(adsbygoogle = window.adsbygoogle || []).push({}); 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?

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# Homework Help: Max distance up incline plane with variable friction

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