Distance up an incline question

[meeklobraca]

Homework Statement

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 coefficient of friction along the incline is 0.22.

Homework Equations

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

The Attempt at a Solution

Im not really sure where to start, I think we can figure out the Ff with the coefficient 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?

I appreciate any help you guys can give me.

Cheers

 

[Sakha]

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

 

[meeklobraca]

Is that m(a + g) ?

 

[Sakha]

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

 

[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 doesn't it?

 

[Doc Al]

In fact, the question implies for some reason that the object is going up the incline doesn't it?

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).

 

[meeklobraca]

Youll forgive me but I am even more confused now. I don't think we ever went over the conservation of mechanical energy. And as I understand it Work is Fnet x d. But I don't get how id know to use that formula.

As for the potential and kinetic energies.

At the bottom we don't 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 ?

 

[Doc Al]

Youll forgive me but I am even more confused now. I don't think we ever went over the conservation of mechanical energy.

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

(FYI, "mechanical energy" just means KE + PE.)

And as I understand it Work is Fnet x d.

There's nothing wrong with that approach: The net work done on the object will equal the change in its KE.

At the bottom we don't have any potential energy, just kinetic energy so for one half of the equation we have 1/2mv2.

OK, assuming you measure PE from the bottom (which makes sense to me).

At the top of the incline we have potenital energy and kinetic energy?

At the highest point, what will the KE be?

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

Almost.

 

[meeklobraca]

Nothing about this makes any sense to me whatsoever.

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, I am just having some difficulty getting my head wrapped around it.

 

[Doc Al]

The formula for conservation of energy that were using is Fad + Ep + Ek = FFd + Ep + Ek.

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

 

[meeklobraca]

Yes, that's Force Applied x distance and Force of friction x distance.

 

[Doc Al]

Yes, that's Force Applied x distance and Force of friction x distance.

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.

 

[meeklobraca]

Why is there no force applied? In this problem the object is going up the ramp isn't it? If that is the case then wouldn't 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?

 

[Doc Al]

Why is there no force applied? In this problem the object is going up the ramp isn't it? If that is the case then wouldn't there be a force pushing it up the incline?

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.)

 

[meeklobraca]

okay fair enough, so that's one half of our equation to sovle this.

Now, I am 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 coefficient (umgcosO) I've determined the Ff is 5.3

-now I need the y height of the ball which is where i think I am lost. Anything I try to figure out the height doesn't fit. The formula i see that I could use is h=v2/2(9.81)?

 

[Doc Al]

okay fair enough, so that's one half of our equation to sovle this.

Now, I am 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

Good.

-using 0.22 as the coefficient (umgcosO) I've determined the Ff is 5.3

Good.

-now I need the y height of the ball which is where i think I am lost. Anything I try to figure out the height doesn't fit.

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 i see that I could use is h=v2/2(9.81)?

The formula you need is the one you already have.

 

[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!