

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Homework Help Overview

The problem involves determining the coefficient of kinetic friction between a box and a patch of dirt as the box slides down an incline. The box has a mass of 35 kg and experiences a change in speed while traversing a distance of 1.50 m across the dirt.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of kinematic equations and forces acting on the box, including the role of gravitational components. There is a focus on the correct formulation of the equations to find the coefficient of kinetic friction.

Discussion Status

Some participants have provided insights into the equations being used and questioned the assumptions regarding velocity components. There is acknowledgment of a potential discrepancy in the calculated coefficient of friction, with participants exploring the implications of using different components of velocity.

Contextual Notes

Participants are navigating the complexities of applying kinematic equations in the context of forces acting on an inclined plane, with specific attention to the effects of friction and acceleration. There is an ongoing discussion about the relevance of negative signs in calculations related to deceleration.

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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