A 10.0 kg block is released from point A in the Figure below. The track is friction-less except for the portion BC, of length 6.00 m. The block travels down the track, hits a spring of force constant k = 2250 N/m, and compresses it 0.300 m from its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between surface BC and block. Answer: 0.328 3.00 m 10.0 kg B 6.00 m C k •୪୪୪୪୪

College Physics
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ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
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Chapter29: Nuclear Physics
Section: Chapter Questions
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KE = ½-mv², Ug = mgh, U₁ = ½kx², E = KE + Ug + Us, Ei = Eƒ,
Wfric=Ffricd,
Ffric = MkN, E₁ - Wfric = Ef
Transcribed Image Text:KE = ½-mv², Ug = mgh, U₁ = ½kx², E = KE + Ug + Us, Ei = Eƒ, Wfric=Ffricd, Ffric = MkN, E₁ - Wfric = Ef
Problem 4:
A 10.0 kg block is released from point A in the Figure below. The track is friction-less except for
the portion BC, of length 6.00 m. The block travels down the track, hits a spring of force
constant k = 2250 N/m, and compresses it 0.300 m from its equilibrium position before coming
to rest momentarily. Determine the coefficient of kinetic friction between surface BC and block.
Answer: 0.328
A
3.00 m
10.0 kg
B
6.00 m
с
k
•^୪୪୪୪
Transcribed Image Text:Problem 4: A 10.0 kg block is released from point A in the Figure below. The track is friction-less except for the portion BC, of length 6.00 m. The block travels down the track, hits a spring of force constant k = 2250 N/m, and compresses it 0.300 m from its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between surface BC and block. Answer: 0.328 A 3.00 m 10.0 kg B 6.00 m с k •^୪୪୪୪
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