Coefficient of kinetic friction of a sliding crate

AI Thread Summary
The problem involves calculating the coefficient of kinetic friction for a 100 kg crate sliding down a 30-degree incline, initially moving at 2.0 m/s and stopping after 1.5 m. The equations of motion and forces acting on the crate are applied, leading to an acceleration of -1.3 m/s². The normal force is calculated to be 848.7 N, and the frictional force equation is set up to solve for the coefficient of friction (μ). The final calculation yields a coefficient of friction of 0.73, confirming the solution's accuracy despite a minor typographical error in the acceleration formula. The discussion emphasizes the importance of clear notation in physics problems.
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Homework Statement


Ugonna stands at the top of an incline and pushes a 100 kg crate to get it started sliding down the incline. The crate slows to a halt after traveling 1.5m along the incline. The initial speed is 2.0 m/s and the angle of incline is 30 degrees. What is the coefficient of sliding friction?



Homework Equations


ΣF=ma
ƒ = μN
aΔx = ½(vf2 – vi2)



The Attempt at a Solution


I have my midterm on Tuesday and am working on problems at the end of the chapters. I am hoping that someone can confirm my work, to make sure everything is okay, or help me through it if anything is wrong. This is what I have.

ΣFx = mgsinθ - ƒ
ΣFy = mgcosθ +N = 0

a = [(1/2)(0-2.0)2] / 1.5 m
a = -1.3 m/s2



Fy = mgcosθ +N = 0
N = (100 kg)(9.8m/s2)cos(30)
N = 848.7 N

ΣFx = mgsinθ - ƒ
Fx = (100 kg)(9.8m/s2)sin(30) - μN
= (100 kg)(9.8m/s2)sin(30) - μ(848.7N)
= 490 - μ(848.7N)


ΣF = ma
490 - μ(848.7N) = (100 kg)(-1.3 m/s2)
- μ(848.7N) = -620
μ = .73
 
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Seems okay to me!
 
Awesome! Thank you!
 
There is one thing; it looks just like a typo though:

This:

a = [(1/2)(0-2.0)2] / 1.5 m

should be written like this:

a = [(1/2)((0 m/s)^2-(2.0 m/s)^2)]/1.5 m

The answer isn't changed, though.
 
Ahh yes, definitely a typo; thanks
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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