(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

When the mass M is at the position shown, it has a speed v0 = 2.17 m/s and is sliding down the inclined part of a slide. The mass reaches the bottom of the incline and then travels a distance S2 = 2.45 m along the horizontal part of the slide before stopping. The distance S1 = 1.11 m and the angle of the incline is θ = 31.1o. Calculate the coefficient of kinetic friction for the mass on the surface.

2. Relevant equations

E= KE + U

E+ W (ext) = 0 -> E = -W (ext) (which is the work done by frictional force)

3. The attempt at a solution

So far I've analysed the x and y components of the mass on path S1 and path S2. This is what I've got

Path S1:

y-dir: N = mgcosθ

x-dir: ƩF = -F(friction) + mgsinθ = -μmgcosθ + mgsinθ

So W1 ( work done on S1) = S1(-μmgcosθ + mgsinθ)

Path S2:

W2 = -μmg(S2)

Since -W (total) = E

W1+W2 = KE + U (at original position)

S_{1}(-μmgcosθ + mgsinθ) - μmg(S_{2}) = - [(1/2)mv^{2}+ mgS_{1}sinθ]

μ(S1gcosθ + gS2) = 2S1gsinθ + v^{2}/2

μ = (2S1gsinθ + v^{2}/2)/(S1gcosθ + gS2)

So far I've got μ = .105 but it's wrong!

I'd really appreciate if anyone helped me out with this cos I totally have no idea how to figure out the right answer!

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

2. Relevant equations

3. The attempt at a solution

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# Homework Help: Coefficient of kinetic friction on 2 different parts

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