# Distance up an incline question

1. Sep 29, 2008

### meeklobraca

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

A 3.0 kg object has a speed of 2.0 m/s at the bottom of a 35 degree incline. How far up the incline will the object slide if the coefficent of friction along the incline is 0.22.

2. Relevant equations

Being a conservation of energy question I am looking at Fad + Ep + Ek = FFd + Ep + Ek

3. The attempt at a solution

Im not really sure where to start, I think we can figure out the Ff with the coefficent of friction, but with this going up an incline I think we need the Fa? If not how does an object go up an incline?

Cheers

2. Sep 29, 2008

### Sakha

You first need to calculate the force that is pulling your object down the incline.

3. Sep 29, 2008

### meeklobraca

Is that m(a + g) ?

4. Sep 29, 2008

### Sakha

Yes, the weight of the object is 3*9.8, but the force pulling it down is the weight*sin(35).

5. Sep 29, 2008

### meeklobraca

OKay, so how does that fit into my equations to figure out the distance travelled? In fact, the question implies for some reason that the object is going up the incline doesnt it?

6. Sep 29, 2008

### Staff: Mentor

It has an initial speed up the incline.

Consider using conservation of mechanical energy in this form:

KEi + PEi = KEf + PEf + Wf

Where Wf is the work done against friction (which is positive).

7. Sep 29, 2008

### meeklobraca

Youll forgive me but im even more confused now. I dont think we ever went over the conservation of mechanical energy. And as I understand it Work is Fnet x d. But I dont get how id know to use that formula.

As for the potential and kinetic energies.

At the bottom we dont have any potential energy, just kinetic energy so for one half of the equation we have 1/2mv2. At the top of the incline we have potenital energy and kinetic energy?

so our equation is 1/2mv2 = mgh + 1/2mv2 + Wf ?

8. Sep 29, 2008

### Staff: Mentor

But in your first post you stated "Being a conservation of energy question..."

(FYI, "mechanical energy" just means KE + PE.)
There's nothing wrong with that approach: The net work done on the object will equal the change in its KE.

OK, assuming you measure PE from the bottom (which makes sense to me).
At the highest point, what will the KE be?
Almost.

9. Sep 29, 2008

### meeklobraca

The formula for conservation of energy that were using is Fad + Ep + Ek = FFd + Ep + Ek. Nothing about conservation of mechanical energy. Nothing about work done etc etc.

When I get a question like this, where do I start? What am I needing to figure out to get the answers I need?

I do appreciate your help with this, im just having some difficulty getting my head wrapped around it.

10. Sep 29, 2008

### Staff: Mentor

Can you tell me what the terms Fad and FFd stand for?

11. Sep 29, 2008

### meeklobraca

Yes, thats Force Applied x distance and Force of friction x distance.

12. Sep 29, 2008

### Staff: Mentor

Ah! So that equation is practically the same as the one I suggested. What you call "FFd", I called the work done against friction (Wf). Same thing. (Work = force x distance.)

In this problem there's no applied force so you can drop that term.

13. Sep 29, 2008

### meeklobraca

Why is there no force applied? In this problem the object is going up the ramp isnt it? If that is the case then wouldnt there be a force pushing it up the incline? OR are we going from the top of the incline to the bottom to sovle this?

14. Sep 29, 2008

### Staff: Mentor

Yes, the object starts out going up the ramp, but no, there's no force pushing it up. All you need is an initial speed and it will go up the ramp--until it stops. Just like when you toss a ball in the air it continues going up, even though the only forces on it act down. (As soon as the ball leaves your hand, there's no more upward force.)

15. Sep 29, 2008

### meeklobraca

okay fair enough, so thats one half of our equation to sovle this.

Now, im left with

1/2mv2 = Ffd + Epf + Ekf

The kinetic energy at the top when the ball stops is 0 so that eliminates that.

1/2mv2 = Ffd + mgh

-using 0.22 as the coefficent (umgcosO) ive determined the Ff is 5.3

-now I need the y height of the ball which is where i think im lost. Anything I try to figure out the height doesnt fit. The formula i see that I could use is h=v2/2(9.81)?

16. Sep 29, 2008

### Staff: Mentor

Good.

Good.

What you really need is the distance the object slides up the ramp, which is "d". With a little trig you can express the height "h" in terms of "d". (Then you can solve for d.)

The formula you need is the one you already have.

17. Sep 30, 2008

### meeklobraca

Yup, got it all under control. I solve for h by making h = dsin 35, then combining the formulas to solve for d.

Thank you very much for your help!