Final Velocity of Mass on Frictionless Incline

[dgl7]

1.
A spring with a spring-constant 3.6 N/cm is compressed 39 cm and released. The 9 kg mass skids down the frictional incline of height 36 cm and inclined at a 20◦ angle. The acceleration of gravity is 9.8 m/s^2. The path is frictionless except for a distance of 0.6 m along the incline which has a coefficient of friction of 0.4 .

What is the final velocity of the mass?

2. RELEVANT EQUATIONS
Ko+Uo+Wnc=U+K
kinematic equations
ma=kx

3. MY ATTEMPT
I tried kx=ma-->
(3.6)(.39)=(9)a
a=.156
I put that into the kinematic equation v^2 = vo^2 + 2a(X - Xo)
v^2=(0)^2+2(.156)(.39)
got v=.3488266045 m/s (note: this would be vo for the overall equation and v only for the compression of the spring)

I put that into Ko in the Ko+Uo+Wnc=U+K equation-->
.5(9)(.3488266045)^2+(9)(9.8)(.36)+[(.4)(9*9.8*cos20)](.6)(cos20)=0+.5(9)v^2
solved for v and got 3.366216946 m/s but that isn't right...I think there is something wrong with the nonconservative work, but I'm not sure what...

 

[Delphi51]

I'm wondering about the 39 cm where the spring pushes the block. Is that 39 cm part of the incline? If so, it makes your kx=ma calc incorrect because a component of mg will also be acting. Might be easier to just include a 1/2kx^2 term in your conservation of energy equation.

Also, the cos(20) on the end of the friction term seems odd to me. What is it for?

 

[dgl7]

Yes that would make a lot more sense to just use Uspring+Ugravitational+Wnc=K. I didn't/don't know how to solve for Wnc, so I just used Ffriction*r*costheta, which brought me to the cos20...

 

[Delphi51]

It says "distance of 6 cm along the incline" so it is just Ff*.06 with no cos(20). Keep in mind the faint possibility that it runs out of energy somewhere in the friction stretch.