What is the coefficient of kinetic friction of the surface?

[DKPeridot20]

I have a homework question (due tomorrow) that I cannot figure out how to answer. It is:

This problem involves a single surface with a single unknown coefficient of kinetic friction that can be found. A box starting from rest, slides down along an incline a distance of 8.8 meters. The incline is at 35.5 degrees above the horizontal. After it reaches the bottom of the incline, it travels horizontally
17. meters. What is the coefficient of kinetic friction of the surface?

In class we derived an equation that I thought would work. I know that
W = Fd or mgh [h being dsin(zeta)] and that
Wtotal = Wgravity + Wfriction.
(I'm going to say that the distance on the incline is di and the distance on the horizontal is dh) So I end up with Wtotal = mgh(di) - Coefficient Fk(Normal force on the incline)di - Coefficient Fk(Normal force on the horizontal)dh which is:

Wtot = mg(di)sin(zeta) - Coefficient Fk(mg)di(cos)(zeta) - Coefficient Fk(mg)dh

Is this even the right equation and if so, how then do I solve for Coefficient Fk?:uhh:

I'm sorry that it's so hard to read.
Thanks very much.

 

[Just some guy]

As the box starts from rest you need to use an energy argument - the gravitational potential energy the box starts with has to completely be dissipated by the frictonal force, therefore the work done against friction (mu*normal reaction force) is equal to the gravitational potential energy and you can solve it to find mu (as the mass of the block cancels out).

Remember the normal reaction force on the block is different when it's on the slope so the work done against friction will be split up into two parts: (mu*normal reaction force on slope + mu*normal reaction force on horizonal plane).

Hope that helps.

[edit]basically you have the right equation - just set Wtot equal to zero[/edit]

 

[DKPeridot20]

Oh, okay, thanks very much. I'm getting the right answer now.

I also have another problem that is exactly the same except that it deals with a frictionless surface, the box goes up another incline, and the only supplied number is the distance it travels down the first incline. Is it correct to use the same equation only setting the coefficient of friction equal to zero? I know that without friction the box will go up the second incline just as far as it came down the first, but I'm not exactly sure how to find that with an equation. I speculate:

(d1 - 1st inlcine
d2 - horizontal
d3 - 2nd incline)

Having set Wtot equal to zero I get d3sin(zeta) = d1sin(zeta) - 0 - 0 the
sin(zeta)'s cancel and there's d3 = d1.

Is this right?

 

[Just some guy]

Oh, okay, thanks very much. I'm getting the right answer now.
I also have another problem that is exactly the same except that it deals with a frictionless surface, the box goes up another incline, and the only supplied number is the distance it travels down the first incline. Is it correct to use the same equation only setting the coefficient of friction equal to zero? I know that without friction the box will go up the second incline just as far as it came down the first, but I'm not exactly sure how to find that with an equation. I speculate:
(d1 - 1st inlcine
d2 - horizontal
d3 - 2nd incline)
Having set Wtot equal to zero I get d3sin(zeta) = d1sin(zeta) - 0 - 0 the
sin(zeta)'s cancel and there's d3 = d1.
Is this right?

That's correct if both slopes are at the same angle, otherwise the sin(zeta) doesn't cancel out

 

[DKPeridot20]

Thanks again. The problem doesn't specify, but the answer is the same distance so I'm assuming they're the same angle...