Max distance up incline plane with variable friction

  • Thread starter PhizKid
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  • #1
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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?
 

Answers and Replies

  • #2
TSny
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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.
 
  • #3
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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
TSny
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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?
 
  • #5
476
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Ohh I just learned this a few minutes ago, lol. Integrate the Friction Force, not multiply Friction Force * displacement.
 
  • #6
TSny
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Yes. Good.
 

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