Max distance up incline plane with variable friction

[PhizKid]

Homework Statement

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.

Homework Equations

F = ma
(1/2)mv^2
mgh

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?

 

[TSny]

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

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.

 

[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?

 

[TSny]

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?

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?

 

[PhizKid]

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

 

[TSny]

Yes. Good.