- #1
Northbysouth
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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 = ukN):
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-ukmgsin(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.