# There is no mass given. How do I solve this?

1. Jan 29, 2014

### djalle12

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

2. Relevant 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.

3. 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 dont know where to go from here since all these forces seemingly depend on knowing the mass.

2. Jan 29, 2014

### 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...

3. Jan 29, 2014

### 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?

4. Jan 29, 2014

### BvU

I agree. This is for the downhill part.

5. Jan 29, 2014

### 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.

6. Jan 29, 2014

### 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 travelled along the level ground. This all seems to cancel out the mass. Does this seem correct to all of you?

7. Jan 29, 2014

### BvU

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

8. Jan 29, 2014

### djalle12

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

9. Jan 29, 2014

### BvU

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

10. Jan 29, 2014

### djalle12

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

11. Jan 29, 2014

### djalle12

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

12. Jan 29, 2014

### djalle12

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

13. Jan 29, 2014

### BvU

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

14. Jan 29, 2014

### djalle12

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

15. Jan 29, 2014

### djalle12

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

16. Jan 29, 2014

### BvU

What is the 2 * cos^2 theta ?

17. Jan 29, 2014

### BvU

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

18. Jan 29, 2014

### 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

19. Jan 29, 2014

### djalle12

Yes, g cancelled out for me.

20. Jan 29, 2014

### 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.