# Ramp With Friction - Help Appreciated

1. Feb 10, 2013

### Nfinley1

Ramp With Friction - Help Appreciated!!

1. The problem statement, all variables and given/known data
When mass M is at the position shown, it is sliding down the inclined part of a slide at a speed of 2.41 m/s. The mass stops a distance S2 = 1.9 m along the level part of the slide. The distance S1 = 1.22 m and the angle θ = 37.10°. Calculate the coefficient of kinetic friction for the mass on the surface.

2. Relevant equations

I'm unsure what to use.

3. The attempt at a solution

So to start off am I looking for the height it starts off at.
So 1.22Sin(37.1)

Then am I looking for Potential Energy? and such. I'm lost from here

I'm not for someone to solve this for me, but someone who could help me get through this would be awesome! :)

Thanks,
Nick

2. Feb 10, 2013

### SammyS

Staff Emeritus
Re: Ramp With Friction - Help Appreciated!!

Please post the figure, and give more detail regarding your attempts at solving this.

While you're at it, please give the complete problem with the exact wording.

3. Feb 10, 2013

### Nfinley1

Re: Ramp With Friction - Help Appreciated!!

When mass M is at the position shown, it is sliding down the inclined part of a slide at a speed of 2.41 m/s. The mass stops a distance S2 = 1.9 m along the level part of the slide. The distance S1 = 1.22 m and the angle θ = 37.10°. Calculate the coefficient of kinetic friction for the mass on the surface.

This is all the information given.

I'll post more of my work on the subject in a moment, as I'm still working on even getting my first full attempt done.

Thus far, I've brainstormed a bunch of ideas...
Tried thinking about the component of gravity going along the horizontal axis until I realized I don't have a mass.
Then I thought about using a = mu * g solving for a and then use Vf^2 = Vo^2 + 2aS2. But I realized im solving for Mu anyways. So I can't exactly plug it in to get another constant.

So in all terms, I'm lost.

I tried
Vf^2 = Vo^2 + 2aS2
0 = (2.41m/s)^2 +2(a)(1.9)
0 = (5.5081) + 2a *1.9
A = 1.4495

Then mu = a/g
1.4495/9.8 = 0.1479 = mu Which is the incorrect answer.

found out 2.41^2 is actually 5.8081, and that would make mu = 0.155964, but still not correct.

Last edited: Feb 10, 2013
4. Feb 10, 2013

### SammyS

Staff Emeritus
Re: Ramp With Friction - Help Appreciated!!

I assume that the coefficient of kinetic friction is the same at all places along the path of the object.

One thing that I find helpful -- I always had my students do this -- is to draw a Free Body Diagram.

Actually, draw one for the object (mass) when it's on the incline and one when it's on the level.

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5. Feb 11, 2013

### lep11

Re: Ramp With Friction - Help Appreciated!!

It might be easier to think it in terms of energy.

6. Feb 15, 2013

### Nfinley1

Re: Ramp With Friction - Help Appreciated!!

Finding time to work on this during the week is hard for me. I'm going to review my work now and keep at it.

Thanks, I'll post more here later. (Just wanted you guys to know I wasn't leaving it for dead).