Work Energy Problem Package released to Spring

[XwyhyX]

Homework Statement

A 2.00 kg package is released on a 53.1 degree incline, 4.00m from a long spring with force constant 120 N/m that is attached at the bottom of the incline. The coefficients of friction between the package and the incline are u{s} = 0.40 and u{k} = 0.20. The mass of the spring is negligible. (a) What is the speed of the package before it reaches the spring? (b) What is the maximum compression of the spring? (c) The package rebounds back up the incline. How close does it get to its initial position?

m = 2.00kg
θ = 53.1°
k= 120 N/m
L1= 4m
u_{s} = 0.40
u_{k} = 0.20

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Homework Equations

Conservation of Energy
K_{1} + U_{1} + W_{o} = K_{2} + U_{2}

The Attempt at a Solution

I know that to solve for a I will only consider the distance it has covered before touching the spring which is 4.0 m

So
K_{1} = 0 because it is released from rest
U_{1} = mgy = (2)(9.8)Lsinθ
W_{o} = W_{f} = -fL1 = -μmgL1 =- (0.2)(2)(9.8)(4)
K_{2} = 1/2mv^{2} = 1/2(2)v^{2}
U_{2} = 0 I'm not sure with this but i assume that at this point it's the reference point

Solving this

v = \sqrt{(2)(9.8)(4sin53.1°) - (0.2)(2)(9.8)(4)}
v = 6.86m/s

Is this correct? I can't continue because I'm not sure. I'll continue solving the b and c after I clarify this part. Thanks!

 

[PhanthomJay]

You have the equations correct but you did not calculate the friction force correctly. You used umg when it should be uN, where N = ?

 

[XwyhyX]

Oh N = mgcosθ. My bad, i'll have to correct that. Thanks!