How Do I Calculate Sliding Distance with Unknown Mass?

[djalle12]

Homework Statement

Children sled down a 41m long hill inclined at 25°. At the bottom, the slope levels out. If the coefficient of friction is 0.12, how far do the children slide on the level ground?

Homework Equations

This question is from the force chapter of my textbook, so I believe on force equations are to be used and not the conservation of energy.

The Attempt at a Solution

I have only gotten as far as drawing a force diagram. I have labelled all the forces to include, the normal forces, force of gravity down the plane, and the frictional force up to the plane. I don't know where to go from here since all these forces seemingly depend on knowing the mass.

 

[BvU]

They certainly do. But perhaps the mass cancels if you calculate how far they get...
And if you can post the diagram...
The trip consists of two parts: down the hill the forces are different from when sliding horizontally. Perhaps TWO diagrams are needed...

 

[djalle12]

I believe I made some progress. According to my force diagram then the following should be true F=mg(sinΘ)-μmg(cosΘ)
if you set that equal to ma because of F=ma, then a=g(sinΘ)-μg(cosΘ).

Right?

 

[BvU]

I agree. This is for the downhill part.

 

[orion]

As a general rule of thumb, if a seemingly needed quantity is not given in the statement of the problem, it usually means that it cancels out somewhere.

It looks like you are making great progress. Now you have to work on the flat part.

 

[djalle12]

Ok now from the equation (v^2)=(vi^2)+2a(Δx), I figured v=√(2g(sinΘ)-μ(cosΘ)x_1)

So I can now equate that to being an intial velocity vector Θ° below the horizontal at the start of level ground. Then solve for for initial velocity in the x direction in relation to the level ground. After that I should be able to use the above equation again to solve x_2, being the distance traveled along the level ground. This all seems to cancel out the mass. Does this seem correct to all of you?

 

[BvU]

Yes ! except that I might miss a bracket around 2g(sinΘ)-μ(cosΘ) int he first line...

 

[djalle12]

My final answer was 101.50 meters. If anyone else did it, did you get the same?

 

[BvU]

Show a bit more detail in the calculation. I slide even further, but I'm also prone to sloppy writing out...

 

[djalle12]

I just started deriving my equation more thoroughly and noticed a mistake. I'll post another answer in a bit.

 

[djalle12]

OK now I got 182.45 meters. How does that match up with you BvU?

 

[djalle12]

OK now my answer is 182.45 meters, how does that match up with you BvU?

 

[BvU]

I was within 10% of what you had. Now you are very far ahead of me. Show some detail...

 

[djalle12]

My final equation was Δx_2=[2(cosΘ)^2(Δx_1)(sinΘ-μcosΘ)]/μ.

 

[djalle12]

OK I believe I finally have it :) Looked through again and found a mistake. Now I got 108.23 meters.

 

[BvU]

What is the 2 * cos^2 theta ?

 

[BvU]

Did you notice that even g cancels out ?
You are still 1.00 meter ahead of me...

 

[djalle12]

Nevermind I got 91.23 half of my 182 answer because I believe I initially forgot the 2 in v^2=vi^2+2ax

 

[djalle12]

Yes, g canceled out for me.

 

[djalle12]

What I have is [(cosΘ^2)(Δx_1)(sinΘ-μcosΘ)]/μ and I've looked through my derivation and redone it and get the same thing each time, and this equation results in 91.23 meters.

 

[djalle12]

Well the 2 is a mistake and I took it out and the cos theta came from finding the initial velocity in the level ground direction from the final velocity of the downhill which was in the direction of theta below the x axis.

 

[djalle12]

Oh and I realized I've been using 26° for theta this whole time. With 25° I actually get 88.08 meters.

 

[djalle12]

without the cos theta^2 I get 107.24, which I believe is what you got, right? From what I see the cos^2 is necessary though.

 

[BvU]

Incidentally, leaving the symbols (as opposed to the numberical values) in the expression until the last moment like you do that, is very good practice. It allows dimension checking and greater accuracy if things happen to cancel. It also allows quicker adaptation if you discover something went wrong on the way.

Thought you had one or two cosines too many in the denominator, but I coudn't reverse-engineer them.
Hence the questions for details.

-- OK, now I understand. I missed page 2 (not refreshed the page) and the 25/26 blurred the numbers a bit.

However,

You continue with the final velocity along the slope and take the horizontal component as the initial speed for the horizontal stretch. Understandably so, but you may well assume that the friction is sufficiently low that hitting the flat does not dissipate kinetic energy. No work is done to change the direction of the motion, because the force that changes the direction is perpendicular to it. It is a slight approximation, but a good one.
So the cos^2 goes away.
If it still troubles you, try to explain where else the 1-cos^2 part of the kinetic energy goes to. Difficult eh ?

Remember that originally you had trouble with missing m ? What do you think of the cancelling g ?

 

[djalle12]

Ohhh that makes sense. Ok so the final velocity of the down hill section should(as far as solving this problem is concerned) equal the initial velocity of the level section. I just realized that it makes no sense to automatically jump back to a slower velocity. Thanks a lot!