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Coeficient of Kinectic friction question

by Shanetm
Tags: accleration, friction, kinetic, mass, static
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Shanetm
#1
Nov3-11, 01:30 PM
P: 9
1. The problem statement, all variables and given/known data

I am having trouble getting the answer for this one. The question is about two students moving a piano. The piano weighs 260kg and one student is pushing with 280N Forward and the other 340N Forward. The piano is accelerating at 0.30m/s^2 Forward.

They are asking for kinetic friction as well as how long it will take for the piano to stop moving after pushing it for 6.2 seconds from rest

2. Relevant equations


3. The attempt at a solution

So far I have tried adding the two students Newtons and then diving the weight of the piano (times gravity) which gives me 2548. Then I divide by the ammount of Newtons the students are applying on the piano (Forward) which is 620N (combined) by 2548

The answer I get is around 0.24 for the coefficient of kinetic friction. The back of the book says 0.21. Could someone please tell me what I am doing wrong, and possibly how I could go about finding the second part of the question. Thanks.
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grzz
#2
Nov3-11, 01:38 PM
P: 950
To find the coefficient of frction you have to use also the acceleration 0.30...
dacruick
#3
Nov3-11, 01:44 PM
P: 1,084
well I got 0.21 as well, I guess make sure that you are using 9.8 for gravity not 10? I'm not sure what else it could be.

Try calculating it again, maybe you swapped 260 kg with 280 N and vice versa.

As for the second part, I believe there is an equation vfinal^2 - vinitial^2 = 2ad

where a is the acceleration and d is the distance.

Shanetm
#4
Nov3-11, 01:46 PM
P: 9
Coeficient of Kinectic friction question

Quote Quote by dacruick View Post
well I got 0.21 as well, I guess make sure that you are using 9.8 for gravity not 10? I'm not sure what else it could be.

Try calculating it again, maybe you swapped 260 kg with 280 N and vice versa.

As for the second part, I believe there is an equation vfinal^2 - vinitial^2 = 2ad

where a is the acceleration and d is the distance.

Ok, thanks. I will look into that!

Edit: Nevermind... I still keep getting 0.2433281 dividing 620 by 2548
grzz
#5
Nov3-11, 01:52 PM
P: 950
Did you use the value 0.30 of tjhe acceleration to find the coefficient of ffriction. I did not see it in your explanation.
Shanetm
#6
Nov3-11, 01:55 PM
P: 9
Quote Quote by grzz View Post
Did you use the value 0.30 of tjhe acceleration to find the coefficient of ffriction. I did not see it in your explanation.
No... I didnt. Where exactly would I use acceleration in the equasion? Thats probably exactly what im missing
grzz
#7
Nov3-11, 01:59 PM
P: 950
Use Fnet = ma for the piano.

First draw a FBD for the piano.
Which are the forces acting on the piano?
Shanetm
#8
Nov3-11, 02:29 PM
P: 9
Quote Quote by grzz View Post
Use Fnet = ma for the piano.

First draw a FBD for the piano.
Which are the forces acting on the piano?
I think its applied force, gravity, normal force and friction
grzz
#9
Nov3-11, 02:32 PM
P: 950
Correct.
Now use Fnet = ma.
then find the coefficient of friction and you get 0.2127.
Shanetm
#10
Nov3-11, 02:57 PM
P: 9
Hmm... I wish I could tell you I got it, but to be honest I am just getting more frustrated. I guess I will have to wait till class and have it taken up.

I know you want me to figure it out on my own using the mathematical steps provided. But, I'm at the point I just want to see the question broken down step by step (like what to do with each number to show exactly how to get it)

But thanks, regardless
grzz
#11
Nov3-11, 03:05 PM
P: 950
Fnet = ma

(280 + 340) - muN = ma

But N = mg

(280 + 340) - mumg = ma

(280 + 340) - mu260 x 9.8 = 260x0.30
Shanetm
#12
Nov3-11, 03:19 PM
P: 9
Quote Quote by grzz View Post
Fnet = ma

(280 + 340) - muN = ma

But N = mg

(280 + 340) - mumg = ma

(280 + 340) - mu260 x 9.8 = 260x0.30
Thank you very much. I appreciate it.


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