Friction force on a 3.35 kg collar sliding down a spring system

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engineering810
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Homework Statement


I've been attempting this problem all day now. My current submissions are 10.6N and I keep getting 7.04N every time I retry it but this is also wrong. Can somebody please help me out so I can figure out what I'm doing wrong? Thanks for any help in advanced!

The 3.35kg collar shown below is attached to a spring and released from rest at A. The collar then travels downward a distance of y = 2.00m . The spring has a spring constant of k = 22.5N/m . The distance a is given as 1.15m . The datum for gravitational potential energy is set at the horizontal line through A and B.

Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at C is 3.52m/s and the spring has an unstretched length of 0.65m .
Express your answer to three significant figures and include the appropriate units.
14.6.jpg


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Homework Equations



mgy = YF+(1/2)mv^2+(1/2)kx^2

F = frictional force

The Attempt at a Solution


(3.35)*(9.81)*(2) = 2F + (0.5)*(3.35)*(3.52^2) + (0.5)*(22.5)*(1.657^2)
65.727 = 2F + 20.7539 + 30.8886
65.727 = 2F + 51.6425
2F = 14.08
F = 7.04 Newtons
 

Discussion

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jack action said:
The spring displacement is wrong. Displacement is the difference between initial and final state.
Yea but unstretched is still 0.65. It starts at 1.15 but isn't unstretched at 1.15
 
It doesn't matter. The only energy added to the spring is done between initial and final state. The energy stored within the 0.65 and 1.15 was already in the spring at the initial state and is still in the spring at the final state.
 
engineering810 said:
Yea but unstretched is still 0.65. It starts at 1.15 but isn't unstretched at 1.15
The spring contained some elastic energy before the collar was released, this should be deducted from the elastic energy figure appearing in your equation.