Finding coefficient of friction

[462chevelle]

Homework Statement

A 3 kg block slides down a 30 degree inclined plane with constant acceleration of 0.5 m/s2. The block starts from rest at the top. The length of the incline is 2 m.

Homework Equations

Vf^2=Vi^2 +2ad velocity is 1.414
Fg=3 x 9.8=29.4
Fn= cos(30) (9.8 x 3) = 25.5
f//=sin (30) (9.8 x 3) =14.7
f=ma = 4.242=3 x 1.414
Ff= F//-f = 10.458=14.7-4.242
Ff=mu x (Fg) = 10.458= mu x 29.4 = .356

The Attempt at a Solution

Ff=mu x (Fg) = 10.458= mu x 29.4 = .356
this is incorrect. I feel like I am missing a simple step. I just can't figure out where I am messing up. any direction here? thanks

 

[barryj]

I am having a hard time following your equations.
These are correct and applicable:
Fg=3 x 9.8=29.4
Fn= cos(30) (9.8 x 3) = 25.5
f//=sin (30) (9.8 x 3) =14.7

Ff = u X Normal Force = u(Fn)

Set up an equation relating Ff, F//, m, and a and then solve for u

 

[SammyS]

Homework Statement

A 3 kg block slides down a 30 degree inclined plane with constant acceleration of 0.5 m/s2. The block starts from rest at the top. The length of the incline is 2 m.

Homework Equations

Vf^2=Vi^2 +2ad velocity is 1.414
Fg=3 x 9.8=29.4
Fn= cos(30) (9.8 x 3) = 25.5
f//=sin (30) (9.8 x 3) =14.7
f=ma = 4.242=3 x 1.414
Ff= F//-f = 10.458=14.7-4.242
Ff=mu x (Fg) = 10.458= mu x 29.4 = .356

The Attempt at a Solution

Ff=mu x (Fg) = 10.458= mu x 29.4 = .356
this is incorrect. I feel like I am missing a simple step. I just can't figure out where I am messing up. any direction here? thanks

Hello 462chevelle. Welcome to PF !

From what you have given it's difficult to know what it is that you are trying find out.

Please state the complete problem, word for word as it was given to you.

 

[462chevelle]

mass = 3
acceleration = .5 m/s
velocity = 1.414
Fg = 29.4
Fn = 25.5
F// = 14.7
Ff= 10.458
and what I got for mu
10.458 = mu x 29.4
solve it and get .356 which is the wrong answer.
and the question I posted is the question. the only thing not stated is that I am trying to find the coefficient of friction.
Sammy s . thanks. most of the time when I am having trouble with a physics problem. if I google it. it always pops this forum up for good explanations.

 

[barryj]

Ff = mu X Fn
mu is what you are trying to find. You cannot find mu until you find the net force on the mass.
You must use net force = ma to find mu
I would like to give you the complete solution but this is prohibited.

 

[462chevelle]

oh. I though it was Ff = mu x Fg so if I need the Fn that's 25.5.
so I have 10.458 = mu x 25.5 I get .41 but that is still wrong. am I doing anything wrong that's obvious or skipping a step?

 

[barryj]

Not quite. You are given the mass and acceleration. What is the net force on the mass? The net force is the component of the weight down the incline minus the friction force. Have you drawn a diagram of this situation? A diagram often helps.

 

[462chevelle]

would that be the F// minus the force of friction. so that is 4.242, and yes I have drawn out a diagram. that helps a lot in understanding what's going on

 

[barryj]

I think yes. The net force is F// - Ff. Where did you get 4.242?
You can get the net force from the given mass and acceleration.

 

[462chevelle]

maybe my force of friction is incorrect. cause if you do f = ma you get 1.5 = 3 x .5

 

[462chevelle]

so I could take 1.5 from 14.7 and get 13.2 as force of friction

 

[barryj]

F// - Ff = ma

You said that F// = 14.7 and that ma = 3 X .5
what is Ff?

Ff = (mu)Fn
and you said that Fn = 25.5

 

[barryj]

so I could take 1.5 from 14.7 and get 13.2 as force of friction . YES

now, use Ff = (mu) X Fn to get mui

 

[462chevelle]

that would be .51 but that is incorrect to. on my lunch break I am going to get a new page and start over to see if I am just writing something down wrong or skipping something

 

[barryj]

Check your precision. .51 might be slightly off.

 

[462chevelle]

FINALLY its .52. when should I decide which to round to when I am doing an equation like this?

 

[462chevelle]

I appreciate the help

 

[barryj]

When you do the calculations, keep the numbers in your calculator if you can to have the highest precision possible. In this problem, since the mass was 3 kg one might think that only one digit of precision is necessary but obviously this is not the case. So, keep as many digits of precision as you can. I assume you are entering this answer into Quest or some other online program, yes? These programs have a small tolerance on the answers but I guess .51 is not close enough.

 

[462chevelle]

yes its thinkquest. I've found that out that sometimes I have to redo the problem a few times rounding to different decimals to get it right.