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
 
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