# Homework Help: Coefficient of kinetic friction of a sliding crate

1. Jul 12, 2008

### ally1h

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
ΣF=ma
ƒ = μN
aΔx = ½(vf2 – vi2)

3. 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

2. Jul 12, 2008

### arildno

Seems okay to me!

3. Jul 12, 2008

### ally1h

Awesome! Thank you!

4. Jul 12, 2008

### Noein

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