Coefficient of kinetic friction of a sliding crate

Click For Summary

Homework Help Overview

The problem involves a crate sliding down an incline, where the original poster seeks to determine the coefficient of kinetic friction based on given parameters such as mass, distance traveled, initial speed, and incline angle.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster outlines their approach using force equations and kinematic relationships, while some participants confirm the validity of the work. A participant also points out a potential typo in the calculation of acceleration without altering the outcome.

Discussion Status

The discussion appears to be progressing positively, with participants providing confirmation and minor corrections. There is no explicit consensus on the final answer, but the original poster's reasoning has been acknowledged as sound.

Contextual Notes

The original poster is preparing for a midterm and is working through problems from their textbook, indicating a time constraint and a focus on understanding the material thoroughly.

ally1h
Messages
61
Reaction score
0

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
 
Physics news on Phys.org
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
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
3K
Replies
14
Views
3K
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
48
Views
8K
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 16 ·
Replies
16
Views
6K
Replies
5
Views
2K