# Max distance up incline plane with variable friction

1. Dec 16, 2012

### PhizKid

1. The problem statement, all variables and given/known data
A block with initial velocity of 10 m/s is sent up an incline at 30 degrees from the horizontal. The coefficient of friction is 0.2x where x is the displacement. Find the maximum distance the block moves up the incline.

2. Relevant equations
F = ma
(1/2)mv^2
mgh

3. The attempt at a solution
I decided to try and use energy for this one.

Initial energy is (1/2)m(10)^2 + friction and final is mgh because the block comes to rest eventually.

So: (1/2)m(10)^2 + friction = mgh

Friction = coefficient * Normal force, so Friction = 0.2x * mgcos(30). Therefore:

(1/2)m(10)^2 + (0.2x*mgcos(30)) = mgh

Masses cancel out:

50 + (0.2x*gcos(30)) = gh

We want to find x because x is along the incline, and sin(30) = h / x, so h = xsin(30) and substitute back in:

50 + (0.2x*gcos(30)) = g(xsin30)

Simplify:

50 + 1.697x = 4.9x
50 = 3.203x
x = 15.6 meters

The solution is 5.31 meters, though. What have I done incorrectly?

2. Dec 16, 2012

### TSny

By "friction" do you mean the force of friction or the work done by the friction force? Note that you are setting up an energy expression.

3. Dec 16, 2012

### PhizKid

Oh...the work done by friction would be 0.2x * mgcos(30) * x * cos(180) I think. So is that what I'm supposed to use for the "friction" part in my equation?

4. Dec 16, 2012

### TSny

Right, you need the work done by friction. But the force of friction is a variable force (it depends on x). So, how do you get the work done by a variable force?

5. Dec 16, 2012

### PhizKid

Ohh I just learned this a few minutes ago, lol. Integrate the Friction Force, not multiply Friction Force * displacement.

6. Dec 16, 2012

Yes. Good.