# Homework Help: Two inclines

1. Jul 27, 2010

### GOPgabe

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

A 2.2kg block slides down a planar incline. The incline makes an angle of thirty degrees with the horizontal and the block starts 10 m above ground. The coefficient of friction between the block and the incline is .1. The block reaches the bottom of the incline and then slides up an identical incline on the other side. To what vertical height will the block slide on the second incline? Ans. 7m

2. Relevant equations
P.E.= mgh
Ff=u mg cos (theta)

3. The attempt at a solution

The potential energy is =2.2*9.81*10=215.82

Here's where I'm a little unsure.
The work done by Ff is u mg cos(theta) * 20 ( the hypotenuse) *2, because you have to account for it up, right?

So 215-74.686=141.34

To account for height I think mgh works so 141.34/(2.2*9.81).

I get 6.53, not 7. Did I do this correct?

2. Jul 27, 2010

### collinsmark

So far so good.
Well, you do have to account for it going back up, yes. But the way you have done it, you are accounting for it going all the way back up, to reach a total height of 10 m (i.e. another distance of 20 m back up along the hypotenuse).

But you can be sure, that due to friction (and conservation of energy), that it's not going to make it that far back up. You can bet that the frictional losses will be greater when going down than back up, because the object doesn't travel as far on the way back up. (For a constant force like friction on a dry flat surface, [itex] W = \mathbf{F_f} \cdot \mathbf{s} [/tex]. But we don't know what s is yet, on the way back up.)

I suggest breaking the problem up into two parts. Find the kinetic energy when it reaches the bottom. Using that interim value, then set up a new equation in terms of an unknown distance back up the incline.

3. Jul 27, 2010

### GOPgabe

Got it! Thanks!