Find the maximum deflection of the spring

[Northbysouth]

Homework Statement

The 3.8-kg collar is released from rest at A and slides down the inclined fixed rod in the vertical plane. The coefficient of kinetic friction is 0.51. Calculate (a) the velocity v of the collar as it strikes the spring and (b) the maximum deflection x of the spring.

I have attached an image of the question.

Homework Equations

The Attempt at a Solution

I was able to find the velocity of the collar with:

U₁ + K₁ = U₂ + K₂ + F

K₁ and U₂ are 0

F is the friction force

0.5mv² = mgh₁ - F

Drawing a FBD of the collar shows that the Normal force (F = u_kN):

N = mgsin(27)

0.5(3.8kg)v² = (3.8kg)(9.81 m/s²)(0.57sin(63)) - (0.51)(0.57m)(3.8kg)(9.81m/s²)sin(27)

Solving for v gives me:

v = 2.7157

I can't seem to find the displacement of the spring though.

I realize that F = -kx and I need to find F to get x.

Looking at the practice question (identical scenario but different numbers) led me to believe that the answer should be:

F = mg - mgsin(27)

x = sqrt[mg-u_kmgsin(27)/2700]

x = 103 mm
But it says this isn't correct and I'm not sure where I'm making my mistake.

Help is appreciated.

 

[TSny]

U₁ + K₁ = U₂ + K₂ + F

Note that force and energy have different dimensions, so they can't be added together.