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I was wondering if anyone could help me out in this question:

A mass M is sliding down the inclined plane at a speed of 2.13 m/s. The mass stops a distance of 2.45 m away from the base of the inclinde, along a flat surface. The mass slides 1.24 m down the incline before reaching the horizontal surface. The inclide has an angle theta of 31.1 degrees above the horizontal. Calculate the coefficient of kinetic friction for the mass on the surface.

I said that

Mechanical energy@ topof the incline = Mechanical energy@bottom

KE(top) + PE(top) = KE(bottom) + PE(bottom)

0.5m(Vtop^2) + mgh = 1/2m(Vbottom^2) + 0 J

Solving for Vbottom = 4.14 m/s

**I said that h = sin(theta) x 1.24 m; h=0.641 m

Then by using Vf^2 = Vo^2 + 2ad, and solving for accleration, given that Vf=0 m/s (when it comes to a stop), and Vo = 4.14 m/s(the velocity at the bottom of the incline), and d = 2.45 m

I got an acceleration of -3.50 m/s^2

then looking at a freebody diagram, I said that the sum of all forces in the x-direction is = ma.

I only have friction acting in the x direction?? Is there another force?

I said. -f=ma

-(mu)FN = ma where mu=coefficient of friction, FN = normal force

-(mu)mg= ma mg=FN, because, FN=Fg

masses cancel, and solving for mu....

mu = -0.357,

but, mu cannot be -?? and 0.357 is not the answer.

If anyone could help me out, it would be greatly appreciated thanks!!

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